P~9a~~--------------------------
PHYSICAL CHEMISTRY
Dr. J.N. Gurtu
Dr. H.C. Khera
M.Sc., Ph.D.
M.Sc., Ph.D.
Former Principal
Reader & Head, Deptt. of Chemistry,
Meerut College, Meerut
loP. College, Bulandshahr.
»
PRAGATI PRAKASHAN
© Authors-Physical Chemistry Vol. I
PRAGATI PRAKASHAN Head Office: Educational Publishers PRAGATI BHAWAN 240, W. K. Road, Meerut-250 001 SMS/Ph. : (0121) 6544642, 6451644 Tele/Fax: (0121) 2640642, 2643636 Visit us at : www.pragatiprakashan.in e-mail:
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Edition 2010
ISBN-978-81-8398-496-6
Published by : K.K. Mittal for Pragati Prakashan, Laser Typesetting : Pragati Laser Type Setters Pvt. Ltd., (Phone: 2661657) Meerut. Printed at: Arihant Electric Presss, Meerut.
CONTENTS
IIUlij' MATHEMATICAL CONCEPTS AND COMPUTER Logarithmic and Antilogarithmic Relations Find out the values of the following Differentiation with Examples Numerical Problems Integration and give important formulae Numerical Problems Terms Permutation and Combination with Examples 24-41. Numerical Problems 42. Probability Definition 43-50. Numerical Problems 51. Logarithmic, Trigonometric Series 52. Maxima and Minima 53-57. Numerical Problem 58. Functioning, Characteristics, Limitations Computer Programing Flow Charts Fortan, Cobol, Basic, Pascal Operating System Exercise Multiple Choice Questions Fill in the Blanks True or False 1. 2. 3. 4-12. 13. 14-23. 24.
1-55 1 5 7 11-13 13 16
22 . 23
29 32 32 33 34 38 49
50 51 52 52 54 54
o o o
111~lij" GASEOUS STATE
5~102
1.
Nature of R and its value in different units
56
2.
Short account of kinetic theory of gases and derivation of kinetic equation
57
3.
(a)
59
4.
Short account of kinetic theory of gases. Derivation of PV = RT and show how the various gas laws are consistent with it? (b) Expression for kinetic energy of one mole of gas Values of Cv and Cp from kinetic equation and variation of CplCv with molecular complexity of the gas
62
(viii)
Distribution of molecular velocities of Maxwell's law 65 (a) Average velocity, root me~n square velocity and most 67 probable velocity and relation among them (b) Calculation of RMS velocity from kinetic theory of gases 67 7. Kinetic equation of gases 68 8. Ideal gas and its difference with a real gas 70 9. (a) Limitations of PV = RT and improvements suggested by 70 vander Waals. Derivation of vander Waals equation. (b) Units of vander Waals constants 70 (c) Show that effective volume of gas molecules is four 70 times greater than actual volume of molecules 10. (a) Critical phenomenon, calculation and determination of 75 critical constants, short note on continuity of state 79 11. Some short questions on vander Waals equation 81 12. Short notes on : (a) Various equation of state (b) Law of corresponding states (c) Mean free path (d) Critical phenomenon and its utility (e) Collision frequency (0 Law of equipartition of energy (g) Specific heat ratio (h) Boyle temperature (i) Continuity of state 13. Methods for producing cold and liquefaction of gases, 86 inversion temperature o Numerical Problems 98 o Multiple Choice Questions 100 o Fill in the Blanks 101 o True or False 102
5. 6.
IIn"j'" CHEMICAL AND PHASE EQUILmRIUM Chemical Equilibrium
I
I
103-171
1.
Law of mass action and eqUilibrium constant
103
2.
Short notes on the following : (i) Work function (ii) Free energy Thermodynamic derivation of law of mass action
104
3.
108
4. 5. 6. 7.
Thermodynamic derivation of van't Hoff isotherm Thermodynamlc··-derivation of van't Hoff isochore or van't Hoff equation Thermodynamic derivation of Clapeyron equation and Clausius-Clapeyron equation Le-Chatelier-Braun principle and applications to different equilibria o Numerical Problems
109 111
Explanation and illustration of phase, component and degree of freedom Phase rule and its thermodynamic derivation Explain: Can all four phases in a one component system co-exist in equilibrium? Application of phase rule to water system Application of phase rule to sulphur system Short notes on : (a) Non-variant system in phase rule studies (b) Triple point (c) Transition point Two component system, graphical representation, reduced phase rule equation, condensed state Application of phase rule to lead-silver system Application of phase rul~ to potassium iodide and water system Determination of number of phases and components of different systems Calculation of degree of freedom Determination of number of phases, components and degree of freedom of different systems Ideal solutions, vapour pressure of such solutions Non-ideal or real solutions, vapour pressure curves of completely miscible binary solutions Theory of fractional distillation of binary solutions (a) Theory of partially miscible liquid pairs, e.g., (i) Phenol-water system Oi) Triethyl-amine water system
126
IPhase Equilibrium I
1.
2. 3. 4. 5. 6.
7. 8. 9. 10. 11. 12. 13. 14. 15. 16.
114 118 122
-129 131 131 134 137
138 140 142 143 143 144 144 147 150 154
(x)
(b)
(iii) Nicotine-water system Influence of impurities on critical solution temperature
154
IDistribution Law I 1.
2.
Nemst's distribution law. limitations. applications 157 Nemst's distribution law. modification when the solute 162 undergoes dissociation or association Numerical Problems 163 o Multiple Choice Questions 169 Fill in the Blanks 170 True or False 170
o o o
COLLOIDAL STATE 1.
2.
3.
4.
172-210
Explain the terms : colloidal state and colloidal solution, 172 methods for preparation and purification of colloidal solutions (a) Difference between true solution. colloidal solution and 177 suspension (b) Types of colloidal systems. 177 Preparation of colloidal solutions of AS2S3. Fe(OH)3 gold. 178 sulphur. silicic acid. carbon. mastic. iodine Short notes on : 180 (i) Lyophilic and lyophobic colloids (ii) Peptisation (iii) Dialysis (iv) Ultramicroscope (v) Tyndall effect (vi) Brownian motion (vii) Electrophoresis (viii) Electro-osmosis (ix) Coagulation (x) Hardy-Schulze law (xi) Protection (xii) Gold number (xiii) Stability of lyophilic colloids (xiv) Iso-electric point (xv) Emulsion (xvi) Gel (xvii)Electrical double layer or Zera potential
(xi)
5.
(a) (b)
6.
Explain the following facts: 199 (a) A sulphur sol is coagulated by adding a little electrolyte, whereas a gelatin sol is apparently unaffected. (b) What happens when a colloidal solution of gold is brought under the influence of electric field? (c) What happens when an electrolyte is added to colloidal solution of gold? (d) What happens when a beam of light is passed through a colloidal solution of gold? (e) A colloidal solution is stabilised by addition of gelatin. (f) Presence of H2S is essential in AS2S3 sol though H2S ionises and should precipitate the sol. (g) Why ferric chloride or alum is used for stoppage of bleeding?
7. 8. 9.
Applications of colloids in chemistry Sol-gel transformation Note on thixotropy Multiple Choice Questions Fill in the Blanks True or False
Origin and significance of charge on a colloidal particle Classification of the sols: Gold, Fe(OH)3, gelatin, blood, sulphur, AS2S3
198 198
200
2C4 206 208 209 209
o o o
III~I'M CHEMICAL KINETICS AND CATALYSIS Chemical Kinetics
I
I
211-276
1. (a)
Explain the terms: rate of chemical reaction, velocity 211 coefficient, molecularity and order of reaction (b) Difference between molecularity and order of reaction (c) Why reactions of higher orders are rare? (d) Factors which affect reaction rates? 2. Zero order reaction, rate expression, characteristics. 215 3. Half order reaction, rate expression, characteristic 216 4. First order reaction, rate expression, characteristics, examples 217 5. (a) (b)
Pseudo-unimolecular reactions Study of kinetics of hydrolysis of methyl acetate
6. Half life period for a first order reaction
219 220
7. Second order reaction, rate expression, characteristics, examples and study of kinetics 8. Half life period for a second order reaction 9. Third order reaction, rate expressions, characteristics and examples 10. nth order reactions, rate equation and characteristic 11. Methods employed in determining the order of reaction 12. Energy of activation and temperature coefficient 13. Activation energy, potential energy barrier and Arrhenius law 14. Collision theory for unimolecular reactions 15. Mathematical treatment of transition state theory, comparison with collision theory. Numerical Problems
o
221 224 225 227 228 230 233 235 236 242
ICatalysis I 1.
2.
3. 4. 5.
Catalyst, catalysis, types and classification of catalysis. 251 characteristics of catalytic reactions Notes on the following: 258 (a) Catalytic promoters (b) Catalytic poisons Theories of catalysis, industrial applications of catalysts 261 Enzyme catalysis, characteristics and examples of enzyme 265 catalysis, kinetics of enzyme catalysis Note on acid-base catalysis 269 Multiple Choiae Questions 270 Fill in the Blanks 274 True or False 275 o Log and Antilog Tables (i)-(iv)
o o o
MATHEMATICAL CONCEPTS AND COMPUTER MATHEMATICAL CONCEPTS Problem 1: Expillin the logarithmic and antilogarithmic reilltions with suitable examples.
[A] Index Multiplication of equal terms: then multiplication will be x n , i.e.,
If the term x is multIplIed
Il
tImes,
x x x x x x x x ... n times = x" Here x is called the base and n is called index.
[8] Laws of Index m
and ~----tn-II
x"
3. xo= 1
4. (Xlll)" = Xlllll
5. (xyt =x"l
6. (xlyr = x"ly"
1 d x -n =1 7.xn =-an -n n X
8.
xlln
X
= n{;
X
[C] Logarithms Definition: If ab = c; then exponent 'b' is called the logarithm of number 'c' to the base 'a' and is written as log.. c = b, e.g., J~ = 81 ~ logarithm of 81 to the base 3 is 4, i.e., log3 81 = 4. Note: a b = c, is called the exponential form and loga c = b is called the logarithmic form, i.e., T3 = 0.125 (i) (Exponential form) (Logarithmic form) log2 0.125 =-3 1 (Logarithmic form) (ii) log64 8 ="2 (64)112 = 8
(Exponential form)
Laws of Logarithms [I] First Law (product law) : The logarithm of a product is equal to the slim of logarithms of its factors. logo (m X 11) = log" III + log" /I
2
PHYSICAL CHEMISTRY-I
loga (m X n xp) = loga m + logan + lo~p Remember: 10& (m
+ n);/:. logam + logan
[II] Second Law (Quotient law) : The logarithm of a fractl~n is equal to the difference between the logarithm of numerator and (he logarithm of denominator.
Remember: loga m
-1-- ;/:. loga m - loga n oga Il
[III] Third Law (Power law) : The logarithm of a power of a number is equal to the logarithm of the number multiplied by the power. loga (mt = n loga m
Corollary: Since .,
Jlr-
-vm = m
l/n
1Jrl/n 1 lo~ -vm =lo&m =-Io~m
n Note: (a) Logarithms to the base 10 are known as common logarithms. (b) If no base is given, the base is always taken as 10. (c) Logarithm of a number to the same base is always one, i.e., loga a = 1; 10glO 10 = 1 and so on. (d) The logarithm of 1 to any base is zero, i.e., loga 1 = 0; logs 1 = 0; 10glO 1 =0 and so on. (e) 10gJO 1 = 0; 10gJO 10 = 1; 10gIO 100 = 2 [.: 10gIO 100 = 10gJO
102 = 2 log 10 10 = 2 xl = 2]
Similarly, 10g]O 1000 = 3; 10gIO 10000 = 4 and so on.
Example: If log 2 = 0.3010 and log 3 = 0.4771; find the value of: (i) log 6 (ii) log 5 (iii) log -V24 Solution: (i)
log 6 = log (2 x 3) = log 2 + log 3 = 0.3010 + 0.4771 = 0.7781
(ii)
10 log 5 = log 2" = log 10 -log 2 = 1 - 0.3010 = 0.6990 (': log 10= 1)
MATHEMATICAL CONCEPTS AND COMPUTE:,:R-'--_ _ _ _ _ _ _ _ _---=::3
(iii) log ..J24 = log (24)1/2 = ~ log (23 X 3)
="21 [3 log 2 + log 3] ="21 [3 X 0.3010 + 0.4771] =0.69005
[0]
Common Logarithms and Use of Four Figure Log Tables
[I] Common Logarithms : Logarithms to the base 10 are known as common Logarithms. If no base is given, the base is always taken as 10. [II] Characteristics and Mantissa: The logarithm of anum ber consists of two parts : (i) Characteristic-It is the integral part of the logarithm. (ii) Mantissa-It is the fractional or decimal part of the logarithm. For exampLe, in log 273 = 2.4362, the integral part is 2 and the decimal part is .4362. Therefore, characteristic = 2 and mantissa = .4362. [III] How to Find Characteristic? (i) The characteristic of the logarithm ofa number greater than olle is positive and is numerically one less than the number of digits before the decimal point. In number 475.8~ the number of digits before the decimal point is three. .. Characteristic of log 475.8 = 2, i.e., (3 - 1 = 2) Similarly, Characteristic oflog 4758 = 3, i.e., (4 - 1 = 3) Characteristic of log 47.58 = 1, i.e., (2 - 1 = 1) Characteristic of log 4.758 =0, i.e., (l - 1 =0) (il) The characteristic of the logarithm of a number less than one in negative and is numerically one more than the number of zeros immediately after decimal point. The number 0.004758 is less t;lan one and the number of zeros immediately after decimal point in it are two. ~ :. Characteristic of log 0.004758 = - (2 + 1) = -3, which is also written as 3. Note: To find the characteristic of the logarithm of a number less than one, count the number of zeros immediately after the decimal point and add one to it. The number so obtained with negati~ sign gives the characteristic. .. Characteristic of log 0.3257 = - 1 = 1 [Since, the number of ~eros after decimal point = 0] Characteristic of log 0.03257 = - 2 = 2 [Since, the number of zeros aftet; decimal point = 1] Characteristic of log 0.0003257 = -4 = 4 and so on.
4
PHYSICAL CHEMISTRY-I
[IV] How to Find Mantissa? The mantissa of the logarithm of a number can be obtained from the logarithmic table. A logarithmic table consists of three parts: (1) A column at the extreme left contains two digit numbers starting from 10 to 99. (2) Ten columns headed by the digits O. 1,2, 3,4,5,6, 7, 8,9. (3) Nine more columns headed by digits 1,2,3,4,5,6.7,8,9. A part of the logarithmic table is given below: (Difference to be added) 0
1
2
3
4
5
6
7
8
9
1 23 456
789
30 4771 4786 4800 4814 4829 4843 4857 4871 4886 4900 134 678 JO 11 13 31 4914 4928 4942 4955 4969 4983 4997 5011 5024 5038 1 34 678 JO 11 12 32 5051 5065 5079 5092 5105 5119 5132 5145 5159 5172 134 678 9 11 12
(a) To find the mantissa of the logarithm of one digit number. Let the number be 3. :. Mantissa of log 3 =Value of the number 30 under O. = 0.477l. (b) To find the mantissa of the logarithm of two digit number. Let the number be 32. :. Mantissa of log 32 = Value of the number 32 under O. = 0.5051. (c) To find the mantissa of the logarithm of three digit number. Let the number be 325. :. Mantissa of log 325 = Value of 32 under 5 = 0.5119 (d) To find the mantissa of the logarithm of a four digit number. Let the number be 3257. Mantissa of log 3257 = Value of 32 under 5 plus the difference under 7 =0.5128 [5119+9=5128]
1. How to find the logarithm of a number from the logarithm table? First. find the characteristic and then mantissa. Suppose the number is 3257. Characteristic of log 3257 =3 Mantissa of log 3257 = 0.5128 Therefore,
log 3257 =3.5128
[Note: The mantissa of the logarithms of all the numbers having the same significant digits is the same. While tinding the mantissa, ignore the decimal point.]
5
MATHEMATICAL CONCEPTS AND COMPUTER
For example:
log 3257 = 3.5128 log 325.7 = 2.5128 log 3.257 = 0.5128 log 0.3257 = 1.5128 log 0.003257 = 3.5128. and so on]
[Note ..:. 3.4682 is equivalent to (-3 + 0.4682) and -3.4682 = -(3 + 0.4682) =-3 - 0.4682 i.e., in 3.4682, the mantissa is positive, while in -3.4682, the mantissa is negative. (2) Remember, the mantissa should always be written positive. (3) To make the mantissa positive, subtract 1 from the integral part and add 1 to the decimal paJ;t, Thus, -3.4682 =-3 - 0.4682 =(-3 - 1) + (l - 04682) (1)
=- 4 + 0.5318 =4.5318] Problem ~ : Find out the values of the following: (i) ~.8321 + 1.4307 (ii) !.9256 - 4.5044 (iii) 1.7544 x 2 (iv) 2.3206 + 3 Solution: (i)
3.8321 1.4307
Since, after adding decimal parts, we have 1 to carry.
1.2628
:.
-
-
1 + 3 + 1 = 1 - 3 + 1 =-1 == 1.
(ii) 1.9256
-4.5044 3.4212
Since,
-
1 - 4 =-3 =3
(iii) 1.7544
x2
-
1.5088 Since,
--
lX2+1=2+1=1
(iv) 2.3206 + 3 2.3206 3 3 + 1.3206 3
To make the integral part 2 divisible by3.
3 + 1.3206 3
Take
-
-
= 1 + 0.4402 = 1.4402
2= 3+1
6
PHYSICAL CHEMISTRY-I
[E]
Antilogarithms If log 5274 = 3.7221, then 5274 is called the antilogarithm of 3.722l. We write, antilog 3.7221 = 5274. To find an antilogarithm, the antilogarithm tables are used. The antilogarithm tables are used in the same way as the logarithm
tables. The only difference between the two tables is that the column at the
extreme left of the log table contains all two digit numbers starting from 10 to 99; whereas an antilog table contains numbers from .00 to .99 (i.e., all fractional numbers with only two digits after decimal) in the extreme left column of it. x
0
1
2
3
4
5
6
7
8
9
123 456 789
0.35 2239 2244 2249 2254 2259 2265 2270 2275 2280 2286 1 1 2 233 445 0.36 2291 2296 2301 2307 2312 2317 2323 2328 2333 2339 1 12 233 445 037 2344 2350 2355 2360 2366 2371 2377 2382 2388 2393 I 12 234 455 0.38 2399 2404 2410 2415 2421 2427 2432 2438 2443 2449 1 1 2 234 455
[Note : (i) Antilog tables are used only to find the antilogarithm of decimal part. (ii) To find the antilog of 2.368 means to find the number whose log is 2.368].
Ex. 1 : If log x = 2.368,jind x. Solution : log x = 2.368 x = antilog 2.368 Antilog 2.368 = the number characteristic of whose log is 2 and mantissa is 368. From antilog table, the value of .36 under 8 is 2333. Since the characteristic of log of the number is 2. :. The number has 2 + 1 =3 digits in its integral part (i.e., 3 digits before the decimal point). Antilog 2.368 = 233.3. Ex. 2 : Find the antilog of 2.3536 Solution: From the antilog table, find the value of .35 under 3 and add to it the mean difference under 6. The number thus obtained is 2257. Now place the decimal point so that the characteristic of its log is 2. .. Antilog 2.3536 = 0.02257 [Note: (i) Antilog 0.5362 = 3.438 (ii) Antilog 2.5362 = 343.8 (iii) Antilog 4.5362 = 34380
Antilog 1.5362 = 0.3438 Antilog 1.5362 = 0.003438 Antilog 5.5362 = 0.00003438 and so on].
MATHEMATICAL CONCEPTS AND COMPUTER
7
Problem 3: Explain differentiation with suitable examples.
[I] Differentiation If y = f(x) , x is the function of x, then there will be a change inf(x) with respect to every change in x. If the increment in x is denoted by cSx , increment inf(x) will be f(x + ax) - f(x) . The ratio of increment in function f(x) and increment in variable x, i.e., f(x + cSx) - f(x)
Ox is called the difference quotient. But, if cSx -+ 0, then this ratio tends to a definite quantity. This definite quantity is called the differential coefficient of f(x) with respect to x and it is denoted by
d1;)
or
~
or f'(x) or y'
Therefore, differential coefficient of f(x) with respect to x,
!!l = !:lfJE = dx
dx
lim f(x x~o
+ cSx) - fix) cSx
The process of finding differential coefficient of f(x) with respect to x is called 'differentiation' If Ox = It, then
!!l =
lim f(x + It) - f(x) It
x~o
dx
[II] Some Standard Derivatives (1)
!
(xn) = nxn -
1,
where n is a real number
For example, differential coefficient of ;?= 5x5 - 1 = 5x4 (2)
.!£(..Jx) = _1_
2Vx
dx
(3) _d (eX) = eX
dx d mx mx (4) .dx (e ) = me For example, differential coefficient of e3x = 3e 3x
d
<
X
(5 ) dx (a') = a loge a For example, differential coefficient of 5x is
.!£ (5") =5x log 5 dx e
8
PHYSICAL CHEMISTRY-I
d 1 (6) - (log x) =dx e x For example, differential coefficient of log (x + 1) d [log (x
+ 1)]
_1_. 1 = _1_ x+1 x+l
dx
d dx
1
(7) --- (log x) = - log e
For d
-
d'(
(log x) 2
a
a
X
example, 1
differential
coefficient
of
log2 x,
i.e.,
=-log~ e X
-
[III] Differentials of Trigonometric Functions
(3)
(2)
d ~ dx (tan x) = sec- x
(4) dx (cot x) = cosec x
2
d
(6) dx (cosec x) = - cosec x cotx
x
~ (sin mx) = m cos mx d
sinx
d
d (5) -d (sec x) = sec x tan x
(7)
~ (cos x) = -
d (SIn . x) = cos x
(l) dx
(8)
~ (cos I1lx) = -
m
sin mx
2
(9) -d (tan IIlx) = 111 sec mx
x
[IV] Differentials of Inverse Functions (1) dd (sin-I x) =
x
bI-x
(2) -d (cos-I) x = - -1 -
{l=7
dx
(3) dd (tan-I x) = _1_ry
x 1+x1 d (-I (5) -d sec x) = _~ x x'lx- - 1
(4 ) -d (cot-I) x = - - -1- , dx 1 +x~ d ( _) 1 (6) -d cosec x =- _~ x 'IX- - 1
x
[V] Differentiation of a Product The differential coefficient of product of two functions is the sum of the product of one function and the differential coefficient of the other. d du du -- (ll . u) dx
=u . - + u . dx
dx
For example, to find the differential coefficient of xe-' with respect to x, we have d t d r x d dx (xe-) =x . d; (e') + e . d-; (x) =x.e-'+ex.l l
=e (x+1)
9
MATHEMATICAL CONCEPTS AND COMPUTER
[VI] Differentiation of a Division If u and v are any two functions of x, then
d
d
~(~)== v .~(u)-u.~(v) dx v
(v/
For example, to find the differential coefficient of xm Iloge x with respect to x, we have
!!.- (~) == loge (x) . d;d(m X ) dx loge x
m d X
•
d; (loge x
)
(loge x)2 loge x (mxm -I) -
==
(loge x)2 m-I
==
x'" . (lIx)
m. 1ogeX. X
m-I
-x
(loge xi m
x -
I
•
(m loge x - 1) ADS.
(loge x)2
[VII] Partial Differentiation We know that the differential coefficient off (x) with respect to x is . f(x + ox) - f(x) 11m Iix~O x provided that limit exceeds and is expressed as,
a
/' (x) or
'
~ [f(x)]
If u == fix, y) be a continuous function of two independent variables x and y, then the differential cofficient w.r.t. x (taking y as constant) is called the partial derivative or partial differential coefficient of u w.r.t. x and is represented by different symbols such as
au ax ' E1 ax ,Ix (x, y) ./x· Symbolically, if u == f(x, y) then · f(x + Ox, y) - f(x, y) 11m Iix ..... O Ox If it exists then. it is called the partial derivative or partial differential coefficient of u w.r.t. x. It is denoted by,
au ax
or
af ax
or
Ix or
Ux
10
PHYSICAL CHEMISTRY-I
Similarly, by keeping x constant and allowing y alone to vary, we can define the partial derivative or partial differential coefficient of u w.r.t. y. It is represented by anyone of the following symbols.
au af oy' Oy,/y (x, y),/y ou = lim f(x, y + By) - fix, y) oy dy-+O By
Symbolically,
provided this limit exists.
then
u = ~ + 2hx2y + bi ou 2· ox = 3ax + 4hxy
and
-=2hx +2by
For example, if
ou oy
2
[I] Rules of Partial Differentiation Rule (1) : (a) If u is a function of x, y and we are to differentiate partially w.r.t. x, then y is treated as constant. (b) Similarly, if we are to differentiate u partially w.r.t. y, then x is treated as constant. (c) If u is a function of x, y, z, and we are to differentiate u w.r.t. x, then y and z are treated as constants. Rule (2) : If z = u ± v, where u and v are functions of x and y, then
oz = ou + ov
ax ax - ax
and
oz = ou + ov
ay ay - ay
Rule (3) : If z = uv, where u and v are functions of x and y, then
oz 0 ov ou -=-(uv)=u-+vox ox ox ox oz 0 ov ou -=-(uv)=u-+vand oy oy oy oy Rule (4) : If z =~, where u and v are functions of x and y, then v oz ox and
=~ (~ ) = v ( ~ ) ox
v2
V
ou.)
u(
f)
(OV)
~; =;y (~)= v ay ~ u ay (
Rule (5) : If z = f(u) , where u is a function of x and y, then
OZ = oz OU and OZ = oz ou ox ou· ox oy ou· oy
11
MATHEMATICAL CONCEPTS AND COMPUTER
Problem 4. Determine thejirst order partial derivatives ofyx. Solution. Let z = yx Taking logs.
log z =x log y
... (i)
az
. . w.r.t. x. -1 -;- = Iog y O I·ftierenttatmg Z aX
az = z Iog y= yx Iog y ax Now differentiate equation (i), w.r.t y, we get
;~~=x(~)=~ az (X) ay=Z y =yx(X) y =xy
x-I
or
Problem S. Find the differential coefficients ofthejollowing: (i) x S (U)
M
(iii) X-SIl
Solution, (i)
!!:.-. (x5 ) = 5~ - I = 5x4
(ii)
!!:.-. U =.!!:.- (x)3/2 =1 x~-l = 1 xl/2-
'1'1') (1
dx
dx d dx
2
dx
(-5/2) X
-5
=T X
-7/2 - I 2
2
-5
=T x
-7/2
Problem 6. Find the differential coefficients of the following: (i) ax4 (ii) 7 loglo x Solution. (i) (ii)
:.x (ax = a :.x (x4) = 4ax3 :.x (7 loglO x) = 7 :.x (log lO x) = 7 4
)
~ loglo (e)
Problem 7. Find the differential coefficients oj the following: 2 (i) 4x + ~ + 10 (ii) tan x 1- sin x 2 (iii) x sin x Solution . . d 2 d 2 d d (I) dx(4x +6x+1O)= dx(4x)+ dx"(6x)
+ dx(1O)=8x+6
..) dx d (tan x sm .) d (tan x ) + dx d (-sm .) (II x = dx x = sec2 x + cos x d (. d (2). ...) dx d (2. (III x sm x ) -::: X .z dx sm x ) + di x . sm x
= "'=' ~ SIn-X
12
PHYSICAL CHEMISTRY-I
Problem 8. Find the differential coefficients of the following: (i)
sin x
x
(ii)
(x+lf
xl
Solution. 2d(.
(i)
)
.
d(2)
~ (Sin x) = x ;t;; sm x - sm x.;t;; x (X 2)2
x2
dx
= =
i
cos x - sin x . 2.x X4
X
cos X
2 sin x
-
x
2
d d 2 (x+l) ;t;;(x)-x.;t;;(x+l) 2
••) - d [ X (II dx (x + 1)2
(x+ 1)4 _ (x+ 1)2. I-x. 2. (x+ 1) (x+l)4 _ (x+ 1) [x+ 1-2.x] _ (I-x) (x+l)4 -(x+l)3
Problem 9. Differentiate the following with respect to x : (i) sin 5x (ii) log sin x (iii) (3x 2 + Il Solution. (i) Let y = sin 5x => y = sin t, where t =5x
(Meerut 2007)
~=cost
..
dt =5 dx dv dv
and
dt d; = di .dx = cos t . 5 = 5 cos 5x. (ii) Let y = log sin x => y = log t, where t = sin x
4r_l
dt - t dt dx =cosx
and
4r = 4r. dx
4r =_._1_ . cos x = cot x
or
(iii)
dt = 1 . cos x dtdx t
1
(3x
dx
2
smx
2
+ 1)2 = 2 (3x + 1)2-1 . (6x) = 2 (3x2 + 1) + (6x) =12x (3x2 + 1)
13
MATHEMATICAL CONCEPTS AND COMPUTER
Problem 10. Calculate the value of ~ from the following: (i) x=acost,y=asint Solution. (i) x = a cos t, y = a sin t dx . dv - = - a sm t ::::.L = a cos t dt ' dt cos t !:!l. =d dt dt = a cos t y . dx - a sin t = - si~ dx
!:!l. = -
or
dx
1
cott
Problem ll.lfy.= tan- [1
~x2}
then calculate the value of
Solution. Let x = tan 8, then .• Y = tan -I [ 2 tan 82 I-tan 8 ..
Problem 12. If y
J= tan-
1 ( tan
~.
28) = 28 = 2 tan - 1 x
dy _ '2 dx - 1 +x2" =:
d . tan-1 (Sin x + C?S x) then calculate the value of f!:1...dx cos x - SID X
SoIu tion. Let y = tan -I (sin x + cos . x~
cosx- smx
:
Dividing ilie :::::~(f;;::n~)~ ~; r:~(: ~Jl~: + ~ ...
!:!l. dx --
1.
Problem 13. Explain integration and give important integral formulae.
[I] Integration The reverse process of differentiation is called the integration and it is represented by a symbol
f.
If differential coefficient of f(x) is
(x), i.e., d - [((x)] = dx
f
then
Therefore,
f
dx =f(x)
(x) dx
14
PHYSICAL CHEMISTRY-I
is called the integration of c1>(x) with respect to x.
[II] Tables 01 Integral Formulae (1) J al(x) dx = a JI(X) dx + C
± c1>(x)} dx = JI(X) dx ± J c1>(x) dx
(2) J If(x)
(3) Jl. dx=x, JOdx=C
n :n;:, (n;t - 1)
(4) Jx dx:%
For example, integration of x5 is given by, 5+1 1 6 5 x dx = ; + 1 =6" x (6) J
f ~dx=IOgeX
(7) f C dx=e
x
(8)faXdx=~ loge a (9) ISin x dx = - cos x + C
(10) Jcos x dx = sinx + C (11)
f
tan x dx = loge sec x + C = - 10& cos u + C
(12) fcotx dx= 10& sin x + C = -loge cosec x + C
(13) fcosec x dx = loge (cosec x - cot u) + C = loge tan 2
(14) f sin xdx= 2
(15) J cos x dx = 2
t t +t
x-t sin x cos x + C x
sin x cos x + C
(16) Jsec x dx = tan x + C
~+C
MATHEMATICAL CONCEPTS AND COMPUTER
(17)
15
f
cosec 2 x dx = - cot x + C 2
(18) ftan x dx = tan x - x + C 2
(19) fcot xdx=-cotx-x+C (20) f sec x tan x dx = sec x
(21) Icosec xcotx dx= - cosec x (22) (23)
f ~1 ~X2
l
l
dx=sin- x=-cos- x
f~ dx = tanl+x
l
l
x = - coe x
(24) ff(y) dx = ff(y) dy :
(25)
f~=.!tan-l~+ C 2 2
(26)
J
x +a
a
dx x- -a
a
1
~=-2
a
x-a 2 2 loge--+C,x >a x+a
1 a-x 2 2 =-2 log--+C,x
(27)
Ih= a -x
sin-
1
~a + C, a > 0
[III] Definite Integral When any functionf(x) is integrated between the lower limit and upper
limit of x, then it is called the definite integral. For f(x), if the lower and upper limits of x are a and b, respectively, then
f
b
f(x) dx =
[F(x)]~ =F(b) - F(a)
a
where
ff(X) dx = F(x).
For example,J(x) =x 5 and a = 2, b =3, then
16
PHYSICAL CHEMISTRY-I
1 [(3)6 -
(2)6] = 729 - 64 = 665 6 Problem 14. Evaluate the following integrals: =
(i)
f
X7
dx
S~lution. (ii)
f
f
(li)
(i)
f
eX dx
7+ 1
7
x dx
8
=; + 1 =~
eX dx = eX
Problem 15. Find the value of the following integrals: (i)
f
3
(x + 2x + 7) dx(ii)
Solution. (i)
f
f
(x
3
2)2 dx
-
f f f 3
3
(x + 2x + 7) dx =
x dx +
4
(ii)
f
2x dx +
2
4
7 dx
=£+~+7x=£+i+7x 4
(x
3
-
2)2 dx =
=
2
f f f f
4
3
6
(x + 4 - 4x ) dx
6
x dx +
x7
4 dx -
3
4x dx
4x3+ 1
=-+4x--7 3+1 x7 4X4 =-+4x--
7
4
7
= ~+4x-x4 7
f f
Problem 16. Evaluate the integral · (.) So Iubon. I
f~-!1-
-
. SIO X
1
(1
-
d~
-smx
.
(l. + sin x) dx . X )sm x). (1 + Sin
17
MATHEMATICAL CONCEPTS AND COMPUTER
f(
=
1 + sin x) dx 1 - sin 2 x
=f(l+Si~X)dX cos x =
f[ cos1 x
cos x
f
f
sin x ]
--2-+~-
=
sec
2
x dx +
dx
tan x . !>ec x dx
=tanx+secx
Problem 17. Find the value of the integral
Solution.
f+-dx=fx4~ 1 + 1 dx=fx:-l dx+f-,_l_dx x- + 1
x- + I
X
3
=--x+tan 3
'Problem 18. Evaluate
f
x- + I
-1
x- + 1
x
sec x + tan x dx sec x - tan x
(Meerut 2006)
Solution. Multiplying numerator and denominator by (sec x + tan x).
f
sec x + tan x sec x - tan x
dx = f
-f -
(sec x + tan x) (sec x + tan x) (sec x - tan x) (sec x + tan x)
(sec x + tanx)2 sec 2 x - tan 2 x
dx
dx
18
PHYSICAL CHEMISTRY-I
f I I I I
(".. sec 2 x - tan 2 x = 1)
=
(sec x + tan X)2 dx
=
see 2 x + tan2 x + 2 sec x tan x dx
=
see 2 x + (see x-I) + 2 sec x tan x dx
2
=
(2 see 2 x-I + 2 sec x tan x) dx
=
2 see x dx -
=2
f
2
2
I I f I . dx +
see x dx - fl. dx + 2
2 sec x tan x dx
sec x tan x dx
=2tanx-x+2secx Problem 19. Evaluate the following integrals by substitution method: (i)
f
(ax + b)' dx (ii)
Solution. (i) In
I
f
cos (ax + b) dx
bf dx, put ax + b = t, so that
(ax +
adx=dt dx= dt a
or
I
7
(ax+b) dx=
I7 II7 dt
t -;=~
=~(~) 1 8 = 8a (ax + b) (ii) In
f
cos (ax + b) dx, put ax + b = t adx=dt
or
dx=dt a
t dt
19
MATHEMATICAL CONCEPTS AND COMPUTER
f cos (ax + b) dx = f cos t .
~ =~ f
cos t dt
=-a1 SIn.t 1 =-a .SIn(ax + b) Problem 20. Evaluate the /oUowing integrals: (i)
f~dx eX + 1
(iii)
f
(ii)
f
log x dx x
eX 2x dx
l+e
Solution. (i) In
f:L e'+1
dx, put eX + 1 = t
= log (eX + 1)
(ii) In
f 10!
e
x dx, put log x = t
ldx=dt x
f IO! =f x dx
t dt =
~
-~ 2
(iii) In
f4 1 +e
r
dx, put eX =t
t f _e_t-dx=f~=tan-I l+x2x 1+t2 =tan- I e'.
20
PHYSICAL CHEMISTRY-I
Problem 21. Evaluate the following integrals:
f_1_ dX
(ii)
(i)
X
log X
Solution. (i) Put x2 + 3x + 2 = t (2x + 3) dx =dt
2x+3 dx= fdt 2 -= log t f x + 3x + 2 t e = log e (x 2 + 3x + 2) (ii) Put log x = t
1 - dx =dt x
or
f
1 --dx= x log x
f
dt -=log t t e
= loge (log x)
Problem 22. Evaluate the following integrals: (i)
(iii)
f f
log X dx
(ii)
(iv)
x log x dx
Solution. (i)
f
f
x
xe dx
f xsinxdx
log x dx = fl. log x dx
= log x f dx - f {
~ (log x) f
= x log x -
f ~ .x dx
=xlogx-
f
dx
= x log x - x = x (log x-I) (ii)
f
x
xe dx = x
f f{~ f eX dx -
(x) .
eX dx } dx
dx } dx
21
MATHEMATICAL CONCEPTS AND COMPUTER
= xe' - fl. eX dx
= eX (x - 1) (iii)
f
x log x dx = log x
f f1fx x dx -
(log x) .
f
x dx } dx
2 x2 1x =--Iogx--2 2 2
1
2
=~
(logX-
2)
i
= 4 (2 log x-I) (iv)
f
x sin x dx = x
=x(-cosx)-
f
f
sin x dx -
f{fx f (x)
sin x dx } dx
1.(-cosx)dx=-xcosx+
f
cosxdx
= - x COSJ + sin x Problem 23. Evaluate the integral Solution. (i) Let 1=
f
eX sin x dx.
e" sin x dx
f f{fx f \. f
= sin x
eX dx -
= e' sm x -
(sin x)
cos x . e''dx
eX dx } dx
22
PHYSICAL CHEMISTRY-I
ff f
= eX sin x - [ eX cos x -
= eX sin x - [ eX cos x
I = eX sin x - eX cos x -
+
sin x . eX dx ]
sin x eot dx ]
(sin x) eX dx
I = eX sin x - eX cos x - I
or
21 = eX sin x - e'\ cos x
or
1=
t
eX (sin x- cos x)
Problem 24: Explain the terms permutation and combination with suitable examples.
[I] Permutation The number of different arrangements which can b~ made by Taking one til1le some or all elements of any set of things, thell such every arrangement is called a permutation. Example: If in the set {a, b, c}, we take two elements in one time, then the different arrangements which can be made are ab,ba,bc,cb,ca,ac Therefore, it is clear that the number of arrangements which can be made by taking any two things at one time out of three things will be six.
[II] Principle of Permutation If one work can be done in m ways and second work in n ways, then both the works can be done together in m X n ways. Important Formulae: The number of arrangements of taking r things out of II different things is "P r = n (11 - 1) (n - 2) ...... (11 - r + 1)
Factorial n : II! = n (II - 1) (n - 2) ... 3.2.1 Here n ! is read as factorial Il and this is also represented by Clearly, if r < 11 "Pr=n!
Also
tip =_II_!_ r
(II - r) !
1lL
23
MATHEMATICAL CONCEPTS AND COMPUTER
When aU things are not different: If out of n things, p things are of one type, q things are of second type and r things are of r type, then Number of arragements =
, n '; , p.q.r.
[III] Combination The number of combination of r things taken from a set of II dissimilar things is given by nc = r
n! =_np_r r! (n - r) ! r !
[IV] Complementary Combinations Important Results : (l)nCn =1 (3) nCr + nCr _ 1= n + IC r
n
(4) 2 =
Co + CI + ... + CII
Problem 25. Prove that 0 ! = 1. Solution. The number of permutations of n different objects taking all of them at a time, is given by, nPn = (n - l)(n - 2) ... (n - n
+ 1)
= n{n - 1) (n - 2) ... 1 = n! nPn = nl.
.... (l)
Also by the formula,
Pp = n
n! nl (n-n)!=O!'
... (2)
From equations (1) and (2), we get II! =
0;n'
or O! = 1.
Problem 26. Find the value of6P3 .
So Iuti· on.
6p
3
6!
6!
= (6 _ 3)1 = 3!
6. 5 .4 . 3! 6 5 4 120 3! =.. = .
Problem 27. Prove that nPr = n x n -lpr _ 1• . S So Iutlon. R.H .. - n x
n- I
n! (n - 1 - r
+ I)!
Pr -
_
n (n - 1) ! 1 _ (r - 1)] !
1 - [n _
,
II.
(II - r) !
="p =LHS r
...
Problem 28. Three persons enter into a car offive seats. In how many ways can they occupy their seats?
24
PHYSICAL CHEMISTRY-I
Solution. First person can sit in five ways as all the five seats are lying vacant at the time of his entrance. The second person can sit on anyone of the remaining four seats (one is already filled by first person). So. the second person can sit in 4 different ways. Similarly, thIrd person can SIt in 3 different ways on anyone of the three vacant seats (two already filled). Hence. all the three can sit in 5 X 4 x 3 = 60 different ways. Problem 29. (i) How many words can be formed with the letters of the word "DELHI"? (ii) How many of these will begin with D? (iii) How many of these will end at D? (iv) How many of these will begin with D or L? (v) How many of these will begin with D and end at L? (vi) How many of these will begin with D or end at L? (vii) How mallY of the vowels "E, I" occupy the even number of places? (viii) How many of these will end at vowel only? (ix) III how many of these vowels come together? (x) How many of these will begin and end at vowel? (xi) How many of these will begin and end with D or L? Solution. (i) There are five letters "D, E, L, H, I" in the word DELHI. These five letters can be rcalTanged among themselves in 5 ! ::; 120 ways, (or the say 5P s• i.e., the number of permutations of these five letters taking five at a time). Hence, we can form 120 different words. (ii) All the arrangements in which D is 'in the beginning can be obtained by fixing 'D' at the first place and then rearranging the remaining four letters. Remaining four letters can be arranged in 4 ! ways. Hence, in 4 ! = 24 arrangements the words will begin with D. Alternatively. First place can be filled by Din 1 way only, the second place in 4 ways as any of the letters L. H, E, I can be put there; 3rd place can be filled in 3 ways, 4th place in 2 ways and 5th place in one way only. Hence, all the five can be filled in 1 x 4 x 3 x 2 x 1 = 4 ! = 24 ways. (iii) The problem is similar to problem (ii) except that now we fix the letter D on the last place. Hence, the remaining four letters can be arranged in 4 ! ways. Hence, the required numb~r of words, ending at D, is 4 ! = 24. (iv) As above there will be 4 ! arrangements starting with D, and also 4 ! an'angements will begin with L. Hence, the total number of arrangements beginning with D or L are 4 ! + 4 ! = 48. AlternaHvely. First place can be filled in two ways as any of the letters D or L can be put there. Then second place can be filled in 4 ways as anyone of the remaining fOur letters can be put there. The third place can be filled in three ways. fourth place in two ways and fifth place can be filled in one way only. Hence, all the five places can be filled in 2 x 4 x 3 x 2 x 1 = 48 ways.
all
MATHEMATICAL CONCEPTS AND COMPUTER
25
(v) In this we fix D at the beginning and L at the end. The remaining three letters E, H, I can be rearranged in 3 ! ways. Hence, 3 ! = 6 arrangements will begin with D and end at L. (vi) The arrangements which begin with Dare 4 ! and which end at L are also 4 !. Hence total number of arrangements which begin with D or end at L are 4 ! + 4 ! = 48. (vii) There are only two even places namely 2nd. 4th. E and I can be rearranged on these places in 2! ways. Further. remaining three letters D, L, H can be put on 1st, 3rd and 5th places in 3 ! ways. Hence. the total number of arrangements in which E, I occur at even places only, is 2 ! x 3! = 12. (viii) The arrangements ending at E or I give the arrangements ending at vowels. Such arrangements are 4 ! + 4 ! = 48. Proceed as in (iv). (ix) We consider the two vowels forming as one letter say (E I). Thus there are four letters D, L, H, (E I). These can be rearranged in 4 ! ways. Further, two vowels E, I can be rearranged among themselves in 2 ! ways. Hence, total number of arrangements in which vowels come together is 2 ! x4! =48. (x) We want to put E or I in the beginning and at the end. For the 3 ! arrangements will begin with E and end at I. Again 3 ! arrangements will begin with I and end at E. Hence, the number of required ways are 3 ! + 3 ! = 12. (xi) As above the required number of ways is 12 (only letters E and I are replac~d by D and L).
Problem 30. How mallY words call be formed from the letters of the word 'DAUGHTER' so that the vowels always come together? (Meerut 2006) Solution: We consider three vowels forming as one letter say (AUE). So, there are six letters DGHTR(AUE). These can be arranged in 6! ways. Further three vowels, A, U. E. can be rearanged among thermselves in 3! ways. So, total number of arrangements in which vowels come together is 6! x 31 = 4320 ways.
Problem 31. How mallY words call beformed with the letters of the word "MEERUT"? In how many of these words vowels occupy only even places? Solution: (i) There are 6 letters in the word "MEERUT", and the letter E is repeated 2 times in this word. Therefore, the total number of words that can be formed with the letters of the word "MEERUT" _ 6! _ 6 x 5 x 4 x 3 x (2!) - 2! 2!
=6 x 5 x 4 x 3 =360 (ii) There are three even places in the word "MEERUT", namely, second. fourth and sixth places, respectively. At these places the vowels E, E and U are to be arranged. So, the number of ways to arrange E, E and U at these three places
26
PHYSICAL CHEMISTRY-I
= 3! =3 2! Now at the odd places, the letters M, R, T are to be arranged; so the number of ways to arrange M, R, T at three odd places =3! =6 Thus, the total number of words in which the vowels occupy even places
=3 x6= 18. Problem 32. How many permutations can be made out of the letters of the word "BUSINESS"? How many ofthese will begin with B and end with
N? Solution: (i) There are 8 letters in the word "BUSINESS", and the letter S is repeated 3 times and others are different. Then the total number of words formed by the letters of the word "BUSINESS" are
8! - 3!
=8x7x6x5x4=6720 (ii) If each word begins with letter B and ends with N, then except these two letters, the remaining letters are 6 and out of 6 the letter S is repeated 3 times. Now in this case, the total words formed by the letters of word "BUSINESS" which begin with B and end with N are
6'
=~=6
3!
x 5 x 4 = 120
Problem 33. Prove that nCr = nCn _ r Solution. R.H.S.
= nCn _ r = (
(Il -
) , [n ! (
n-r . nn!
r) ! (II -
II
Il-f
-)] , .
+ r) !
n! (n - r) ! r! =
nCr = L.H.S.
Problem 34. Find the value of the following 24 24! Solution. C4 = 4 ! (24 - 4) !
24 C4
.
[
24 ! 4! 20! _ 24 x 23 x 22 x 21 x (20!) 4 x 3 x 2 x I x (20!) 24 x 23 x 22 x 21
4x3x2xl
.,'
nC _ r-
,
n.
r! (n - r) !
]
27
MATHEMATICAL CONCEPTS AND COMPUTER
= 6 x 23 x
11 x 7
= 10626 Problem 35. Find the value ofr if20Cr _ 1 = 20Cr + l' Solution. Since. we know that if nCx = nCy. then either x = y or x + y = n
2OCr _ I = 20Cr + I and r - 1 "* r + 1
Here.
(":x+y=n) r+ 1 + r- 1 = 20 2r=20 => => r=lO Problem 36. In how many ways can a cricket eleven be chosen out of 15 players? How many of them will always (i) include a particular player? (ii) exclude a particular player? So.
Solution. Number of ways of selecting cricket eleven = Number of ways of selecting 11 players out of 15 = lSC 11
= ~ = 15 . 14 . 13 . 12 . 11! = 1365
11! 4!
11! 4. 3. 1
.
(i) Now since a player is to be included always. we are to select remaining 10 players out of the rest 14 players. This can be done in 14CIO = 1001 ways. (ii) Again since a player is never to be included. i.e .• always excluded. we are selecting 11 players out of 14 only. This can be done in IO ClI = 364 ways.
Problem 37. How many triangles can be made by joining 12 points in a plane, given that 7 are in one line? Solution. The triangles can be formed by joining any three points. But 7 points are in one line. Hence. with three points out of these 7 points in one line. we cannot form a triangle. Hence. the required number of triangles is 12C3 - 7 C 3 = 185. Alternative. Triangles can be formed in the following ways : (i) Three points are taken from the five non-collinear points. Number of ways is 5C3 = 10. (ii) One point is taken from 5 non-collinear and 2 out of 7 collinear points. This can be done in SCI x 7C2 ways = 105 ways. (iii) Two points are taken from 5 non-collinear and 1 from 7 collinear points. This can be done in sC2 x 7C1 ways = 70 ways. Therefore. the required number of triangles = 10 + 105 + 70 = 185. (Note that possibilities (i). (ii) are (iii) are mutually exclusive.) Problem 38. Find the number of diagonals that can be drawn by joining angular points. of a sixteen sided figure? Solution. In a sixteen sided figure there will be 16 angular points (vertices).
28
PHYSICAL CHEMISTRY-I
Total number of lines which can be drawn by joining any two angular points = 16 C2 = 120. But out of these lines, 16 will be sides. Hence. the required number of diagonals = 120 - 16 = 104. Problem 39. In how many ways can 12 things be divided equally among 4 persons? Solution: 12 things are divided equally among 4 persons, so each person gets three things. Therefore. the total number of ways 12! 3! 3! 3! 3! 12xUxlOx9x8x7x6x5x4x3x2xl = 6x 6 x 6 x 6
=
47900 1600 = 369600 6x6x6x6
Problem 40. There are six points on the circumference of a circle. How many straight lines can be drawn through these points? Solution: Here no three tJoints lie on a line and through any two points, a straight line can be drawn. Therefore. the total number of lines that can be drawn from 6 points =6 C2
=~= 6x5 2! 4!
2 xl
= 15
Problem 41. Out of 6 teachers and 4 students, a committee of 5 is to be formed. How many such committees can be formed including (i) at least one student (ii) 3 teachers and 2 students? Solution: (i) In this case we have to select at least one student; this means that from one to all students are to be selected. So. there can be the following formations, of a committee of 5 persons : (a) 1 student and 4 teachers (b) 2 students and 3 teachers (c) 3 students and 2 teachers (d) 4 students and 1 teacher Tota I ways
.
III
(4C case a) = I
X
6
C4= I!4!3! x 4!6!2!
= 4 x 15 = 60
Total ways in case (b) = 4C2 x 6C3 4! 6! ::: 2! 2! x 3! 3! = 6 x 20 = 120 Total ways in case (c) = 4C3 X 6 C2 4!
=
6!
3!1! x 2! 4! = 60
~M~AT~H~E~M=A~T~IC=A~L~C~O~N~C~E=P~T~S~A~N~D~C~O~M~P~U~T=E~R___________________
29
Total ways in case (d) = 4C4 x 6C 1 =lx6=6 Hence, the total ways = 60 + 120 + 60 + 6 = 246. (ii) In this case the committee has 3 teachers and 2 students. So the total ways to form this committee
=6C3 x 4C2 6!
4!
=--x-3! 3! 2! 2! =20x6= 120.
Problem 42 : What is probability? Give its definition also. In our daily life, we generally come across with the following statements: (i) Most probably Amit will stand first in his class. (ii) It is quite probable that Amit may stand first in his class. (iii) It is least expected that Amit may stand first in his class. (iv) It is impossible that Amit will stand first in his class. In all these above statements we have tried to express the chances of Amit for standing first in his class qualitatively. This is an event which may and may not happen. But we are predicting the result of the event with some uncertainty. This uncertainty associated with the event may be lesser or greater, i.e., it may vary. In mathematics we measure this uncertainty in terms of number quantitatively which we call probability or chance. With the help of probability we can predict the outcome of any random experiment by associating some probability to that outcome. Probability: If an event E can happen in m ways and fails (cannot happen) in n ways, all the ways are equally likely to occur, then the probability of happening of the event E, denoted by pee), is given by P(E)=~ ... (1) m+n Note that 0 :5 peE) :5 1. If we denote the event of "not happening of event E" by symbol E or by E', then according to the above definition n pee) = pcE') = ... (2) m+n From equations (1) and (2), we get
or and
peE) + peE) = 1 peE) + peE) = 1 - peE) peE) = 1 - peE)
In equation (1) note that (m + II) are the total number of ways (outcomes) in which a trial or an experiment may end. Out of these (m + n) ways in m
30
PHYSICAL CHEMISTRY-I
ways, event E happens or say m ways are favourable to the event E. Therefore, the probability of event E is also given by P(E) = Favourable number of ways to event E Total number of ways of the experiment Problem 43 : Find the chance of throwing more than 4 in one throw of cubic dice marked 1 to 6 its six faces, Solution: Here are 6 equalJy likely cases of which only 2 are favourable because we want 5 or 6 on the upper face of the cubical dice. Hence, the required probability of throwing more than 4 in one throw with one dice 2 1
=6=3"
Problem 44. In a single throw with two dices, what is the probability of throwing 9? Solution. The number of the first dice may appear in 6 ways. Similarly, on the second dice also the number may appear in 6 ways. Hence, the two dices may appear in 6 x 6 ways namely, (1, 1), (1,2), (1,3), (1,4), (1,5), (1,6), (2, 1), (2,2), (2, 3), (2, 4), (2,5), (2,6), (3, 1), (3,2), (3, 3), (3, 4), -(3,5), (3,6), (4, 1), (4,2), (4,3), (4,4), (4,5), (4,6), (5, 1), (5, 2), (5, 3), (5, 4), (5,5), (5,6), (6, 1), (6, 2), (6, 3), (6, 4), (6,5), (6,6). Out of these 36 ways, those which give desired sum of 9 are (3, 6), (4, 5), (5, 4) and (6, 3), i.e., only 4 favourable ways. :. The probability of throwing 9 = 4/36 = 119. Problem 45. A bag contains 5 white, 8 black and 3 red balls. If three balls are drawn at random from the bag, then find the probability of the event, (i) that all the balls may be white, (ii) that one ball may be black and the other two white. Solution. Total number of balls = 5 + 8 + 3 = 16. Total number of ways of drawing three balls from the 16 balls in the bag
C3 = ~ = 16. 15 . 14 = 560 3!13! 3.2.1 (i) Total number of ways of drawing three white balls out of 5 16
= 5 C3 = 10. . .. The probability of drawing 3 white balls together 10
1
= 560 = 56'
MATHEMATICAL CONCEPTS AND COMPUTER
31
(ij) Total number of ways of drawing two white balls out of 5 and one black ball out of 8 = 5C x 8C =80 . 2
J
So, number of ways of drawing three balls of which one is black and the other two white 80 =-=560 7" Problem 46. Three cards are drawn from a pack of 52 cards. Find the probability that: (i) aU the three will be kings, (ii) the cards are a king, a queen and ajack. Solution. (i) Total number of ways of drawing 3 cards from a pack of 52 cards C =~52X51X50=22100. 3 3! 49! 3x2x 1 Number of kings in the packet is 4, so the favourable number of ways of drawing three kings =52
4!
4
= C3 = 3! I!
= 4.
So, the required probability 4 1 = 22100 = 5525· (ii) The favourable ways are 4CJ x 4C J X 4C1 = 4 x 4 x 4 =.64 . ... Required probability 64 16 :::: 22100 = 5525· Problem 47. Two cards are drawn from a pack of 52 Find the probability that one may be queen and the other a king. Solution. Total number of ways of drawing 2 cards out of pack of 52. =~= 52 x51 x50! = 1326 2! 50! 2 x 1 x 50! Now, number of favourable ways of drawing 2 cards together, of which one is a queen and the other a king =52C 2
=4 C1 ..
X
4
C 1 =4 x4 = 16.
. d probab·l· 16 = 663· 8 I Ity = 1326 the reqUIre
Problem 48. Two cards are drawn from a full pack of 52 cards. What is the chance that (i) both are aces of different colours (li) one is red and other is black?
32
PHYSICAL CHEMISTRY-I
Solution. Total number of ways of drawing two cards from 52 cards =52 52! 52x51 x50! =26x51 C = 2 2!(52-2)! 2xlx50! . (i) Now there are 2 black and 2 red aces. So, two aces can be drawn in = 2 C , x 2C , ways = 2 x 2 ways = 4 ways.
Required probability
6 2 26 x 51 = 663" (ii) There are 26 red and 26 black cards. So, one red and one black card can be drawn in 26C, x 26 C , ways = 26 x 26 ways. => Required probability 26 x26 26 26 x 51 51"
Problem 49. A pair of dice is thrown. What is the probability of getting a totalofn Solution. Total number of possible outcomes = 6 x 6 = 36 Let E be the event of getting a totai of 7, then E = [(1, 6), (2, 5), (3, 4), (4, 3), (5, 2), (6, 1)] n{17'\ 6 1 Pea total of7) = ~ = -- =-. II (S) 36 6 Problem 50. Find the probability that a leap year selected at random will contain 53 Sundays. Solution. A leap year contains 366 days, i.e., 52 weeks and 2 days. The different possibilities for the remaining two days are: (1) Monday and Tuesday (2) Tuesday and Wednesday (3) Wednesday and Thursday (4) Thursday and Friday (5) Friday and Saturday (6) Saturday and Sunday (7) Sunday and Monday. So, we see that the last two cases are favourable to the happening of 53 Sundays, out of a total of seven equally likely cases. So, the required proba.l. 2 bllty = 7.
Problem 51: Explain with examples the following: (a) Logarithmic series (b) Trigonometric series
(a) Logarithmic Series
MATHEMATICAL CONCEPTS AND COMPUTER
33
(b) Trigonometric Series 3
5
' x x (1) smx=x-'3!+5"!......
00,
(-oo<x
where x is very small and measured in radians. 2. 4 X X x6 (2) cosx= 1 - -, +-4' - - , + ...... 00, (-oo<x
Problem 52: Define the terms maxima and minima and also mention the conditions lor finding them.
[I] Definition Let y =f(x) be some given function and x = a is some given point. Suppose on the left hand side of x = a, there exists some nearest point x = a -/z and on the right hand side of x = a there exists some point x = a + h, where h is very small. At x =a f(x) =f(a) At x =a - h f(x) =f(a - h) At x = a + h f(x) = f(a + h) Maximum: At x = a, the function is maximum, if 1(0 - h) I(a + h) Minimum: At x = a, the function is mimmum, if I(a - h) > I(a)
[II] Conditions for Finding Maximum and Minimum At any point x = a, the function f(x) has maximum or minimum value if the following two conditions are satisfied. (a) Necessary condition: For the existence of both maximum or minimum, the necessary condition is
=0 ~=O
f'ex) or
dx
34
PHYSICAL CHEMISTRY-I
(b) Sufficient condition: (a) At x
= a,
if
d2~ is negative, then we get maximum.
dx2
(b) At x = a, if d ~ is posItive, then we get minimum. dx(c) At x = a, if
d
2
~=
3
d °and ~ *- 0, then neither we get maximum nor dx
dx minimum. (c) To determine maximum or minimum with the help of differential coefficient : (i) For maximum: If for nearest small value of x = a ~
dx
.. = posItIve va Iue
and for greater value of x which is nearest to x
= a,
~ = negative value Then there exists maximum of y =f(x) at x =a (ii) For minimum : If for nearest small value of x
= a,
2:
== positive value
and for greater value of x which is nearest to x
=a
dv .. I d; = posItIve va ue
then there exists minimum of y
=f(x) at x = a.
(iii) For neither maximum nor minimum :
If for small and great nearest values of x = o'!:!x has same sign (either (X L both positive or both negative), then at x = a, y :::: f(x) has neither maximum nor minimum.
Problem 53 : (a) Find the maximum and minimum values of x 3 - 3x 2 - 9x. (b) Show that the function y = X S - 5x 4 + 5x3 - 10 has a maximum at x = 1, while minimum at x = 3 (Meerut 2007) 2 Solution: (a) f(x) = x 3 - 3x - 9x y = x3 -
or
3.l - 9x
Differentiating with respect to x
!!.l. = 3x2 dx
6x - 9
35
MATHEMATICAL CONCEPTS AND COMPUTER
Differentiating again with respect to x d2 E..l'.2 = 6x - 6 dx For maximum or minimum value of x
4l. =0
dx 3x - 6x-9 =0 x2-2x-3=0 (x + 1) (x - 3) = 0 x = - 1,3 2
=> or or
6=6(-1)-6 2
Atx =-1
dx
= - 6 - 6 = - 12 (negative)
:. Maximum at x = - 1. 2
~=6(3) -6 2
Atx= 3
dx
= 18 - 6 = 12 (positive)
:. Minimum atx= 3. (b) y=x5 -5x4 +5x3 -10 Differentiating with respect to x,
i=
5x
4
3
2
20x + 15x
-
Differentiating again with respect to x,
6dx
2
= 20x'3 -
2
60x + 30x
For finding out the maximum and minimum value of x, put
4l.=0 dx or or or At x-I
x
2
-
4x + 3 = 0
(x - 3) (x - 1) = 0
=0, i.e., x = 1, d
2
~ = 20 - 60 + 30 = - 10 (negative)
dx :. Maximum at x = 1 At x - 3 = 0, i.e., x = 3, 2
~2 = 20 (3/ dx
60 (3)2 + 30(3)
= 540 - 540 + 90 = + 90 (positive)
:. Minimum at x = 3
36
PHYSICAL CHEMISTRY-I
Problem 54. Show that in all the rectangles of the. same area, the sum of the sides of the square is the least. Solution. Suppose x and y are the length and breadth of a rectangle, then its, (a) Area
=xy =A
(given)
... (i)
(b) Perimeter" P = 2 (x + y) = 2 ( x + ~ ) ..
[Using equation 0)]
P is a function of x, ~o for maximum and minimum value of P, we must
have,
Now Now
2
x"'-{;\,
At
d P
-> 0 2 dx
Therefore, the perimeter (P) of the rectangle is least when x =-{;\ 1
x- =A =xy.
or
2
x =xy =H (x - y) == 0
*
x == Y ( ... x 0) So, all the rectangles with given area are square if the perimeter of rectangle is least.
=}
Problem 55. What is the maximum area of a rectangle of perimeter 176 em? Solution. Suppose x and yare the length and breadth of the rectangle, respectively. Then, Area,
2(x+y)== 176 or x+y=88 A == xy = x(88 - x) dA == 88 - 2x dx
and
dA =O=}x=44 dx x=: 44 is a point of maximum.
Maximum area
== 44 (88 - 44) = 44
x 44 = 1936 cm2•
37
MATHEMATICAL CONCEPTS AND COMPUTER
= 1, then find the minimum z = x + y = x +.1 [. : xy = 1]
Problem 56. If x > 0 and xy Solution. Let
value of x + y,
x
dz = 1 _ ~ and d2~ = 23 dx x2 dx- x dz =0 dx 1
means that 1 - -:; = 0 or x = ± 1 xThus, x = 1, because x > 0 (given), so 2 dz 1= 2 > 0
Clx2 )x=
:. x = 1 is a point of minimum. :. Minimum value = 1 + 1 = 2.
Problem 57: What is a linear graph? How will you calculate the slope of a linear curve? Solution: If a graph is plotted on two axes, say X and Y and we get a straight line, then it is known as a linear graph. The equation of a straIght line is y = mx + c, where m = slope which the line makes on X-axis and c is :::..........
',"
.... .
A Intercept
vt B~--------------~--
-X
m = slope = tar, e = (~) c = Intercept
=AB
o
-x m = slope = e c Intercept = 0
=
the intercept (or portion) cut by the straight line on Y-axis. This can be represented by the curves shown in figures (1) and (2).
Problem 58. Write the equation of a straight 2 an d Y-axIs . .Intercept IS . 6. - -3
line whose slope is (Meerut 2007)
38
PHYSlCAL CHEMISTRY-I
= mx + c Y-axIs. Here m = - ~ and C = 6, so the
Solution. The general equation of a straight line is y where m
= slope and C = intercept on
equation of the straight line will be given by,
y=( or or
-~)x+6
3y=-2x+ 18 2x + 3y= 18
COMPUTER Problem 58: Explain the terms functioning, characteristics, limitations and components of a computer. (Meerut 2005. 04)
[I] Introduction to Computer The word computer comes from the word 'compute' which means to calculate. So, normally a computer is considered to be a calculating device that can perform arithmetic operations at enormous speed. Computer is an invention that has taken more than a hundred years to take the shape it is in today. The way the nature of computers have manifested in our lives, is slightly less than the effect that the discovery of fire or the Illvention of the wheel had on evolution of mankind. Launching of a spacecraft, making an airline reservatIOn, scheduling year, your day or keeping a child occupied with some games is like a game for this wonder machine. The essence being, computer can no longer be termed as a mere invention-it is a revolution. Computer can be defined as an electronic machille which processes raw data to give meaningful if!formation. In other words, the term computer is used to describe a device f//ade up of combination of electronic and electromechanical components. Data : Data are collection of facts, figures, statistics which can be processed to produce meanlllgful informations. Processing: Proce~sing IS the manipulation of the letters, numbers or graphic symbols that constitute data. It includes calculation, comparison, decision making, logic. Information: Information is the term given to meaningful form of data. In other words, it is processed data.
IInput Data 1--7 IProcess 1--7 ~tput (Informatic;;J [II] Characteristics of Computer By itself, a computer has no intelligence. A computer or computer system does not come to hght until it is connected to the other parts of its system. A computer is generally used due to the following attributes:
MATHEMATICAL CONCEPTS AND COMPUtER
39
1. Speed: A computer can work very fast. It can perform in minutes/seconds the tasks that would take a person long time to complete. Present day computer can perform 100 million computation in 1 second. 2. Accuracy: A computer always give accurate results. 3. Reliability: Computers consistently provide the same accurate results under all operating conditions. 4. Storage capacity: A computer can store large amounts of data. 5. Diligence: Computer can keep on working for long hours and never gets tired.
[III] Limitations of Computer Unlike human brains, a computer cannot think on its own. It cannot detect flaws in logic input to it. Computer cannot draw a conclusion without going through all intermediate steps. It does not learn from experience.
[IV] Components of a Computer A computer system is a combination of three elements : (1) Hardware (2) Software {1) Person (1) Hardware: Hardware as the term indicates is anything which is hard. Those parts of a computer which we can see, touch or feel are called hard wares. The physical components and equipments which make up a computer system are called hardware. Hardware consists of the following components:
Memory Control Unit
Arithmetic
and logical urut
Now let us study each of the components: C.P.u.: The central processing unit is also referred as 'brain' of the computer. It controls the flow of data through the system, directing the data to enter the system, placing data in memory, reentering them when needed and directing the output of information. It consists of : (a) Arithmetic Logic Unit (A.L.U.): It performs all the arithmetic calculations and takes logical decisions. It can perform addition, subtraction etc.
40__________________________________~P~H~Y~S~I~C~A~L~C~H~E~M~IS~T~R~Y~-I
(b) Control Unit: This unit controls and coordinates the activities of all other parts of a computer system. It performs the following functions: (i) It can get instructions out of the memory unit. (il) It can determine the storage from which it is to get the net ll1structions after the previous instruction has been executed. (iIi) It can decode the instructions. (c) Memory: All the data, intermediate and final results are stored in memory. It is used to hold the instructIOns to be carried out and the data to be processed. It is c1as~ifled mto two types: (i) Primary memory Oi) Secondary memory. (i) Primary memory: It is also called the main memory or the central memory. It is again of two types RAM and ROM. RAM is essentially a 'ReadlWrite MemO/y'. Information can be written into and read from a RAM. It is volatile in nature, i.e., it retains the stored information as long as the power supply is not switched off. ROM is a permanent type memory. ROM retains the data in it even in the absence of power and IS thus non-volatile storage. ROM does not allow user to write data onto it. It stands for 'Read Only Memory'. (ii) Secondary memory: It is found outside the CPU box and hence ~ometimes called external memory or the external storage. Examples are floppy disks, t~lpe~. compact disks (C.D's) etc. (2) Input Devices: These devices are required to get raw data into the computer. It is considered as 'interface' between the user and the system. Various types of input devices are as follows: (i) Keyboard: It is just like a typewriter. But there are additional keys that control functions. These are alpha-nulIleric keys, special keys and/ullction keys. (ii) Mouse: It is one of the most popular type of specialized input deVIce for computers. It is used as a pointing device. (iii) OMR: The 'Optical Mark Reader' is a device which is capable of recognising pre-specified type of mark made by pencil or pen. (iv) OCR: The 'Optical Character Reader' cannot only detect a mark but also recognizes its shape and identify characters directly from source documents. (v) Light pen: Light pen uses a light sensitive photo-electric cell to a single screen position to the computer. (3) Output devices: Output devices are those which output the processed information. These devices are as follows: (i) Visual display unit: It is used for interactive processing. It is commonly known as monitor. (ii) Printers: The most common form of computer output is printed output is also called hard copy. Printers are classified as/to how they print and how fast they operate. Character printer, page printer and line printer are different types of printers. (iii) Storage: It has storage areas, often referred to as memory. The memory can receive, hold and deliver data when instructed to do so. This
41
MATHEMATICAL CONCEPTS AND COMPUTER
fonn of storage stores the data permanently in the given media, examples of which are floppy diskettes, magnetic discs, C.D's etc.
Problem 2 : Explain the binary system used in computers. Define the terms, bit, nibble and byte. Solution : Binary System : A computer interprets infonnation composed of only zeros and ones. So, instructions and data processed by a computer must be in the fonn of zeros and ones. In other words. the data are stored and processed as strings of two symbols or two state devices. A switch is for example, a two state device, it can be either ON or OFF. As decimal system uses total ten digits to make any number, similarly the binary system uses only two digits (0 and 1) to represent any large number. Each character in computer is represented by a number of decimal system and further converted into 0 and 1 (binary number). Since only two digits 0 and 1 are used in the system, it is called, binary number system. The each digit of this number system is called binary digit or bit. This system of numbers can also be said as base 2 number system. In this system also each digit has its place value as in decimal number system. The digit written at extreme left is called 'Most Significant Bit' (MSB) and the digit written at extreme right is called 'Least Significant Bit' (LSB). Note the place ofLSB and MSB in the example given below.
o
1
o
o
o
1
1
f
Most Significant Bit Least Significant Bit (1) Bit. As it has already been stated above that each digit of binary number system is known as 'Billat:v Digit'. To make its name short, the B has been taken from Binary and IT is derived from Digit, thus making its name as BIT. A bit represents the smallest part of memory locations in computer, which stores either 0 or 1. Refer figure (3) for bit.
I
Binary number Power of base Decimal equivalent
I
1
()
0
I
0
2'
.,6
25
24
23
22
21
128
64
32
16
8
4
:2
I,
ON
-
Binary digits are sho\',.11 in the form of switches. If any switch is ON. It . represents digit L otherWise (} in it's OFl· condItion
/
~~~~~~~~ OFF I
1
0
0
I
0
I
Fig. 3. Binary digits
0
Blllary digit or bit. which represents the smallest part of computer memory
=
20210
I
42,___________________________________P~H~Y~S~I~C~A~L~C~H~E~M~IS~T~R~Y~-I (2) Nibble. The combination of 4 adjacent bits is known as a nibble. (3) Byte. The combination of 8 bits is called byte. The byte represents one character in computer memory. It is now clear that to store one character in computer memory we need at least the space of 1 byte or 8 bits. Now the question arises here that why should there be only 8 bits to represent one character in computer memory? The answer to this question is that the computer works with the help of only 256 characters generated by its keyboard. These characters are A to Z, a to z, 0 to 9 and some special characters such as punctuation marks, arithmetic operator symbols, colon. semicolon, questiQn mark, = , >, <, etc. Every character is represented by a numeral code. These numeral codes are converted into binary numbers and stored in computer memory. The codes starting from 0 to 255 are represented by a group of binary digits. Now we have to find out the total number of digits required in a group to represent 256 combinations using only two digits, i.e., 0 and 1. We find that 8 is the number which can make 256 combinations using two digits. 28 or 2 x 2 x 2 x 2 x 2 x 2 x 2 x 2 = 256 Decimal Group of binary numbers equivalent using 8 bits 0 0 0 0 0 0 0 0 0 1 0 0 0 0 0 0 0 1 2 0 0 0 0 0 0 1 0 3 0 0 0 0 0 0 1 1 4 0 0 0 0 0 1 0 0 5 0 0 0 0 0 1 0 1 6 0 0 0 0 0 1 1 0 7 0 0 0 0 0 1 1 1 8 0 0 0 0 1 0 0 0 9 0 0 0 0 1 0 0 1 10 0 0 0 0 1 0 1 0 11 0 o 0 0 1 0 1 1 12 0 0 0 0 1 1 0 0
1 1 1 1 1 1 1 1 1 1 1 1
0 1
255 256
Problem 3 : Discuss the conversion of numbers from one system to another with examples. Solution: Conversion of Numbers: We know that computer understands only binary numbers, so the methods of converting the numbers from one system to another, has been explained below : (1) Decimal to binary conversion. The base of binary number system is 2 because it uses only two digits to represent the number. The base of decimal number system is 10 because total 10 digits are used to represent the
43
MATHEMATICAL CONCEPTS AND COMPUTER
decimal numbers. Each decimal number is converted to binary before it is stored in computer memory. For converting the decimal number to binary, we divide the decimal number by 2 (because the base of binary system is 2) unless we get remainder less than 2. This can be understood by the following problem. 2
7
4
2
2
3
7
1
0
2
1
8
5
1
2
9
2
1
2
4
1
0
2
2
0
1
2
1
0
0
2
5
0
2
2
1
1
0
(LSB)
I Remainders
(MSB)
Problem 4. Convert decimal number 742 to binary. Solution. After dividing the number by 2, the remainder is noted in each step. The last remainder will make most significant digit and first remainder as least significant digit. The binary number will be as follows : (0 1 0 0 1 0 1 Oh = (742ho (2) Binary to decimal conversion. As the numerals are converted to binary system for storing into computer memory, these numbers are converted back into decimal form for output purpose in the same way. The conversion of these numbers is done automatic by the electronic parts fitted in computer. For converting, binary numbers into decimal, following steps are to be followed. (a) Find the least significant digit and most. significant digit in the number. (b) Mark the place position of each digit starting from LSB to MSB. The first position will be zero position. (c) Find the place value of each digit with the help of the following formula, Digit X Base number place pOSItIOn of digit. It is important to note that the base value in binary system is 2. (d) Add all the place values to get the equivalent decimal number. This can be understood by the following problem. Problem 5. Convert the binary number (1 0 0 1 0 1
equivalent.
Oh to its decimal (Meerut 2006)
44
PHYSICAL CHEMISTRY-I
Solution. Binary digit MSB 1 0 0
Place position 6 5 4
Place value 1 x 26 Ox25 1 X24 1 x23 Ox2 2
64 0 0 8 0 2 0
=
1 3 2 0 lX2' 1 1 Ox2° LSB 0 0 The equivalent decimal number will be : + 0 + 0 + 8 + 0 + 2 + 0 = 74 64 (3) Decimal to octal conversion. Like binary number system, octal system is also used for coding of decimal numbers. As its name clarifies, octal means 8, so total 8 digits are used in octal system. These digits are 0 to 7. The highest number that can be derived from the group of octal digits without any conversion is equivalent to (777)8 of octal number system. For converting decimal number to binary number, we divide the decimal number by the base value of binary number system (which is 2). Similarly, in octal system also we divide the decimal number by base value of octal system which is 8. This can be understood by the following problem.
Problem 6. Convert a decimal number (3 4 8 9 2)10 to its octal equivalent. Solution. 4
834 892
(LSB)
84364 8
5 4 5
8
6 8
8
Remainders 4
8 4
o
o (MSB)
Thus, the octal equivalent of (34892)10 decimal number is (1041 14)8. Just like binary conversion here also we keep dividing the number by 8 unless we get remainder less than 8. (4) Octal to decimal conversion. We know that only 8 digits (0 to 7) are used in octal system. It is clear now that digits used in octal system cannot be more than 7. For converting an octal number to decimal. same steps are applicable as from binary system to decimal. (a) Find the most significant and least significant digits. (b) Mark the place position of each digit considering LSB at zero place and MSB at highest place. (c) Find the place value of each digit by applying the following formula, · ·t x B ase num be r Place positIOn of dIgit D Igl
45
MATHEMATICAL CONCEPTS AND COMPUTER
(d) Add all the place values to get the decimal equivalent number. This can be understood by the following problem.
Problem 7. Convert (6 5 3 4 2)8 octal number to its equivalent decimal number. Solution. (MSB)
i
Octal digit 6
Place value 4 6x8 24576 5 x 83 2560 ::: 192 3 x 82 4 x 8' 32
Place position 4
5
2
3
2
4 2
1
(LSB) 2 x 8° 2 o :. Decimal equivalent number is: 24576 + 2560 + 192 + 32 + 2::: 27362 So, (6 5 3 4 2)g::: (2 7 3 6 2)10 (5) Decimal to hexadecimal conversion. An in binary number system only two digits (0 and 1) are used and in octal ~stem 8 digits (0 to 7) are used to represent the numbers, similarly in hexadecimal number system total 16 digits are used to store and represent the numbers. The digits are 0 to 9 and A to F, where A is used for 10, B for 11, C for 12, D for 13, E for 14 and F for 15. The reason for using these alphabets in place of numbers is that each digit should be represented by single entity. If we write 10 (i.e .. one zero) to represent eleventh digit making it a combination of two digits i.e., 1 and 0, which have already been used for representing the digits 1 (one) and 0 (zero) independently, it will create confusion to recognise among the numbers 0, 1 and 10. So, t;:> avoid this confusion, alphabets from A to F are used to represent numbers from 10 to 15. As shown above, total 16 digits are used in hexadecimal system. While converting decimal number to hexadecimal number, we divide decimal number by 16 unless we get remainder less than 16 because the base of hexadecimal system is 16. This can be understood by the following problem.
Problem 8. Convert (186275 2ho decimal number to hexadecimal. Solution. 16
8
16
1
16 16 16
6
2
7
5
2
6
4
2
2
0
7
2
7
0
2
4
5
4
6
2
8
6
(LSB)
l~' 1
(MSB)
46
PHYSICAL CHEMISTRY-I
So, (1 8 6 2 7 5 2)10 =(1 C 6 6 2 0)16 (6) Hexadecimal to decimal conversion. The system of number conversion from hexadecimal to decimal system is same as from binary to decimal and octal to decimal. The difference is only of their base value because binary number system has its base value as 2. octal has 8 and hexadecimal has 16. This can be understood by the following problem. Problem 9. Convert (5 7 A C ~16 hexadecimal number to decimal number. Place position Solution. Hexadecimal Place value digit ,f MSB 4 5 X 164 5 327680 7 X 163 = 7 3 28672 Ax 162 A 2 2560 = 1 ex 16 C 1 192 LSB 2 x 16° 2 0 2 The decimal equivalent number will be : 327680 + 28672 + 2560 + 192 + 2 = 359106 or (5 7 A C 2)16 = (3 5 9 1 o 6)10 (7) Hexadecimal to binary conversion. The conversion of hexadecimal to binary number is very simple just by converting each digit of hexadecimal to its binary equivalent. To make it simpler. let us know the binary equivalent of each decimal number from 0 to 15 as shown in table-I. Table-1. Relationship between the four number systems
I
Decimal
Binary
Octal
Hexadecimal
0
000 0
0
0
I
000 I
I
I
2
0
I 0
2
2
o
3
001 I
3
3
4
010 0
4
4
5
0 I 0 I
5
5
6
o
I I 0
6
6
7
0 I I I
7
7
8
I
0 0
10
8
9
100 I
II
9
10
I 010 I o 1 1
12
A
II
13
B
12
1 1 0 0
14
C
o
13
1 1 0 1
15
D
14
1 I 1 0
16
E
15
1 I 1 I
17
F
This can be understood by the following problem.
MATHEMATICAL CONCEPTS AND COMPUTER
47
Problem 10. Convert (7 3 0 A D Eh6 hexadecimal number into its binary equivalent. D E o A Solution. 3 7 1101 lll0 01ll 0011 0000 1010 The equivalent binary number will be : (0111 0011 0000 1010 1101 11lOh = (730ADE)16 (8) Binary to hexadecimal conversion. We know that the hexadecimal numbers are represented in a group of 4 bits because the largest digit of hexadecimal system is F, which is equivalent to decimal number 15 and binary number 1111. As the largest digit in hexadecimal system is represented by group of four bits, so each hexadecimal digit will be represented by a group of four bits. To convert a number from binary to hexadecimal, we have to make the group of each four digits starting from LSB (Least Significant Bit) to MSB (Most Significant Bit). Add the zeros to left side of MSB if required to make the combination of four bits. Now write the hexadecimal equivalent of each group of four bits starting from MSB to LSB. This can be understood by the following problem. Problem 11. Convert (0010 0110 1001 1110 0101h to its equivalent hexadecimal number. Solution. Let us make the group of each four bits starting from LSB to MSB. 10 0110 1001 1110 0101 (These two zeroes have been added to make the MSB as a combination of four bits) While making the group of each four bits we find that only two bits i.e., 1 and 0 (one and zero) are left in group ofMSB, so we have added two zeroes at the left most side to make the combination of four bits which in our problem now shows 00 10. Now we write the hexadecimal equivalent of each group of four bits. 0010 0110 1001 1110 0101 2 6 9 E 5 So, the hexadecimal equivalent number will be (2 6 9 E 5)16' (9) Binary arithmetic. (a) Addition: Now let us consider addition in binary. The following scales of binary addition are to be remembered : 0+0=0 0+1=1 carry 1 to next column to the left 1+1=0 carry 1 to next column 1 + 1 + 1 =1
48
PHYSICAL CHEMISTRY-I
Example: 1111
11011 + III
LaOOlO (b) Subtraction: This method is also known as complementary subtraction. We will be doing the following three steps to perform subtraction: (i) We have to find the complement of the number we are subtracting. (ii) To the complement of number obtained in step 1, we add the number we are subtracting from. (iii) If there is a carry 1, add the carry to the result of the addition, else recomplement the sum and attach a negative sign. Example: Number Complement 10001101 01110010 Consider the following example of subtraction. Example: 1010101 - 100100 Step 1. Find the complement of 1001100 11011 Step 2. Add the number you are subtracting from
(c) Multiplication:
1110111 Carry 1010101 + 0110011 0001000 (Since there is a carry of 1) +1 -7 Result 0001001 The table to be remembered is :
Oxl=O lxO=O 1 x 1 == 1 Example:
10101 x 11001
10101 11001 10101 00000 00000 10101 10101 1000001101 -7 Result (d) Division: The table 'for binary division is Oil =0 1/1 = 1 The steps for binary division are: (i) Start from the left of the dividend.
MATHEMATICAL CONCEPTS AND COMPUTER
49
(ii) Perform subtraction in which the divisor is subtracted from the dividend. (a) If subtraction is possible put 1 in quotient and subtract the divisor from the corresponding digits of the dividend. (b) Else put a 0 in the quotient. (iii) Execute step 2 till there are no more digits from left to bring down from the dividend. Example: 1000011110 (Quotient) Then 0101 (Dividend) (Divisor) 110) 100001 Step 1 110 Step 2b 1000 Step 2a ~ Step 2b 100 Step 2a .-WL 1001 Step 2b Step 2a 110 11 (Remainder)
Problem 12 : Explain the terms (i) computer programming and (ii) flow charts.
[I]
Computer Programming
In order to solve a problem on a computer, we have to develop an algorithm. An algorithm is a set of instructions which if strictly followed, will give a solution to the problem. In case an instruction is obeyed, we say it is executed. The following activities are involved when we solve a problem on a computer.. (a) Define the problem. (b) Analyse the problem. (c) Develop an algorithm or a method for solving the problem. (e) Test and debug the program. (f) Document the program. There is normaIly some overlap of the above activities. For example, with a large program, a portion may be written and tested before another portion is written. For any problem, there will nonnaIly be more than one algorithm (method) to solve it. Each method will have its own advantages and disadvantages. The user will have a choice of algorithms and it will be his/her job to decide which algorithm is the best and why this is so.
[II] Flow Charts The algorithm uses a numbered list of instructions, the instructions themselves are English-like statements of what must be done. Another method which is useful for specifying smaIl algorithms is the flow chart. A flow chart consists of 'flow chart 'Symbols , connected by arrows. Each symbol will have information about what must be done at the point and the arrows indicate the
50
PHYSICAL CHEMISTRY-I
flow of execution of the algorithm, i.e., they indicate the order in which the instructions must be executed. Figure below shows the symbols commonly used in flow charting.
C)
Oval, indicates Start or Stop: tenninator symbol)
[J
Parallelogram, used to specify an Input or Output operatIon, ... g., getting data or printing data. (liO symbol)
o
Rectangle, used to specify an operation or process, e.g., 'find average' or . set F to 32 I C (process symbol)
~"
Rhombus, used to specIfy a condition. This usuall)' takes the lorm of a quesllon with possible answers Yes or No (or True or False). These arrows lead to the reqUIred action corresponding to the answer to the question (the decision symbol. more commonly called the decIsIOn box).
o
Cirde, used as a connectmg point for arrows coming from different directions (conncctor symbol).
Flow chart symbols.
Problem 13 : Explain the following computer programming languages: (i) FORTRAN (ii) COBOL (iii) BASIC (iv) PASCAL (1) FORTRAN. It means 'Formula Translation' and is the oldest high level language. It was developed in 1956-57. It was designed to solve scientific and engineering problems and is currently the most popular language among the scientists and engineers. After 1957, it appears in different variations such as FORTRAN II, FORTRAN IV and FORTRAN 77 (a version of 1977). The updated version of FORTRAN 77 is FORTRAN 90. (2) COBOL. It means 'Common Business Oriented Language' and was designed to solve the data processing problems of the business community, e.g., problems relating to pay roll, stock control, accounts received and paid, cheque analysis etc. The activities involve little numerical work but a large amount of data has to be processed. The processing consists in creating and updating large files of data. Unlike FORTRAN, the COBOL does not have all the commonly used mathematical functions such as square root, sine, cosine, tangent, logarithms and exponentials etc. In FORTRAN, matrices encountered in science and engineering can be manipulated easily using the FORTRAN arrays, whereas in COBOL this cannot be done. (3) BASIC. It means 'Beginner's All-purpose Symbolic Instruction Code' and was developed by John Kemeny and Thomas Kurtz (1963). Basic is easy to learn and use. It was mainly designed to be used as an interactive language. Almost every computer manufacturer provides, BASIC on its machines. BASIC is probably the most widely used programming language in the world because of the widespread use of microcomputers. (4) PASCAL. It was designed by NikIaus Wirth and named after the famous French mathematician, Blaise Pascal (l7th century). The first Pascal compiler appeared in 1970. The language was chiefly designed as a tool for
MATHEMATICAL CONCEPTS AND COMPUTER
51
teaching structured programming concepts. Most of the colleges and universities use it to teach computer programming. Pascal has the facilities to manipulate, not only numbers, but also vectors, matrices, strings of characters, records and files etc. PASCAL is more readable and thus self documentary as compared to BASIC.
Problem 14 : Discuss the operating systems as used in computers. Operating system is a software loaded into computer memory for ensuring the control and proper working of computer hardware. No computer can work without an operating system. The reason for it is that the system of working of each part of computer is defined in operating system program.
We know that many programs can be loaded in computer memory. Different parts of computer are used to operate these programs on computer. There are chances of clashing of instructions issued through different programs. It may create a halt to operation of the computer. If two different programs need the service of some part of the computer, it may create a state of confusion that whose service is to be completed first. Operating systcm software has been assigned this responsibility to decide the priority and conro) the entire system. Operating system is required in a computer in the same way as a principal is required in a college or a vice-chancellor in a university, who takes the decision above all the teaching staff. Operating system attends to the request of different programs running in computer memory and decides their working through the hardware. For example, suppose you want to read the data from a floppy and execute the command for this purpose. The operating system will read this command and check the availability of floppy drive before reaching the data. Similarly, if you execute the comma-v-d for printing a file through printer, the operating system program will check the availability of printer. The complete control of hardware is done through operating system program.
[I] Personal Computer Operating Systems The examples of operating systems are MS-DOS, MS-WINDOWS, UNIX, ZENIX etc. The main mode of use of a personal computer (PC) is hi
•
52
PHYSICAL CHEMISTRY-I
a single user. Thus, operating systems (OS) for PCs have been constructed as a single user single task operating system, i.e., it is assumed that only a single user uses the machine and runs only one program at a time. The operating system of PCs consists of two parts, viz.: (i) Basic Input Output System (BIOS) which is stored in a floppy disc or a hard disc and (ii) Microsoft Disc Operating System (MS DOS). This type of OS is widely used. The BIOS provides basic low level services, whereas DOS provides many user-level services. MS-DOS provides services such as editing, filing and other utility programs. MS-Windows is an improved operating system for more powerful personal computers which have a large memory (4 to 8 MB) and disc (200 to 400 MB). This gives a very good 'Graphical User Interface' (GUI) which simplifies the use of the computer. It also helps us; to allow the multiple programs to be simultaneously stored in memory and executed. (1) UNIX operating system. This system is a very powerful operating system used in personal computers and supercomputers. UNIX is written in C, so it makes it portable. It is organised as a layered operating system. The innermost portion is known as a kernel. It provides low-level services, e.g., device drivers and memory management. The next portion is known as a shell. It is a command interpreter. The outermost portion is known as utilities portion. It provides miscellaneous services such as text processing, text setting, games, calender, graphics, language compilers etc. The user interacts with the kernel using commands and utilities.
EXERCISES [I]
Numerical Problems
1.
' d the d''''' Illerentl'al coe ff"IClent 0 f,log x . FIII x
2.
Calculate the value Of: if Y! = tf-'y.
3.
Evaluate
4.
Find the value of
5.
If 211 + I PII _ I : 211- I Pn = 3 : 5, find the value of n. Convert (11011.101)z into decimal numbers. Fmd the binary equivalent of (146)10' Convert (39ho into binary equivalent. Convert (O,1l5ho to octal equivalent. Convert (537)8 into binary equivalent.
6 7.
8. 9. 10.
f5
dx,
f
..Jax + b , dx.
[II] Multiple Choice Questions 1.
If nCIO = nC 15 • then value of 11 is, (a) 5
(b) 15
(c) 20
53
MATHEMATICAL CONCEPTS AND COMPUTER
2.
(d) 25 Values of 8C3 and 9P2 are respectively: (a) 28, 72 (b) 56, 72 (d) 28,36
f~
(c) 56, 36
3
3.
The value of
is :
1
(a) log 3 (d) log 10 x 4.
(b) log 2
~ b is :
The value of fax
I ax+b
(b) iog--
(a) log(ax+b)
(c)
bI log (ax + b)
I (d) -log (ax + b)
a
5.
The probability of selecting a girl from a class consisting of.s boys and 7 girls is: 5 (b) ! (c) (a) ~ 6 7 7 (d)
6.
7.
8.
12
Value of
!x
72 / is :
(a)
'2 x 9/2
(d)
~ log x 7/2
7 5/2 (b) "2X
2
The slope of straight line 2x + 3y = 4 is : (a)
-~
(d)
-~
(b)
Th
I f 12! . e va ue 0 3! x 8! IS
(d) Every statement is wrong
:
(b) I
(a) 2900
2
(d) 198
10.
3 2
(c) -
An equation of a straight line y = 2x means that: (a) Une passes through the OIigin (b) Une does not pass through the origin (c) The slope of line is ~
9.
~
The value of 5
f+x +4
(a) log (x + 4) (d)
*
5
log (x + 4)
dx is :
(b)
(c) 1980
54 11.
12.
13.
14.
PHYSICAL CHEMISTRY-I
The (a) (d) The (a)
unit used to read the exernal data and instructions in a computer is : Input unit (b) Output unit (c) CPU None of the above computer memory is measured in terms of : Units (b) Bytes (c) Binary (d) ALU The decimal equivalent of (IOOlllh is : (a) (19)10 (b) (29)10 (c) (39)10 (d) (49)10 I second consists of ...... microseconds : 3 (a) 10 (b) 106 (d) 1012
15.
The following is an output device: (a) Keyboard (b) Arithmetic logic unit (d) Scanner
(c) Monitor
[III] Fill in the Blanks 1.
Differential coefficient of a constant function 'c' is .............. .
2.
d The value of dx
3.
The value of
4. 5. 6. 7. 8. 9. 10.
The value of OJ is ....................... . The value of 6 P3 is ......... . A mouse is ................ device generally used in graphical interface software. COBOL is an example of ................ level language. 8 bits are equal to .................. byte. The components which we can see and touch are known as ........... . 15 In decimal system is equivalent to .......... in hexadecimal system.
f
ifl
(x)
'
.
·12 (x)] IS ............... .
(ax + b)n - I dx is ................ .
[IV] True or False State whether the following statements are true (T) or false (F) ? 1.
The number of arrangements of n distinct objects taken all at a time is (n - I)!.
2.
2n It· p 3 = 2xn P4. then n =7 .
3.
f2 x 3dx =
3l
I
4. 5. 6. 7. 8. 9.
The value of probability varies from - I to I. For a straight line 2y = 3x - 3. the intercept on Y-axis is - 3. The speed of computer is very fast as compared to a calculator. A computer consists of CPU only. The computer memory is measured in terms of bytes. In computer memory, 0 and I are called binary digits or bits.
55
MATHEMATICAL CONCEPTS AND COMPUTER
10. 11. 12.
Laser printers are coloured printers. The octal equivalent of (10)10 is 12. The binary equivalent of (0)10 is 1000.
ANSWERS [I] Numerical Problems 1.
l-logx
x2
2.
logx (I + logx)2
S. 4
6. (27.101)10
9. (0.0727024)8
10. (100101111 lh
3.
-k
2x2
2
4. 3a (ax + b)3/2
7. (IOOIOOIOh
8. (IOOlllh
3. (a)
4. (d) 8. (a) 12. (b)·
[II] Multiple Choice Questions 1. (d) S. (d) 9. (e) 13. (e)
2. (b) 6. (b) 10. (d) 14. (a)
7. (a) 11. (a)
15. (e)
[III] Fill in the Blanks I
1. 0
2. j(x)J'2 (x) +hex) .f'(x)
3. - (ax + b)n all
4. I
S. 120
6. input
7. high
8. one
9. hardware
10.
2. (F) 6. (T) 10. (F)
3. (T)
nc
[IV] True or False 1. (F) S. (1') 9. (T)
7. (F) 11. (T)
4. (F) 8. (T) 12. (F).
000
GASEOUS STATE Problem 1 : What is the nature or significance of R in general gas equation PV = nRT? Calculate its value in different units. Nature or significance of gas constant R. From general gas equation, PV=nRT R
= PV = IlT
Pressure x Volume Mole x Temperature
... (1)
Force Force Pressure =- - = Area (Length)2 Volume = (Length)3
and
So, substituting the values of pressure and volume in equation (1), we get R=
x
Force
(Length)2
(Length)3
Work
Mole x Temperature
Mole x Temperature
C-.. Work = Force x Length)
The work can be expressed in different units and so R can have different units. (1) Value of R in litre atmosphere deg-1 mole-
1
1 mole of every gas at N.T.P. occupies a volume of 22.4 litres. ..
P = 1 atmosphere, V = 22.4 litres, T = 273K
..
=Lx 22.4 R = PV T 273
= 0.0821 litre atmosphere deg-1 mole-1 (2) Value of R in erg deg- 1 mole- 1
P = 76
x 13.6 x 981dyne/sq cm, V= 22400 ml, T= 273 K
PV 76 x 13.6 x 981 x 22400 R=T= 273 = 8.314
x 107 erg deg- 1 mole-1 (56)
57
GASEOUS STATE
(3) Value of R in calorie deg-1 mole-I. We know that, 1 calorie = 4.18 x 107 erg 7
1.99 cal deg- I rnole- I
R = 8.314 x 10
7
4.18x10
"" 2 cal deg-1 mole- I (4) Value of R in joule deg- I mole-I. We know that, 1 joule = 107 erg 7
..
. I d -I I-I R = 8.31410x7 10 JOU e eg rno e = 8.314 joule deg-1 mole-I.
Problem 2 : Write the postulates oj kinetic theory of gases. Derive the kinetic equation oj gases. Or, Prove the equation PV =
1
mnu
2
•
Postulates of Kinetic Theory Kinetic theory of gases was given by Kronig, Clausius, Maxwell etc. to explain the behaviour of gases theoretically. This theory is applicable only to a perfect or an ideal gas. The main postulates or assumptions of the kinetic theory are: J. A gas consists of a large Ilwllber of small particles, called molecules. The molecules are so small that their actual volume can be neglected in comparison to the total volume oJthe gas. 2. The molecules are ill a state of collstant rapid motion in all possible directions, colliding with each other and with the walls of the containing vessel. i':::::'::.: : .": .' ..... .. .'. :-':::-:""'::::: '::'.: .::.: ..:.'.: ::'.:.., "'::: 3. The l110lecules are f..~ y
4.
5.
perfectly elastic. Thus, there is no loss ofkinetic energy wizen they collide with each other or with the walls of the containing vessci. The molecules do not exert any appreciable lIlutual force of attraction.
~;u~~~;/~;~:~l~a~~:
.. :. .:
/: /
::-:
I
/
f--~-----~
....
km A
.' ..'
-ux
.r ~
:::: '.' ..
.. ",'0------------/---.. Z
z:~'l=·
..
R...
r""'~""'~=f""~a"'"~"".~" "oZ:'I: ;:e~" '~" i~" '~:;:('
+H;:"-:.:?.. g=,l=.=Ki='n::;:e""tl"'c""p=jc=tu"' " "
=."'-
X·.·.
.... ..............
ment of the moving molecules 011 the walls of the colltaillillg l'essel,
58
PHYSICAL CHEMISTRY-I
6. 7.
The kinetic energy of the molecules is directly proportional to the absolute temperature of the gas. There is no effect of gravity on the gas molecules.
Derivation of Kinetic Equation From the above postulates, the kinetic equation of gases can be derived easily as follows: Consider a gas enclosed in a cubical vessel of volume V, each side of the cube being I cm. Let the number of gas molecules be II, mass of each molecule be m and the velocity be u cm/sec. The velocity can be resolved into three components ux' uy and U z along the three axes X, Y and Z, mutually at right angles to one another. These components will be related to the velocity (u) by the expression, 2 2 2 u2 =ux+uy+u .•• (1) z For the sake of simplicity, consider one molecule moving with a velocity U x along the X-axis striking the faces, A and A'. On striking the wall A, it will rebound in opposite direction with the same velocity. Therefore,
Momentum of the molecule before collision = mux Momentum of the molecule after collision = m (- ux ) = -mux :. Change in momentum per collision
=(mux) -
(-mu x ) = 2mux-
After striking the wall A, the molecule will have to travel a distance of I cm before striking the wall A' the molecule will strike the wall A in II Ux sec. In other words, the number of collisions suffered by the molecule per second will be ux/i on the walls A and A'. :. Change in momentum per molecule per second along X-axis ry
2mu; =2mu x - = - x I I Similarly, we can derive expressions along Y- and Z-axes. :. Change in momentum per molecule per second along Y-axis Ux
_ 2mu~ - I Change in momentum per molecule per second along Z-axis 2
= 2muz
:. Total change in momentum per molecule per second 2 2 2mil:2 _ 2m 2 ~ _ 2mux2 :::!!!!!2 2. 2 _ 2111uI + I + I - l (u x + uy + uJ - I _Total change in momentum per second for all the molecules 2mu2 2mnu 2 =-l-xn=-l-
... (2)
S9
GASEOUS STATE
According to Newton's second law of motion, the rate of change of momentum (i.e., change of momentum per second) is the applied force. So, we have, 2mnu 2 Force=-1 The gas pressure, by definition, is the force per unit area. Thus, 2
2
Pressure, P = Force = 2mnu / 1 =! . mnu Area 6/2 3 P 2
1 m1lu 3 P="3v (As volume of cube, V=!)
or
1 2 PV=3"mIlU
... (3) ... (4)
This equation is known as kinetic equation of gases. Equation (3) gives the pressure exerted by an ideal gas. For 1 mole of a gas, equation (4) can be written as, 1 2 PV=-mNu 3
where N = Avogadro's number.
Problem 3 : (a) Write a short account of kinetic theory of gases. Derive the equation PV = RT from kinetic considerations. Show how the vllrious gas laws are consistent with it? (b) From the kinetic equation derive an expression for the kinetic energy of one mole of a gas.
(a) [I] Short Account of Kinetic Theory of Gases See problem 2.
[II] Derivation of Equation PV = RT
(t
According to the postulates of kinetic theory of gases, the kinetic energy mni) of the molecules and absolute temperature (n of the gas are directly
proportional to each other. Thus,
"21 mnu2 oc T
or
"21 mnu 2 = kT (where, k = constant).
lx!mnu 2 =kT or !mnu2=~kT · to2k·metIc 3 . equatIOn . 0 f gases, 3 PV = 3" mnu 2 Accordmg
or
i
PV=~kT or
PV = ~ k = constant T 3
... (1)
60
PHYSICAL CHEMISTRY-I
For 1 g mole of a gas, the value of constant is equal to R, i.e., gas constant. Therefore, PV =R or PV=RT. T
[III] Consistency of Various Gas Laws (i) Boyle's law: According to it, 'at constant temperature, the volume of a given mass of a gas is inversely proportional to pressure', i.e.,
Voc or
p1 , at constant
T
PV = constant, at constant T.
From kinetic equation, we have, 1 2 2 1 2 PV=-mnu =- X-mnu 3 3 2
The kinetic energy of the gas, E = ~ mnu 2 PV=-·E ... (2) 3 At constant temperature, kinetic energy (E) of the gas is constant. 2
PV = constant. This is Boyle's law. (ii) Charles' law : According to it, 'at constant pressure, the volume of a given mass of a gas is directly proportional to its absolute temperature, ' i.e., Voc T, at constant P As already derived, from equation (2), we have
2
2 E
PV =3 . E or V =3 . P At constant pressure, V = constant X E or VocE But we know that E oc T, where T is absolute temperature. .. Voc T This is Charles' law. (iii) Avogadro's hypothesis: According to it, 'equal volumes of all gases under same conditions of temperature and pressure contain equal number of molecules '. For any two gases, the kinetic equation can be written as, 1 2 2 1 2 PI VI
="3 mllllul ="3 X 2" mlll\U\
1 2 2 1 z Pz V2 = "3 11121l 2U 2 ="3 X 2" m21l 2UZ
=G~AS~E~O~U~S~ST.~A~~=E
___________________________________ 61
When pressures and volumes of the gases are the same, i.e., PI = P2 and VI = V2 , it follows that,
2"I mInIuI2 = 2"I m2n2u22
••• (3)
When the gases are also at the same temperature, their mean kinetic energy will also be the same, i.e.,
121
2
2" mIuI = 2" m2u2
... (4)
Dividing equation (3) by (4), we get, ni =n2
This is Avogadro's hypothesis. (iv) Graham's law of diffusion:
According to it, 'the rate of diffusion (r) of a gas is inversely proportional to the square root of the density (d) of the gas, at constant pressure', i.e., r oc
I Td ' at constant pressure
From kinetic equation, we have
pv=imnu2 or
u=~=~
mn _ Total mass of the gas - d
(As V -
Vi I oume
. f th - enslty 0 e gas,
d)
Therefore, at constant pressure,
The rate of diffusion (r) of the gas is directly proportional to the mean velocity (u) of the molecules, i.e., u oc r. I
rocTd' This is Graham's law of diffusion. (v) Dalton's law of partial pressures: Suppose ni molecules each of mass m) of a gas A are contained in a vessel of ,:,olume V. Then, according to the kinetic theory, the pressure, PA of the gas A will be given by, 1 mInIuI
PA=3"-VNow, suppose n2 molecules, each of mass m2 of another gas Bare contained in the same vessel at the same temperature and there is no other gas present at that time. The pressure, PB of the gas B is then given by, 2
1 m2n 2u2
PB=3"-V-
62
PHYSICAL CHEMISTRY-I
If both gases are present in the same vessel at the same time, the total pressure, P is given by, 1 mllllUr 1 m2n2u~ P=3-V-+3-V-=PA+PB Similarly, if three, four or more gases, e.g., A, B, C, D, etc are present in the same vessel, the total pressure is given by,
P=PA +PB+PC+PD+ ... This is Dalton's law of partial pressures.
(b) Kinetic Energy of One Mole of a Gas According to kinetic equation of gases, for one mole of a gas, 1 2 PV = 3 mNu (where N = Avogadro's number) 2 1 2 PV=-x-mNu 3 2 The kinetic energy (E) of one mole of a gas is given by, or
1
E=-mNu 2
... (5)
2
Thus, from equation (5),
2 PV=3. E=RT
m
3 E=jRT
For n mole of an ideal gas, the kinetic energy will be given by (312)nRT. Problem 4 : From kinetic equation, calculate the ratio of specific heat at constant pressure to specifIC heat at constant volume in case ofmonoatomic gases. How does this ratio vary with molecular complexity of the gas ? Or, Explain with reason: The value ofy is 1.67 for a monoatomic gas. (Meerut 2(05)
Or, Explain specific heat of gases at constant volume Or, Why does the value of Cp differ from that of Cv? Or, Show that for an ideal gas, Cp - Cv = R.
(Meerut 2(04)
(Meerut 2(02)
"[1] Molar Heat* of a Gas at Constant Volume (Cv) It is defined as, 'the quantity of heat in calories required to raise the ~ temperature of 1 mole of a gas through 1° at constant volume'.
*
We will use the term molar heat instead of specific heat, as in chemistry it is customary to use the term molar heat. Morwver, molar heat = specific heat x molecular weight.
63
GASEOUS STATE
In this case, the volume of a gas is kept constant, so there is no external work done by the gas when it is heated. In other words, all the heat supplied to the gas is used up in increasing its kinetic energy. If the temperature of one mole of a gas is raised through 1°, the increase in its kinetic energy is equal to the molar heat at constant volume. For 1 mole of a gas, the kinetic equation is written as, PV = 1. mNu 2 = ~ .1. mNu 2 = ~ . E = RT
332
3
where, E = kinetic energy per mole at temperature, T =~ mNu2 E=lRT 2
Similarly, kinetic energy (E1) per mole at temperature (T + 1). 3 E1=-R(T+I) 2 Increase in kinetic energy per mole for one degree rise in temperature is given by, 3 3 3 E1-E=2 R (T+ 1)-2 RT =2 R
C v = 1 R '"
2
12 x 2 '" 3 calories
Thus, the molar heat of a monoatomic gas at constant volume should
be nearly equal to 3 calories. [II] Molar Heat at Constant Pressure (Cp) It is defined as, 'the amount of heat required to raise the temperature of 1 mole of a gas through 1° at constant pressure. ' When the gas is heated at constant pressure, there will be an increase in its volume, i.e., the gas will expand and do some external work. Thus, some extra heat (in addition to the heat required to raise the kinetic energy of the gas molecules) must be supplied to the gas to enable it to do the external work. Thus, at constant pressure, the heat supplied to the gas is utilised in two ways: (a) In increasing the kinetic energy of the gas molecules, which is equal to Cv. (b) In doing external work during expansion. So, the value of Cp is always greater than Cv•
Therefore,
Cp = Cv + Work done
... (1)
The work done by the gas during expansion is calculated as follows :
*
When a gas expands at constant pressure, the work done (W) is given by. W = Pressure x Change in volume
64
PHYSICAL CHEMISTRY-I
Suppose 1 mole of a gas* is enclosed in a cylinder containing frictionless piston. Let its pressure, volume and temperature be P, V and T, respectively. For 1 mole of a gas, ... (2) PV=RT When the temperature is raised to (T + 1) at constant pressure P, let the volume increases to VI' Thus, PVI =R(T+ 1) ... (3) From equations (2) and (3), we get
PVI or
-
PV = R (T + 1) - RT = R
P (VI - V)
=Pressure x Change in volume = R
:. Work done, W From equation (1), therefore,
=R
3 5 Cp = Cv + R = -2 R + R = -2 R ::::: 5 calories. Thus, the molar heat of a monoatomic gas, at constant pressure is nearly 5 calories. The molar heat ratio (y) is defined as the ratio ofmolar heat at constant pressure and molar heat at constant volume. For a monoatomic gas, molar heat ratio is given by, 5 Cp 2R 5 y=-=-=-= 1.66 Cv lR 3
2
[III] Variation of y With Molecular Complexity of a Gas : In monoatomic gases, the heat supplied to the gas is utilised only in increasing the kinetic energy of the molecules. But in case of polyatomic gases (whose atomicity is more than one) the heat supplied is utilised in not only increasing the kinetic energy of the molecules but also in increasing their rotational and vibrational energies (intramolecular energy), say x. Consequently, more beat is needed to raise the temperature of the gas through 1°. Thus,
Cp =
%R + x
and
Cv =
5 2R+x Y=i!=-3-v -R+x
~ R +x
C
2 The value of x varies with the complexity of the gas molecules. For monoatomic gases, * x = 0
*
Inert gases, vapours of elemenlS,
65
GASEOUS STATE
For diatomic gases,** x
= R, 5
7
2R+R 2R 7 Y=
= - = - = 1.40
lR+R 1R 5 2
2
. . gases *** x = -3 R For tnatomlc , 2
5
y=
3
2R+2R 4
=-= 1.33
lR+1R 2
2
3
For polyatomic gases, the value of y is still less. In fact, y decreases as the molecule becomes more and more complex.
[IV] Difference Between the Values of Cp and Cv At constant pressure, the heat supplied to the gas is used in increasing the kinetic energy of the molecules as well as in doing work during expansion of the gas against external pressure. Moreover, at constant volume, the heat supplied is used only in increasing the kinetic energy of the molecules. Thus, at constant pressure the gas requires more heat to raise its temperature by 1° than at constant volume. So, the value ojCp is always greater than C v by an amount equivalent to work done by the gas. In other words, Cp = Cv + Work done = C v + R ~-~=R
or
Problem 5: Explain the distribution of molecular velocities. How will you verify the distribution of molecular velocities experimentally? Or, Write a short note on Maxwell's law. We know that all the gas molecules do not travel with the same velocity. This is because the molecules are colliding with one another quite frequently and so their velocities keep on changing. Maxwell worked out the distribution of molecules between different possible velocities by using probability considerations. According to him. the distribution of molecular velocities is given by, dn r _ 4 (~)312 - 1t 2- D • e n JIJ\.T,
-M/12RT . c2dc
***
C02, NzO, H20 vapours etc,
... (1)
66
PHYSICAL CHEMISTRY-I
where, dnc is the number of molecules with velocities in the range c and c + dc, n is the total number of molecules of the gas, M is the molecular weight of the gas, T is the absolute temperature, c is the lower value of velocity for the selected velocity range, dc is the velocity interval for the velocity range, e is a mathematical constant, base of natural logarithm, whose value is 2.71. DiVIding equation (1) by dc, we get
~ . dll c _ n
::
(JL) -
dc - 4n 21tRT
3/2
.e
M/I2RT
.c
2
... (2)
'"
"3 u
'.'
'0
:::8
.....0
r--.
..
:;':
;';.
~
"
.
E
.'. : :
.'. .... .'.
~
·1 ~
ii . . . . . . . . . . . . . . .
:::.
~. 2. '"""b;~.:M:::~ V"'~ty.;if.;;..;t!
This is the usual form of the equation for Maxwell's law of distribution of velocities. The left hand side of expression (2) gives the probability, p of finding molecules having the velocity c. This is virtually (not exactly) the fraction of the molecules having the velocity c. Knowing the molecular weight of the gas, we can calculate the fraction of the molecules having any particular velocity c, at any desired temperature. The general form of distribution of molecular velocities is shown in figure 2. Fractions (percentages) of the molecules having various velocities are plotted along Y-axis and velocities are plotted along X-axis. The actual curve depends upon the molecular weight of the gas and temperature, but the general form of the curve is the same for every gas. From the curves, it is seen that the fraction of molecules having velocities greater than zero goes on increasing with increase in velocity, reaches a maximum value and then begins to fall towards zero again for very high velocities.
67
GASEOUS STATE
The validity of the Maxwell's distribution law can bt: checked by an apparatus shown in figure 3. A beam of meta'i atom is created by heating a metal like silver or bismuth in an' oven. The two discs are mounted on a rotating axis. The slits in the discs are displaced by a known angle so that the slit in one disc comes in the path of vapour ~olecules coming out of the slit m second disc after an interval of time determined by the rate of rotation of the axis. Particles moving with different velocities can be collected in the detector by adjusting the rotation rate of the diSCS. On plottmg a function of number of particles collected in the detector and the velocities of the partic\e~ as given by rotation rate of discs, we obtain curves as shown in figure 2.
:','y..;.:.::;.... ::;.:::':.:;.:'::::.;:-::-::': :. :':-. '':
Pin-Hole
.. . ::.,:-:'::... .:.....
.:.::.:-......:':. ::-:-
'\..
r---------------,-------, :.:'
0
• • I:~-v~--~
~ol~c,:t!a! - -~;:o~~-----Velocity ! :Detector Selector
:::;
I
I
.:;'
I
::::
Evacuate} Chamber
C:~:.::-:::::?::.;::: .. ·......
. ....
....
......... "'.::.:.:::.y::':.:
'"
.:.:.....:.:.:.:.:.:.:... ::: Fig. 3. Apparatus for testing Maxwell's distribution law. .. :: : .....
.::::.:'::::;:::;:;:;::.::;::; ... ... .........
:. '::'.;':;: :.;. :<:.
Problem 6 : Define the terms average velocity, root mean square velocity and most probable velocity of molecules and how are they related to each other? (Meerut 2006) Or, Find out a relation between RMS velocity and average velocity. (Meerut 2005)
Or,
Define root mean square velocity a1ld average velocity.
(Meerut 2007)
(i) Average velocity (v): It is defined as, 'the average of the velocities of all molecules at allY time'. If UI, u'}., U3, '" Un are the velocities of individual molecules in a gas and n is the total number of molecules con tamed in a gas, then the average velocity is given by, V
= UI + U2 + U3 + ... + Un ~
Il
Root mean square velocity or R.M.S. velocity (u)*: It IS defined as, 'the square root of the mean of squares of all velocities of the molecules '. (ii)
*
.
Mean square velocity. u
2
. . IS given by, i
=
l(uh 111 + II~ + ... + u~ I! Il
68
PHYSICAL CHEMISTRY-I
.
_
RM.S. velocIty -
~(
2+ 2+ 2+ ... + 2)
UI
U2
U3
Un
11
(iii) Most probable velocity (a. or u): It is defined as, 'the velocity possessed by the l1laximum number of molecules of the gas '. The relationship of most probable velocity to the average velocity and root mean square velocity can be developed from the following expressions:
Average velocity, v = " ( : - )
u
3~TJ
Most probable velocity, a. =
~( 2~TJ
Root mean square velocity,
= "(
Thus, 'v = 0.9213 X RM.S. velocity, u
Average velocity,
Most probable velocity, a. = 0.8164 x RM.S. velocity, u From the above equations, it can be seen that the ratio of the three velocities are given by :
a : v : u = I : 1.128 : 1.224 Problem 7 : From kinetic equation oj gases, how will you calculate the root mean square velocity oj gas molecules under different conditions? From kinetic gas equation, we can calculate the root mean square velocity of gas molecules under different conditions of temperature, pressure, etc., as shown below.
[I] When only temperature is given For 1 mole of a gas we have, from kinetic equation
PV=.l mNu 2 = RT 3
u="C::)=~(3~T)
or where,
\
mN = M =molecular weight
8.31~X 10
7
..
or
u = "(3 X
4
II
= 1.58 X 10 X " (
X
T)
~)
Knowing the value of T, we can calculate u.
(': PV= R7)
69
GASEOUS STATE
[II] When both pressure and temperature are given From kinetic equation, we have for 1 mole of a gas, 1 2 J " PV=-mNu =-Mu(M= molecular weight) 3 3
u=-V(3~VI
or
... (1)
The value of V can be determmed !rom N.T.P. considerations, i.e.,
PV _ PoVo
T-r;
(where, Po, Vo and To correspond to N.T.P. values) or
PoVo T V=--xTo P
Thus, the value of V can be calculated and substituted in equation (1).
[III] When both pressure and density are given For 1 mole of a gas, the kinetic equatIon is written as,
PV=~mN1l2=~MU2
or
u~-vC~V)=-V(~)
[As d Knowing the values of P and d, we can calculak u.
= MIV = density]
Problem 8 : What are real and ideal gases? In what respect does a real gas differ from an ideal gas? (Meerut 2004)
Ideal or Perfect Gas A gas which strictly obeys Boyle's and Charles' laws at all temperatures and pressures is known as an ideal or perfect gas. The characteristics of an ideal gas are as follows : (i) The compressibility (Z) of an ideal gas is unity, i.e., Z= PV = 1 nRT
(ii) The product of pressure and volume of an ideal gas at constant temperature is constant. (iii) If an ideal gas is cooled at constant pressure, its volume decreases continuously, till at -273°C its volume becomes zero. (iv) If it is allowed to expand without doing any external work, it shows no thermal effect. (v) There is no force of attraction between molecules of an ideal gas.
Differences between an Ideal and Real Gas (i) A real gas does not obey various gas laws particularly at low temperatures and high pressures. Only at high temperatures and low pressures, real gas tends towards ideal behaviour.
70
PHYSICAL CHEMISTRY-I
(ii) The volume of a real gas does not become zero at -273°C. In fact, when cooled sufficiently, a real gas is suddenly converted into liquid state. Problem 9 : (a) What are the limitations of the equation PV = RT? Show in what respects vander Waals equation is an improvement over the simple gas equation. Derive vander Waals equation and discuss it or write the drawbacks of this equation. (Meerut 2004, 03, 01, Garhwal 2006, 02, 2000, Kanpltr 2005; Agra 2006, 01.2000)
Or, Discuss vander Waals equation of state in detail. (Meerut 2005) Or, Explain vander Waals equation along with volume correction and (Meerut 2007) pressure correction. (b) What are the units of vander Waals cOllstants. (Meerut 2003, 2000) (c) Show that effective volume of the gas molecules is four times (Meerut 2002) greater thall the actual volume of the molecules.
(a) Limitations of the Equation PV = RT The general gas equation, PV = RT derived from the postulates of the kinetic theory is valid for an ideal gas only. Real gases obey this equation only approximately and that too at high temperature and low pressure. The higher the pressure and lower the temperature, the greater are the deviations from the ideal behaviour. In general, the more easily liquefiable and highly soluble gases exhibit larger deviations. Thus, gases like CO 2, SO:! etc. show much larger deviations than N:!, H2o O2 etc. -.
..
i ...'" (1)
>,
(1)
A
c:>.
'"
J (1)
~
~
>-Po.
.. 15
200
400 P-Atmosphere _
600
800
GASEOUS STATE
71
Regnault and Amagat studied the effect of pressure on volumes of gases like H 2 , He and CO 2 etc. at constant temperature and the curves are as shown in figure (4). According to them, if Boyle's law IS obeyed, the values of PV for a given mass of a gas should be constant at all pressures and the graph should be a straight line parallel to the pressure axis, as shown by dotted line. But, it is only an ideal behaviour which no real gas will exhibit. Hydrogen and helium at all pressures are less compressible than Boyle's law requires, i.e., PV increases with increase in pressure at constant temperature right from the very beginning. On the other hand, CO 2 and other gases at low pressures are more compressible than Boyle's law requires. i.e., PV decreases wIth increase in pressure at constant temperature. This continues with an increase in pressure till PV passes through a minimum at a certain stage. With further increase in pressure, the compressibility is less than expected, i.e .. PV increases with increase in pressure and this contInues thereafter. At low temperatures, the deviations are much more pronounced than at high temperatures.
[I] Improvements by vander Waals In order to explain the deviations of real gases from ideal behaviour vander Waals suggested that it is necessary to modify the kinetic theory of gases. The following two postulates of the kinetic theory, according to him. do not appear to hold good under all conditions. (i) The volume occupied by the gaseolls molecules is negligibly small as compared to the total volume 0/ the gas. This postulate can be justified only under ordinary conditions of temperature and pressure. At low pressure or high temperature, the volume of the gas is comparably large and so the small volume occupied by the gas molecules can be neglected without producing any appreciable error. However, at high pressure or low temperature, the volume of the gas becomes small and now even the small volume occupied by the molecules cannot be neglected, as the molecules are incompressible. Hence, under conditions o/high pressure and low temperature, the above postulate is flOt valid. (ii) The mutual force of attraction between gaseolls lIlolecules is negligible. This postulate also holds good at low pressure or hIgh temperature. because under these conditions, the volume of the gas is large and the molecules lie far apart from one another. But at high pressure or at low temperature, the volume of the gas IS small and the molecules lie c\o:o.er to one another. Thus, the interlllolecularforces ofattraction canllot be neglected. So, at high pressure and 101V temperature, the above postulate is also not valid. So, it is necessary to apply suitable corrections to the ideal gas equatIOn to make it applicable to real gases. vander Waals introduced two correction terms in the ideal gas equation to account for the errors introduced as a result of neglecting the volume of the molecules and intermolecular forces of attraction.
72
PHYSICAL CHEMISTRY-I
(i) Volume correction: For an ideal gas, PV = RT, where V is the total volume occupied by 1 g mole of the gas. As the molecules are incompressible, the volume occupied by them remains the same irrespective of pressure. vander Waals suggested that a factor b should be subtracted from V, the total volume in order to get the ideal volume, which is compressible. Thus, the volume which is compressible is not V but (V - b). So, Ideal volume = V-b. The factor b is constant and is characteristic of each gas. It is known as co-volume, effective volume or excluded volume. It has been shown that the co-volume is nearly 4f2 times the actual volume occupied by all the molecules contained in the gas. (ii) Pressure correction: Consider a molecule lying somewhere in the middle of the vessel as shown in figure (5). It is attracted uniformly on all sides by other molecules. These forces neutralise each other and hence there is no resultant force of attraction on the molecule. However, as the molecule approaches the wall of the containing vessel, it experiences force of attraction which tends to drag it backwards. Hence, it will strike the wall with a lower velocity and will exert a lower pressure than it would
have done if there were no forces of attraction at all. Thus, it is necessary to add a certain quantity to the pressure P of the gas in order to get the ideal pressure. So, Ideal pressure = P + p' where, p' is the pressure due to intermolecular forces of attraction. The value of p' depends upon (i) the number of molecules per unit volume in the bulk of the gas, i.e., directly on the density of the gas and (ii) the number of molecules striking the wall at any given time which in turn also depends upon the density of the gas. Thus, correction factor p' is proportional to the square of the density (d) of the gas or is inversely proportional to the square of volume (V) [As, density oc l/Volume]. So, a or p = -2 I
V
where, a is a constant depending upon the nature of the gas and is known as coefficient of attraction Thus, a Ideal pressure = P + 7: V Introducing the above two corrections in the gas equation PV = RT, we get the following equation:
GASEOUS STATE
73
(p+ ;2)(V-b)=RT
... (1)
This is known as vander Waals equation. The constants a and bare also known as vander Waals constants. For n moles of a gas, the vander Waals equation becomes :
n:~ }V-nb) =nRT naJ ':poc nd oc n v2 =7
i
p+
2
... (2)
2 \
2 2
[II] Discussion of vander Waals Equation or Drawbacks of vander Waals Equation We can now explain the departure of real gases from ideal behaviour at different pressures and temperatures as shown in figure (4) on the basis of vander Waals equation. (i) At low pressures: When the pressure is low, the volume will be sufficiently large and b can thus be easily neglected in its comparison. Thus, the vander Waals equation (1) becomes, (
p + ~) V = RT 2
or
V
a
PV + !!:.. = RT
PV = RT- V = (YV}J(leal -
V
a
V
The product PV i~,less than the ideal value by an amount equal to a/V. As P increases, V decreases or a/V increases and so PV becomes smaller and smaller. This explains the dip in the curve of CO2 etc. at low pressures. (ii) At high pressures: When the pressure is very high, the volume V is quite small. It is now not possible to neglect b. But as P is large, the factor a/V can be neglected in comparison to the high value of P. Thus, equation (1) becomes, P (V-b) =RT or PV- P. b=RT or PV = RT + P.b = (PV)ideal + P . b Thus, PV is greater than the ideal value of PV by an amount equal to P.b. As P increases, the product P.b increases and so PV increases. This explains why the value of PV after reaching a minimum increases with further increase of pressure. fiii) At high temperatures: If at a given pressure, the temperature is very high, the volume of the gas will be sufficiently large to make the value of a/V2 negligibly small. Under this condition, the value of b can also be neglected in comparison to V. So, vander Waals equation approaches ideal gas equation. This explains why the deviations are less at high temperatures. (iv) Exceptional behaviour of hydrogen and helium: As both H2 and He have comparatively small masses, the attractive forces of attraction
74
PHYSICAL CHEMISTRY-I
between the molecules become too small. So, the correction term alVz, due to attraction factor is negligible even at ordinary temperature or low pressure. Therefore, equation (1) becomes, P (V - b) =RT or PV =RT + P.b =(PV)ideal + P.b This explains the rise in the curves of Hz and He with an increase of pressure right from the very beginning even at ordinary temperatures or low pressures.
(b) Units of vander Waals' Constants
From equation (3), the constant a is expressed by PVzII?, i.e., pressure x (volume)2/moI2. If pressure is expressed in atmospheres and volume in cm 3 the value of a will be atm cm6 moCz. If volume is expressed in litre (or dm 3 ), the value of a will be atm L2 mol-2 or atm dm 6 mol- 2. In Sl system. the unit of a will be 4-2 (Nm-2) (m 3)2 ry or Nm mol. (molt
The constant b is incompressible volume per mole of a gas so, it will have the same units as volume per mole, e.g., cm3 moC I , L mor) or dm 3 mol-I. In SI units, the unit of b will be m 3 mol-to
(C) Effective Volume of Gas Molecules vander Waals suggested that a correction term nb should be subtracted from the total volume V in order to get the ideal volume which is compressible. In order to understand the meaning of the term lib we consider two gas molecules as unpenetrable aM incompressible spheres each of which has a diameter, 1", as shown in figure (6). It is clear that the centres of the two spheres cannot approach each other more closely than the distance r. For this pair of molecules. therefore, a sphere of radius r and of
Excluded Volume
,, I
.... ..... .; ;::,:":':~ •.. ::;.;::: .•.;::: •.•.:;:::-.::;: ••••;•. ;:;: .. ;: .• ;::: .•.;:: .•;:;: .. +=:':::::::::::C":::::.:-:::'·.:·':'?:·:·(t.\~
;: :;:;: :/: : ,: : : : : : :;. . ~.i~: .~. . .r:::::-:':::}:.:':':':' ..:
. vo Iume 3"4 1[r3 constItutes wh at 'IS
known as excluded volume or co-volume. The excluded volume per molecule is thus half the above volume, viz., ~ 1[,.3. The actual volume of one gas · r I 'IS "3 4 /1 S 0, mo IeeuIe 0 f ra dIllS 1[r·. 4
':1
31[r'
4
(r}3 =61[1" 1
=31[l2
3
75
GASEOUS STATE
2
:. Excluded volume per molecule == 31tr
3
1 3 == 4 X -1tr
6
== 4 x Actual volume of the gas molecules.
Problem 10: What do you understand by critical temperature and critical pressure in relation to vander Waals' equation and critical phenomenon. Calculate the values of critical constants in terms of vander Waals' constants. How are the values of critical constants determined experimentally? (Meerut 2002; Agra 2005,03,01; Rohilkhand 2005,01; Kanpur 2006)
Or,
Starting from vander Waals' equation find the values of critical constants. Or, Write a short note on continuity of state. (Meerut 2002, 2000) Or, Define critical temperature. (Meerut 2(06) The curves which are obtained by plotting pressure against volume at various constant temperatures are known as isotherms (isos == equal; therm:::: heat). Andrews obtained isotherms of carbon dioxide at different temperatures which are shown in figure (7). First consider the isotherm ABCD at 12.1 Dc. The point A represents CO 2 in the gaseous state occupying a certain volume under certain pressure. On increasing the pressure, the volume of the gas decreases along AB in accordance with Boyle's law until at a certain pressure (at B) liquefaction occurs. Further, decrease of volume is not accompanied by any change of pressure, as shown by horizontal portion BC, until the vapour has been condensed completely at C. The liquid is only slightly comp;'essible, so further increase of pressure produces only a very small decrease in volume. This is shown by a steep line CD which is almost vertical. Thus, along the curve AB, CO 2 exist as a gas. Along BC, it exists partly as gas and partly as liquid. Along CD, it exists completely as liquid. At higher temperature (21. 1°C). a similar isotherm EFGH is obtained. It differs from the first isotherm ABCD in two respects. viz .• (i) liquefaction starts at a higher pressure and (ii) length of the horizontal portion of the curve becomes shorter. With further rise in temperature, the horizonal portion gradually decreases until it is reduced to a point J and the isotherm becomes continuous as seen in the isotherm UK at 31.1°C. So. there is no horizontal portion of the curve and no sudden change from the gaseous to the liquid state. At this temperature (31.1 ° C), the gas passes into the liquid state without any visible separation of one phase from the other. This is called continuous transition of state. The idea of continuity from the gaseous to liquid state can be explained from the isotherm UK. The change at J shows no sharp discontinuity but a continuous transition occurs during the conversion. Above 31.1°C, the isotherms are continuous and there is no evidence of liquefaction at all.
76
PHYSICAL CHEMISTRY-I
Andrews observed that if the temperature of CO2 was above 31.1°C it cannot be liquefied, whatever the pressure may be. Other gases also behave similarly. So, for every gas there is a limit of temperature beyond which it cannot be liquefied, whatever high pressure may be. This limit of temperature is known as critical temperature (Tc) of the gas. The pressure required to
liquefy a gas at critical temperature is called critical pressure (Pc)' The volume occupied by one mole of a gas at critical temperature and critical pressure is known as critical volume (Vc)' The point of inflexion (1) is called the critical point. The isotherm passing through J, is called the critical isotherm. This phenomenon is called critical phenomenon.
vander Waals' Equation and Critical State or Calculation of Critical Constants For one mole of a gas, vander Waals' equation, (p+
~2)cV-b)=RT
may be simplified as, a ab PV-Pb+-V- v2 -RT=O Multiplying throughout by V2 and dividing by P, we get
77
GASEOUS STATE
V3 _ b V 2 + a V _ ab _ p p
RW == 0 p
Arranging in descending powers of V, we get 2 V3 -(b+ RT) p V +aV p _ ab p == 0
... (1)
This equation is cubic in V and as such there may be three real roots or one real and two imaginary roots of V for each value of P and T. In other words, for given values of P and T, there will be either three real values or one real and two imaginary values of V. This behaviour is not shown by isotherms of CO2 in figure (7). At 51° and 31.1°, there is only one volume for each pressure. At 12.1°, there are two different values of V corresponding to points Band C for the same pr~ssure. However, the third volume predicted by equation (1) is missing. By substituting the experimental values of a and b in equation (1), Thomson (1871) calculated the values of V for different values of P and T. He plotted these calculated values of V against P and the isotherms obtained by him are shown by dotted curves in figure (7). The isotherms for temperature 31.1° and above are exactly of the same form as obtained by Andrews. However, the theoretical isotherms below the critical temperature differ from experimental isotherms (of Andrews). They have no sharp breaks and the horizontal portIons of the curves are replaced by wave like portions. For example, the experimental isotherms ABCD and EFGH are replaced by theoretical isotherms ABB 1B 2B 3CD and EFF j F2F 3GH, respectively. In these, there are obviously three volumes represented by B, B2 and C (and F, F2 and G), corresponding to one pressure as predicted by equation (1). As the temperature rises, the wave portion of the curve becomes smaller and smaller and the three values of volume get closer and closer until they merge into one point J at 31.1 ° (critical temperature). At J, the three roots of V (say x, y and z) of equation (1) are identical. Since the temperature is critical, the value of V represents the critical volume of the gas, i.e., V == V,.. The three values of V can be represented as, (V -x) (V- y) (V - z) == 0, At critical point, x==y==z== Vc (V-
vi ==0
Expanding and writing in decreasing powers of V, we get • .3 V -
3 vy2 + 3 V~2 V - Vc3 == 0
... (2)
This equation must be identical with vander Waals' equation (1) at critical temperature and pressure, which may be written as
78
PHYSICAL CHEMISTRY-I
,.3 ( V - b
RTc)
2
aV
ab
+ Pc V + Pc - Pc = 0
" .(3)
Since equations (2) and (3) are identical, the coefficients of equal powers of V in the two equations must be equal to one another. Therefore, RTc 3V(=b+ p ... (4) r
3V2=~ c Pc
... (5)
V3 = ab
... (6)
c
Pc
Dividing equation (6) by (5), we get Vc= 3b Substituting the values of Vc in equation (5), we get
P =~=
3~
c
a
a
3 X (3b)2 = 27b 2
... (7) ... (8)
Substituting the values of Pc and Vc in equation (4), we get
=~. PrY,. =~. (a/27b ) x 3b 2
T
('
3
R
or
3
R
Sa
... (9)
T{,'= 27bR
Determination of Critical Constants (a) Critical temperature and critical pressure : These values can be determined by a simple method which is based on the principle that at the critical temperature, the surface of separation, i.e., meniscus between the liquid and the vapour phase disappears. It is generally used when the substance is in liquid state at ordinary temperatures. The experimental liquid is taken in a vessel V enclosed in a glass jacket J. The temperature of J can be varied gradually by circulating a suitable liquid from a thermostat. The vessel V is connected to a mercury manometer M containing air. The temperature is first lowered so that the vessel is cooled and the surface of separation between the liquid and its vapour becomes sharp. The temperature of the jacket is raised slowly. This rise is continued till the meniscus be-
ig.8. Apparatus for determining T.;.:
t;:l-_ _ _
79
GASEOUS STATE
tween the liquid and its vapour just disappears. This temperature say t~ is noted. The jacket is then cooled slowly till cloudiness due to the condensation of vapours appears again. This temperature t~ is again noted. The critical
temperature will thus be
C' ; ( 2
).
The mean of the pressures read from the manometer M, corresponding to temperatures t~ and t~ ; gives the value of critical pressure. (b) Critical volume: The determination of Vc takes the advantage of the observations made by Calliatet and Mathias that when mean values of the densities of liquid and saturated vapour of a substance are plotted Density against the corresponding temperatures, a straight line is obtained. In figure (9), the curves VC and LC show the plots of the densities of saturated vapours and those of liquid against the corresponding temperatures. The point C, where the two curves meet gives the critical temperature. This point is not sharp as the curve in this range is rather flat. Therefore, the mean densities are then plotted against different temperatures when a straight line MC is obtained. The point C where this line cuts the curve VCL, gives the critical temperature as the density of the liquid now becomes identical with that of vapour. The point C gives the critical density of the substance. The critical volume is obtained by dividing the molecular weight of the substance by the critical density.
i
Problem 11 : (a) Show in what aspects vander Waals equation is an improvement over the simple gas laws ? (b) Calculate the values of vander Waals constants a and b and gas constant R in terms of critical constants.. (c) Calculate the values of constants (a, b) of vander Waals equation in terms of critical constants and show that, 9 1 RTc a=-RTV and b = - (Meerut 2(}()2) 8 C C 8 Pc
80
PHYSICAL CHEMISTRY-I
(a) (i) vander Waals' equation explains the deviations of gases from Boyle's law. Regnault and Amagat showed that gases at low temperatures and high pressures do not obey the simple gas laws. (ii) vander Waals' equation explains the critical phenomenon, whereas simple gas laws do not. (b) In problem 10, we had derived the following three equations:
RT,. 3V,.=-+b Pc
... (1)
V~ = ab
... (3)
Pc Dividing equation (3) by (2), we get V,
Vc::: 3b or b=3 From equation (2), we get a = 3P,.v;
... (4) ... (5)
Substituting the values of a and b in equation (1), we get,
RTc Vc 3V = - + c Pc 3 RTc V,. 8 -=3V"--3 =-3 Vc P" 8 Pc V,. R=-·3 Tc
or or
... (6)
(c) From equation (6),
Squaring both the sides, 64 :z R2~ -V=-
9
c
p;c
Substituting the value of we get
or
V; from equation (2), in the above equation,
9R2r;
a
3P = 64P; c
or
27 R2r; a = 64P,.
... (7)
Substituting the value of V from equation (4) in equation (1), we get,
RI;. 3x3b=-+b Pc
81
GASEOUS STATE
RTc 8b=Pc From equations (4) and (8),
... (8)
or
or
or
PC=8l1
3RTc c
Substituting the value of Pc in equation (7), we get 27 x 8R2~Vc 64 X 3RTc
a = --:--=-==-=--
9
or
a =gRT,yc
and
b =- . 8 Pc
1 RT,.
[From eq. (8)J
Problem 12 : Write short notes on the following:
(1) (2)
Various equations of state. Or, Kammerling Onnes' equation.
(Meerut 1989)
lAw of corresponding states and its applications. (Meerut 2004, 01; Kanpur 2005,02, Rohilkhand 2004,02, Garhwal2006, 03, 2000)
(3)
Mean free path.
(4) (5)
Critical phenomenon and its utility. Collision frequency.
(6) (7)
lAw of equipartition of energy. Specific heat ratio.
(8) (9)
Boyle temperature. Continuity of state.
(Meerut 2004, 02) (Meerut 2(01) (Meerut 2002, 2000)
1. Various Equations of State vander Waals equation has been found to be accurate over a wide range of pressure and temperature. Near the critical point, it does not gives satisfactory results. Several other equations have been proposed to express P-V-T relationship in accordance with the experimental observations. These equations of state are: [I] Clausius equation: This is vander Waals equation modified by Clausius to account for the variation of vander Waals constant a with-temperature. It is given by,
[
p+
a
T(V + c)
2](V-b)=RT
82
PHYSICAL CHEMISTRY-I
where. c is a new constant. This equation is fairly satisfactory but does not hold good for all gases. rIl] Dieterici equation: Dieterici introduced an exponential factor to account for the effect of molecular attraction on the pressure. His equation is given by. P (V - b) = RT. e-aiRTV
where. e is the base of natural logarithm. This equation gives more satisfactory results at high pressures than the vander Waals equation. [III] Berthelot's equation: Berthelot proposed the following empirical equation to explain the behaviour of real gases.
(p+ ;2)(V-b)=RT
[IV] Kammerling Onnes' equation or virial equation: Kammerling Onnes gave an empirical equation which gives the product PV as power series of the pressure at any given temperature. The equation is represented as :
cr
3
PV = A + BP + + Dp +... The factors A. B, C and D etc. are known as first, second, third, fourth etc. virial coeffreients (Greek: virial = force). At low pressures, the coefficient A (which is equal to RD is important and others cancel out. At increasing pressures, other coefficients become significant. [V] Beattie-Bridgemann equation: This empirical equation is given by:
PV=RT+~+~+ ~ +... where,
p, y, 0 are also called virial coefficients.
2. Law of Corresponding States vander Waals showed that if P, V and T of a gas are expressed in terms of critical pressure, critical volume and critical temperature of a gas, we obtain another important generalisation, known as law of corresponding states. Let, and
P
V
-=n -=cj> P(' , V('
•
L=9
1'..
where n, cj> and 9 are known as reduced pressure, reduced volume and reduced temperature, respectively. Therefore, P=nPc ; V=cj>V(' and T=9T,,, Substituting the values of P, V and T in vander Waals equation, we get,
83
GASEOUS STATE
(p+
we get,
(npc +
0
}(V-b)=RT
~ 2) «Wc - b) = RaTc
V~
Substituting the values of Pc, Vc and Tc in the above equation, we get. 8a n . + a ( • 3b - b) = R a . 27 b R [
~ 2 2] 27b (3b)
or or
2~~2 (n + :2)<3<1> - 1) =Ra . 2~~R (n+
:2}<3<1>-I)=8a
... (1)
The above equation known as vander Waals reduced equation of state does not involve vander Waals' constants as well as gas constant, hence it is a general equation applicable to all substances. From equation (1), if two substances have the same reduced temperature (a) and the same reduced pressure (n), they will have the same reduced volume (<1». This statement is known as law of corresponding states. In other words, two or more substances having the same reduced temperature and same reduced pressure and thus having the same reduced volume, are said to be in corresponding states. Importance : While studying the relationship between physical properties and chemical constitution of various liquids, their properties should be studied at the same reduced temperature as pressure has practically no effect on liquids. It is seen that the boiling point of a liquid on absolute scale is nearly two-thirds of its critical temperature. Various liquids at their boiling points are thus very nearly in corresponding states and to study their physical properties at the same reduced temperature, these should be studied at their boiling points.
3. Mean Free Path The average distance travelled by a molecule between two successive collisions is known as its mean free path. It is denoted by the symbol I or A. If v be the average velocity of the molecule and z be the number of collisions suffered by a single molecule per second, then the average distance covered by a molecule be~ween two successive collisions, i.e., I, is given by 1= !'..
z
The value of z has been shown to be equal to (-.1"2) nv 0'2111 ; where, 0' is the collision diameter and nl is the number of gaseous molecules per cm 3.
84
PHYSICAL CHEMISTRY-I
v
1=
..J(2). nva2nl
..J(2).na2nl
... ( 1)
For n g mole of an ideal gas PV=nRT=!!J..RT N where, N = Avogadro number.
Therefore, from equation (1) 1=
RT ... (2) ..J(2). na2 PVN Thus, at constant pressure, the mean free path is directly proportional to absolute temperature. Thus, higher the temperature, the greater is the mean free path. Similarly, at constant temperature, the mean free path is inversely proportional to the pressure. Thus, lower th~ pressure, the greater is the mean free path. The mean free path is also inversely proportional to the viscosity of the gas. The mean free paths for H2, O2 and He are 11.2 X 10-6,6.43 X 10-6 and 18.0 x 1O-6cm., respectively.
4. Critical Phenomenon and its Utility For critical phenomenon, see problem 10. It is utilised in the liquefaction of gases.
5. Collision Frequency The number of collisions between molecules per second per unit volume (ml) of a gas is known as the collision frequency.
From kinetic theory of gases, it can be seen that if the number of molecules per millilitre of a gas be N, then the number of molecules with which a separate gas molecule will collide with be given by, z' = fin v a 2N where v = average velocity of the molecules in cm s-I; a = collision diameter incm. :. Total number of molecules colliding per millilitre per second, z =z' xN= (..J2)nv~N2 As each collision involves two molecules, the number of molecular collisions per millilitre per second will be z/2 i.e., Collision frequency,
z 1 2 2 2 '-12 So, collision frequency of a gas increases with increase in temperature, molecular size and the number of molecules per millilitre. N
c
=-=~·nvaN
85
GASEOUS STATE
6. Law of Equipartition of Energy According to Maxwell and Boltzmann, the law of equipartition of energy can be defined as, 'the energy given to a gas is equally distributed amongst each degree of freedom. The degree of freedom is the system to represent the probable velocity of a molecule. For mono-atomic gases: In case of mono-atomic gases like He, Ar etc. the molecules have only translational motion, i.e., they move in one direction only. This velocity can be resolved along three axes at right angles, so the system will have three degrees of freedom. The kinetic energy of the gas will be
~ mu2 which can be resolved along 3 -
axes, i.e., x, y, z.
1212121 2 =2"mux +2"muY +2"muz
"2 mu
According to law of equipartition of energy,
1.2 mux2 = 1.2 mti.y = 1.2 mu z2 = 1.2n RT (AS total kinetic energy =
t mi =~ t
For each degree of freedom of velocity, the energy will be mole. The total energy is
RT)
RT per
~ RT. so the molecular heat capacity will be ~ R.
For di-atomic gases: A gas where molecule has more than one atom will have rotational energy and vibrational energy besides the translational or kinetic energy. For di-atomic molecule, this velocity will be only in one direction, i.e., parallel to the axis of the particle. So, such a molecule possesses two degrees of freedom, one for potential energy and the other for kinetic energy. Now consider the rotational motion. Due to collision with other molecules, the molecule will rotate about its own centre of gravity. Assuming the velocity to be in one plane, it can be resolved along two axes and each axis will have one degree of freedom. Therefore, for a di-atomic gas molecule, the translational energy will have four degrees of freedom, two due to rotation and two due to vibration. If the molecule is rigid, there will be no degree of freedom due to vibration. So, the energy will be
%RT.
Molecular heat capacity = %RT "" 5.0 calories The above value is valid at ordinary temperature when the vibrational energy is very less. When vibration is more at a high temperature, then the molecular heat capacity becomes nearly equal to 7.0 calories/mole.
86
PHYSICAL CHEMISTRY-I
7. Specitic Heat Ratio See problem 4.
8. Boyle Temperature The deviation of real gases from ideal gases is best explained by means of compressibility factor (Z). It is defined as, Z=
PV _ PV (PV)ideal nRT
The value of Z = 1 for an ideal gas at all temperatures and pressures. When for a gas Z is less than or greater than 1, it shows less or more deviation from ideal behaviour.
So, the temperature of the gas at which the value of PV remains constant, so that Boyles' law is fully obeyed, is known as Boyle's temperature (Tb)' If the temperature is less than Boyles' temperature, the value of Z first decreases and then reaches a minimum value and finally as the pressure is gradually increased, the value of Z starts increasing. Different gases have different Boyle temperatures. For hydrogen and helium, Boyle temperatures are -80oe and -240oe, respectively. It means that at -80oe, hydrogen obeys Boyles' law within a maximum range of pressure. For a real gas, Boyle temperature is given by the equation, a
Tb= bR where 'a' and 'b' are vander Waals constants.
9. Continuity of State See problem 10.
Problem 13 : Mention the various methods that are adopted for producing cold and what is the lowest temperature hitherto attained? How has it been reached? Show how these have been used in the liquefaction of gases ? Mention the importance of liquefaction. What is inversion temperature. (Meerut 2002,2000; GarhwaI2001; Rohilkhand 2002)
For liquefaction of any gas it is necessary to cool it below its critical temperature and then apply high pressure. Following are the methods used for producing cold and for liquefying gases.
1. By using freezing mixture and volatile refrigerants:
This method is based on the principle that when a volatile compound evaporates, it absorbs its latent heat of vaporisation from the surroundings and consequently the temperature becomes low. A mixture of solid e0 2 (dry ice) and others can produce a temperature of -IOOoe by evaporating under reduced
87
GASEOUS STATE
pressure. Difluorodichloro methane or freon (CF2CI 2) is also used for producing low temperatures. 2. By Joule-Thomson effect: When a compressed gas at a particular temperature is allowed to expand through a porous plug the temperature falls. During expansion, work is done in overcoming the intermolecular forces of attraction. This work is done at the expense of heat energy and the temperature of the gas falls. Thus, when a gas under high pressure is allowed'to expand into a region of low pressure, it suffers a fall in temperature. This phenomenon is known as louie-Thomson effect. Experiments have shown that gases become cooler during louleThomson's expansion only if they are below a certain temperature known as inversion temperature (T,). The inversion temperature is characteristic of each gas and is given by, 2a
T,= b.R where, a and b are vander Waals' constants. At inversion temperature, there is no louie-Thomson effect. In other words, if a gas under pressure passes through a porous plug and expands adiabatically into a region of low pressure, then at inversion temperature, there is neither a fall nor rise in temperature. If the expansion occurs below the inversion temperature, there is a small fall in temperature and if it occurs above the inversion temperature, there is a small rise in temperature. In most cases, this temperature lies within the range of ordinary temperatures. Hydrogen and helium have very low inversion temperatures, i.e., -80°C, and -240°C, respectively. Thus, at ordinary temperatures, these gases get heated up instead of being cooled in loule-Thomson's expansion. But, if hydrogen is first cooled below -80°C and helium below -240°C, then these gases also get cooled down on louIe-ThomA~,.--_-, son expansion. louleThomson effect was used by Linde in the liquefaction of the gases. The Linde's process is described below: The gas or air freed from impurities and moisture is compressed to a high pressure and enters the
g~rntj[F!i~g:.~l~O:::~~!J~:;~mm@tj
inner tubes of concentric pipes at 0 as shown [ in figure (10) under a
88
PHYSICAL CHEMISTRY-I
pressure of 200 atm. The valve V in J is then opened to allow it to expand suddently into a wide chamber C. It thus gets cooled and its pressure is reduced to about 50 atm. The cooled gas is then made to pass through the outer tube B and cools the incoming gas passing through the inner tube A. Thus, the incoming gas gets further cooled on expansion into the wide chamber C. The gas is compressed again to 200 atm, and made to pass through the inner tube and again expanded into chamber C. This process continues till the gas or air is liquefied and liquid gas comes out of the jet J. 3. By performance of external work: When a gas is allowed to expand against an external pressure, it does work against pressure. If the system is a closed one and is so insulated that neither heat can enter nor leave the system, it is said to undergo adiabatic expansion. During such an expansion, work is done by the molecules against external pressure at the cost of their own kinetic energy and so the temperature falls. This principle combined with Joule-Thomson effect has been used by Claude for the liquefaction of air. Thus, Claude's process (fig. 11) is as follows: Air dried and freed from CO 2 is compressed to about 200 atm, and passed througn the pipe ABC which divides at C. Some of the air enters the cylinder D provided with aIL air j tight pIston. Here the gas expands and does external work in moving the piston outwards. I The temperature of the gas thus falls. The cooled air enters the chamber E from below and then goes up as shown. Thus, the gas cools the portion of the compressed air passing down the coil CEo This chilled gas then passes through a jet or throttle J and is further cooled by JouleThomson effect on account of expansion. This process goes on till the gas is converted into the liquid state. 4. By adiabatic demagnetisation: It is based on the principle that if a paramagnetic substance is magnetised and then demagnetised adiabatically, the temperature falls. On applying the magnetic-thermal effect to the gas, Giauque produced as Iowa temperature as 0.0033 K. By the method of nuclear magnetls~tion it has been possible to attain a temperature of 0.001 K.
li/:":::"I
I y
89
GASEOUS STATE
Importance of liquefaction : Liquefied gases or gases compressed under high pressure are finding important uses in industries. These are : (i) Liquid CO 2 is used in soda fountains. (ii) Liquid NH3 and liquid S02 are used as refrigerants. (iii) Liquid CI 2 is used for bleaching and disinfecting purposes. (iv) Compressed He is used in airships. (v) Liquid air is an important source of O2 in rockets and jet-propelled planes and bombs. (vi) Compressed O 2 is used in welding. IMPORTANTFORMULAE1~______________~
1. Kinetic energy of an ideal gas, E = ~ nRT where n = number of moles, R =gas constant, T =absolute temperature. (Note : E will have the same unit, in which R is expressed). 2. Molecular velocity 4
(a) When only T is given: u = 1.58 X 10 x
%
where, M = molecular weight of the gas. (b) When both P and T are given: u =
~ 3~V
PV PoVo The value of V is calculated from, T =
To
where, Po, Vo and To refer to N.T.P. values. The values are Po =76 x 13.6 x 981 dyne cm-2, Vo =22400 cm 3, To = 273 K. (c) When both P and d are given: u =
~
3. Molar heats of gases (a) Molar heat = Specific heat x Molecular weight (b) Cp=Cv+R (c) y = ~ = 1.66 (mono-atomic), 1.40 (di-atomic), 1.33 (tri-atomic) v
4. For n moles of a gas, vander Waals' equation is
(p + n~ ) (V - nb) = nRT. The units of a and bare lit2 atm mole-2 and lit mole-1, respectively. 5. Critical constants, V" = 3b, P" = 2;b2 ' T" = 2~b
90
PHYSICAL CHEMISTRY-I
The value of R is generaIly expressed as 0.0821 lit atm deg- I mole-I.
6. Inversion temperature, Tj = ~
7. Diameter of a molecule is given by, b = 4N (~nr3) where, r = radius of the molecule :. Diameter = 2r.
NUMERICAL PROBLEMS Ex. 1 : (a) Calculate the kinetic energy of two moles of CO 2 at 300 K, assuming it to be an ideal gas. (b) Calculate in ergs the kinetic energy of a mono-atomic gas (molecular weight 20.2) molecule, moving with a velocity of 5 x 1(/ cm/s. Solution: (a) The kinetic energy (E) of 2 moles of an ideal gas is given by:
E=lnRT 2 3 E="2 x 2 x 8.314 x 300 = 7482.6 J. (b) The kinetic energy
1 2 = -2 InU
42 =-21 X 6.02320.2 ~3 x (5 X 10 ) x 10-
= 41.92 x 10-5 erg molecule-1
Ex. 2 : Calculate the velocity of amelecule of oxygen at N.T.P. and 500·C. Solution: The velocity of a molecule is given by (When only temperature is given) :
where, M = Molecular weight of O 2 = 32. (a) At N.T.P., T = O·C U
= 0 + 273 = 273 K.
-\113"21
= 1.58 x 104 x - F273) = 46,149.4 cm per sec.
(b) At 500·C, T= 500 + 273 = 773 K U·
1.58 X [0' x
"C3
7 ;) • 77,655.4 em per sec.
91
GASEOUS STATE
Ex. 3 : The density of hydrogen at O·C and 760 mm pressure is 0.00009 g / cm3• Find the RMS velocity and the average velocity ofhydrogen molecules (density of Hg = 13.6 g/cm 3). Solution: In this case, the values of density and pressure are given. Therefore, the RMS velocity (u) is given by
u=v'e:) p g/cm 3•
= 760 mm = 76 cm = 76 x u
13.6 x 981 dynes/sq cm and D
= 0.00009
=-
'{3 x 76 x 13.6 x 981} = 183 844 2 / -" 0.00009 ' . em s
The average velocity (v) is given by : v = 0.9213u = 0.9213 x 183,844.2 = 169,375.6 em/so
Ex. 4 : (a) Calculate the RMS velocity of oxygen molecules at 2S·C. (Meerut 2005)
(b)
Calculate the RMS velocity of hydrogen gas at 27·C. 4
~(~)
4
v'(
Solution: (a) We know that, u = 1.58 X 10 x
u = 1.58 X 10 x = 4.8 X 104 ems-
8 2;2 )
I
7
U
_ - '(3 x 8.314 x 10 x 298)- 4 8 104 -I - -" 32 - . x em s T = 25 + 273
(b)
u
= 298 K; M = 32
= 1.58 X 104 x ~(~) = 1.58 X 10
4
x
~(3~)
= 19.35 x 104 em S-I
Ex. 5 : (a) Cal~te the RMS velocity ofoxygen molecules at a temperature of 27·C and 740 mm pressure. (b)
Find out the RMS velocity for CO 2 gas at 12·C and 78 cm pressure. Solution: (a) The RMS velocity is given by ... ( 1) -2
p= 74 x 13.6 x 981 dyne em ;
M=32
92
PHYSICAL CHEMISTRY-I
The value of V can be calculated from NTP concept, i.e.,
u =~3 x 78 x
13.6~ 981 x 22785 =40145.22 emfs PV
PoVo
T=r; P= 74 cm, T= 273
+ 27 = 300 K, V=?; Po= 76 cm,
Vo == 22400 C.c., To = 273 K.
74 x V
300=
76 x 22400 273
22400 x 300 == 252806 V == 76 x 74 x 273 • . C.c. Thus from (1).
U
==
~l3 x 74 x 13.6 ~;81 x 25280.6)
== 48372.6 em/so
(b) Calculate as in part (a) For C02>M==44, P==78cm, T== 12+273=285K V == 76
x 22400 x 285 78 x 273
== 22785
cc.
u = - /3 x 78 x 13.6 x 981 x 227850 == 40145.22 cmfJ "
44
Ex. 6 : Calculate the temperature at which root mean square velocity of S02 molecules is the same as that of O2 molecules at 27°C. Solution: The RMS velocity (Ut) of O2 molecules at 27°C. 4
== 1.58 x 10
~(33020J
The RMS velocity (lI2) of S02 mblecules at T K == 1.58
. 1.58 X 10' or
10
~F)
~( ~)~ 158 X 10' ~em
L = 300 64
4
X
32
or
T == 300 x 64 == 600 K
32
=327°C
Ex. 7 : Calculate the RMS velocity of O2 molecules at NTP (Density of O2 =0.001429 g/c.c. at N.T.P., Density of Hg =13.6g/c.c.; g =981 cm.lsec2). Solution: The RMS velocity of O 2 molecules when P and density are given is expressed as :
93
GASEOUS STATE
=U
'(3P) =_ d -"
'(3 x 76 x 13.6 x 981) =46137 6 -1 0.001429 • em s
-"
Ex: 8 : The density of O2 at O'C and at 1 atmosphere pressure is 1.429 g/cm 3• Calculate the mean square velocity of the gas at O·C. S I ti
• o u on.
= - '(3P) =u" d
'(3 -"
X
76 X 13.6 X 981) = 145899 -1 1.429 • ems
Ex. 9 : Calculate the RMS velocity of a carbon dioxide molecule at 1000·C. (Meerut 2006, 05) 4
~(~)
4
~C~3}=84,985.5 em
Solution: RMS velocity, u = 1.58 X 10
= 1.58 X 10
s-1
"" 8.5 X 104 em s-1 Ex. 10 : Calculate the root mean square velocity and average velocity in em S-1 of H2 gas at O·C and 760 mm pressure, if its density is 9 x 10-5 g/ml. · u = ~(3P) · : RMS veIoClty So Iu t Ion - = ~(3 x 76 x 13.65x 981) d 9 X 104 = 18.38 X 10 em S-1 Average velocity =0.9213 xu = 0.9213 x 18.38 x 104 em S-I = 16.93 X 104 em S-I
Ex. 11 : Calculate the pressure exerted by 1023 gas molecules each of mass 10-22 g in a container of volume 1 Utre. The RMS velocity is Iff cm/sec. 22 Solution : Weight of 1023 gas molecules = 1023 x 10- g = 109 23 23 10 x 6.023 X 10 . :. Weight of 6.023x 10 gas molecules = 23 = 60.23 10 :. Molecular weight, M = 60.23, = P cm Let pressure
u=~3~V 5
/
10 cm see or
_
=- '[3 x (P x 13.6 x 981 dynes/cm2) x 1000 -"
(105)2 X 60.23
60.23
_
P - 3 x 13.6 x 981 x 1000 -15,040 em.
ml]
94
PHYSICAL CHEMISTRY-I
Ex. 12 : The specific heat at constant volume of argon is 0.075 and its molecular weight is 40. How many atoms are there in its molecule? Solution: Molar heat at constant volume (CI') = Specific heat x Molecular weight = 0.075
x 40 = 3
Cp =Cv +R=3+2=5 Cp Y= C
v
5
=3" = 1.66
Argon is thus monoatomic. i.e., there is one atom in its molecule. Ex. 13 : The specific heats ofa gas at constant pressure and constant volume
are 0.125 and 0.075 cais, respectively. Calculate the molecular weight of the gas. The specific heat and molar heat are related as : Molar heat = Specific heat x Molecular weight ..
Cp=cpxM andCv=cvxM andCp-Cv=R
Therefore, Cp =0.125 M and C v = 0.075 M .. or
0.125M - 0.050M = R 2 or m = 0.05 = 40
0.050M = 2
Ex. 14 : The specific heat of a gas at constant volume is 0.152 and that at constant pressure is 0.215. Calculate the atomicity and molecular weight of the gas. Also name the gas.
Solution: Molar heat = Specific heat x Molecular weight .. ..
Cp=cpxM Cp =0.215M
and and
Cv=cvxM C v =0.152M
Cp -C v =2 0.215M - 0.152M =2 or
or
0.063M =2
2
M = 0.063 = 31.74
Y= Cp
Cv
= 0.215 = 1.41 0.152
.'. Molecular weight of gas = 31.74 Atomicity = 2 Name of the gas = Oxygen.
Ex.1S : The specific heats ofa gas at constant volume and constant pressure are 0.090 and 0.45 calories. If the atomic weight of the gas be M, calculate its molecular weight.
95
GASEOUS STATE
We know that, Molar heat = Specific heat x Molecular weight Cp
0.125
Y= C = 0.090 = 1.38:::: 1.40
v So, the gas is diatomic and so its molecular weight = 2 x M .. Cp =0.125 x 2M =0.250M and Cv =0.090 x 2M =0.180M Cp-Cv=R 0.250M - 0.180M = 2
or 0.070M = 2 2 or M =0.070 = 28.57 :. Molecular weight of the gas = 28.57 Ex. 16 : Calculate the constants a and b,
if Tc =31"C, Pc = 72.8 atm and
R = 0.082 lit atmldeg. The values of Tc and Pc are given by:
P =_a_ c
... (1)
27b 2
8a Tc= 27bR
... (2)
Dividing (2) by (1), we get Tc 8b 304 8b - = - or - - - - Pc R 72.8 0.082 .. b = 0.0428 lit mole-1 Substituting the value of b in equation (1), we get,
a 27 X (0.0428)2 a = 3.6 Iit2 atm mole-2
72.8 = or
Ex. 17 : The critical temperature and critical pressure of ethane are 305.1 K and 48.1 atm respectively. Calculate its vander Waals constants 'a' and
'b'. Solution: As in Ex. 16, we can write, Tc 8b 305.1 8b - = - or - - = - Pc R 48.1 0.082 or
x 0.082 = 0 065 I·t 1-1 b = 305.1 48.1 x 8 • I mo e
Substituting the value of b in the equation Pc = ~ , we get 27b
or
a 48.1 = 2' 27 x (0.065) 2 2 a = 5.48 Iit atm mole-
96
PHYSICAL CHEMISTRY-I
Ex. 18 : The critical temperature and critical pressure of oxygen are -U8.8°C and 49.7 atm. Calculme its vander Waals constants. The values of Tc and Pc are given by Sa r..= 27bR
... (1)
P =_a_
... (2)
27b2
c
Dividing (1) by (2), 2
Tc Sa 27b Sb -=--x--=Pc 27bR a R Tc = -l1S.S + 273 = 154.2 K; Pc = 49.7 atm. or b = 154.2 x 0.OS2 0.OS2 49.7 x S = 0.0318 lit mole-1 Substituting the value of b in equation (2), we get
154.2 49.7
=~
49.7 = a 2 49.7 x 27 x (0.031S)2 27 x (0.031S) :::;!
=1.357 8tm Iit2 mole-2 Ex. 19 : The vander Waal\' constants in lit atm/mole of CO 2 are : a = 3.6, b = 4.28 X 10-2• Calculate Tc and V(Jor the gas. The values of Vc and Tc are given by, ... (1)
Vc= 3b Sa Tc = 27bR From (1),
Vc = 3 x 4.2S
From (2),
Tc = 27 x
X
... (2)
2
10- = 3 x 0.042S = 0.1284 lit
0.~2~·: 0.OS2 = 303.9 K = 30.9°C
Ex. 20 : Two gases obey vander Waal\" equation, their constants al and a2 are equal but b I and b2 are different. Which of the gases under identical conditions wiD occupy more volume? Solution: Under similar conditions of temperature and pressure, vander Waals' equation for two gases 1 and 2 can be written as : P
+!!l.. (VI ~ (V2 -
b 2)
=RT
VI - b l = \/2 - b2 The gas which has higher value of b will occupy more volume. When
al
= a2 , then
b l ) = RT
97
GASEOUS STATE
Ex. 21 : One mole oj NH3 is enclosed in afive litre flask at 27°C. Calculate the pressure oj NH3 using vander Waals equation. For NH3, a = 4.17 atm lif mole-2 and b = 0.0371 lit mole-to
Solution: vander Waals' equation is given by,
(p+ ~ )cV-b)=RT Substituting the values, we get
. (p + 4;~7) (5 - 0.0371) or
or
= 0.0821
x 300
(25P ;54.17 )<4.9629) = 0.0821 x 300 1 [0.0821 x 300 x 25 ] P = 25 0.9629 - 4.17 = 4.71 atmosphere
Ex. 22 : From vander Waals equation, show that
3
PcVc=SRTc
Solution: vander Waals equation under critical conditions is written as: pc +
or
~ )(V- b) = RT('
a V--=O ab V3 - (RTc b+- V2 +-
Pc Pc Pc \ At critical point, the three values of volume are identical and equal to Vi.' Thus, . (V-
vi=o
or V3 - 3 Vc y2 + 3 ~V - V~ = 0 As (1) and (2) are identical. therefore. 3Vc =b+
Rt
pc
... (2)
... (3)
... (4) ... (5)
Dividing (5) by (4), we get Vc=3b
b= Vc 3 Substituting the value of b in equation (3), we get Vc RTc Vc RTc 3V = - + - or 3V - - = c 3 Pc c 3 Pc
or
... (6)
98
PHYSICAL CHEMISTRY-I
9Vc - Vc 3
or
RTc
= Pc
RTc
8
or
"3 Vc = Pc
3
or
PcVc=iRTc
Ex. 23 : If vander Waals' equation of a gas is given by :
(p + 0.~;86) (V - 0.00224) =0.0041 (273.16 +
t)
Calculate the values of Tc and Pc of the gas.
Solution : The values of a, b, Rand T are given by : a =0.00786, b =0.00224, R =0.0041, T= (273.16 + t) The values of Pc and Tc are given by: (i)
P c
(1·1·)
T c
=_a_2 = 27b
0.00786 27 (0.00224)2
8a
=58.01 atm.
8 x 0.00786
= 27bR = 27 x 0.00224 x 0.0041 =
253 5 K.
.
Ex. 24 : 2 moles of CO2 at 27°C is filled in a five litre fliJsk. Calculate its pressure using the following: (i) Gas equation, (ii) vander Waals equation. For CO2 : a lit mole-I.
=3.6
atm l;r mole-Zj R
=0.082 lit atm,
b
=4.28 x 10-z
Solution : (i) Pressure from gas equation:
PV=nRT or P=
~T
P = 2 x 0.082 5 x 300 =9.84 a tmosph eres. (ii) Pressure from vander Waals' equation:
(p + n;)(V - nb) = nRT
n
(where = No. of moles = 2)
(p + 4 ~;.6) (5 _ 0.0428 ~ 2) = 0.082 x 300 x 2 or
or or
(25P + 14.4) = 0.082 x4~9~~ 25 x 2
25P= 0.082 x3oox25 x2 -144 4.9144
P = 235.8 25
=9.43
Pressure = 9.43 atmospheres.
.
.
99
GASEOUS STATE
Ex.2S : Calculate the pressure using (i) vander Waals' equation and (ii) ideal gas equation of 10 g ammonia contained in a 1 litre flask at O·C. (a = 4.17 atm litr; mole-", b = 0.0371 litre mole-I) Solution: (i) Using vander waals' equation
(p + n~: ) (V - nb) =nRT p or
[
+
~~ 2 (4.17)] 12
[1-(!~)<0.0371)]=(~~)<0.0821) (273)
or
(P + 1.442) (1 - 0.0218) = 13.1843
or
(P+ 1.442) (0.9782) = 13.1843 P = 13.1843 0.9782
1442 . = 12.036 atm (ii) Using ideal gas equation PV=nRT
or
or
P x 1=
(~~) (0.0821) (273)'
P = 13.18 atm
or
Ex. 26: 0.5 mole of CO2 was filled in a vessel of volume 0.6 litre at 47·C. What pressure would be expected on the basis of vander Waals equation? (a =3.36 atm luJ mole-", b =4.27 X 10-2 lit mole-I; R =0.082 lit atm deg- I mole-I. From vander Waals equation,
(p + n;) (V - nh) = nRT.
or
[p + (0.5)2(0.6)x ;.36] (0.6 _ 0.5 x 0.0427) = 0.5 x 0.082 x 320
or
(P + 2.33) x 0.57865 = 13.12
or
P
13.12 =0.57865 -
2.33 =20.34 atmospheres
Ex. 27 : At 27°C, the volume of 1 mole of CO is 1.5 litre. If R is taken in litre atmosphere and values of a and bare 1.5 atm lit mole-2 and 0.04 lit mole-I, respectively, then what will be the pressure of the gas according to (a) ideal gas equation and (b) vander Waals equation? Solution: (i) From ideal gas equation:
100
PHYSICAL CHEMISTRY-I
PV==nRT p==nRT == 1 x 0.082 x 300 V 1.5
or (ii)
P == 16.4 atmospheres From vander Waals equation: (p+
or or or
(
~2)(V-b):::::RT
p + ~21 (l.5 - 0.04) == 0.082 (1.5) )
lp /5) x -I-
x 300
1.46::::: 0.082 x 300
P == 0.082 x 300 _ _ 1 == 16.183 1.46 1.5 Pressure == 16.183 atmospheres.
Ex. 28 : For NO gas, vander Waals' constant, b == 0.02788 lit/mole, calculoJe the diameter of NO molecule. (N::::: 6.023 X Itf.1) Solution: We know that b == 4N [ 1nr3 ]
or
27.88 == 4 x 6.023
X
10 x ~ x 3.14 x r3 23
(b == 27.88 cc/mole).
or r3 == 2.76 X 10- cm or r == 1.4 X 10-8 cm. :. Diameter of NO molecule == 2 x 1.4 X 10-8 == 2.8 X 10-8 em. 24
3
Ex. 29 : For hydrogen the vander Waals' constants a and bare 0.246 and 2.67 x 10-1 in lit atmosphere units. Calculate the inversion temperature for hydrogen. Solution : We know that.
2a Ti== b.R
or
2 x 0.246 T, == 0.0821 x 0.0267 ::::: 223.6 K Inversion temperature. T; == 223.6 K or - 49.4°C
MULTIPLE CHOICE QUESTIONS 1.
A gas cannot be liquefied if for a gas the temperature is greater than: (i) Critical temperature (ii) Critical pressure (iii) Critical volume (iv) Critical density
2.
At constant pressure the specific heat of a gas (molecular weight::::: 80) is 0.125. It is :
101
GASEOUS STATE
(i) 3.
5.
(ii) Di-atomic
(iii) Tri-atomic
The first virial coefficient of Kammerling Onnes equation at low pressure is : (i)
4.
Mono-atomic
PV
(ii) RT
The unit of vander Waals constant 'Q' is : (i) mole atm- I (iii) Iit2 atm mole-2
(iii)-b
(iv) ~T
(ii) atm mole-I (iv) atm 1it-2 mole 2
If the rms velocity of a gas is 100 cm s-I, its average velocity is : (i) lOOcms- 1 (ii) 0.01 cm s-I (iii) 92.l3cms- 1
(iv)81.64cms- 1
6.
The conpressibility factor of an ideal gas is : (i) Zero (ii) Infinite (iii)2.0
7.
The temperature at which the second virial coefficient of a real gas is zero. is known as:
8.
(i) Critical temperature
(ii) Boiling point
(iii) Boyle temperature The virial state equation for an real gas is : 8' C' (i) PV=RT+-y+ V2 + ...
(iv) Freezing point
(iii) PV=RT+Rr+R'f3 + ...
9. 10. 11.
12.
(iv) 1.0
. B' C' (Iv)PV=RT+-y+-y+ ...
At 2TC the kinetic energy of 2 moles of an ideal gas is : (i) 1800 cal (ii) 900 cal (iii) 600 cal (iv) 3600 cal The following is called a permanent gas : (iv) CO2 (i) NH3 (ii) N2 (iii) H2 The ratio of most probable, average and root mean square velocity is : (i) I: 1.224 : 1.128 (ii) 1 : 1.128 : 1.224 (iv) 1 : 1.178: 1.524 (iii) 1 : 1.128: 1.524 The vander Waals equation for n moles of a real gas is : (i)
(p + ~) (V-b) =nRT
(ii)
(p + ~) (V- nb) =nRT
(iii)
(p + n:~) (V -nb) = nRT
(iV>(P + n:v2) (V -nb) = nRT
Fill in the Blanks 1. 2. 3. 4. 5. 6.
In general, the more easily liquefiable gases have a .............. Boyle temperature. The average energy of translational motion of gas molecules is proportional only to the .................. . Real gases behave ideally at .................. temperatures and .................. pressures. At 274 K, rrns velocity 'of methane molecules would be .................. ms- I . . vander Waals replaced the volume term, V in the ideal gas equation by ........ . The intermolecular forces which exist between the neutral molecules are known as ................. .
102 7.
8. 9. 10. 11.
PHYSICAL CHEMISTRY-I
The temperature at which the average velocity of O2 molecules would be twice that at 20·C is .................. K. Average kinetic energy of gas molecules is .................. to absolute temperature. The Cp/Cv value for a gas is 1.47. If its atomic weight is x, its molecular weight is ............ . The unit of vander Waals constant a is .................. and b is ................. . The pressure of real gases is less than that of ideal gases because of .............. ..
True or False State whether the following statements are true (T) or false (F) ?
±mNi is an expression for the real gases.
1.
PV =
2. 3;
The molecules of real gases have both volume and mutual attraction. Pressure depends on the number of molecules and temperature.
4.
The equation
5. 6.
The observed pressure of a real gas is less than the pressure of an ideal gas. There are large number of particles in any sample of a gas for which the average velocity must be zero, whereas the average velocity will always be final. Gases which are difficult to liquefy have a low Boyle's temperature. The vander Waals constant b is the actual volume of gas molecules. Root mean square velocity of gas molecules is always greater than the average velocity under similar conditions. The ratio of the average molecular kinetic energy of CO2 to that of SF6 both at 300 K is one.
7. 8. 9.
10.
(1t + :2) (341- 1) = 89 is known as the reduced equation of state.
ANSWERS Multiple Choice Questions 1. (i), 2. (i), 3. (ii), 4. (iii), 5. (iii), 6. (iv), 7. (iii), 8. (i), 9. (i), 10. (iii) 11. (ii) 12. (iii) Fill in the Blanks 1. high 2. absolute temperature 3. high, low 6. vander Waals forces 4.652 5. (V-b) 7.586 K 8. directly 9. directly proportional 10. x 1l.lit2 atm mole-2, lit mole- 1 12. mutual attraction between gas molecules True or False 1.
(F)
2.
(1)
7.
(1)
8.
(F)
..
3. 9.
(1)
4.
(T)
(T)
10.
(1)
5.
(T)
6.
(1)
000
CHEMICAL AND PHASE EQUILIBRIUM CHEMICAL EQUILIBRIUM Problem 1: Exp1o.in 1o.w of mass action and equilibrium constant. Guldberg and Waage (1867) found the effect of reacting substances on a reversible reaction and gave a law known as the law of mass action. According to this law, "The rate at which a substance reacts is directly proportional to its active mass and the rate ofa chemical reaction is proportional to the product of the active masses of the reactants." The term active mass of a reactant is the molecular concentration, i.e., the number of gram moles per unit volume. It is expressed by enclosing the formula of the substance within two square brackets, e.g., [H2], [HI] etc. Consider a reversible reaction, A + B ~ C + D According to the law of mass action, Rate of forward reaction oc [A][B] = kJ [A][B] where kJ is the velocity constant for the forward reaction. Rate of backward reaction oc [C][D] = k2 [C][D] where k2 is the velocity constant for the backward reaction. At equilibrium: Rate of forward reaction = Rate of backward reaction or or
kJ [A][B]
= k2 [C][D]
[C][D] _~_ [A][B] - k2 - Kc
... (1)
where Kc is known as equilibrium constant in terms of concentration. In a gaseous reaction, the partial pressure (P) of the gas is proportional to its molecular concyntration. Therefore, the equilibrium constant in terms of partial pressures, will be
K =PCXPD P
PA XPB
... (2)
where PA,PB,PC andpD represent the partial pressures of the respective constituents. (103)
104
PHYSICAL CHEMISTRY-I
Equations (1) and (2) are the mathematical forms for the law of mass action. In general, for a reaction n 1A+n2B + ... ~ m 1C+m2D + '" we can write,
K C
= [C]m. [D]mZ ... [A]n. [B ]n2 ..•
m.
Similarly,
X
m,x
K = !....P"'-C_P!....D/.!.-_ _ • '-' P
n. X n2 X
PA
PH
•••
Problem 2: Write short notes on the following: (i) Workfunction (ii) Free energy
[I] Work Function We know that the concept of entropy is the fundamental consequence of the second law of thermodynamics. There are two other functions, which utilize entropy in their derivations. These two functions are 'free energy function' and 'work function.' These functions like the internal energy, heat content and entropy are fundamental thermodynamic properties and depend upon the states of the system only. We know that energy can be converted into work, but it is not always
necessary that all the energy may be converted into work. So, any kind of energy which can be converted into useful work is called 'available energy', such as the operation of an engine or a motor. But the energy which cannot be converted into useful work is known as 'unavailable energy'. :. Total energy = Isothermally available energy + Isothermally unavailable energy The isothermally available energy present in a system is called its free energy. It is denoted by the symbol G. Free energy is also known as thermodynamic potential or Gibbs' function. It has also been explained before, that a part of internal energy of a system can be utilized at constant temperature to do useful work. So, this fraction of internal energy which is isothermally available is called the 'work function' of the system. It is denoted by the symbol A. The exact nature of the functions will be clear as follows : The function, G is defined by the equation, G = E - TS + PV ... (1) where, P, V, T as usual refer to the pressure, volume and temperature of the system, E is the energy content of the system whose entropy is S. As E. T and S depend only on the thermodynamic state and not on its previous history, it is clear that the same views must apply to G, the free energy
105
CHEMICAL AND PHASE EQUILIBRIUM
function. Therefore, G may be regarded as single valued function of the state of the system. Hence dG, will be a complete differential. As E and S are extensive properties, therefore. G will also be extensive in character or in other words, the value of G will depend on the quantity of matter specified in the system. As E + PV = H, Equation (1) reduces to. G=H- TS ... (2) The work function, A is defined by, A=E-TS ... (3) where, all the symbols have their usual significance. Like the free energy, the work function A is also seen to be a single valued function of the state of the system. Hence, likewise dA is also a complete differential. Similarly. as E and S are extensive properties, A will also be extensive in nature.
[II]
Physical Significance of Work Function
In order to obtain some understanding of the physical significance of the work function, consider an isothermal change from the initial state to the final state. The two states are represented by the subscripts 1 and 2, respectively. In such a case, (3) can be written as, Al = EI - TS I ••• (4) and
... (5)
The temperature remains the same as the process is an isothermal one. Subtracting equation (4) from (5), we have, A2 - AI = (E2 - EI) - T (S2 - SI) or
M=AE-T6.s
... (6)
where, M is the increase in the work function, AE is the corresponding increase in internal energy and AS is the increase in entropy of the system. Suppose the change is brought about reversibly at a constant temperature T, then the heat absorbed will be equal to Qrev (say). Thus, the increase in entropy is given by,
AS= Qrev T
Substituting this in equation (6), the result is M = AE- Qrev
... (7)
From first law of the thermodynamics, AE=Qrev- W or
- W=AE-Qrev
Combining this equation with equation (7), we have W=-M
... (8)
106
PHYSICAL CHEMISmY-1
Since the process is carried out reversibly, W represents the maximum work. It is clear from equation (8) that the decrease in work function (- M) gives the maximum work that can be obtained from the system at a constant temperature. It is this fact, which justifies the use of the concerned term as
work function.
[III]
Physical Significance of Free Energy
Consider an isothermal change from the initial state to the final state at constant temperature T. The two states are represented by the two subscripts 1 and 2, respectively as before. Equation (2), therefore, reduces as follows for both the states: G 1 = HI - TS, ... (9) G 2 =H2 - TS2 Subtracting equation (9) from (10), we have (G2 - G,) = (H2 - H,) - T(S2 - S,)
... (10)
... (ll) or !lG = !lH - T !lS where, !lG and !lH represent the change in free energy and change in heat content, respectively. At constant pressure, we know that,
!lH=M+P!lV ... (12) !lG = !lE + P !l V - T !lS Combining equations (6) and (12), we have !lG=M +P!lV or -!lG = W - P!l V [From equation (8)] ... (13) Since P!l V represents the quantity of work done by gas on expansion against the external pressure, therefore, - !lG gives the maximum work which can be obtained from the system other than that due to change of volume, at constant temperature and pressure. The work other than that due to change of volume is caned the net work. Hence, Net work = W - P !lV Therefore, -!lG is a measure of the net work. In other words, the decrease in free energy (- !lG) is a measure of maximum net work that can be obtained from a system at constant temperature and pressure. G can also be defined as, 'the fraction of total energy which is isothermally available for converting into useful work'. This decrease in free energy is of great importance in chemistry, especially in physical chemistry.
[IV] Variation of Free Energy With Pressure and Temper· ature From equation (1), we have
107
CHEMICAL AND PHASE EQUILIBRIUM
G=E+PV-TS On differentiation, we have, dG =dE+PdV+ VdP-SdT-TdS From first law of thermodynamics, we have dE=dQ-dW If the work is only due to expansion, then dW = P dV Hence, dE =dQ - P dV or dQ=dE+PdV dS=
Now we know that, or
... (14)
d~rev
T dS =dQrev =dE + P dV
Combining (14) and (15), we have dG= VdP-SdT
... (15) ... (16)
A constant temperature, dT = 0 dG T = VdP =V (OG) oP
or
.... (17)
T
Equation (17) represents the variation offree energy with pressure. When pressure is constant, dP = 0 ..10 such a case (16) reduces to, dGp=":'-SdT or
(~~)p = -
S
... (18)
Equation (18) represents the variation of free energy with temperature.
[V] Change of Free Energy With Pressure for a Perfect Gas For an isothermal change, dT= 0, hence equation (16) reduces to dG=VdP On integration, we have,
(G2 dG= J,P2 VdP
JGI
PI
RT For 1 mole of perfect gas, V = P
1,G2 dG= J,~2 G.
PI
RT dP
P P2
G2 -G 1 = I1G =RTloge PI
108
PHYSICAL CHEMISTRY-I
For n moles of perfect gas, ... (19) Equation (19) applies to a change occurring at constant temperature for a perfect gas. It also applies, whether the process is reversible or otherwise.
Problem 3: Give thermodynamic derivation of law of mass action. Consider the gaseous reaction A+B~C+D
Suppose there are two large vessels at the same temperature, in each of which the four substances A. B, C and D are always in equilibrium. Suppose each of the four walls of the vessels is permeable to only one of the substances. LetPA,PB,pcandpD be the partial pressures of A, B, C, D in the first vessel, whereas P'A, P' B, P' c and P'D that of A, B, C, D in the second vessel. The vessels are supposed to be so large that the transfer of relatively small quantities of the material from one vessel to the other results in no appreciable change in the concentrations or partial pressures of the substances. It is also assumed that an substances are ideal gases. First a reversible and isothermal process is carried out. Suppose 1 mole of A is transferred isothermally and reversibly from the first vessel to the second. This is done by means of the wall permeable to A. If PA and P'A represent the initial and final pressures, respectively, then the change in free energy (~G) is given by, llG=
P'
J
A
VdP
PA
SO, increase of free energy of the substance A is given by, llGA =
J~
RT J~ -~ VdP= J~ -dP=RT
PA
PA
P
PA
P
=RTlOgp'A IPA
Similarly, for the transfer of 1 mole of B from PB to P'B in the same direction, the change in free energy is given by, llGB = RT log P' B IPB During the time A and B are transferred in one direction, 1 mole of C and 1 mole of D are transferred isothermally and reversibly in the opposite direction. Thus, the free energy change for the transfer of 1 mole of C from P' c to Pc is given by, llGc=\RT logPc1p'c For the transfer of 1 mole of D, we have, llG D = RTlog PD Ip'D
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CHEMICAL AND PHASE EaUILIBRIUM
The total change of free energy is given by. IlG = IlGA + IlG B + IlG c + IlG D 'P'A P' B Pc PD = RTlog- + RTlog-+ RTlog -,- + RTlog -,- ... (1) PA PB PcP D As the vessels are large and the system is in equilibrium at constant remperature T. the total change of free energy at constant temperature is zero. i.e .• IlG =0.
,
,
log P A + log P B + log P,C + log P,D = 0 PA PB PcP D
,
or or or
PA log PA
,
,
,
PB
Pc
PD
(From equation 1)
PD + log -PB = log Pc - + log-
P'AP'B I P'CP'D Io g - - = o g - PAPB
PCPD
=
P'A P'B P'c P'D PAPB
PCPD
, ,
PCPD PcPD - - = -,-,- = Constant (Kp) ... (2) PAPB PAPB As the partial pressure is proportional to the molar concentration. equation (2) becomes.
or
CC' CD
CA' CB
[C] [DJ _ [AJ [BJ - K("
... (3)
The equations (2) and (3) are different forms of law of mass action which have been deduced thermodynamically.
Problem 4: Gil'e the thermodynamic deril'ation of maximum work obtainedfrom gaseous reactions or l'an't Hoff isotherm. When a small change is considered occurring in a reversible reaction. it is assumed that the system is at equilibrium. So. the change in free energy (1lG) can be put to zero. If. however. the constituents are not at eqUilibrium. but at certain arbitrary concentrations. there is a definite change of free energy•. as the reactions get converted into the products. This change in free energy can be determined by the usual equilibrium box. Imagine a large box in which the various gases taking part in the reaction A + B ~ C + D are present at equilibrium. Let the P' terms represent the equilibrium pressure. It is assumed that all gases are perfect. Then the following steps are carried out : (i) Transfer 1 mole of A isothermally and reverSibly at arbitrary pressure PA. into the equilibrium box through the wall permeable to A. Let the final pressure be p'A- Then the change in free energy of the substance A is given by.
110
PHYSICAL CHEMISTRY-I
P'A JPA
JP' PA
dP-RTIogP'A P PA (ii) Similarly, the change in free energy for the transfer of 1 mole of B isothermally and reversibly into the box, from pressure PB to final pressure P' B is given by, AGB = RTlogp'B IpB (iii) With the passage of the reactants A and B into the box, it is assumed that (he products are removed or withdrawn. . The free energy change in the transfer of 1 mole of C from the box isothermally and reversibly, when the pressure of the gas C is reduced from P'c ro Pc is given by, AGe = RTlogpc Ip'c (iv) For the transfer of 1 mole of D, the free energy change when the pressure is reduced from P'D to PD is given by, AGD= RTlOgPD Ip'D AGA =
VdP=RT
A
The rotal change of free energy (AG) for the whole process is given by, AG=AGA +AGB+AGc+AGD P'A P'B Pc PD + RT log - + RTlog -,- + RT log -,PA PB Pc PD
= RTlog -
,
,
I
,
PA PB Pc PD + RT log - - RT log - - RT log PA PB Pc PD
= RT log -
or
aG = RTlog Pc ,PD _ RTlog '!:c ,P:D PA,PA
, ,
PA,PB
... (1)
AG-RTI PCPD - RTIo g PC,PD og -,-,-PAPB PA,PB PC,PD ... (2) or - AG =RTlog Kp - RT l o g - PA,PB where, Kp is equilibrium constant, as the fraction in the second term of equation (1) on R.H.S. is obviously equal to Kp. The first term on R.H.S. is also of the same form, but it involves arbitrary partial pressures instead of 'equilibrium partial pressures. Consider the general equation, aA + bB + cC + .. , ~ pP + qQ + rR + ... for which equation (2) can be written as, -u
- AG =RTlog Kp - RTlog
or
p'f,POPR ... a b ,.
PAPBPC ... - aG = RTlog Kp - RTI.lIlogp
... (3)
whe:-e. I.n log P represents the algebraic sum of all the II log P terms, those for products being taken as positive and for reactants as negative. Equation (3)
111
CHEMICAL AND PHASE EQUILIBRIUM
is one of the forms of reaction isotherm deduced by van't Hoff (1885). If all the partiaJpressure terms are replaced by concentration (c) terms, then equation (3) can be written as, -I1G = RTlog K,. - RTLn loge. . .. (4) Similarly, if partial pressure or concentration terms are replaced by mole fraction (x) tenns, then equation (3) or (4) becomes at constant pressure, -I1G = RT log Kx - RT Ln log x, ... (5) So, van't Hoff isotherm gives the increase offree energy change accompanying the transfer ofreactants at any arbitrary concentration (or partial pressure) to products at any arbitrary concentration (or partial pressure). Equations (4) and (5) are other fonns of van't Hoff isothenn. The maximum work obtained is also given by equations (3), (4) and (5), and often -I1G is called the affinity of the reaction.
Problem 5: Give the thermodynamic derivation ofvan't Hoffisochore or van't Hoff equation. Mention its applications also.
[I] Thermodynamic Derivation The vant Haff isotherm is given by, -I1G = RTlog Kp - RT Ln log p where all letters have their usual significance. For a homogeneous gaseous reaction in which the products and reactants are at unit pressures, the last equation becomes I1G = - RTlog Kp ... (1) Differentiating equation (1) with respect to temperature, at constant pressQre, the result is
a(I1G)] = _ R Iog Kp _ RT d log Kp dT [ aT p On multiplying equation (2) by T, we get T
a(I1G)] [ aT p = -
Combining equations (1) and (2), the • T
or
...2.
RTlog Kp - Rl res~lt
a (I1G)] _ _ r [ aT p -I1G R
d log Kp dT
... (2)
is,
d log Kp dT
Rrdl~~Kp=I1G-T[a~~G)t
... (3)
From Gibb's-Helmholtz equation, we have Standard free energy: Standard free energy of a reaction gives the increase of free energy when the reactants, all in their standard states of unit pressure or unit concentration are converted at constant temperature (1) into the products, which are also in their standard states. The standard free energy (AG"T) is given by the expressions,
AGOT = - RTiog K,J
112
PHYSICAL CHEMISTRY-I
AG
1
=AH + T[d ~ATG)
..
(4)
From Gibbs-Helmholtz equation (4), equation (3) reduces to,
or
Rrdl~~Kp
AH
dlogKp dT
AH Rr
... (5)
Equation (5) represents the variation of equilibrium constant with temperature at constant pressure. This equation is referred to as van't Hoff reaction isochore (Greek: isochore = equal space), as it was first derived by van't Hoff for a constant volume system. Since AN is the heat of reaction at constant pressure, the name isochore is thus misleading. Therefore, equation (5) is also called as van's Hoff equation. An alternative form of equation (5) showing the variation of K(. with temperature and involving the heat of reaction at constant volume is obtained as follows: We know that, ... (6)
where, An is the increase in the number of molecules in the gaseous reaction. Taking logarithms of equation (6) and differentiating it with respect to temperature, we have, log Kp = log K(. + An log RT
d log Kp d log Kc + An dT dT T Combining equations (5) and (7), we have dlogKc AN An dT = Rr-T' or
d log Kc dT
or
dlog Kc dT
... (7)
AN - An . RT
Rr
!J.E
=Rr
... (8)
[c.f. thermodynamics, AE = AH - An R71 where, !J.E is the increase of energy or heat of reaction at constant volume. Equation (8) is another form of van't Hoff isochore or equation.
Integrated Form of van't Hoff Isochore If AH remains constant over a range of temperature, then on integrating equation (5), the result is,
113
CHEMICAL AND PHASE EQUILIBRIUM
f
log., Kp =
till t1H Rr dT =- RT + constant
t1H 10gIO Kp = - 2.303 RT + constant
" .(9)
... (10)
From equation (9), it is clear that there is a linear relation between log Kp and liT. On integrating equation (5) between two temperatures T j and T2 at which the equilibrium constants are Kp and K,:. we have, 10geK'p-loge Kp = -
~
(;2 -;1)
On converting loge to 10gIO, we get IOglO K'p -IOglO Kp = - 2.:R
(;2 -;J
..
(11)
Equations (9), (10) and (11) are integrated forms of van't Hoff
isochore. [II] Applications of van't Hoff Isochore (i) To fmd the heat of reaction (MJ): It can be obtained by either of the following two methods : (a) If values of IOglO Kp are plotted agairist liT, we get nearly a straight line. The slope of this line at a given point will be equal to - MJI2.303R. The value of till can thus be measured by measuring this slope. (b) The second way to calculate t1H is from equation (11). Knowing the equilibrium constants at two given temperatures, we may find the value of t1H. (ii) van't Hoff isochore is also applicable to find the heat of solution of a substance from the temperature coefficient of the solubility, the heat of dissociation of a weak electrolyte from the temperature coefficient of the dissociation constant etc. According to equation (5) the equilibrium constant of a gaseous reaction increases with temperature, if till is positive. An increase in the equilibrium constant means an increase in the proportion of the resultants, so that if heat is absorbed in the reaction, an increase in temperature favours the formation of the resultants. On considering the reverse case, if heat is evolved in the reaction, an increase of temperature displaces the equilibrium in the direction ofthe reactants. The behaviour is again in accordance with the Le-Chatelier's principle.
Problem 6: Give the thermodynamic derivation of Clapeyron equation and Clausius-Clapeyron equation. Discuss their applications also.
114
[I]
PHYSICAL CHEMISTRY-I
Clapeyron Equation
Suppose a single substance exists in two phases A and B in equilibrium with each other at constant temperature and pressure. If one mole of a substance is transferred from one phase A to the other phase B without altering the temperature and pressure, then there will be no work done other than that of expansion. So, from equation (dG = V dP - S d1), we get, dG = 0, as dP = 0 and dT = 0 i.e., In other words, the molar free energy of a substance is the same in the two phases which are in equilibrium. In a phase change, dG = VdP - SdT can be written as ... (1) dGA = VA dP-SA dT and
... (2)
Subtracting equations (1) from (2), we get, dGB-dGA = (VBdP-SBdn - (VA dP-SA dn
or or VBdP-SBdT= VAdP-SAdT or or
(VB - VA) dP = (SB - SA) dT
l::.VdP=l::.SdT dP -l::.S or dT l::.V· where, l::.S and l::. V have their usual significance.
Recognising further that at equilibrium we have the change in entropy (t::.S), given by l::.S=l::.H
Therefore,
dP dT
T t::.H
[As l::.G = 0, at equilibrium]
Tl::.V
... (3)
Equation (3) was first derived by a French engineer Clapeyron in 1834. It gives the variation of the equilibrium pressure (P) with temperature (n for any two phases of a given substance.
[II] Application of Clapeyron Equation Equation (3) is ~pplicable to various equilibria such as solid-liquid equilibria, liquid-vapour equilibria and equilibria between two solid modifICations. The Clapeyron's equation for these various equilibria can be easily obtained as follows : (1)
Solid-liquid Equilibrium
We know that solid and liquid forms of a substance can exist in equilibrium only at the freezing or melting point. Hence, in equation (3), Twill
115
CHEMICAL AND PHASE EQUILIBRIUM
be the freezing point and P will be the external pressure exerted on the system. So, equation (3) can also be written in a reversed form as : dT T~V dP= MI ... (4) If Vs and VL represent the molar volumes of the solid and the liquid phases, respectively at temperature T and pressure P, then, ~V=VL-VS
where, ~ V represents the increase in volume in transferring 1 mole from solid to liquid phase, MI the amount of heat absorbed may be replaced by ~Hf' i.e., molar heat of fusion. So making these changes in equation (4), we get the following Clapeyrons's equation.
dT T(VL - Vs) dP= Ml
... (5)
f
Liquid-vapour Equilibrium In this equilibrium, the increase in volume (~V) accompanying the transfer of one mole of liquid to the vapour state will be equal to Vv - VL. where Vv and VL represent the molar volumes of the vapour and liquid, respectively.I1H may then be replaced by i1Hv, i.e., molar heat of vaporisation. From the above considerations, equation (4) becomes: dT _ T(Vv - VL) ... (6) dP Mlv (2)
It is a well known fact that the boiling point of a liquid is the temperature at which its vapour pressure becomes equal to the external pressure. So, equation (6) represents the variation of boiling point of a liquid with pressure P. Thus equation (6) can also be written as : dP Mlv ... (7) dT T(V v - VJ Equation (7) represents the rate of change of vapour pressure (P) of the liquid with temperature. Note: The volume measured at the same temperature and pressure, i.e., Vvand VL are sometimes known as 'orthobaric volumes'.
[III] Clausius-Clapeyron Equation If the temperature of the liquid is not too near the critical point then we can easily neglect the volume of the liquid (VL), compared to that of the vapour (Vv). In such a case equation (7) reduces to equation (8) on replacing Pby p. So, ... (8)
116
PHYSICAL CHEMISTRY-I
Furthermore, under such conditions the vapour pressure is very samll, so it may be assumed that the vapour behaves as an ideal gas to which the equation PVv = RT is applicable. So, dp AHv dT=P· RT-
:. From equation (8), 1 dp
AHv
or
p. dT= RT-
or
dlogp AHv ( i f = RT-
... (9)
Equation (9) is sometimes known as Clausius-Clapeyron equation and is generally spoken to as first latent heat equation. It was first derived by Clausius (1850) on the thermodynamic basis of Clapeyron f?quation. Equation (9) is valid for evaporation and sublimation processes, but not valid for transitions between solids or for the melting of solids. ClausiusClapeyron equation is an approximate equation because the volume of the liquid ~as been neglected and ideal behaviour of the vapour is also taken into account.
[IV] Integrated Form of Clausius-Clapeyron Equation : Assuming the heat of vaporisation to be independent of temperature, if we integrate equation (9) between the limits TI to T2 (for temperature) and PI to P2 (for vapour pressure) we get,
f
Pz
1
d log P = T A ~ dT z
TI
PI
or or
Rr
r
log P2 = AHv Tz dT e PI R JTI TP2 10glO PI
AHv (T2 - TI) =2.303R T) T2
... ( 10)
If AHv is expressed in cal/mole and R = 1.987 cal/degree/mole, then equation (10) becomes pz AHv (T2 - TI) loglo PI = 4.576 TIT2
... (11)
Equation (11) is an integrated form of Clausius-Clapeyron equation. If the integration is carried out indefinitely (without limits) then we can write the vapour-pressure equation (9) as, AHv
10& P = - RT + C (constant)
... (12)
117
CHEMICAL AND PHASE EQUILIBRIUM
[V]
Applications of Clausius-Clapeyron Equation
(i) Latent heats from vapour pressure data: Suppose the vapour pressures PI and P2 are determined at two temperatures T. and T2 respectively; then from (11) it is possible to calculate the molar or specific heat of vaporisation. It is also possible to calculate the value of Hv graphically. It is clear from equation (12) that if log P is plotted against the reciprocal of the absolute temperature, i.e., liT, then a straight line is obtained. The slope of this line will be equal to - AHIR. So, knowing the value of R, the value of AH can thus be easily determined after measuring the slope. (ii) Deduction of Trouton's law: Equation (9) can be written in the form,
1 dp
AHv
p' RT= Rr
For liquid-vapour equilibrium ANv = Le , i.e., molar heat of evaporation. Hence, the last expression becomes,
1 dp
Le
p' dT= Rr 1..!!P- _ Le
or
... (13)
p dT- RT
We know that,
TITc = e and PIPe = 1t, where Pc, Tc represent the critical pressure and critical temperature, respectively and 1t, are the reduced pressure and reduced temperature, respectively. So (13) \>ecomes,
e
e .d1t - =Le 1t
de
... (14)
RT
vander Waals (1888) suggested the empirical relationship,
P ( I -Tc) log-=k Pc T where k is a constant which is equal to 3 for many substances. Hence in reduced terms it can be put as, log 1t = 2.3k (1 - lie) Differentiating equation (?) with respect to
... (15)
e, we get,
dlog1t =2 3~ de . e2
or or
.!. . d1t =2.3 ~2 1t
de
e
d1t k de =2.3 9
n'
e
... (16)
118
PHYSICAL CHEMISTRY-I
From equations (14) and (16), we conclude that, Le k RT=2.3 or
e Le T=6.g e R
[As k = 3]
If the temperature is taken to be the boiling point Tb , then Tb/Tc is nearly equal to 0.6 and, therefore,
e, which
is
Le -:::: 23 [As R = 1.99/degree/mole] ... (17) Tb So, the molar heat of vaporisation of a liquid divided by its boiling point 011 absolute scale, i.e., Le/Tb is constant and is approximately equal to 23, provided the latent heat is expressed in calories. This fact was first observed by Pictet (1876) and rediscovered by Ramsay (1877) and Trouton (1884) and is commonly known as Trouton's law. Problem 7: State and explain Le-Chatelier-Braun principle and mention its applications to different equilibria. (Meerut 2006) Le-Chatelier and Braun (1888) put a generalisation which helps us to predict the effect of changes of pressure, temperature and concentration on the course of two opposing processes at equilibrium. It is known as Le-Chatelier-Braun or simply Le-Chatelier's principle. It is widely applied to physical and chemical processes. According to this principle, 'If a system is in equilibrium and one of the factors, i.e., pressure, temperature or concentration, involved in the equilibrium is altered, the equilibrium will shift so as to tend to annul the effect of the change.' (1) Effect of temperature change on the position of equilibrium In reactions which proceed entirely in the gas phase, phosphorous pentachloride dissociates to phosphorous trichloride and chlorine or hydrogen iodide dissociates to hydrogen and iodine in a reversible reaction as follows: PCI5(g)~ PCI 3 (g)+CI 2 (g); DoH = Qkcal. 2HI (g)
~
H2 (g) + 12 (g);
DoH = x kcal.
In the above, forward reaction (dissociation of PCI 2 or HI) is accompanied by the absorption of heat. If the temperature is increased, the equilibrium will be disturbed. Le-Chatelier's principle requires the reaction to respond to oppose this change, that is to lower the temperature. This can be achieved if the forward reaction which is endothermic, is allowed to predominate over the backward reactIon, which is exothermic. In such a case, the position o~ balance of the reaction is disturbed and we say that the position of equilibrium has been shifted from left to right, In other words, the dissociation of PCI5 or HI increases.
119
CHEMICAL AND PHASE EQUILIBRIUM
We may summarise the effect of temperature on a chemical equilibrium as follows: Forward reaction (left to right)
Change in temperature
Exothermic
Increase Decrease
Endothermic
Increase Decrease
Effect on position of equilibrium New equilibrium has more of substances on left (reactants in forward reaction). New equilibrium has more of substances on right (products in forward reaction). New equilibrium has more of substances on right (products in forward reaction). New equilibrium has more of substances on left (reactants in forward reaction).
(2) Effect of pressure change on the position of equilibrium Consider the gas phase reaction involving the decomposition of dinitrogen tetra-oxide into nitrogen dioxide. N 20 4 (g) ~ 2N02 (g) Other reaction is: PCIs (g) ~ PCl 3 (g) + Cl2 (g) At equilibrium, the mixture will contain the two compounds in a definite proportion. If the pressure is increased, Le-Chatelier's principle demands that the equilibrium position of the reaction should change in order to restore the balance and this can occur by a decrease in volume (since the total capacity of the reaction vessel is fixed, a decrease in volume of the gases is equivalent to a decrease in pressure). An increase of pressure will thus shift the eq~ilib rium to the left, i.e., dissociation of N20 4 or PCls is decreased. However, pressure will have no effect on those reactions in which there is no change in the number of'molecules as a result of the reaction, i.e., in the reaction 2HI ~ H2 + 12, The effect of pressure on an equilibrium system may be summarised as follows: Type of reaction I.
2.
3.
Effect of increase in pressure
Increase in number of Position of equilibrium molecules, left to right, e.g., moves to the left, i.e., less dissociation of Pe1 5. PCI 5 ~ Pel 3 + C12· Decrease in number of Position of equilibrium molecules left to right, e.g., moves to the right, i.e., more NH3 will be formed. N2 + 3H2 ~ 2NH 3· No change in number of No effect molecules, left to right, e.g., Position of maintained. H2+12 ~ 2HI.
Effect of decrease in pressure Position of equilibrium moves to the right, i.e., more dissociation of Pe1 5.
Position of equilibrium moves to moves to the right, i.e., more NH3 will be formed. No effect. equilibrium Position of equilibrium maintained.
I
(3) Effect of concentration change on the position of equilibrium If the concentration of one of the substances present in an equilibrium reaction is changed without change in any of the other conditions, then by
120
PHYSICAL CHEMISTRY-I
Le-Chatelier's principle, the position of equilibrium will move to decrease the concentration of the added substance. Thus, in the reaction N2 (g) + O2 (g) ~ 2NO (g), at a given temperature, adding N2 or O2 would shift the equilibrium from left to right, i.e., more nitric oxide will be formed. The effect of changes in concentration of substances on the position of equilibrium in a chemical reaction may be summarised as follows: Change in concentration of substance
EtTect on eqUilibrium position of reaction A+B~C+D
Increase in concentration of A or B Proportion of C shifts to right Decrease in concentration A or B Proportion of C shifts to left Increase in concentration C or D Proportion of A shifts to left Decrease in concentration C or D. Proportion of A shifts to right.
and D increased, i.e., equilibrium and D decreased, i.e., equilibrium and B increased, i.e., equilibrium and C decreased, i.e., equilibrium
Applications of Le-Chatelier's Principle (1) Physical Equilibria (a) Melting point of ice: Ice melts with decrease in volume ,as well as absorption of heat, e.g., H 20 (s) ~ H20 (I). It is represented as : Ice
Water
(More volume)
(Less volume)
- Heat
Increase of pressure or temperature will shift the equilibrium from left to right. In other wards, melting point of ice is lowered by an increase of pressure or temperature. (b) Vaporisation of water : The equilibrium between water and steam is represented as : Water ~ Water vapour - Heat (Less volume)
(More volume)
On increasing the temperature, the equilibrium will shift in that direction in which heat is absorbed, i.e., forward reaction. So, more steam will be produced. Similarly, on increasing the pressure, the equilibrium will shift in that direction in which volume is decreased, Le., backward reaction. So, steam will condense into liquid. In other words,formation of steam will befavoured by increase of temperature and decrease of pressure. (c) Solubility o/substances: Certain substances like sugar, NaCI etc. dissolve with an absorption of heat, e.g., Sugar + aq ~ Sugar (aq) - Heat So, increase of temperature will shift the equilibrium to the right. So, the solubility of such substances increase on increasing the temperature.
121
CHEMICAL AND PHASE EQUILIBRIUM
Certain substances like Ca(OHh etc. dissolve with an evolution of heat,
e.g., Ca(OH)z + aq ;=::: Ca(OHh (aq) + Heat So, increase of temperature will shift the equilibrium to the left, i.e., direction in which heat is absorbed. So, the solubility of such substances
decrease on increasing the temperature. (d) Solubility of gases in liquilfs: Consider the solution of a gas in equilibrium with the gas. The equilibrium can be represented as : Gas + Solvent ~ Solution of gas I I (Less volume) (More volume)
If pressure is increased, volume will be reduced without affecting the pressure and some of the gas will dissolve in the solvent. Thus, the solubility of the gas increases on increasing the pressure. [2] Chemical Equilibria (a) Synthesis of ammonia by Haber's process: Haber's process involves the reaction Ml=-22 kcal 1 vol.
3 vol.
2 vol.
(i) Effect oftemperature,' If the temperature of the reaction is lowered, the equilibrium must shift so as to tend to raise the temperature again (LeChatelier's principle). That is, heat must be liberated by the production of ammonia. That is, low temperature favours the formation of ammonia. But
lowering of temperature reduces the rate of reaction, so it is necessary to use a catalyst which will give a sufficient reaction rate inspite of a relatively low temperature. (ii) Effect of pressure. Ammonia is produced from its elements with reduction of volume. Therefore, if the system is in equilibrium and the pressure is then raised, the equilibrium must shift so as to tend to lower the pressure (Le-Chatelier's principle). To do this, the volume must be reduced by the production of more ammonia. That is, high pressure favours the formation of ammonia. (iii) Effect of concentration. If the system is in equilibrium and more N2 is added to increase its concentration, then according to Le-Chatelier's principle, the equilibrium will shift so as to tend to reduce the N2 concentration. That is, more ammonia will be produced to use up N 2. This increases the yield of ammonia relative to H 2, and vice versa if the H2 concentration is increased. The formation of ammonia is favoured by : (i) Low temperature (ii) High pressure, and
122
PHYSICAL CHEMISTRY-I
(iii) High concentration of the reactants. (b) Formation of sulphuric acid by the contact process: The first step in the production of sulphuric acid is the conversion of sulphur dioxide into sulphur trioxide according to the reaction 2S0 2 (g) + O 2 (g) ~ 2S0 3 (g); tJ.H = - 47 kcal 2 vol.
1 vol.
2 vol.
This reaction is just similar to the synthesis of ammonia described above. So, the effect of pressure, temperature and concentration will oe the same as mentioned in the synthesis of ammonia. Low temperature, high pressure and increased concentrations of S02 and O2 wiU favour the formation of sulphur trioxide. The S03 is removed from the equilibrium mixture by dissolving it in fairly concentrated sulphuric acid, forming oleum which is then diluted to get the acid of the required concentration. (c) Formation of nitric oxide: The reaction is represented as N 2(g) + 02(g) ~ 2NO(g); tJ.H = + 43.2 k. cal. I vol.
I vol.
2 vol.
(i) Effect of pressure: As no change of volume occurs during the formation of nitric oxide, there will be no effect ofpressure on the equilibrium. Oi) Effect of temperature: If the temperature is increased then the equilibrium will shift in that direction in which heat is absorbed, i.e., in the forward direction. So, high temperature favours the formation of nitric oxide. (iii) Effect of concentration: If to the system in equilibrium N2 is added, the equilibrium will shift in that direction so as to reduce the concentration of N2• So, more nitric oxide will be formed. Similar is the effect of adding oxygen. So, the formation of nitric oxide is favoured by (i) high temperature and (ii) high concentrations of N2 or O2,
Problem 8: Explain with reasons that high pressure and low temperature are favourable for the high production ofammonia gas by Haber's process. (Meerut 2006)
See Problem 7.
NUMERICAL PROBLEMS Problem 1: In the formation of silver chloride from its elements under normal conditions, tJ.G is - 26.3 k .. cals. and m is -30.3 k. caL per mole at 18°C. What is the corresponding entropy change? Solution: From the following, we have equation
tJ.G = tJ.H - TtJ.S or
-26.3 = -30.3 - 291 x IlS
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CHEMICAL AND PHASE EQUILIBRIUM
or
AS = -30'~9~ 26.3
=0.0137 k. cal mole-1
Problem 2: Calculate the change infree energy (in cals.) which occurs when 2 g moles ofa perfect gas expands reversibly and isothermally at 37°C from an initial volume of 55 litres to 1000 litres. Solution: The change in free energy for an isothermal expansion is given by, VI
I1G = 2.303 nRT log 10 V
2
VI = 55 litres; V2 = 1000 litres;
T= 310 K; R = 1.987 cal. deg-I mole-I; n =2. 55 I1G = 2.303 x 2 x 1.987 x 310 loglO 1000
= 2.303 x 2 x 1.987 x 310 (log 55 -log 1000) = - 3.574 cals.
Problem 3: At 300 K and 1 atmosphere pressure N 20 4 is 20% dissociated to N02 • Calculate the standard free energy change for the reaction. Solution: N20 4 1 (1 - 0.2)
~
2N02 0 0.4
[Initially) [At equilibrium)
Total number of moles = 1 - 0.2 + 0.4 = 1.2.
and On applying the law of mass action, 2
K = PNo! p
PN 0
2 4
= 119 =1.. 2/3
6
Standard free energy change (I1GO) is given by, I1Go = - RT loge Kp = - 2.303 RT loglO Kp
=- 2.303 x 1.98 x 300 log 116 =1075.22 cals. Problem 4: At 1000 K water vapour at 1 atmosphere pressure has been found to be dissociated into hydrogen and oxygen to the extent of 3 x 10-5%. Calculate thefree energy decrease of the system in this reaction. (R = 1.98 callmole/degree). Solution: The partial pressures in the equilibrium mixture are thus given as,
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PHYSICAL CHEMISTRY-I
7 7 PeHp = 1; PeH 2 = 3 X 10- ; PeO! = 3/2 X 10as one molecule of water yields one molecule of hydrogen and half a molecule of oxygen. For the reaction, 2H2 + O2 = 2H20 2
We have,
PeH,O 1 K ----:-;------= 14 7 p - P;H! x PeO! - 9 X 10- x 1.5 X 10-
1 =----= 1.35 X 10-20 The change in free energy is given by,
AG = - RT loge Kp = - 2.303 RT log 10 Kp = - 2.303
x 1.98 x 1000 log
1 20 1.35 x 10-
= - 2.303 x 1.98 x 1000 (log 1020 -log 1.35) = - 2.303 X 1.98 X 1000 X 19.8697 = -90,610 cals. :. Decrease in free energy of the system is 90,610 cals.
Problem 5 : The equilibrium constant of the reaction 2S02 + O2 ~ 2S03> at S28°C is 98.0 and at 680°C is 10.5. Find out the heat of the reaction. Solution: We know that, loge Kp (I)
-
loge Kp (2) = -
~
UI -~)
where, Kp is the equilibrium constant and All is the heat of reaction. log Kp (I) = 98.0 Kp (2) = 19.5 = 10.5 TI = 528 + 273 = 801 K T2 = 680 + 273 = 953 K or loge 98 -loge 10.5 = -
1~~7 (8~1 - 9~3)
953 X 801) All =- 1.987 X 2.303 (loglO 98 -loglO 10.5) ( 953 _ 801 = - 1.987 X 2.303 X 0.97 X
953 X 801 152
=- 22,290 cals. Problem 6: The vapour pressures of water at 9S0C and 100°C are 634 and 760 mm., respectively. Calculate the latent heat of evaporation of water per gram between 9S0C and 100°C. Solution : or
[1-_ 1-]
log P2 = Le W PI R TI
T2
P2 Le [T2 - TI] 2.303 log]() PI = Ii TI T2
= Le [T2 - TI]
R
TI T2
CHEMICAL AND PHASE EQUILIBRIUM
125
where, PI and P2 are the vapour pressures at temperature TI and T2, respectively and L. is the latent heat of evaporation/mole. It is given that: PI =634 m.m. P2 = 760 m.m. TI = 95 + 273 = 368°K T2 = 100 + 273 = 273°K So,
760 Le [373 - 368] 2.303 IOglO 634 = 1.987 368 x 373
or
Le = 9886 cals/mole = 9886/18 = 549.2 caVg.
Problem 7: At what height must the barometer stand in order that water may boil at 99°C? Given that latent heat of vaporisation of water per gram is 536 cal As known, the integrated form of Clapeyron-Clausius Solution equation is, P2 -Le[T2 TI] -- Iog e PI - R TIT2
=76 cm. mercury P2 =? TI = 100 + 273 =273°K T2 =99 + 273 = 372°K
PI
Le = 536 cals/g = 536 x 18 cals/mole P2
2.303 IOglO 76 =
536 x 18 [372 - 373] 1.987 372 x 373
536 X 18 xl 1.987 x 372 x 373 536 x 18 xl loglo P2 -loglo 76 = - 1.987 x 372 x 373 x 2.303
=
or
536 x 18 xl loglo P2 = loglo 76 - 1.987 x 372 x 373 x 2.303
= 1.8808 - 0.0159 =1.8659 Taking antilog we get, P2 = 73.43 cm. of mercury Problem 8: The dissociation pressure of magnesium sulphate hydrated is 35.6 m.m. at 35°C and 47.2 m.m. at 40°C, the hydrated salt being in equilibrium with the anhydrous salt. Calculate the heat of dissociation of the hydrate.
Solution: We know that, dlogp _JL. dT - Rr where, Q is the heat of dissociation per mole of salt. Integrating this between proper limits, we find,
126
PHYSICAL CHEMISTRY-I
or
or
Substituting the values. we get. 47.2) 313x308 Q = 2.303 x 1.987 ( loglO 35.6 x 5 = 13,610 cals. Problem 9 : Cakulate the equilibrium constant for a reaction in which AGo value is - 22 k cal at 25° C. (Meerut 2006,2007) Solution. From vant Haff isotherm. in the standard state
= - RTlog K K=_~G· =_
AGo
10
or
or
g
RT
(-22 k cal) (2 x 10 - 3 k cal)(298)
22 = 36.9 2 x 10- 3 x 298 Taking anti log. K = 0.7943 X 1037 36 K = 7.943 X 10
PHASE EQUILIBRIUM Problem 1 : Explain and illustrate the terms phase, component and degree offreedom. (Meerut 2006, 2005) In order to deal effectively with the heterogeneous equilibria, W.J. Gibbs (l873J gave a generalised rule in the form of phase rule. Its utility was not immediately known until several scientIsts like Ostwald. Roozehoom applied the rule to several well known heterogeneous equilibria. Before defining phase rule one must clearly understand the definitions of three basic terms, which are frequently used In this connection. The three terms are : (1) Phase, (2) Number of components, (3) Degree of freedom.
[I] Phase A phase is defined as "any homogeneous, physically distinct part of a system which is mechanically separable and bounded by a definite surface. A phase can exist in either state of matter. viz .• solid. liquid or gas. A system may consist of one phase or more than one phases.
127
CHEMICAL AND PHASE EQUILIBRIUM
Examples: 1. Pure substances: A pure substance (s, / or g) made of one chemical species only has one phase, e.g., oxygen (02), ice (H20), alcohul (C 2H50H) etc. 2. Mixture of gases: A mixture of gases, say H2, N2 and O2 contributes one phase only as all gases mix freely to form a homogeneous mixture. 3. Miscible liquids: Two or more completely miscible liquids give a uniform solution. e.g., a solution of water and ethanol has one liquid phase. 4. Non-miscible liquids: A mixture of non-miscible liquids forms as many number of liquid phases as that of liquids, because on standing they form separate layers, e.g., a mixture of water and chloroform forms two liquid phases. 5. Aqueous solutions : An aqueous solution of a solid substance such as NaCI is uniform throughout. So, there is only one liquid phase. 6. Mixture of solids : A mixture of two or more solid substances contains as many phases. Each of these substances have different physical and chemical properties and form a separate phase. Thus, a mixture of calcium carbonate and calcium oxide has two solid phases. 7. In the dissociation of calcium carbonate there will be three phases, viz., two solid phases (CaC03 and CaO) and one gaseous phase (C02), 8. In water system, there are three phases, viz., ice (solid), water (liquid) and vapours (gas). Similarly, in sulphur system, there are four phases, viz., rhombic sulphur and monoclinic sulphur (solids), liquid sulphur (liquid) and vapour sulphur (gas).
[II] Components The number of components of a system is defined as, "the smallest number of independently variable constituents by means of which the composition of each phase can be represented by means of a chemical equation." Constituents can either be elements or compounds. While writing the chemical equations, we can use zero as well as negative quantities of the constituents, besides the positive ones (as is the convention). Examples: 1. Water system: We know that water system consists of three phases, viz., solid (ice), liquid (water) and gas (water vapours). Each of the three phases is nothing else but water. Hence, all the three phases can be represented in terms of the composition of only one constituent, water i.e., by the formula H20. Ice (s) = H 20; Water (/) = H 20; Vapour (g) = H 20 So, water system is a one component system. 2. Dissociation of calcium carbonate: The case of dissociation of calcium carbonate is rather complicated. Its equilibrium can be represented as : CaC0 3(s) ~ CaO(s)
+ CO2(g)
The composition of all the three phases can be expressed in terms of either of the two components. Any two out of three substances can be chosen as the two components. This is clearly understood as follows :
128
PHYSICAL CHEMISTRY-I
(a) When CaC03 and CaO are the two components : Phase Components CaC03 = CaC03 + O.CaO CaO = O.CaC03 + CaO CO 2 = CaC0 3 - CaO (b) When CaC03 and CO2 are the two components : CaC03 = CaC03 + O.C0 2 CaO = CaC03 - CO2 CO2 = O.CaC03 + CO2 (c) When CaO and CO2 are the two components : CaC03 = CaO + CO2 CaO = CaO + O.C02 CO2 = O.CaO + CO2 Hence, from the above three cases, it is crystal clear that only two constituents are needed to express the composition of each of the three phases. Hence, it is a two component system. 3. Dissociation of ammonium chloride : The case of dissociation of ammonium chloride is very interesting. It dissociates as follows : NH4Cl(s) ~ NH 3(g) + HCI(g) There are two phases, viz., one solid and one gas. The system wiII be a one component or two component system depending upon the relative quantities of HCl and NH3 formed. (a) When NH3 and HCI are in equivalent quantities : Gaseous phase Component NH3 + HCI = NH4CI Solid phase = NH4CI NH4CI So, we see that the composition of both the phases has been expressed in terms of only one substance, viz., ~Cl. So, the system is a one component
system. (b) When NH3 and HCI are not in equivalent quantities : Suppose: ~CI(s) ~ x NH 3(g) + y HCI(g) (x> y) The composition of the solid phase can only be represented by ~CI, but the composition of the gaseous phase cannot be represented by ~CI, but in terms of NH3 and NH4CI as follows : Gas phase: xNH3 + yHCI = yNH3 + yHCI + (x - y) NH3
= y~CI + (x - y) NH3
Solid phase :
~CI = ~CI
+ O.NH3
CHEMICAL AND PHASE EQUILIBRIUM
129
Hence, in this case the two components are NH4CI and NH 3. If HCI is present in excess over NH 3 , then the two components will be NH 4CI and HCI. However, the system remains a two component system.
[III] Degree of Freedom or Variance There are three variable factors, vi;::., temperature, pressure and concentration, on which the equilibrium of a system depends. In some cases, we have to mention only one factor to define the system completely, sometimes two or three. So, the degree of freedom (or variance) of a system is defined as, the least number of variable factors such as temperature, pressure or • concentration which must be specified so that the remaining variables are fIXed automaticaUy and the system is completely defined. System having degrees of freedom three. two, one or zero are known as trivariant, bivariant, univariant (or monovariant) and non-variant systems, respectively. Examples: 1. For ice-water-vapour system, F =0 : In the system, ice ~ water ~ vapour, the three phases co-exist at the freezing point of water. As the freezing temperature of water has a fixed value, the vapour pressure of water also has a definite value. The system has two vaJiables (Tand P) and both these are already fixed. So, the system is completely defined automatically and there is no need to specify any variable. So, it has no degree of freedom, i.e., F = O. 2. For saturated NaCI solution, F = 1 : The saturated solution of sodium chloride in equilibrium with solid NaCI and water vapour, i.e., NaCI(s) ~ NaCI (solution) ~ Water vapour is completely defined if we specify temperature only. The other two variables, i.e., composition of NaCI solution and vapour pressure have a definite value at a fixed temperature. So, the system has one degree of freedom. 3. For a pure gas, F = 2 : For a sample of pure gas, PV = RT. If the values of P and T are specified, volume (\I) can have only one definite value or that the volume, i.e., third variable is fixed automatically. Any other sample of the gas under the same pressure and temperature as specified above will be identical with the first one. So, the system containing a pure gas has two degrees of freedom. Problem 2: (a) Define phase rule. (b) Give the thermodynamic derivation of phase rule.
(Meerut 2(04)
[I] Statement of Phase Rule Phase rule as given by W.J. Gibbs, is defined as follows: "If any heterogeneous system in equilibrium is not affected by electrical or magnetic forces or by gravity, then the degrees of freedom (F), number of components (C) and number of phases (P) are connected by means of the equation, F=C-P+ 2."
130
PHYSICAL CHEMISTRY-I
The mass of the phase does not enter into the equation. as it has no effect on the state of equilibrium.
(II] Thermodynamic Derivation of Phase Rule Consider a heterogeneous system in equilibrium consisting of C components distributed in P phases. The degree of freedom of the system is equal to the number of independent variables which must be fixed arbitrarily to define the system completely. The number of such variables is equal to the total number of variables minus the number of variables which are defined automatically because of the system being in equilibrium. At C(quilibrium. each phase has the same temperature and pressure. so there is ,one temperature variable and one pressure variable for the whole system. So. these variables total two only. The number of composition (or cOl\~ntration) variables. however. is much more. In order to define the composition of each phase. it is necessary to mention (C - 1) composition yariables. because the composition of the remaining last component may be . obtained by difference. Thus. for P phases, the total number of concentration or composition variables will be P (C - 1). Total number of variables = P (C - 1) + 1 + 1 for composition for temperature for pressure
= P (C -1) + 2
According to thermodynamics, when a heterogeneous system is in equilibrium. at constant pressure and temperature, the chemical potential (~) of any given component must be the same in every phase. Therefore. if there is one component in three phases x. y and z and one of these phases. say x is referred to as standard phase. then this fact may be represented in the form of two equations : ~l (x)
=
~l (y)
~l (x)
= ~l (z)
So. for each component in equilibrium in three phases. two equations are known. In general, therefore, for each component in P phases. (P - 1) equations are known. For C components, thus the number of equations or concentration variables that are known from the conditions of equilibrium are C (P - 1). Since chemical potential is a function of pressure. temperature and concentration. it means that each equation represents one concentration variable. Therefore. the number of unknown variables (which should be fixed) or degree of freedom. F = (Total number of variables) - (Number of concentration variables which are already fixed)
or or
F=[P(C-l)+2]-[C(P-l)] F=C-P+2 This equation is the phase rule as given bv Gibbs.
131
CHEMICAL AND PHASE EQUILIBRIUM
Problem 3 : Explain with reason: Can all the four phases in a one component system co-exist in equilibrium? No, all the four phases in a one component system cannot co-exist in equilibrium. Reason : In a one component system, C =1 and, therefore, the phase rule equation (F = C - P + 2) becomes: F = 1 - P + 2 =3 - P. The minimum degree of freedom in any system can be zero. i.e. F= 0 and so, 0=3 - Pmax or Pmax = 3. Thus, not more than three phases can co-exist in equilibrium in a one component system. Problem 4 : Discuss the application ofphase rule with a neat and labelled diagram of water system. . (Meerut 2006, 2005, 2002, 2(00) Water exists in three phases, viz. (a) solid-ice, (b) liquid-water, (c) gas-vapour. The system is a one component system, as the composition of all the three phases can be expressed in terms of only one constituent, H 20. The equilibrium diagram for the water system is as shown in figure (l) A
218
f atm. ~
~~
Ice
a..
4.58 mm
Vapour
0.0075 100 375 Temperature (OC)_
Fig.l The salient features of the phase diagram are as follows : (1) The curves AO, OB, oe.
132
PHYSICAL CHEMISTRY-I
(2) The triple point, O. (3) The areas AOB, BOC, AOC. The significance of each of these features is discussed below. (i) Curve AO : It is known as vapour pressure curve or vaporisation curve !!f liquUJ water. The two phases in equilibrium along AO are water and water vapour. Hence, the curve is univariant. It also follows from the phase rule,equation, F=C-P+2= 1-2+2= 1. From the curve it is also clear that in order to define the system along it, we have to mention either temperature or pressure. This is because for one value of temperature there can only be one value of pressure. The curve AO terminates at 0 the critical point of water (374°C). (ii) Curve OB : It is known as the vapour pressure curve or sublimation curve of ice. The two phases in equilibrium along OB are solid ice and water vapour and so the curve is univariant (F = C - P + 2 1 - 2 + 2 = 1). This also follows from the curve, because there is only one value of pressure for any value of temperature. The curve OB extends to B, which is nearly absolute zero where no vapour exists. (iii) Curve OC : It is known asfreezing point curve of water orfusion curve of ice. The two phases, ice and water, are in equilibrium along OC. So, the curve is univariant (F - C - P + 2 = 1 - 2 + 2 = 1). This curve shows the effect of pressure on the melting point of ice* or freezing point of water. As the curve OC slopes towards the pressure axis, the melting point is lowered as the pressure is increased. This fact can also be predicted by Le-Chatelier's principle, as ice melts with decrease in volume. (iv) Triple point: The curves OA, OB and OC meet at point 0, where all the three phases, viz., liquid water, ice and water vapour exist in equilibrium. So, it is known as a triple point. Applying phase rule to triple point, we have
=
F =C - P + 2 = 1 - 3 + 2 =o. So, this point is non-variant, i.e., in order to define the system at 0, we have not to mention any variable factor, i.e., the system is self-defined. The triple point, posseses fixed values for pressure and temperature,
i.e., 4.58 mm. of Hg and +O.OO75°C, respectively. (It is clear that 0 is not the actual melting point of ice, i.e., OoC. Its value has been increased due to the fact that OoC is the normal melting point of ice at 760 mm of Hg and decrease of pressure will increase the melting point of ice. Since a decrease of pressure by 1 atmospheric pressure or 7(fJ mm of Hg increases the melting point by O.OO8°C, therefore, a decrease of pressure to 4.58 mm wilt raise the melting point to +O.OO75°C.)
*
Normal melting point of a substance is defined as the temperature al which the solid and liquid are in equilibrium at atmospheric pressure.
133
CHEMICAL AND PHASE EQUILIBRIUM -
(v) Areas: The diagram consists of three areas viz., AOB, BOe and AOe, which show the conditions of temperature and pressure under which a single phase-water vapour, ice and water respectively, is capable of stable existence. Applying phase rule to different areas, we have F = C - P + 2 = 1 - 1 + 2 =2, i.e, the systems are bivariant. This also follows from the diagram, because in order to define the system completely at any point within the area, we have to express both the variable factors-pressure and temperature-making the system bivariant. Metastable Equilibrium : Fahrenheit observed that under certain conditions, water can be cooled to -9°C, without the separation of ice at oDe. Similarly, every liquid can be cooled below the freezing point without the separation of the solid phase. So, water cooled below its freezing point is known as supercooled water. But as soon as the equilibrium is disturbed either by stirring or by adding a small piece of ice, supercooled water immediately changes into ice. Therefore, it can be said that such water is in itself stable, but becomes unstable on disturbing the equilibrium. Such an equilibrium is known as metastable equilibrium. It can be defined as, "an equilibrium which in itself is stable but becomes unstable on being disturbed by stirring or adding a piece of the solid phase." From the diagram, it is clear that if water is cooled along AO undisturbed, then at 0, no ice separates out, as should have happened. In such a case, the curve AO merely extends to A'. The phases along OA' will be water and vapour, in metastable equilibrium, as water normally should not exist below O. The curve OA' which is known as metastable curve will be univariant. As soon as the equilibrium is disturbed by stirring or adding a piece of ice, the curve OA' immediately merges into OB with liquid water changing into solid ice. The curve OA' is in accordance with the general principle that the vapour pressure of the metastable phase is always higher than the stable phase. This is evident because the curve OA' lies above the curve OB. Below 0, ice is the stable phase and water is the metastable phase. Salient Features of water System
Curve/Regionl Point
Name
Phases : Solid (S), Liquid (L) and Vapour (V)
Variance I
I.
Curve OA
Vaporisation curve of water
L --" V
2.
Curve OB
Sublimation curve of ice
S --" V
I
3.
Curve OC
Fusion curve of ice
L --"S
I
..-
..-
4.
Region AOB
-
Vap;;;;;:-
2
5.
Region AOC
-
Liquid
2
6.
Region BOC
-
Solid
2
134
PHYSICAL CHEMISTRY-I
Curve/Regionl Point
Phases: Solid (S), Liquid (L) and Vapour (V)
Name
Variance
7.
Point '0'
Triple point
S~L~V
0
8.
Curve OA'
Metastable vaporisation curve of water
L~V
I
Problem 5 : Apply phase rule to system of one component comprising of more than one solid phase or apply phase rule to sulphur system. (Meerut 2004, 2001)
Sulphur exists in two crystalline forms, viz., rhombic and monoclinic. Normally, sulphur exists in rhombic form which is octahedral in shape. When sulphur is heated to 95.6°, the rhombic sulphur changes into monoclinic variety which is prismatic in shape. Above 95.6°, monoclinic sulphur exists as the stable phase. If monoclinic sulphur is cooled then at 95.6°, it changes into rhombic type. Therefore, it is clear that below 95.6°, rhombic sulphur exists as a stable phase, whereas above 95.6°, only monoclinic variety occurs. At 95.6°, these two crystalline forms are in equilibrium with one another. Hence, 95.6° is the transition temperature of SUlphur. 95.6'
Rhombic sulphur ~ Monoclinic sulphur. Besides the above two crystalline forms, sulphur also exists as : (1) Liquid sulphur and (2) Vapour sulphur. Monoclinic sulphur when heated to 120°, melts into liquid form which is pale yellow. On heating gradually the colour of the liquid sulphur changes and the viscosity increases. At 444°, liquid sulphur begins to vaporise. The complete equilibrium is represented as follows :
J
95.6"
Rhombic sulphur (SR)
120"
~Monoclinic
sulphur---7 Liquid sulphur
(SM)
Vapour sulphur (Sv)
(SL) Changes in colour and viscosity
444'
As evident, not more than three phases can co-exist in equilibrium at anyone time in a one component system. In sulphur system, there are four phases, therefore, all the phases can never co-exist in equilibrium. Only three out of the four phases can exist in equilibrium at anyone time. Equilibrium Diagram : The salient features of phase diagram represented in figure (2), are as follows : (i) The six curves, AB, BC, CD, BE, CE and EP. (ii) The three triple points B, C and E. (iii) Tlte four areas ABCD, ABEF, BCE and DCEP. .....", .... ,.
135
CHEMICAL AND PHASE EQUILIBRIUM
r"""''''''''''''''''''''''''''''''''''''''''''''''''''''""""""".,.".""""".,.,', "" """"::"""""'~"""'\I:I
I: ..••
i
SR
o
I$~ . ~~~
i
r
I T.:~~mW~'~~O::~~) -
,I:
::;:;';;';;~;·;:;~i·;;;r~~·~1;;;;;;;;;;;;!
The significance of the features is discussed below. (1) Curve AB : It is known as vapour pressure or sublimation curve of Sib as it gives the vapour pressure of solid SR at different temperatures. Two phases in equilibrium are SR and Sy. The system is, therefore, univariant as,
F=C-P+2 F=I-2+2=1. At B, SR changes reversibly into SM' (2) Curve BC : It is known as vapour pressure or sublimation curve of SM as it gives the vapour pressure of solid SM at different temperatures. The two phases in equilibrium are SM and Sy. The system is again univariant. At C, i.e., at 120°, SM changes into SL' (3) Curve CD: It is known as vapour pressure curve ofSL, as it gives the vapour pressure of SL at different temperatures. The two phases in equilibrium are SL and Sy. The system is univariant, as
F=C-P+2=1-2+2=1.
i
136
PHYSICAL CHEMISTRY-I
(4) Curve BE : It is known as transition curve and the two phases in equilibrium are solid SR and solid SM' The system is univariant. The curve indicates the effect of pressure on the transition temperature of SR to SM' As the curve BE slopes away from the pressure axis, an increase of pressure will increas;! the transition temperature. The curve BE terminates at E. beyond which SM disappears. (5) Curve EF : It is known as melting point or fusion curve of SR' The two phases co-existing in equilibrium are SR and SL' The system is univariant. (6) Curve CE : It is known as melting point or fusion curve of SMThe two phases in equilibrium along it are SM and Sv The system is univariant. The curve CE shows the effect of pressure on the melting point of SM' As this curve slopes slightly away from the pressure axis. the melting point of SM increases with an increase of pressure. This is in accordance with LeChatelier's principle. as melting of SM is accompanied by a slight increase of volume. (7) Triple Points (B, C and E). (i) Triple point B .' It is the meeting point of the three curves AB, BC and BE. Three phases, viz., solid SR' solid SM and Sv exist in equilibrium at point B. Thus, it is a non-variant point (95.6°C).
F=C-P+2=1-3+2=O At B, SR is changed to SM and the process is reversible. (ii) Triple point C.' It is the meeting point of the three curves BC, CD and CEo Three phases, viz., solid SM' SL and Sv exist in equilibrium at point C. So, it is a non-variant point (l20°C). (iii) Triple point E .' It is meeting point of the three curves BE, CE and EE Three phases, viz., solid SR' solid SM' and SL exist in equilibrium at point E. It is a non-variant point (150°C). (8) Areas or Regions (ABCD, DCEF, BCE and ABEF) : Any point within the area ABCD, DCEF, BCE and ABEF gives the conditions of temperature and pressure for the stable existence of only one phase, i.e., Sv, SL solid SM, solid SR, respectively. The areas have two degrees offreedom as shown below. F = C - P + 2 = 1 - 1 + 2 = 2. (9) Metastable Equilibria: The change of SR to SM occurs very slowly. If enough time for the change is not allowed and SR is heated rapidly, it is possible to pass well above the transition point (B) without obtaining SM' In that case, the curve AB extends to O. The curve AD is known as metastable vaporisation curve of SR' The phases SR and Sv will be in metastable equilibrium along this curve. It is a univariant system. On super-cooling along DC, the curve CO is obtained. It is, infact, the back prolongation of DC. The curve CO, known as vaporisation curve of supercooled Sv represents the metastable equilibrium between supercooled SL and Sv. It is also univariant. Similarly, the two metastable phases SR and SL exist in equilibrium along EO, which is known as metastable fusion curve of SR' It is also univariant.
137
CHEMICAL AND PHASE EQUILIBRIUM
At point 0, three metastable phases SR' SL and Sy are in equilibrium. It is known as metastable triple point. It is non variant, as
F=C-P+2=1-3+2=O Salient Features of Sulphur System System Curves I Regions! Triple points
Name of the system
Variance (F=CP+2)
Phases in equilibrium
I.
Curve AB
Vapour pressure curve of SR
SR~Sy
I
2.
Curve BC
Vapour pressure curve of SM
SM~ Sv
1
3.
Curve CD
Vapour pressure curve of SL
SL ----" Sy
1
4.
Curve BE
Transition curve
SR~ SM
I
5.
Curve CE
Fusion curve of SM
SM----" SL
I
~
~
6.
Curve EF
SR~ SL
I
7.
Region ABCD
-
Sy
2
8.
Region DCEF
-
SL
2
Fusion curve of SR
9.
Region BCE
-
SM
2
10.
Region ABEF
-
SR
2
II.
Point B
First triple point (95.6", 0.006 m.m.)
SR~SM~SV
0
12.
Point C
Second triple point (\20", 0.04 m.m.)
SM
' SL~Sy
0
13.
Point E
Third triple point (15 J", 1288 cm)
S~S~SL
0
14.
Curve BO
Metastable sublimation curve of SR
SR~SV
I
15.
Curve CO
Metastable vaporisation curve of SL
SL~Sy
I
16.
Curve EO
Metastable fusion curve of SR
SR~SL
I
17.
Point 0
Metastable triple point (114°,0003 m.m.)
SV~SL<
'Sv
0
Problem 6 : Write short notes on the /oUowing : (i) Non-variant system in phase rule studies. (ii) Triple point. (iii) Transition point. (i) Non-Variant System in Phase Rule Studies A system for which degree of freedom is zero is called non-variant system. Under such conditions, there can be no change in temperature, pressure and concentration, because if any change is made in either of them, one or more phases may disappear. Such systems are also called self defined systems.
138
PHYSICAL CHEMISTRY-I
In water system, the three phases viz., ice (s), water (/) and vapours (g) remain in equilibrium at a fixed temperature (O.0075°C) and fixed pressure
(4.58 mm). At no other temperature and pressure, all the three phases can coexist in equilibrium. From phase rule equation (F = C - P + 2), we can also show that value of F of the system is zero, i.e., F= 1-3 +2=0.
(2) Triple Point Triple point is that point where all the three phases in a one component system exist in equilibrium. At this point, both the variables, e.g., temperature and pressure are fixed, i.e., they have definite values. Thus, triple point is a non-variant point, i.e., degree of freedom is zero. It is also clear from phase rule equation i.e., t' = c - P + 2 = 1 - 3 + 2 = O. The system at triple point is self defined. If any change is made in either of the variable factors, one of the phases of the system disapppears. In water system (figure I), 0 is the triple point where ice, liquid water and water vapours exist in equilibrium. In sulphur system (figure 2), there are three triple points namely B, C and E. (3) Transition Point It is defined as that temperature at which one allotropic form of a substance is converted into another allotropic form of the same substance. The point where this change occurs is called the transition point. Example: (i) At 95.6°C, rhombic sulphur is changed into monoclinic sulphur. Below 95.6°C, rhombic sulphur can exist and above 95.6°C monoclinic sulphur exist. Thus, 95.6°C is the transition temperature of sulphur (See problem 3). (ii) On cooling, white tin is converted into grey tin at 18°C (transition temperature). Problem 7 : What is a two component system and how it is graphically represented? Define reduced phase rule equation and condensed state.
[I] Two Component System A two component system is that system in which the composition of each phase present in it can be represented in terms of two substituents, e.g., lead-silver system, potassium iodide-water system etc. Phase rule when applied to a two component system becomes
F=C-P+2=2-P+2=4-P Since the minimum number of phases in any system is one, it is evident from the above equation that the maximum degree of freedom in a two component system is three. Therefore, in addition to temperature and pressure, a third variable, namely composition, has also to be taken into account. In order to represent such an eqUilibrium graphically, it is, therefore, necessary to have three coordinate axes at right angles to one another. This will lead to
139
CHEMICAL AND PHASE EQUILIBRIUM
three dimensional or space models which cannot be easily represented on paper (figure 3-a). Therefore, it is customary to choose any two of the three :
~
/'
:
T
P
t
t
C constant
P
-T
-P
I
C
(a)
:
T
t
(b)
P
P constant
-c
T constant
t P
(c)
(d)
: '::':':':':':':':':':';'::}}}}i
:::::::::::: .:.:.:.
P
-c Fig. 3 :
d :::::::
:::::::::::::::::::::::::::::::::.:::::::::::::::::::::::
variables for graphical representation, assuming the third variable to remain constant. Thus, we get three types of curves as shown in (figure 3 b, c, d).
[II] Condensed Phase Rule Equation When in a system, one variable out of pressure, temperature or concentration is taken as constant, the number of degrees of freedom is reduced by one. In that case, phase rule equation (F = C - P + 2) becomes, F=(C-P+2)-1 or F=C-P+ 1 This equation is known as condensed or reduced phase rule equation.
[III] Condensed State or Phase In solid-liquid system, a negligible change in pressure produces no change in equilibrium as the vapour pressure of solid is negligible. Therefore, in such a system, the pressure variable may be taken as nearly constant. So, a system in which vapour phase is ignored is known as a condensed system. For such a system, we apply the condensed phase rule equation, F=C-P+l.
140
PHYSICAL CHEMISTRY-I
Problem 8: Apply phase rule to lead-silver system. Mention the effect of cooling and also explain the Pattinson's process for desilverisation of lead ores. The two metals lead and silver mix together in the liquid form but do not form any chemical compound. The phase diagram is as shown in figure (4). It is explained as follows : 1. Curve AC : Pure lead melts at 327°, while silver at 961°. Point A represents the melting point of pure lead. By the gradual addition of silver, the freezing point of lead decreases along AC. The curve AC is thus known as freezing point or melting point curve of lead. The phases present along AC are solid lead and its solution with silver. Thus, it is a univariant curve as, F=C-P+l=2-2+1=1 2. Curve BC : Similarly, B represents the melting point of pure silver. By the addition of lead to it, the freezing point of silver decreases along BC and we get a solution of it with lead. The curve BC is known as thefreezing point or melting point curve of silver. The two phases present along it are solid silver and a solution of it with lead. It is, therefore, a univariant curve as,
Unsaturated solution
'X
I
1
:::
~ ~~
~
I
Liquid
Liquid + Solid silver
of': y
~----~~----------------~D~'
Solid lead + Eutectic
3030
Solid silver + Eutectic
F=C-P+l=2-2+1=1. 3. Point C : The two curves AC and BC meet at a common point C. Therefore, C gives conditions of temperature and composition under which three phases, viz., solid silver, solid lead and solution co-exist in equilibrium. It is thus a non-variant point as, F = C - P + 1 = 2 - 3 + 1 = O.
CHEMICAL AND PHASE EQUILIBRIUM
141
The point C is known as eutectic point of the system. The temperature and composition corresponding to eutectic point are known as eutectic temperature and eutectic composition, respectively. The eutectIc temperature and composition are 303 and 2.6% Ag and 97.4% Pb, respectively. 4. Areas: In the area above the curves AC and BC, the two components are present in the form of one homogeneous liquid phase. The system becomes bivariant as, F = C - P + 1 = 2 - 1 + 1 = 2. In other areas the phases are present as shown in the figure. Effect of cooling: Consider the phase changes which occur on cooling a liquid mixture. Suppose a liquid mixture of composition x is cooled at constant temperature. The temperature will fall without any change in the composition until the point y on the curve yC is reached. At this point, I&d will begin to separate out. The system becomes univariant, as it consists of two phases. The temperature will now fall with a change in the composition of the liquid mixture along AC. As cooling continues, Pb keeps on separating out while the solution becomes richer and richer in silver. When the eutectic temperature (303~ is reached, the second solid phase, viz., silver begins to crystallise out. The system thus becomes non-variant at C. The two solids lead and silver will separate out together in a fixed ratio on further cooling, so that the composition of the solution remains constant as shown by point C. The temperature also remains constant. When the solution phase has completely disappeared as solid, the system consists only of a mixture of solid Ag and Pb. The system becomes univariant and further cooling will lower the temperature below solidus DD'. In area below DD' two solid phases Pb and Ag will co-exist. If a liquid solution of composition represented by a point x' is cooled its temperature will fall without change in the composition along x'y'. At y', solid Ag begins to crystallise out and the system becomes univariant. Further cooling will shift the equilibrium along y'C, when Ag goes on separating out and the solution becomes richer and richer in Pb. When the eutectic temperature is reached, Pb also begins to crystallise out. Further cooling will not change the temperature as well as the composition as long as three phases are present at C. When solution phase solidifies, only then the temperature falls below solidus DD' within which two solid phases Pb and Ag co-exist. Pattinson's process of desilverisation of lead : This process consists of increasing the relative percentage of silver in ores of lead called argenti/erous lead ores containing very small amounts of silver, say 0.1 %. Its relative content can be increased by taking a liquid solution of the ore and increasing it to a high temperature. It is then cooled and the temperature falls along xy. At y, solid lead begins to crystallise out which can be removed by ladels. Further cooling will shift the equilibrium along ye, making the solution richer and richer in Ag. Along ye, lead will go on separating out which is -:ontinu0
142
PHYSICAL CHEMISTRY-I
ously removed. At C, the percentage of silver increases to 2.6% (starting from 0.1%).
Problem 9: Apply phase rule to potassium iodide and water system. Explain the effect of cooling on the system. A careful study has shown that water and potassium iodide system is a simple eutectic system. The phase diagram is as shown in figure (5) (One essential feature of salt-water system is that the melting point of the salt is usuaUy very high, even higher than the critical temperature of water. It is, therefore, not possible to represent the melting point of the salt in the equilibrium diagram). The phase diagram is discussed as follows : :
Unsaturated solution
x
,
0
I
B
X
i ~
~ Q)
SolidKI
t!
Solution
a. E
+ Ice + Solution
-230 Ice + Eutectic
0
Solid KI + Eutectic
52 Composition (% KI) _
100
1. Curve A C : Point A represents the melting point of ice or the freezing point of water which is 0° at 1 atmospheric pressure. With the gradual addition of KI, some of it dissolves in water and remains in contact with ice. The melting point of ice decreases along the curve AC by the addition of KI. The curve AC is known as freezing point curve of water or fusion curve of ice. Along this curve, solution of KI in water is in contact with ice, hence it is a univariant curve. It is also seen from condensed phase rule equation:
F=C-P+l=2-2+1=1 2. Point C : At point C, a new solid phase, potassium iodide also separates out. Thus, three phases, viz., ice, potassium iodide and liquid, exist at C. Hence, it is a non-variant point as,
F=C-P+l=2-3+1=0
CHEMICAL AND PHASE EQUILIBRIUM
143
The point C is known as the eutectic point or cryohydric point of the system. It gives the lowest temperature which can be attained in the system, i.e., _23°. At the cryohydric point the solution freezes at constant temperature without change of composition. The eutectic composition is 52% KI and 48% ice. 3. Curve CB : If the system is heated at C, ice will melt and potassium iodide will pass into the solution in the same ratio in which it is already present in the solution, so that the composition of the solution remains unchanged. The temperature and composition do not change, as long as three phases are present at C. The heat supplied to the system is utilised in transforming ice . into water. If heating is continued and potassium iodide is present in excess then ultimately all ice will disappear. Then there will be two phases and the system will become univariant. At this stage, addition of more potassium iodide will bring about a change in the temperature also and then the curve CB is traced. The curve BC is also traced in an alternative way. If we add ice to solid potassium iodide then we get a solution of potassium iodide in water in equilibrium with solid potassium iodide. In other words, the addition of ice to potass!um iodide decreases the melting point of solid potassium iodide along Be. The curve BC is known as the solubility curve ofpotassium iodide, as KI is in equilibrium with its solution. The nature of the curve BC shows that the solubility of potassium iodide increases slowly with rise of temperature. At point C, a new solid phase, viz., ice begins to separate out. Effect of cooling : If we cool a solution of composition x, then the temperature will fall along xy without any change in the composition, as the system in bivariant. At y, solid ice will separate out. The system becomes univariant. The change in temperature will now be followed by a change in composition. Hence, the temperature will fall along yC when at C, potassium iodide will separate out as a new solid phase. Similarly, we can predict the effect of cooling a solution of composition x'.
Problem 10 : Determine the number of phases and components in the foUowing systems: (i) Water in a beaker at room temperature. (ii) A solution of ethanol with water which is in equilibrium with vapours. (iii) NH4CI(s) ~NH4Cl(vap) ~NH3(g) + HCI(g) (iv)HCI gas is passed from outside in equilibrium (iii). (i) Components = 1; Phases = 2 (ii) Components = 1; Phases = 2 (iii) Components = 1; Phases = 2 (iv) Components = 2; Phases = 2. Problem 11 : What will be the degree of freedom in the foUowing closed systems? (i) liquid water and water vapours.
144
PHYSICAL CHEMISTRY-I
(ii) Liquid water and water vapours at 2S·C. (iii) Dilute solution of NaCI fiUed partially in a closed vessel. (iv) Saturated solution of Naafilled partially in a closed vessel. (v) A gaseous mixture of N:z, O2 and H2o One(F=I-2+2=1) (ii) One (F = 1 - 2 + 2 = 1) (iii) Two (F 2 - 2 + 2 = 2) (iv) One(F=2-3+2=1)
(i)
=
(v)
Four(F=3-1+2=4).
Problem 12 : Determine the number ofphases, components and degree of freedom of the following systems: (i) Sodium chloride and water (ii) A mixture of nitrogen and hydrogen gases in a vessel. (iii) 2HzS(g) ~ 2HzO(g) + Sz(g) (iv) H20(s) ~H20(g) (v) Na~04. 10HzO(s) ~Na~Ois) + 10HzO(I) (vi) CaCOis) ~ CaO(s) + CO2(g) (i) Sodium chloride-water system Phases = 2; Components Degree of freedom = 2
= 2 (NaCI and H 20)
(ii) System of Nz and Hz gases Number of phases = 1; Components = 2 (H2, N+) Degree of freedom = 3 (iii) Phases = 1, Component = 1, Degree of freedom = 2 (iv) Phases = 2, Component = 1, Degree of freedom = 1 (v) Phases 3, Component = 2, Degree of freedom = 1 (vi) Phases = 3, Component = 2, Degree of freedom = 1
=
SYSTEMS OF LIQUIDS IN LIQUIDS Problem 13: What are ideal solutions? Explain the vapour pressure of ideal solutions. Before defining ideal solutions, we must understand Raoult's law. Raoult measured the vapour pressures of a number of binary solutions of volatile liquids and made the important generalisation, known as Raoult's law. It can be stated as follows:
The partial pressure of any volatile component of a solution at any temperature is equal to the product of the vapour pressure of the pure component and the mole fraction of that component in the solution. Consider a binary solution made of nA moles of a volatile liquid A and ns moles of another volatile liquid B. If PA and Ps be the partial pressures of the two liquid constituents, then according to Raoult's law, PA =XAPAo Ps=xsPso
145
CHEMICAL AND PHASE EQUILIBRIUM
where, XA is the mole fraction of the component A, given by ~ , IlA +IlB
XB
is the mole fraction of the component B, given by
~,
+ liB PA° and PB° are the vapour pressures of the pure components A and B. respectively and n term represents the number of moles. If vapour behaves as an ideal gas then according to Dalton's law of partial pressures, the total pressure (P) is given by nA
P=PA +PB o or P=XAPAo+XBPB ... (1) Experiments show that Raoult's law is obeyed only approximately for a number of binary solutions. It is obeyed perfectly only in case of ideal solutions. So, a solution of two or more components is said to be ideal if it obeys Raoult's law perfectly at all temperatures and concentrations. Similar liquid pairs are generally found to form ideal solutions, e.g., binary mixtures of ethylene bromide and ethylene chloride, benzene and toluene, n-heptane and n-hexane etc.
Vapour Pressures of Ideal Solutions The vapour pressure of an ideal binary solution of two components A and B is shown in Fig. 6. It is clear from the graph that the curve of the partial pressure of each component against its mole fraction in the solution is a straight line and the total vapour pressure of the solution for a given concen-
~,
Total ~ ..... , ..... ,
apOur pressur; e of solution
..... ,
..... ,oq
, ..... :'1<1/ A A + XaPa IJ ..... , IJre8s :.II ~.t'''''' lire
I1IJ..qo'''''',~-1 ,
o\~"
, , , "
O.4PA + O.6PB
P
P =X pO
sS\l{e " , 0 ~&tU.&\ ~{~, , ~ , ::. 'J..-o'Y-O -0 , , ",
,
..... ...
,,.""
, ... ' "
... ..."'........ ,
"
---. Mole fraction
..... ,
, , ,
..... ,
.....
, , , " , ,
..... ,
..... ,
..... ,
,~
..... ,
146
PHYSICAL CHEMISTRY-I
tration is equal to the sum of the partial vapour pressures of the two constituents. The total vapour pressure (P) of the solution is given by equation (1). Thus, when the mole fraction of B is say 0.60, the total vapour pressure (P) of the solution is given by P = 0.4 PAc + 0.6PBo
Activity of a Component in an Ideal Solution Consider a binary solution of two components A and B, forming an ideal solution. Let aA and aB be the activities, and XA and XB be the mole fractions of the constituents A and B, respectively. According to Raoult's law, the partial vapour pressure (PA) of constituent A is given by PA =XAPAo PA -=XA PAc
or
As vapour pressures of solid or liquid components of a solution at ordinary temperatures are usually low, lying well within the range in which the fugacity if) may be taken as equal to pressure, therefore, As known from thermodynamics, PA fA -=-=XA PAc lAO
As we know from thermodynamics,
IA
PA -=-=aA PAc lAO
where aA is the activity of the component A. So, XA =aA and similarly x--B = aB, where aB is the activity of the component B. In other words, activities of components forming ideal solutions are equal to their mole fractions. Thus, ideal solution may also be defined as follows:
aB aA
=1
<
=1
m '0
'0
f
~
llC..-______---""t XA = 1 XB=O
Mole fraction
= =
XA 0 XB 1
147
CHEMICAL AND PHASE EQUILIBRiUM
A solution in which the activity of each component is equal to its mole fraction under all conditions of temperature, pressure and composition, is said to be an ideal solution. The relation between activities and mole fractions of the constituents is shown graphically in figure (7) So, If Xi is the mole fraction of a component i, then its activity a, is given by ai=xi
Problem 14 : What are non-ideal or real solutions? Explain the vapour pressure curves of completely miscible binary solutions. Those solutions which show appreciable deviations (positive or negative) from the ideal behaviour are known as non-ideal or real solutions. Real solutions may be divided into the following three types : Type I. Non-ideal solutions of this type show small deviations from ideal behaviour and total pressure remains always within the vapour pressures of the pure components, as shown in figure (8), in which the dotted lines represent ideal behaviour. It i& observed that the total pressure of each component shows a positive deviation from Raoult's law. However, the total pressure remains within the vapour pressures of the pure constituents A and B. Example of such a type of solution is furnished by carbon tetrachloride and cyclohexane system.
800~---------------.
~
~
f/)
400
~
~ 300
&200
~
100 ,,
o "------,---..--r---r--..... o 0.2 0.4 0.6 0.8 1.0 Ethanol Chloroform disulphide - - . Mole fraction
148
PHYSICAL CHEMISTRY-I
Type II. Solutions of this type show large positive deviations and the total vapour pressure curve rises to a maximum which is above the vapour pressure of either of the two pure constituents A and B as shown in figure (9). Mixture of ethyl alcohol and chloroform is a system which belongs to this type. Type III. Solutions of this type show large negative deviations and the total vapour pressure curve dips to a minimum, for some of the concentrations. The total vapour pressure of the mixture may be below the vapour pressure of either of the two pure constituents as shown in figure (10). Mixture of acetone-chloroform is a system which belongs to this type.
~;'::":-:-:':""\I
• 1:
I~ :: ;:::::-.
1
0 0.2
0.4 0.6 0.8 ,.01
A~ Mole tractionChlorotorm
J:
:;:;:;:;:;:;:;~: ;: :;:;:;:;:;:;:;:;: :;:;: :;:;:;:;: : ~ !~: .~:~: : ;:;: : :;: :;:;:;: : : : : : : : : :~:;:;: : : : : Vapour Pressure (or Boiling Point) Composition Curves of Completely Miscible Binary Solutions Consider a binary solution of two components A and B which are completely miscible with one another. On heating under constant pressure, it will boil at a temperature at which its total pressure becomes equal to the atmospheric pressure. If PA and PB represent the partial pressures of the two components A and B, then conditions for boiling is that P=PA +PB where, P is the atmospheric pressure.
149
CHEMICAL AND PHASE EQUILIBRIUM
Solutions of different compositions have different vapour pressures, so they will boil at different temperatures. Therefore, a solution of lower vapour pressure will boil at a higher temperature and vice versa. This helps us in drawing boiling temperature-composition curves from the corresponding vapour pressure composition curves as shown in figure 6. As illustrated, there are three types of mixtures. Type I. Mixtures in which the vapour pressure changes continuously with composition of the mixture. T = Constant
P = Constant ~A
::l
1ij
Type I
Q; a.
E
$
B
OJ
~
·0
co
1
0
0
Mole fraction of B Composition (i)
Mole fraction of B Composition (iv)
T=Constant
P = Constant
C
~
~
::l
::l
I/) I/)
~
B
a.
Type II
~A Q) a.
E
B
$
OJ
~
·0
co
0
0
Mole fraction of B Composition (ii)
Mole fraction of B Composition (v)
T= Constant
P= Constant
D
~
~
::l
::l
~ Q)
B
~ ~ a. SA
Type III
g-A $ OJ
8.
B
~
~
·0
co
Mole fraction of B Composition (iii)
0 Mole fraction of B Composition (vi)
150
PHYSICAL CHEMISTRY-I
Type II. Mixtures in which the vapour pressure shows a maximum in the vapour pressure-composition curve. Type III. Mixtures in which the vapour pressure shows a minimum in the vapour pressure-composition curve. In type I, the vapour pressure of pure A is the lowest and that of pure B is the highest, therefore, the boiling point of pure A will be highest and that of pure B will be lowest. Since the vapour pressures of the mixtures of A and B lie in between the vapour pressures of pure components, their boiling points will also lie in between as shown in figure 11 (iv). At a given temperature, the vapour pressure will be richer in the more volatile component B, the composition of the vapour phase will be always richer in B than that of the liquid phase. Thus, the vapour composition curve will lie above the liquid composition curve. In type II, the vapour pressure shows a maximum for a definite composition, say C, as shown in figure 11. The solution of that composition will boil at the temperature. Thus, there will be a minimum in the boiling temperature-composition curve, as shown in figure 11 (v). In type III, the vapour pressure shows a minimum for a definite composition, say C, as shown in figure 11 (iii). The solution of that composition will boil at the highest temperature. Thus, there will be a maximum in the boiling temperature-composition curve, as shown in figure 11 (vi).
Problem 15: Discuss the theory offractional distillation of binary solutions. As the boiling point curves of the three types of solutions are different. the behaviour of these solutions on distillation will be different.
1. Solutions of type 1. Mixtures which shows neither a maximum nor a minimum on the vapour pressurecomposition curve or boiling temperature-composition curve are known as
zeotropic mixtures. Consider the given figure (12). Suppose a solution of composition X is heated. The boiling will start when temperature T is reached. At this temperature, the vapour coming off will
o
- - . Mole fraction of B
1
151
CHEMICAL AND PHASE EQUILIBRIUM
have the composition X since it is richer in B, than the composition of the residual liquid will become richer in A. Let it be shown by Y. This liquid, having different composition cannot boil at T. Instead, it will require a higher temperature, say T I • The vapour coming off now will again be richer in B, as represented by point X2• Thus, the composItion of the residual liquid will be further enriched in the liquid A. The temperature of this residual liquid will have to be increased again to make it boil again. If this process is continued, the boiling point of the solution will go on increasing from Tto TA , the boiling point of pure liquid A. The composition of the residual liquid will also become richer in A and ultimately only liquid A remains in the; liquid phase. Regarding the vapour phase, it has been stated t~at the vapour coming off from the original solution at its boiling point T will have the composition shown by point XI. If these vapours are condensed and the liquid distilled again, the new boiling point will be T2 and now the composition of the vapours distilling over will be given by Z. Therefore, the distillate is now richer in component B than before. If this process of condensing the vapours and redistilling the liquid obtained is continued, the distillate obtained ultimately will consist entirely of pure component B. In other words, by carrying out fractional distillation of solutions of type I, it is possible to isolate 'both the constituents from each other. 2. Solutions of type II. The boiling temperature-composition curves of the liquid and vapour phases meet at a maximum boiling point fig. (13). In other words, the liquid and vapour phases at this point have the same composition. Thus, the liquid mixture represented by the point M, will boil at a constant temperature and will distil over completely without change of composition. Such mixtures which like pure chemical compounds, boil at :
:
:
100 B
90
U 0
e. !~
-e
80
~ 70
E ~
r
50 10 20 30 40 50 60 70 80 90 100
--+
Percent A
152
PHYSICAL CHEMISTRY-I
a constant temperature and distil over completely at the same temperature without change in composition, are known as constant boiling mixtures or azeotropic mixtures. On applying phase rule to azeotropic mixtures, we should remember that there is only one restriction, viz., compositions in liquid and vapour phase should be the same. H~c~ F=C-P+l=2-2+1=1 i.e .• azeotropes should behave as univariant systems, so that the boiling point is constant, if pressure is fixed. This is what we will see in the next cases. The characteristics of azeotropic mixtures are shown in figs. (14) and (15). The boiling temperature composition diagram is shown in figure (14). The constant boiling mixture has the maximum boiling point, i.e., it is least volatile. The vapour phase for any mixture lying between A and M will, therefore. be richer in A and any mixture lying between z :y I I Band M, will be richer
--+ Composition
boiling mixture M. Suppose a mixture of composition X is distilled. The first fraction distilled will have the composition indicated by Xl' Evidently, it is richer in A. The composition ofthe residual liquid, thus shifts towards constant boiling mixture M. As the distillation proceeds. the composition of the distillate changes towards A and that of the residue towards M. Ultimately, a distillate of pure A and a residue of constant boiling mixture M will be obtained. Similarly, a mixture of composition lying between Band M, say Y will ultimately provide on distillation, a distillate of pure B and a residue of constant boiling mixture M. It is, therefore, evident thaI any binary solution of this type, on complete fractional distillation, can be separated into a residue of composition M and a distillate of either A or B depending upon whether the initial composition lies between A and M or between Band M, respectively. Thus, it is not possible to completely separate such a binary mixture into pure components A and B on distillation.
153
CHEMICAL AND PHASE EQUILIBRIUM
The mixture with the maximum boiling point is called maximum boiling azeotrope and behaves as if it is a pure chemical compound of two components, because it boils at a constant temperature and the composition of the liquid and vapour is the same. But the azeotrope is not a chemical compound, because its composition is not constant under conditions and rarely corresponds to stoichiometric proportions. Pure water and hydrogen chloride boil at 100° and -85°. while their constant boiling mixture (azeotropic mixture) containing 20.25% of hydrogen chloride boils at 108.5°, under a pressure of 1 atmosphere. If a solution containing less than 20.25% of HCl is distilled, (i.e., between points A and M), water will pass over as the distillate and the residue left behind in the flask will consist of 20.25% solution of HCl in water. Thus, pure HCl can not be obtained. Similarly, if a solution containing more than 20.25% HCl is distilled, then pure HCl will pass over as distillate and the residue left behind in the flask contain a mixture of the same constant composition, viz., 20.25% HCl in water. 3. Solutions of type III. The boiling temperature-composition curve for the liquid and vapour phases for such type of solutions is shown in figure (15). The constant boiling mixture in this case has the minimum boiling point, i.e., it is highly volatile. Consider the distillation of composition represented by X (fig. 15). Then the A first fraction collected will have the composition X \. It will be richer in the constant boiling mixture. The : (.'9(.' composition of the reCl "'et ~ ------r-------T---sidualliquid will shift '0 !XI towards A. As distillaI I tion continues, the x: Z: :Y1: Y composition of the distillate and residual 100% liquid changes towards M and A, respectively. By repeating this process, the mixture of minimum boiling point of composition M will be obtained as distillate, while the residue left over in the distillation flask will contain only pure liquid A. If we distil a liquid of composition represented by Y, then the composition of the first fraction will be represented by Y \. Evidently, it will be richer in the constant boiling mixture. The composition of the liquid will become richer in B. As the distillation proceeds, the distillate and the residual liquid
m;:========z======mrm
I
I I
i
I
I
I I
I I
I
I
I
154
PHYSICAL CHEMISTRY-I
will become richer and richer in constant boiling mixture and pure B. respectively. Finally. the distillate will contain only the constant boiling mixture and the residual liquid in the distillation flask will contain only B. There will be no pure A in this case. If the mixture has the azeotropic composition (say Z), it will distil unchanged. In the system of water-ethanol. the point M corresponds to a minimum boiling temperature of78.13° and a composition of95.57% ethanol by weight. If any solution of composition of pure water and 95.57% ethanol is distilled, then we get a residue of pure water and a constant minimum boiling mixture of 95.57% alcohol in the distillate. No pure ethanol can be recovered. On the contrary. if a solution of composition between pure alcohol and 95.57% ethyl alcohol is distilled. then we get a mixture containing 95.57% ethanol and pure . alcohol. No pure water will be recovered.
Solubility of Partially Miscible Liquid Pairs In the study of miscibility of partially miscible liquid pairs, the exterpal pressure i kept constant and, therefore, the vapour phase is Ignored. The mutual solubilities ar( epresented by means of temperature-composition diagram. On the basis of mutual solubility of two liquids, there are three types of liquid pair such as : Type I: Tlwse pairs whose mutual miscibility increases with increasing temperlature. Type II: Those pairs whose mutual miscibility decreases with increasing tem'perature. Type Ill: Those pairs wlwse mutual miscibility both increases and decreases with increasing temperature.
Problem 16: Discuss the theory of partially miscible liquid pairs with special reference to : (i) Phenol-water system (ii) Triethylamine-water system (iii) Nicotine-water system. Define the upper and lower critical solution temperature. What is the (Meerut 2(04) effect of impurities on them? If we consider two liquids A and B and shake together, then some of A dissolves in B. while some of B dissolves in A. We then have two saturated solutions-one of A in B and the other of B in A. On increasing the temperature. the solubility of A increases in B and also drat of B in A. in this particular case. As an example. we take the familiar phenol-water system.
[I] Phenol-Water System Phenol and water are partially mi'scible at ordinary temperature. When we add phenol to water phenol gets dissolved in water. till its concentration reaches 8%. The addition of more phenol will give rise to two liquid layers. One layer will consist of water in phenol system and the other that of phenol in water system. Such solutions of different compositions co-existing in equilibrium with one another are known as conjugate solutions.
155
CHEMICAL AND PHASE EQUILIBRIUM
As the temperature is raised, the solubility of phenol in water increases, Homogeneous whereas that of water in phenol also increases. Ultimately at a certain tempera,, , ture, the two conjugate solutions change into one :Tie line b ----t-----homogeneous solution. This temperature is known as critical solution temperaI Heterogeneous ture or consolute tempera200 A : C ture. The value of consolute temperature for this system B is 68.3°, and the composi203340 60 80 100 o tion is 33% phenol and 67% --+ Percent phenol water. Above 68.3°, the two liquids are completely miscible in all proportions. The variation of mutual solubility of water and phenol with temperature is shown in fig. (16). The solubility of phenol in water increases with rise of temperature along the curve AB, while the solubility of water in phenol increases along CB. The two curves do not intersect each other, but meet 'at a certain point B, known as C.S.T. In between the area ABC, the system will be heterogeneous. If we prepare a mixture of phenol and water of composition and temperature represented by any point within the area ABC, it will separate into two liquid layers or binary conjugate solutions. Any mixture of composition represented by a point outside ABC will give a homogeneous solution. At any fixed temperature say to, the composition of each layer is fixed as shown by the points a and b, a gives the composition of the aqueous layer, while b gives that of phenol layer. The line ab is known as tie line. The amounts of the conjugate solution may be easily calculated. For example, suppose we have a mixture of phenol and water, whose total composition is 50% phenol at a temperature of 40°C. Then the data of the table of fig. (14) indicates that the two liquids formed have the respective compositions 7.5% and 67% phenol. Let x be the weight of aqueous layer in 100 g of the mixture and, therefore, (loo-x) g will be the weight of phenolic layer. Then,
XU~)+(l00-X>(I~ )=50 x=28.5 Therefore, lOOg of total mixture give rise to 28.5 g of aqueous layer and 71.5 g of phenolic layer, or 285 x 0075 = 213 g of phenol is present in aqueous layer and 715 x 067 = 4790 g of phenol is present in the phenolic layer. Other examples are (i) Water-aniline (167°), (ii) Benzene-aniline (59.so) (iii)Methyl Alcohol-cyclohexane (45.5°), (iv) Bi-Zn (Metallic .\ystem) (85.0")
156
PHYSICAL CHEMISTRY-I
[II] Triethylamine-Water System There are some cases, when solubility of one liquid in another decreases with the rise in temperature. 9prrIT:2TI:ITrrillIT:2TI:ITTIillEJ] The temperature I::: composition curve for tri~ C .a ethylamine and water sysI!! CD tem is shown in fig. (17). Co E Triethylamine and water ~ mix together in all proportions below 19 0 but on raising the temperature above 19°C -------"'---, 19°, the system separates B into two liquids layers. One phase The curve AB shows 0% Amine . . 100% Amine 100% Water -+ ComposItIon 0% Water the decreasing solubility of
i
triethylamine in water ':::JWB~i42~~ii14m.2.22EWJ while curvers CB shows the ~~ilij[lli2iliili~~;;;±ilill2lligili] decreasing solubility of Ii;; water in triethylamine. The two curves meet at B, which is the lower critical solution temperature or lower consolute temperature of the system. Any point within the area ABC will gi ve two liquid layers, while any point outisde the area ABC will give a homogeneous solution. Methyl ethyl ketone and water has a lower consolute temperature of
lO°e. [III] Nicotine-Water System In some cases, it BIZJJZ:SEEZZ=ITEEZZ:==81 has been observed that One phase the mutual solubility curve is a closed curve with both the types of consolute temperatures. Two The temperature phases composition curve for this system is shown in fig. (18). The system of 60.BOC ----nicotine and water beOne phase longs to this class. At 100% Nicotine temperatures below 0 60.8° and above 208 , the two liquids are com- Wg¥.Ig¥.[!4¥~~~@g¥.Ig4.4.4~ pletely miscible in all &lli22lli2llib~~ti2iliJ22bBd
i
157
CHEMICAL AND PHASE EQUILIBRIUM
proportions giving a homogeneous solution. Between 60.8° and 208°, they are only partially miscible and give two layers. Hence, the solubility curve is a closed curve. The upper and lower consolute temperature are 208° and 60.8°, respectively.
[IV] Influence of Impurities on Critical Solution Temperature The critical solution temperature has a fixed value for a given system and is completely defined. Its value is very sensitive to the presence of impurities present in either or both the components. Hence, the determination of C.S.T. gives us an accurate method of determining the presence of impurities qualitatively as well as quantitatively. The effect of dissolved impurities on C.S.T. was observed by Crismer. The C.S.T. of ethanol petroleum system was raised by 17°, by the presence of only 1% of water in ethanol. The C.S.T. of methanol and cyclohexane system is 45.55°. The presence of 0.0 1% of water in methanol raises the C.S. T. of the ssytem to 45.65°. The presence of armoatic hydrocarbons in petrol can be detected and estimated by determining the C.S.T. of petrol-aniline system. Similarly, the amount of ceresin in wax can be determined. The biological importance of C.S.T. is in testing the functioning of kidney. A kidney producing urine which raises the C.S.T. of urine-phenol system by 8° is in good order. The kidney is exceedingly well if the C.S.T. is raised by 12° to 16°. In general, the C.S. T. of the system is raised if the impurity is soluble in only one of the two components and the C.S. T. is lowered if the impurity is soluble in both the components.
Advantage is taken of the above principle in the preparation of well known disinfectant lysol. Lysol is a system of cresols and water. These two components do not mix completely at ordinary temperatures, but the addition of soap to the given mixture-soap is soluble in both cresol and water-lowers the C.S.T. to such an extent that the two components readily mix with one another at ordinary temperature to form homogeneous solution.
DISTRIBUTION LAW Problem 1 : State and explain Nernst's distribution law. What are the limitations of this law? Give some important applications of distribution law with special reference to extraction process. (Meerut 2004, 20(H, Kanpur 2(05)
Or
Define and discuss distribution law.
(Meerut 2(05)
[I] Distribution Law Nemst found that if a substance is present in two immiscible solvents in a system at equilibrium then the solute distributes itself between the two immiscible solvents in such a way that at constant temperature, the ratio of its concentrations in the two solvents is constant, whatever the total amount of the solute may be.
158
PHYSICAL CHEMISTRY-I
If C 1 and C2 represent the concentration of a solute in the two solvents,
then
C1
-C = constant = K • 2
where K is known as distribution or partition coefficient. Nemst found that the ratio C 1/C 2 is constant only when the solute has the same molecular species in both the solvents. If a solute associates to form double molecules in one solvent and not in the other, the equilibrium cannot exist between double molecules present in one solvent and single molecules present in the other. The law is valid only if the ratio of concentrations of single molecules in the two solvents is taken into consideration. The distribution law as enunciated by Nernst does not hold good in cases where the solute undergoes dissociation or association in any of the solvents. For example, if a solute remains unaltered in one solvent and undergoes partial dissociation in another, the ratio of total concentrations in the two solvents will not be constant, but the ratio of concentration of undissociated molecules in the two solvents would be constant. Thus, the distribution law in its proper form may be stated as, 'When a solute distributes itself between two immiscible solvents in contact with one another, there exists for similar molecular species at a given temperature, a constant ratio of concentration between the two solvents, irrespective of any other species which may be present.' Thus, in the above equation, C 1/C 2 == K, the terms C 1 and C2 refer to concentrations of similar molecular species in the two liquids at constant temperature. As the solubility of a solute changes with temperature, and as the magnitude of the change in the two solvents may not be the same, the distribution coefficient (K) is found to vary with change in temperature.
[II] Limitations of Distribution Law The important conditions to be satisfied for the application of the distribution law are as follows : (i) Constant temperature: The temperature should be kept constant throughout the experiment. (ii) Same molecular state : The molecular state of the solute should be the same in both the solvents. The law fails if the solute dissociates or associates in one of the solvents. (iiii) Non-miscibility of solvents : The two solvents should be nonmiscible or only slightly soluble in each other. The extent of mutual solubility of the solvents remains unchanged by addition of solute to them. (iv) Dilute solutions : The concentration of the solute in the two solvents should be low. The law fails when the concentrations are high.
159
CHEMICAL AND PHASE EQUILIBRIUM
(v) EquiHbrium concentrations : The concentrations of the solute should be noted after the equilibrium has been established.
[III] Applications of Distribution Law Distribution law helps us in calculating the degree of dissociation or association of a solute. Distribution indicators also involve the principle of distribution law. 1. The process of extraction : The most common and important application of distribution law is the solvent extraction of substances by solvents. Organic compounds are more soluble in organic solvents like CCl4 • C6Hc; etc., than in water and so in the laboratory, this principle is used for the removal of a dissolved substance from aqueous solution by using organic solvents. Since organic compounds have their distribution ratio largely in favour of organic phase, "most of the organic substances would pass into non-aqueous layer. Finally, this non-aqueous layer is removed and distilled to obtain the pure compound. In solvent extraction, it is advisable to use a given volume of an extracting liquid in small lots in successive stages rather than in one single operation at a time. The greater the distribution ratio is in favour of the organic solvent, the greater wiU be the amount extracted in anyone operation. Suppose a solute A is present in 100 c.c. water and that 100 C.c. of benzene is used for its extraction. Let the distribution coefficient of A between benzene and water be 4.
..
K =Concentration of A in benzene Concentration of A in water
4
(i) When the whole of benzene (100 c.c.) is used at a time for extraction, suppose Xl g of solute pass into benzene layer and x2 g is left in aqueous layer, so xI/lOO x2/ 100
i.e.,
=4;
Xl XI 4 -=4 or - - = X2 XI +X2 5
In other words, 100 c.c. benzene has separated 4/5 or 80% of the solute originally present. (ii) Now let us use 100 c.c. benzene in two successive extractions, using 50 c.c. each time. Then in the first extraction, Xl/50 X2/ l00 =4;
i.e.,
160
PHYSICAL CHEMISTRY-I
In other words, in the first extraction (2/3)rd, i.e., 66.6% is extracted. Hence, (l/3)rd or 33.4% of the original amount is still retained in aqueous medium. In the second extraction using 50 C.c. of benzene, we shall further extract (2/3)rd, of (l/3)rd, i.e., (2/9)th of the original amount. So, in both the extractions, using 100 c.c. benzene we can separate (2/3 + 2/9) = 8/9 or 88.9% of the original amount of the solute. Thus, a two stage extraction is more efficient. If we use four or five extractions, the operation wiIi still be more efficient. Derivation of General Formula .It is possible to derive a general expression for the amount remaining unextracted after a given number of operations. Let V C.c. of a solution containing W g of the substance be extracted with v C.c. of a solvent. Let WI g of substance remain unextracted in aqueous layer. Then Concentration of substance in solvent W-WI
v Concentration of substance in water
WI - V
:. Distribution coefficient
K=
or
WI =
WI/V W-WI
v KV(W- WI) V
KV
=W·-KV+v
If W2 be the amount remaining unextracted at the end of the second extraction with v C.c. of the solvent, then
w2 -W KV -W KV KV _W(~)2 I KV + v KV + v . KV + v KV + v Similarly, after nth extraction, the amount remained unextracted will be given by, KV Wn = W ( KV + vIn ... (8) If the entire quantity of the extracting solvent is used in one lot, then unextracted amount (W) will be given by,
W=W(K~~V)
...(9)
Since the quantity within the bracket is less than unity, (8) is smaller than (9) and Wn will be smaller, the greater the value of n. Hence the efficiency of extraction increases by increasing the number of extractions using only a
CHEMICAL AND PHASE EQUILIBRIUM
161
small amount of the extracting solvent each time. It must be remembered that the value of K, the partition coefficient in equations (8) and (9) is that of the substance between the solvent (A) in which the substance is already dissolved and the solvent (B) which is used for extraction. 2. Determination of association: When a solvent is associated in solvent 1 and exists as single molecules in solvent 2, the distribution law is written as,
'vc;- = K Cz
where n = number of molecules which combine to form an associated molecule. Thus, knowing the values of C 1, Cz and K, we can calculate the value ofn. 3. Determination of dissociation: Suppose a solute is dissociated in aqueous solvent 1 and exists as single molecule in solvent 2. If a is the degree of dissociation of the solute, the distribution law is written as :
Thus, a. can be calculated. 4. Confirmatory test for bromide and iodide: The salt solution is treated with chlorine water, when a small quantity of Br2 or Iz is liberated. This solution is then shaken with CCl 4 or CHCI 3. On standing, the CCl 4 or CHCl3 layer forms the lower layer. The free Br2 or 12 being more soluble in CCl4 or CHCl 3 concentrates into the lower layer, making it brown for bromine and violet for iodine. S. Determination of solubility : If the concentration of a solute in solvents 1 and 2 be C 1 and C2, then distribution coeffici~nt (K) is given by, K=C1/C z· As concentration and solubility (S) are proportional to each other, we can write, K=S I /S2 · If solubility (SI) of a solid in one solvent is known then solubility (S2) in other solvent can be calculated. 6. Distribution indicators: In iodine titrations, the end point is indicated by adding starch solution which turns blue. A greater sensitivity is obtained by adding a distribution indicator. A few drops of an immiscible organic solvent say CCl4 is added to the solution. The bulk of any iodine present passes into CCl 4 layer making its colour more intense. Besides, distribution law is applicable in deducing the formula of a complex ion, in desilverisation of lead, partition chromatography etc.
Problem 2 : State and explain Nernst's distribution law. How it is modified
when the solute undergoes association or dissociation in one o/the solvents?
162
PHYSICAL CHEMISTRY-I
[I] Nernst distribution law See problem 1.
[II] Dissociation of the solute in one of the solvents Let A represents the normal formula of the solute. It is not dissociated in solvent I, but dissociates into X and Y in the second solvent II (figure 19). Let C 1 and Cz be the concentrations in solvents I and II, respectively. The distribution law is valid only for the ratio of concentrations of similar molecular species in both the solvents. The equilibrium is represented as: A~X C2 (1 - a) C2a
Solvent I No dissociation Conc. = C1
t~
--->.
X + y Solvent II Dissociation Occurs
Total conc.
=C2
+ Y
C2 a where, a is the degree of dissociation of solute A in solvent II. According to modern distribution law.
K = Concentration of A in solvent I Concentration of A in solvent II
=
C1 C2 (1 - a)
Such a case is observed in distribution of oxalic acid between water and ether.
[III] Association of the solute in one of the solvents Let A represents the normal formula of the solute. It does not associate in solvent I, but associates in solvent II to give molecules of the type An (figure 20). Let C 1 be the concentration of solute A in solvent I and C2 be its total concentration in solvent II. In solvent II, the associated . molecules exist in equilibrium with single molecules, viz., An~nA
According to the law of mass action,
Solvent I No association Conc. = C1 I~ nA~A"
Solvent II Association occurs
Total conc.
<:2
163
CHEMICAL AND PHASE EaUILIBRIUM
or
... (1)
If the solute exists largely as associated molecules in solvent II, the concentration of the associated molecules (An) may be taken to be equal to C2, i.e., [An] = C2 Therefore, from equation (1), we get [A] = constant x (C2)lIn = kC 2 l1n . Applying the distribution law to similar molecular species, we have, K = Concentration of A in solvent I Concentration of A in solvent II K=
or
Cj k (C2)1I2
C1 -vi" = K . k =constant.
or
(C2) Such a case is observed in the distribution of benzoic acid between water and benzene.
NUMERICAL PROBLEMS Problem 1 : The following data shows the distribution of phenol between water and chloroform : Concentration (cJ) in water 0.436 Concentration (C2) in chloroform
0.094
0.163
0.254
0.254
0.761
1.85
5.43
Calculate the partition coeffr.cient between water and chloroform. What conclusions can be drawn from these results concerning the molecular condition of phenol in chloroform layer? Solution : Phenol in chloroform may be present either as normal molecules or in the form of associated molecule», In the former case, Cl/c2 should be constant, whereas in the latter cl/c211n should be constant, where n is the number of molecules of phenol associate to give a single associated molecule. The value of c 1/c2 in the four cases are given as follows: (i)
~: = ~:~~: =0.3701,
(ii)
~~ = ~:~~i = 0.2142,
= 0.254 = 0 1373
(iv)
~: = ~~36 = 0.0833.
( "') Cj
111
C2
1.85
.
,
Since values of cil C2 are not constant, so phenol does not exist as single molecules in chloroform.
164
PHYSICAL CHEMISTRY-I
CI
Now, the values of t= are given as follows : VC2
.
CI
0.094
(I) -:rc;=";(0.254) (
... ) 11l
CI 0.254 -rc;=";(l.85)
CI 0 163 0.1865, (ii)-;rc;=,,;(O.761) =0.1868,
0 1867 . ,
(.) IV
CI 0.436 -rc;=";(5.43)
01870 . ,
Since c l /";C2 values are constant, therefore, phenol exists as double molecules in chloroform. Problem 2 : In the distribution ofsolute between water l ) and chloroform (C2 ) the following data were obtained: CI C2 0.163 0.761 0.436 5.43 What information do you gather regarding the molecular state of solute in chloroform? Solution: When the given solute exists as simple molecule in both the phases, then C I /C 2 = K.
eC
C1 (In water)
C2 (In chloroform)
CI C2
-=K
0.163
0.761
0.163 = 0 214 0.761 .
0.436
5.43
~~36 =0.083
As the values of K are not constant, the distribution law in its simple form is not applicable. It means that the molecular state of the solute in both the phases is not the same. It may be either associated or dissociated. Assuming the association of the solute in chloroform as double molecules, then C j /";eC2) = K:
C2
C :.rc;=K
0.163
0.761
~(0.761) = 0.1868
0.436
5.43
0.436 ~(5.43) = 0.1870
Cj
I
0.163
As the values of K are constant, the solute exists as double molecules in chloroform layer.
Problem 3 : (Jllculate how much succinic acid would be extracted from 100 C.c. of water containing 5 g of succinic acid if extracted with 50 C.c. of
165
CHEMICAL AND PHASE EQUILIBRIUM
ether. The partition coefficient of succinic acid between water and ether is 5.5. Succinic acid has normal molecular weight in both the solvents. (Meerut 2005) Solution: Let x g of succinic acid passes over into 50 c.c. of ether, then the amount left unextracted in 100 c.c.water will be (5 - x) g. :. Concentration of solute dissolved in ether
=x g per 50 c.c. Concentration of solute left in water = (5 - x) g per 100 c.c. From the distribution law,
CI
K=-
C2
5 .5
or
= (5 -x)/lOO
x/50
or x =0.4166 g. Amount of succinic acid extracted = 0.4166 g
Problem 4: At 20·C, S02 was permitted to distribute between 200 C.c. of chloroform and 75 c.c. water. At equilibrium, the chloroform layer contained 0.14 mole ofS02 and water layer 0.05 mole. Calculate the distribution coefficient of S02 between water and chloroform. ' SI o ution :
Cw = --:ys Q~ mo Ies per c.c.
0.14 I CCHCI3 = 200 mo es per c.c.
K=~= 0.05 x 200 =0.953 CCHCI,
75 x 0.14
Problem 5 : If the distribution coefju:ient of benzoic acid between water and C6H6 is 0.304 at 20·C, calculate the number of moles of benzoic acid which may be extractedfrom 100 C.c. of 0.02 molar aqueous solution by 10 C.c. of C6H6' Solution: Let x moles of acid remain unextracted. The concentration of the acid in benzene will be
0.02-x 10
perc.c.
Also the concentration of the acid in the original aqueous solution is x 1100. x/100 K = (0.02 - x)/l0 x 10 x or 0.304 = 100 x (0.02 - x) 0.02 - lOx x = 0.015 mole
166
PHYSICAL CHEMISTRY-I
:. Number of moles of acid extracted = (0.02 - 0.015) =0.005 Problem 6 : The solubility 0/ a substance is twice as great in ether as in water. Compare the quantities extracted/rom 100 C.c. 0/ aqueous solution by using:
(i)
100 c.c. ether in single operation. (li) 50 c.c. ether in successive operations. Solution: (i) By using 100 c.c. ether at a time. Let 1 g of substance be present in 100 c.c. aqueous layer and let by using 100 c.c. ether at a time, x g of it passes over into ethereal layer. Then from distribution law,
CI C2
K=-
2=_X_ I-x
or
(': K=2 given)
x= 0.66 g (ii) By using 50 c.c. ether in two instalments. Illfirst instalment: Let by using 50 c.c. ether Xl g of substance passes over from 100 c.c. aqueous layer. Concentration of substance dissolved in ether = Xl g per 50 c.c. Concentration of substance left in aqueous layer = (1 - Xl) g per 100 c.c.
Hence from distribution law,
CI C2
K=-
Xl
2::::
or
Xl
50 (1- Xl) 100 =O;Sg
III second instalment: The amount of substance now left in aqueous layer is 1~.5 g, i.e., 0.5 g. Consider by using next 50 c.c., X2 g of substance passes over from 100 c.c. aqueous layer. Then. concentration of substance dissolved in ether = X2 g per 50 c.c.
Concentration of substance left in aqueous layer = (0.5 - x~ g per 100 c.c.
167
CHEMICAL AND PHASE EQUILIBRIUM
Hence, from distribution law, K = C I /C 2 X2
or
2=
50 (0.5 -x2)
100 or X2 = 0.25 g So, total amount of substance extracted in two instalments = 0.5 + 0.25 = 0.75 g Thus, it is clear that it is more advantageous to use 50 c.c. ether in two instalments (as it extracts 0.75 g) rather than 100 c.c. 'in one instalment (as it extracts only 0.66 g).
Problem 7: From the following data for the distribution of C~sCOOH between water and benzene at 20·C : (a)
Show that C~sCOOH is associated as double molecules in benzene, and
(b)
Caku1ate the partition coefficient on the basis ofthis assumption. The dissociation of C6H sCOOH in water may be neglected. CHp = 0.015, 0.0195, 0.0289 Cell, = 0.242, 0.412, 0.970. Solution: If benzoic acid remains normal in both the solvents, then, K = CCH20 = 0.0 15 = 0.0619 C.H6 O. 24 2
= 0.0195 =0 .0461 0.412
... (1) ..,
(2)
= 0.0289 = 0.2980 0.970
(3) .., Since the values of K are not constant, it is clear that benzoic acid does not exist as normal molecules in both the solvents. Suppose benzoic acid associates as double molecules in benzene, then K = Cwater/'l/(Cbenzene) (i) K = 0.0151"(0.242) =0.304 (ii) K = 0.0195/'1/(0.412) = 0.303 (iii) K =0.0289/"(0.970) =0.293 .. ff' . 0.304 + 0.303 + 0.293 S0, partitIon coe tClent = 3
=0.300
Problem 8 : 1 g of 12 is in 50 ml of CS2• 1000 ml of water is added into it.
Cakulate the amount of h extracted into water. Partition coe/ficiellt .?( h in CS2 and H 20 is 200. Solution : Suppose x g of h is extracted into water. Then, amount of 12 left behind in 50 m1 of CS 2 will be (1 - x) g.
168
PHYSICAL CHEMISTRY-I
Conc. of 10 In CSo K == - -_._--._':"
~ 50
=: - -
Conc. of 12 In H20
X
1000 ~~
200==~0--
or
x
x == 0.0909 g
1000 So, amount of 12 extracted in waler == 0.0909 g
Problem 9 : At 25°C, the distribution coefficient of iodine between CCl4 and water is 85. If at 25°C, the solubility of iodine in water is 0.33 gllitre, determine the solubility of iodine in CCI,Jo Conc. orI z In CCI 4 Solubility Ofl2 in CCI 4 (S]) Solution: K - - - - -=---- Conc. of 12 in H 20
Solubility Ofl2 In H 20 (S2)
Sl
or
85=---
or
Sl .= 85 x 0.33 g/litre
0.33 g/litre
...: 28.05 g/Iitre
MULTIPLE CHOICE QUESTIONS 1. 2.
3.
4.
5.
6.
7.
The vapour pressure of water at lOOT I~ 760 mm. The molar heat of vaporisation of water is 41.27 kJ mar I The vapour pressure of water at 95"C is : (a) 63.43 mm (b) 760 mm (c) 740.6 rum (d) 634.3 rum The maximum work done In a reactIOn is given by: (a) van't Hoff isotherm (b) van't Hoffisochore (c) Clausius-Clapeyron equatIOn (d) None The variation of equilibrium constant with temperature is given by : (a) van't Hoff isotherm (b) van't Hoff isochore (c) Law of mass action (d) Le Chatelier's principle The variatIOn of vapour pressure with temperature is given by : (a) van't Hoff isotherm (b) van't Hoff isochore (c) Law of mass action (d) Clausius-Clapeyron equation In the equilibrium A + 2B ~ C + D, increase in concentration of A will give more: (a) B (b) A (c) C (d) No effect In the equilibrium 2A + 3B ~ 3C 1 2D, an increase of pressure will : (a) shift the equilibrium towards the left (b) shift the equilibrium towards the nght (c) have no etl'ect (d) may shift the equiliblium In both directions In a one component system, the maximum number of phases which can exist in equiliblium at a point is (a)
8.
0
(b) I
(c) 2
In water system, the triple potnt has the values of :
(d) 3
169
CHEMICAL AND PHASE EQUILIBRIUM
9.
10. 11. 12. 13. 14.
15.
(a) O·C, 1 atm (b) O·C, 4.58 mm (c) 0.OO75"C, 1 atm (d) 0.OO7S·C, 4.58 mm At congruent melting point, the system is : (a) Non-variant (b) Monovariant (c) Bivariant (d) Trivariant At eutectic point, a system has: (a) Only two phases (b) Highest melting point (c) Minimum melting point (d) Uncertain composition At congruent melting point, Zn - Mg system is a : (a) One component system (b) Two component system (c) Three component system (d) Three phase system Henry's law is not applicable to : (a) H2 (b) O2 (c) CO2 (d) He In lead-silver system, the percentage of silver at the eute~tic point is : (a) 0% (b) 2.6% (c) 97.4% (d) 100% The dissociation of CaC03 is a : (a) One component system (b) Two component system (c) Three component system (d) Two phase system For an ideal solution: (a) Mlmix -70
16.
17.
18.
20.
21.
(c) ilHmix = 0
(d) ilHmix "? 0
A mathematically constant ratio exists between concentrations of similar molecular species in any two phases in contact with each other at a constant temperature. This law is known as : (a) Boyle's law (b) Charle's law (c) Distribution law (d) Dilution law Which one of the statements is false for distribution law? (a) Temperature must remain constant (b) The two solvents must be mutually immiscible (c) The concentration of the solute in two solvents must be kept very high (d) The concentration of the solute in two solvents must be very low When a solute is shaken with water and a solvent X in which it forms a dimer. the concentrations were found to be CIV and C., in water and solvent, respectively. Which one of the following expression will be correct between water and X ? (a)
19.
(b) M-Imix < 0
C,
(b) Cw
rc;
(c) C.,.
(d)
-rc;
Cw C" C.,. Distribution law cannot be applied for the system in which 12 is distributed between: (a) H20 and CS 2 (b) H20 and CCl 4 (c) H20 and ether (d) H20 and alcohol In a pair of immiscible liquids a common solute dissolves in both and eqUilibrium is reached. then the concentration of the solute in the upper layer is : (a) In fixed ratio with that in the lower layer (b) Same as in lower layer (c) Less than in the lower layer (d) None of these A 9S·C, an aqueous solution of iodine containing 0.0156liC! is in equilibrium with a CCl4 solution containing 4.412 g lie!. If the solubility of 12 in water at 9O"C is 0.34i lie l , then its solubility in CCl4 is : (a) 4.412 x 0.0516 (b) 0.0516 x 0.34 0.34 4.412
170
22.
23.
24.
PHYSICAL CHEMISTRY-I
() 4.412 x 0.34 (d). 0.0516 c 0.0516 4.412 x 0.34 The solubility of iodine in water is 0.8 gllitre. Ifthe partition coefficient of iodine between water and carbon tetrachloride is 82, the solubility of iodine in carbon tetrachloride is : (b) 65.6 gllitre (c) 0.009 gllitre (d) 81.2 gllitre (a) 102.5 g/Iitre The condition under which Nernst distribution law will not hold true is : (a) Temperature is constant (b) Molecular state of the solute is same in both solvents (c) The solute does not cause any change in the mutual solubility of the two solvents (d) None of them You are given 100 mL of CCl4 to extract iodine from 200 mL of its aqueous solution for extracting maximum amount of iodine. Which one of the following processes would you use ? (a) All 100 mL of CCl 4 (b) 50 mL of CCl 4 twice (c) 10 mL of CCl 4 ten times (d) 25 mL of CCl 4 four times
Fill in the Blanks 1.
2.
van't Hoff equation gives the variation of ........ with temperature. The equilibrium constant ........... with increase of temperature.
3.
dP
4.
Kp and K,. are related by the equation .............. . Phase rule was given by ............... . In ferric chloride-water system, the number of hydrates formed are ..... in number. Dissociation of NH4Cl in a closed vessel is a .......... component system. In Zn-Mg, the intermetallic compound has the composition ............ . Phase rule equation is given by .............. . In a two component system, the maximum degree of freedom is ......... . In a system of H 2, 02, N2 gases, the number of phases will be ............. . The eutectic temperature of Pb-Ag system is ............. . The solubility of triethylamine in water ., ....... with increasing temperature. Benzoic acid ............... as double molecules in benzene. The temperature during the distribution law experiment should be ......... . Distribution law was given by ........... . When iodine is distributed between water and CCI 4 , the solute exists in the ......... . molecular state in both the liquids.
5.
6. 7.
8. 9.
10. 11. 12. 13. 14. 15.
16. 17.
MI'
.
dT = T!1V IS k nown as ........... equatIOn.
True or False State whether the following statements are true (T) or false (F) ? P2
!1Hv
(I 1"
1.
10glO PI = 2.303 R
2.
equation. Clapeyron equation can be applied to an equilibrium between any two phases of the same substance.
T2 - TI represents integrated formofClausius-Claperyon
171
CHEMICAL AND PHASE EQUILIBRIUM
3.
4. 5. ~.
7. 8. 9. 10. 11. 12. 13. 14. 15. 16. 17. 18. 19.
van't Hoff isochore gives the temperature dependence of the vapour pressure of a substance. In the equilibrium N2 + 02 ~ 2NO, !l.H = x kJ, an increase of temperature will shift the equilibrium towards the right. [At' [Btl
In the reaction, nlA + n~B ~ mlC + m2 D, K~ = . , [C]m, [D]m 2 Nicotine-water has only one consolute temperature. There is only triple point in sulphur system. Henry's law is applicable to the solubility of H2 in water. The eutectic point in KI - H 20 system is also known as cryohydric point. In a two component system, three phases exist in equilibrium at the eutectic point. NaCI - H20 system belongs to a system in which the compound has an incongruent melting point. For non-ideal solutions, !l. Vmix is greater or less than zero. HCl - H20 system has a minimum boiling azeotropic mixture. Critical solution temperature does not change with pressure. The two solvents should be soluble in each other, if distribution low is valid. The solutions should be concentrated in distribution experiment. The process of extraction of a solute is based on distribution law. The degree of dissociation of NaCI can be calculated from distribution law experiment. If K = -rc;;ICB,then solute exists as dimer in solvent B.
ANSWERS Multiple Choice Questions 1. (d), 2. (a), 3. (b), 4. (d), 5. (c), 6. (c), 7. (d), 8. (d), 9. (a), 10. (c) 11. (a) 12. (c) 13. (b), 14. (b), 15. (c), 16. (c), 17. (c), 18. (b), 19. (d), 20. (a),
21. (c), 22.
(c)
23. (d) 24. (c)
Fill in the Blanks 1. equilibrium constant 2. increases 4. Kp = Kc (RT)'~N 5. Gibbs 7. one 8. MgZn2 10. Jhree 11. one 14. associates 13. decreases 16. Nemst 17. same True or False (T) 3. (F) 4. I. 2. (T) (T) 9. (F) 10. 7. 8. (F) (T) 14. (T) IS. (F) 16. 13. 19.
3. Clapeyron 6. four 9. F=C-P+2 12.303"C 15. constant
(T)
5.
(F)
II. (F)
6. (T) 12. (F)
(F)
17.
18.
(F) (T)
(F)
(T)
000
(OllODIAl STATE Problem 1 : Explain the terms colloidal state and colloidal solution. Describe various methods used in the preparation of colloidal solutions. Also mention the different methods used for the purifICation of colloidal solutions. (Meerut 2002; Agra 2005,2003,2001; Kanpur 2005,2000) Thomas Graham (1861) divided soluble substances into two classes. (i) Crystalloids, i.e., salt, urea, sugar and other crystalline substances which could rapidly diffuse through vegetable and animal membranes. Such solutions are called true solutions in which the diameter of the dispersed particles ranges from 1 Ato 10 A. (ii) Colloids, i.e., starch, gelatin, gum, proteins and other amorphous substances which could not diffuse through vegetable and animal membranes (Greek; kola = glue and eiodos = like). In a suspension, as sand stirred into water, the dispersed particles are aggregates of millions of molecules. The diameter of these particles is of the order of 2000 A or more. A colloidal solution is regarded as an intermediate between a true solution and coarse suspension. When the diameter of the particles of a substance dispersed in a sol vent ranges from about 10 A to 2000 A, the system is called colloidal solution, colloidal dispersion or simply a colloid. The material with particle size in the colloidal range is said to be in the colloidal state. CoUoidal solution is a heterogeneous system consisting of two phases. (a) Dispersed phase: The substance whose particles are distributed in a medium is called the dispersed phase. (b) Dispersion medium: The medium in which the particles of the dispersed phase are distributed is called the dispersion medium.
Preparation of Colloidal Solutions Colloidal solutions (also known as sols) can be prepared by different methods depending on the nature of the substances. Many substances., e.g., gelatin, starch etc. form colloidal solutions by merely dissolving them in water. Metals and inorganic substances are brought into the colloidal state by special methods. Two types of methods are mainly used: (172)
173
COLLODIAL STATE
Dispersion methods. Condensation methods. 1. Dispersion methods are those in which the particles of bigger size are broken into smaller colloidal particles. The different ways are : (i) Mechanical dispersion : The substance to be dispersed is suspended in dispersion medium to form a coarse suspension. This suspension is passed through a colloid mill which consists of two metal discs rotating in opposite directions at a speed of about 7000 revolutions per minute. (ii) Electro-dispersion (Bredig's arc method) : This method is generally employed in preparing hydrosols of metals, i.e., Ag, Au, Cu. Pt etc. Two rods of metal to be dispersed are kept immersed in cold water containing some KOH and a direct electric arc is struck between them [Fig. (1)]. The vapours of the metal condense to form colloidal particles. The traces of alkali act as stabilising agent. (i) (ii)
/+----
r-----
::;;E-=--~:ij:~t--
Metal Electrodes De-Ionised Water Traces of Alkali
+
Spark
(iii) Peptisation: If a freshly prepared Fe(OHh precipitate is treated with a small quantity of FeCl3 solution, it immediately forms a dark reddish brown colloidal solution of Fe(OHh. Similarly, a colloidal solution of AI(OHh is obtained when freshly precipitated Al(OHh is treated with a small quantity of dil. Hel, the amount of acid added being insufficient to convert the hydroxide completely into chloride. The process of transferring back a fresh precipitate into colloidal form is known as peptisation. It is the converse of coagulation or in other words, it is the re-dispersion of a coagulated sol. Substances like FeCl 3 and HCl which bring about I-~P~ tisation are known as peptising agents.
174
PHYSICAL CHEMISTRY-I
Cause of peptisation : The peptising action is due to the preferential adsorption of one of the ions of the electrolyte. which then gives to the colloidal particle a positive or negative charge. according to the charge on the adsorbed ion. For example. Fe(OHh adsorbs Fe3+ ions from FeCI 3 (peptising agent) and thereby gets a positive charge on the surface. Similarly. charged particles get separated yielding smaller sized colloidal particles of the type [Fe(OHhJ Fe 3+. Similarly. AS 2S3 precipitate obtained by pasing H2S through arsenious oxide solution peptises on treatment with excess of H2S and yields a negatively charged sol of [AS2S3J S2-. 2. Condensation methods are those in which the substances forming a true solution are converted into colloidal form. These are as follows: (i) By chemical reaction: The methods are : (a) Double decomposition: It is used in the preparation of sols of sulphides. e.g.• sols of AS 2S3 and HgS are obtained by passing H2S through their salt solutions. A solution of 1.0 g AS20 3 in 500 ml or'boiling water is obtained. It is cooled and H2S gas is allowed to pass slowly. When the intensity of yellow colour of the sol does not increase further. the passing of H2S is stopped.
ASP3 + 3H2S -7 As2S3 + 3H20 The excess of H2S is removed either by boiling the colloidal solution or by bubbling H2 through it. when a yellow sol of AS2S3 is obtained. (b) Hydrolysis: A colloidal solution of ferric hydroxide is obtained by adding a freshly prepared saturated solution of ferric chloride (2 or 3 c.c.) drop by drop to 500 ml of boiling distilled water. Ferric chloride hydrolyses to give a brown sol of ferric hydroxide. FeCl 3 + 3H20 -7 Fe(OHh + 3HCl The hydrochloric acid formed and unhydrolysed ferric chloride are separated by dialysis to get pure Fe (CHh sol. (c) Reduction: A colloidal solution of gold (or purple of Cassius) is obtained by reduction of auric chloride with stannous chloride or by HCHO. 2AuCl 3 + 3SnCl2 -7 2Au + 3SnCl4 2AuCl 3 + 3HCHO + 3H20
-7
2Au + 3HCOOH + 6HCl
A trace of K2C03 is added to neutralise the acid formed whose presence renders the sol unstable. (d) Oxidation : A colloidal solution of sulphur is obtained by the oxidation of HzS with dil. HN03 or sulphurous acid. 3H zS + 2HN03 -7 2NO + 4H20 + 3S 2H2S + H2S03 -7 3HzO + 3S The colloidal solution of sulphur so formed is treated with saturated salt solution when sulphur is precipitated. It is filtered. washed with water to
175
COLLODIAL STATE
remove salt and then washed further when the precipitate of sulphur peptises and forms a colloidal solution. (ii) By exchange of solvent, i.e., by lowering of solubility : Colloidal solution of a substance can be prepared by dissolving it in one solvent and pouring this solution ill another solvent in which it is less soluble, e.g., a sulphur or phosphorus sol can be prepared by adding their saturated alcoholic solution in water. (iii) Passing vapours of an element into a liquid: When the vapours of a boiling element are passed through a liquid, condensation takes place sometimes with the formation of a soluble sol, e.g., mercury sol can be prepared by passing a stream of vapours from the boiling element into cold water containing a suitable stabilising agent.
Purification of Colloidal Solutions The colloidal solutions prepared by the above methods contain impurities. These impurities must be removed in order to make the sol stable. This is done by either of the following methods. 1. Dialysis : The purification of colloidal solution by this method is based on the inability of the sol particles to pass through an animal membrane or a parchment paper which allows only the molecules or the ions to pass through. The vessel in which dialysis is carried out is known as dialyser [Fig. (2)]. A dialyser consists of a special type of vessel open at both the ends. To the lower end a membrane is stretched. This membrane allows only the solvent and other molecules to pass through it, but it is impermeable to the colloidal particles. The dialyser is then suspended in a larger vessel
lied to stand
::+---1:
r--- Dialysis bag
~
~------~ ~ ~~-------~
e ~
Fresh water.--
e
/·e \
,'. •
~Sol particle
"e " "e +' '. ~ ~.
:e • ·e-
.'
Molecule
R' +
·.x·"--1--/~," ~
Ions
~ -e
.~
~----------------------------~ Fig. 2
I
+ Water +,Ions + Molecules
176
PHYSICAL CHEMISTRY-I
containing pure distilled water. The distilled water is periodically renewed by allowing water to flow into the outer larger vessel and removing it by means of a syphon. Dialysis is continued till no further contamination of the washing liquid occurs. 2. Electrodialysis: In this process, dialysis is carried out under the influence of electric field (figure 3) Potential is applied between the metal screens supporting the membranes. This speeds up the migration of ions to the opposite electrodes. So, dialysis is greatly accelerated.
~n
DiaIYSiS bag
Impure colloidal solution
Sol particle " + '. + ~,'-.\ (f) /"e - \ Positive-ion '. -
e ",/
Gi....-..
-
e
I-&---
Anode
...e. ,
I
'. Negative ion " rD, ~ ' . . ....... + •__ -...... ~~ \
Olecule
I
+ Water + Ions + Molecules
e
Cathode
I
I
Fresh water.
e
Water ------------------------~ Fig. 3
Electrodialysis is not meant for the removal of non-electrolytic impurities like sugar, urea etc. 3. Hot dialysis: In hot dialysis, a temperature of 65-90°C increases the rate of dialysis by about three times. Heating is done by placing an electrical heating unit in the dialyser. All hydrophobic sols cannot withstand the high temperature and the membranes may give off undrained colloidal impurities. 4. Ultraf"dtration : Sols pass through an ordinary filter paper. Its pores are too large to retain the colloidal particles. However, if the filter paper is impregnated with collodion and then dipped in HCHO or glacial acetic acid, the pore size is much reduced. Such a modified filter paper is called an
ultra..Jilter*. Preparation of ultrafilters : The preparation invol ves the imp~gnation of special filter paper with collodion (5% solution of nitrocellulose in a mixture of alcohol and ether) and subsequently hardening by dipping it in a solution of HCHO and finally drying. The size of pores· of ultrafilter may be altered by varying the concentration of collodion solution. The pore size of ultrafilters can be determined by the rate of penetration of water or by forcing air through the wetted membrane. During ultrafiltration, the sol particles are left behind on the ultrafilter, which can be redispersed in the pure dispersion medium.
177
COLLODIAL STATE
The separation of the sol particles from the liquid medium and electrolytes by filtration through an ultrafilter is called ultrafiltration. It is a slow process. Gas pressure or suction is applied to speed it up. The colloidal particles are left on the ultrafilter in the form of slime. The slime is then stirred into fresh dispersion medium to get the pure sol. By using graded ultrafiIters, the technique of ultrafiltration can be used to separate sol particles of different sizes.
Problem 2 : (a) Mention the differences between a true solution, colloidal solution and suspension. (Meerut 2004)
(b) Explain the types of colloidal systems with examples. (a) The differences between a true solution, colloidal solution and suspension are given in the table shown below. S.N. Characteristic property
Colloidal solution
Suspension
True solution
I.
Nature
Heterogeneous
Heterogeneous
Homogeneous
2.
Size range
Particles are greater than 10-5 cms (or
Particles are between 10-5 and 10-7 ems (or 0.1 i.! to I m~) in diameter.
Particles are less than 10-7 ems (or I m~) in diameter.
0.1 ~ Of 100 in diameter.
m~)
3.
Visibility
Particles are viSIble under a microscope or even with a naked eye.
Particles are generally VIsible under Ultramicroscope.
Particles are not visible even under ultramicroscope.
4.
Diffusibility
Do not diffuse
Diffuse slowly
Diffuse rapIdly
5.
Filtrability
Can be filtered Can be filtered even by an through an animal ordinary fi Iter paper. membrane, through which the colloidal particles do not pass.
6.
Molecular weight
7.
Osmotic pressure (o.P. oc
Low
High
Cannot be filtered
High
Low
Low
High
Depends upon the shape and size of the particle
Depends upon the nature of the ion.
M~W.J
8.
Colour
9.
Tyndall effect
-
Do not exhibit Exhibit
Do not exhibit
178
PHYSICAL CHEMISTRY-I
S.N. Characteristic property 10.
Colloidal solution
Suspension
True solution
Brownian movement
Do not exhibit Exhibit
II.
Electrophoresis
Do not exhibit Exhibit
12.
Coagulation
-
Can be coagulated by adding electrolytes.
Can be precipitated by adding suitable electrolytes.
13.
Presence of electric charge
-
Particles carry either positi ve or negative charge.
Particles do not carry any charge.
Exhibit Exhibit
(b) A colloidal system is a biphasic system consisting of (i) dispersed phase and (ii) dispersion medium. Therefore, the two phases can either be solid, liquid or gas. Thus, different types of colloidal system depending upon the nature of the dispersed phase and dispersion phase are possible, which can be summarised in the following table with examples. Name of the colloidal system
S.N.
Dispersed phase
Dispersion medium
I.
Solid
Solid
Solid sol.
Gems, precious stones etc., alloys.
2.
Liquid
Solid
Gels or jellies
Jelly, butter, cheese.
3.
Gas
Solid
Solid foam
Pumice stone, cork, foam rubber.
4.
Solid
Liquid
Colloidal solution, sol Sols of metals, sulphur etc., paint, ink.
5.
Liquid
Liquid
Emulsion
Milk.
6.
Gas
Liquid
Foam
Soap lather, shavfng cream, whipped cream.
7.
Solid
Gas
Aerosol
Smoke, dust.
8.
Liquid
Gas
Liquid aerosol
Fog, cloud, mist.
Examples
A gas dispersed in a gas forms a homogeneous mixture and so do not form a colloidal solution. Problem 3: How will you prepare the colloidal solutions o/the/ollowing: Arsenic sulphide,/erric hydroxide, gold, sulphur, silicic acid, carbon, iodine, mastic. (Meerut 2000 ) 1. Arsenic sulpbide sol: A solution of 1.Og AS 20 3 in 500 ml of boiling water is obtained. It is cooled and H2S gas is allowed to pass slowly. When
179
COLLODIAL STATE
the intensity of yellow colour of the sol does not increase further, the passing of H 2S is stopped. AS 20 3
+ 3H2S ~ AS 2S 3 + 3H20
The excess of H 2S is removed either by boiling the colloidal solution or by bubbling Hz through it, when a yellow sol of ASZS3 is obtained. 2. Ferric hydroxide sol: A colloidal solution of ferric hydroxide is obtained by adding a freshly prepared saturated solution of ferric chloride (2 or 3 ml) drop by drop to 500 ml of boiling distilled water. Ferric chloride hydrolyses to give a brown sol of ferric hydroxide sol. FeCl3 + 3HzO ~ Pe(OHh + 3HCl The hydrochloric acid formed and unhydrolysed ferric chloride are separated by dialysis to get pure Fe (OH)3 sol. 3. Gold sol: A colloidal solution of gold (or purple of Cassius) is obtained by reduction of auric chloride with stannous chloride or by HCHO. 2AuCl3 + 3SnCz ~ 2Au + 3SnCl 4 2AuCl3 + 3HCHO + 3H20
~
2Au + 2HCOOH + 6HCl
A trace of KZC03 is added to neutralise the acid formed whose presence renders the sol unstable. 4. Sulphur sol : A colloidal solution of sulphur is obtained by the oxidation of HzS with dil. HN0 3 or sulphurous acid. 3H zS + 2HN03 ~ 2NO + 4H 20 + 3S 2H2S + H ZS03 ~ 3HzO + 3S The colloidal solution of sulphur so formed is treated with saturated salt solution when sulphur is precipitated. It is filtered, washed with water to remove salt and then washed further when the precipitate of sulphur peptises and forms a colloidal solution. 5. Silicic acid sol: It is obtained by adding a solution of sodium silicate to a solution of 4N HCI with constant stirring till the acid is in excess. The mixture is diluted to about twice its volume and then dialysed against running water till it is free from HC!. 6. Carbon sol : It is obtained by rubbing lamp black with water in presence of little protective colloid like gum arabic, when a colloidal solution of carbon (Indian ink) is formed. 7. Iodine sol: It is prepared by oxidising hydroiodic acid with iodic acid, when a purple colloidal solution of iodine is obtained. 5HI + HI0 3 = 3HzO + 312 8. Mastic sol: It is a resin type hydrophilic colloid. Its colloidal solution is obtained when it is dispersed in only pure distilled water. As such it carries a very little charge, but when dispersed in acidic solution and alkaline solution, a positively charged and negatively charged sol, respectively is obtained.
180
PHYSICAL CHEMISTRY-I
Problem 4 : Define and discuss or write short notes on the following: Lyophilic and lyophobic colloids· (Meerut 2001) (U) Peptisation (Meerut 2002) (iii) Dialysis (Meerut 2002) (iv) Ultramicroscope (v) Tyndall effect (vi) Brownian motion (vii) Electrophoresis (viii) Electro-osmosis (ix) Coagulation (Meerut 2006, 2001) (x) Hardy-Schulze law (Meerut 2007,2006,2000) (xi) Protection (Meerut 2007) (xii) Gold number (Meerut 2007, 2005, 2003, 2000) (xiii) Stability of lyophobic colloids (xiv) Iso-electric point (xv) Emulsions and characteristics (Meerut 2006, 2004, 2003) (xvi) Gels (Meerut 2002) (xvii) Electrical double layer (Meerut 2007,2005,2002,2000) Or Zeta potential (Meerut 2006) (i)
1. Lyophilic and Lyophobic Colloids A substance which passes into the colloidal state, simply by bringing it in contact with water is known as hydrophilic colloid (hydro = water, philic = loving). But if any solvent like organic liquid is used instead of water, then the more general term lyophilic colloid (lyo = solvent; philic = loving) is used. Gum, starch, soap are lyophilic colloids. These colloids when once precipitated can again be brought back directly into the colloidal state. Hence, they are also known as reversible colloids. (Colloidal state precipitate). Reversible colloids are also termed as resolllble or non-electrocratic colloids. Substances like metals, metal sulphides cannot be brought into the colloidal state simply by bnnging them in contact with water and, therefore, special methods are devised for the purpose. Hence, they are known as hydrophobic colloids (hydro = water; phobic = hating). In case of solvent other than water, the general term lyophobic is used. Further, if these colloids are precipitated, then it is not very easy to reconvert the precipitate directly into the colloidal state. Hence, they are termed as irreversible colloids (colloidal state ~ precipitate),'irresoluble or electrocratic colloids.
Difference between lyophilic and lyophobic sols S.N.
Lyophilic sols
Lyophobic sols
L
Prepared by direct mixing wIth dispersion medium,
Not prepared by direct mixing with the medium,
2.
Viscosity higher than dispersion medIUm. set to a gel.
Vi,cosity almost the same as of medium. do not set to a gel.
181
COLLODIAL STATE
3.
Little or no charge on particles.
Particles carry positive or negative charge.
4.
Do not show Tyndall effect.
Show Tyndall effect.
5.
Particles migrate to anode or cathode or not at all.
Particles migrate to either anode or cathode.
6.
Particles are generally solvated.
No solvation of particles.
7.
Precipitated by higher concentration of electrolytes.
Precipitated by low concentration of electrolytes.
8
Reversible.
Irreversible.
2. Peptisation See problem 1.
3. Dialysis See problem 1.
4. Ultramicroscope Zsigmondy and Siedentopf (1903) devised ultramicroscope to study Tyndall effect. It consists of an ordinary microscope with a special arrangement for focussing a beam of light. The light from an arc lamp S is allowed to pass through a projection lens Ll and objective lens L z and focussed into a vessel containing a colloidal solution C. The light scattered from colloidal particles now enters the microscope M placed at right angle to the path of light. The colloidal particles look as pin points of light. We do not see the colloidal particles in this ultramicroscope but see the light scattered by them. As each particle can be detected as 'spot' of light, therefore, slit ultramicroscope helps to determine the number of particles in a definite volume. The great value of the slit consists in the ability by its use to suction off a portion of the fluid which is being examined. Zsigmondy (1912) devised an immersion ultramicroscope, which was used with strongly coloured liquids. In it, illumination objective and microscope objective are brought as close as possible, and the liquid to be investigated can be used as a drop between the two lenses. The ultramicroscope does not give any information regarding the shape and size of colloidal particles.
5. Tyndall Effect When a strong beam of light is passed through a true solution, it cannot be seen unless the eye is kept directly in the path of light, but when the same beam of light is passed through a sol and viewed at right angles, the path of the light becomes illuminated. This is due to the fact that sol particles absorb
182
PHYSICAL CHEMISTRY-I
f!F""""":;':'"''''''''''''''''''''''''''''''''''''''''"""""""""""""""""""""""","','""""""'""""""""""'''''''''''''''''''''\!J
M
Microscope Knife Objective Projection edge Projection lens lens slit lens
o o
C:>Arc ~Iight
I
;;};;;:;;:~~::~~;i':;;':!
light energy and then emit it in all directions in space. This scattering of light illuminates the path of the beam in the colloidal solution. The phenomenon of the scattering of light by the colloid particles is called Tyndall effect. The illuminated beam or cone formed (figure 5) by the scattering of light by the sol particles is often referred to as Tyndall beam or Tyndall cone. If the sol particles are large enough, the sol may even appear turbid in ordinary light. as a result of Tyndall scattering. As ions or solute molecules are too small to scatter light, the beam of light passing through a true solution is not visible when viewed from the side. So, Tyndall effect can be used to distinguish a colloidal solution from a true solution. The blue colour of sky, hazy illumination of the light beam from the film projector in a smoke filled theatre or light beams from the head-lights of a car on a dusty road are common examples of Tyndall effect.
A
M~SCOpeO
Halo around sol particle
*
• ~ scattered
~
View under microscope
',', '.'
183
COLLODIAL STATE
6. Brownian Motion
• .• ?"""".".,."' • . """"'."'., ,""""'·'·'·"'··""?·ll::
R. Brown (1827) observed that when a sol is examined with an ultrami:::, croscope, the suspended particles are .... seen as shining specks of light. By fol:;:; lowing an individual particle, it is observed that the particle is undergoing a constant rapid motion in straight lines in all possible directions [Fig. (6)]. The continuous rapid zig-zag movement executed by the colloidal particles in a dispersion medium is known as Brownian movement. Suspensions and true solutions do not exhibit Brownian movement. It was seen that Brownian motion was independent of the nature of the colloidal particles but was more rapid when the particles are smaller and the solution is less viscous. Explanation: It was proved that the motion of the particles is due to the unequal bombardment of the suspended particles by the molecules of the dispersion medium in which they are dispersed [Fig. (7)]. On this basis, Einstein suggested that colloidal particles must behave like dissolved molecules and gas laws should apply to these systems, just as they were applied to true solutions. As the particles increase in size, the probability of unequal bombardment diminishes and eventually the collisions on different sides ~
~
equalise each other. As the size is reduced, the probability of unequal bombardment rapidly increases, whereby the Brownian movement becomes more and more violent. Applications and Importance (i) Confirmation of kinetic theory: It is a direct demonstration of ceaseless motion of molecules as pictured by kinetic theory of motion. (ii) Determination of Avogadro's number: With the help of ultramicroscope, the number of particles in a given mass of the gas can be counted and the Avogradro's number, i.e., the number of molecules in one mole of gas can be calculated. (iii) In stabiliSing colloidal solution: The zig-zag motion prevents the
184
PHYSICAL CHEMISTRY-I
setting of colloidal particles by gravity and thus helps in stabilising the colloidal solutions to some extent.
7. Electrophoresis The sol particles carry a charge either positive or negative. It was first observed by Linder and Picton (1892) and is used to indicate the migration of colloidal particles in an electric field. When the particles move towards the cathode, the phenomenon of migration of particles is known as cataphoresis and migration towards the anode is known as allapllOresis. But in practice, a more general term electroplzoresis is used. So, electrophoresis is defined as the
:{"""";.'."""'X."""'' ' ' ' ' ' ';. ,.,""""""""""""""""'\:.•'.•; :::; '.'
", .::: ::::
',' .... ..'
Water
::;: '.'
movement of colloidal particles :::;::-t·?~S7:;zz:::::::::~h::Gs:d towards oppositely charged electrode under the influence of electric field. Electrophoresis can be studied by the simple apparatus [Fig. (8)] which consists of a U-tube fitted with a funnel shaped reservoir and a stop cock. A small amount of water is placed in the V-tube and then some quantity of the sol is taken, so as to form a layer under the pure dispersion medium. An electric current is then applied, by connecting the two electrodes dipped in the solution. It will be seen that the particles begin to migrate towards the oppositely charged electrode. If the sol is coloured, then the movement of the particles can be observed directly by naked eyes. But if the sol is not coloured, then the electrophoretic migration can be observed by microscope or ultramicroscope. When the colloidal particles reach the electrode they lose their charge and generally coagulate into coarse particles. Applications of Electrophoresis (i) In determining the charge : The nature of the cha~ge on a colloidal particle can be ascertained by its migration in an electriC field. (ii) In electro-deposition of rubber: The latex obtained from the sap of certain trees is an emulsion of negatively charged rubber particles dispersed in water. This can be deposited on any substance which is made the anode during electrolysis. (iii) In the removal of colloidal smoke from outgoing gases : The removal of negatively charged carbon particles from smoke can be done by passing the smoke between positively charged metal electrodes. The principle has been utilised in the construction of Cottrell precipitator.
COLLODIAL STATE
185
(iv) Electrophoresis has proved to be of great importance in medicine, industry etc. e.g., in characterising proteins causing diseases as well as in the isolation of enzymes. Flavin enzyme was similarly isolated from yeast by Theorell (1932). (v) Lyophobic sols show electrophoresis: This is because the colloidal particles are electrically charged, which move under the influence of electric field.
8. Electro-osmosis A sol is electrically neutral. So, the dispersion medium carries an equal but opposite charge to that of the dispersed particles. Thus, the medium will move in opposite direction to the dispersed phase under the influence of applied potential. When the dispersed phase is kept stationary, the medium is actually found to move to the electrode of sign opposite than Plug of its own. moist clay The movement ofthe dispersion medium under the influence of applied potential is known as electro-osmosis. ~~~~~~ Electro-osmosis is a di- WB2ill22t::z:::z:::z:z:±2Bili211iU rect consequence of the existence of zeta potential between the sol particles and the medium. When the applied potential exceeds the zeta potential, the diffused layer moves and causes electro-osmosis. The phenomenon of electro-osmosis can be studied by using aU-tube [fig. (9») in which a plug of moist clay (a negative sol) is fixed. The two limbs of the tube are filled with water to the same level. The platinum electrodes are dipped in water ar,d potential applied. It is observed that water level rises on the cathode side, while it falls on the anode side. This motion of the medium towards the negative electrode, shows that the charge on the medium is positive. Similarly, for a positively charged sol, electro-osmosis will occur in the reverse direction.
Applications of Electro-osmosis (i) In the preparation of pure colloids. The case in point is the preparation of colloidal silicic acid of low molecular weight, i.e., in a state of fine sub-di vision.
186
PHYSICAL CHEMISTRY-I
(ii) In the tanning of hides and impregnation of similar materials. (iii) In the manufacture of gelatin for photographic emulsions. The object is to get a gelatin free from fat, minerals and reducing constituents. This phenomenon is also applied in the manufacture of high grade glue. (iv) In the dying of peat, a process due to von Schwerin (1903).
9. Coagulation It has been known for some time that on the addition of a very small amount of suitable electrolytes to a sol, the corresponding substances are thrown out of the solution and form a precipitate. If a suitable quantity of sodium chloride is added to a ferric hydroxide sol, the sol becomes turbid and ultimately precipitates. The electrolyte thus causes the colloid particles to coalesce. So, the process by means of which the particles of the dispersed phase in a sol are precipitated is known as coagulation or flocculation. It has been observed that : (i) All iOIl having a charge opposite to that of the sol is responsible for coagulation. This ion is known as active ion, e.g., ill the coagulation of negatively charged sols the cations are the active ions and anions are the active ions in the coagulation of positively charged sols. Oi) The coagulating or.flocculating power of the active ion increases with increase in the valency of the active ion.
Hardy-Schulze law Hardy and Schulze gave a law, after observing the regularities concerning the sign and valence of the active ion. This law is known as Hardy-Schulze law. It can be defined as, 'Greater the valency ofthe active ion, greater is the coagulating power of the active ion. ' Thus, in the case of a positively charged sol the coagulating power of anions is in the order of PO~- > SO~- > CC whereas the coagulating power is in the order of A1 3+ > Bi+ > Na+ in the case of negatively charged sols. The precipitate after being coagulated is known as 'coagulum'. The amount of electrolyte required to produce coagulation depends upon the total surface exposed by the colloidal particles. Hence, more concentrated sols require more electrolyte for coagulation. The minimum concentration of an electrolyte required to bring about coagulation or flocculation of a sol is known as coagulation or flocculation value. It is evident, therefore that polyvalent ions are the most effective active ions in causing coagulation. The coagulation of one sol may be effected by mixing an oppositely charged second sol. Thus, when a positive sol of Fe(OH)3 is mixed with an equi valent amount of a negative sol of AS 2S3• both get coagulated and separate out. This process is known as mutual coagulation of sols.
187
COLLODIAL STATE
10. Hardy-Schulze Law See part 9 of this problem.
11. Protection Lyophilic sols are more stable towards electrolytes. It is due to the fact that these particles are highly solvated and the electrolyte does not penetrate easily to coagulate lyophilic sols. On the other hand. lyophilic sols are readily
~
Hydrophilic sol particle
:. _ :
H2 0 Layer
~j
.0-
-
Dehydrating
---,s-,-ub-,s-,-ta_nce,,--_~..
--: . 0 j 0 _
Hydrophobic _ sol particle
Sol particle
precipitated by small amounts of electrolytes. However. these sols are often stabilised by the addition of lyophilic sols. So, the property of lyophilic sol to prevent the ceaguiation of a lyophobic sol is called protection. The lyophilic substance which is used to protect a lyophobic sol from coagulation is called aprotective colloid. This protection is due to the fact that the particles of the lyophobic sol adsorb the particles of the lyophilic sol. Thus. the lyophilic colloid forms a coating around the lyophobic sol particles The lyophobic colloid thus behaves as a lyophilic sol and is precipitated less easily by electrolytes [Fig. (10)). Thus. if a little gelatin (protective colloid) is added to a gold sol the latter is protected. The protected gold sol is no longer precipitated on the addition of an electrolyte. say NaCl.
12. Gold Number Different protective colloids have different protective powers. Zsigmondy showed that protective power of a protective colloid can be measured in terms of its gold Ilumber. Gold number may be defined as 'the number of milligrams of a protective colloid which will just stop the coagulation of 10 ml of a given gold sol on addillg 1 ml of 10% sodium chloride solution'.
188
PHYSICAL CHEMISTRY-I
If no protective colloid is present in the gold sol, it will turn from red to blue. The smaller the gold number of a protective colloid, the greater is its protective power. Gold Number
Colloid
Gold number
Casein
0.1
Gelatin
O.OOS-O.OI
Dextrin
6.20
Starch (potato)
2S.0
Colloidal Si02
oc
Gum arabic
0.IS-D.2S
Thus, starch has a high gold number, which shows that it is an ineffective protective colloid, while gelatin has a small gold number and is thus an effective protective colloid. (problem : Gold number of starch is 25. Calculate its amount to be added intd 100 ml of gold sol so that it is not coagulated in presence of 10 ml of 10% NaCI solution. Solution : Goid number of starch is 25 means 25 mg of starch IS added to prevent coagulation of 10 ml of gold sol by I ml of 10% NaCI solution. For 100 ml of gold sol, 250 mg of starch is needed to prevent coagulation by 1 ml of 10% NaCI solution. But for 10 ml of 10% NaCI solution, only
251~ 1 mg or 25 mg of starch is
I[Cquired.
13. Stability of lyophobic Colloids Preventing coagulation of a colloidal solution by any means is known as stabilising the colloid. Bancroft (1915) gave the position with regard to stability. He said that 'allY substance may be blVugllt into a colloidal state, plVvided the particlt:.s of the dispersed phase are so small that the Brownian movement keeps the particles suspended and plVvided the coagulation ofllie particles is prevented by a suitable sUlface film '. It was observed that lyophobic sols are stable due to their electric charge. As most of the lyophobic sols are prepared by preferential adsorption of ions, the mutual repulsion of particles is responsible for their stability. The sols become unstable, as soon as they are robbed off their charge. However, in case of lyophilic sols, the stability is due to electric charges as weIl as solvation-a phenomenon in which the colloid particle is surrounded by a thin film of the solvent. The layer of the solvent forms an envelope around each colloidal particle and thus pre-forms an envelope around each colloidal particle and thus prevents the particles from coming together. Groups like
COLLODIAL STATE
189
-COOH and -NH2 in proteins and -OH in hydroxides and polysaccharides can bind water molecules. Lyophilic sols are coagulated when first the solvent layer and then the charge of the ion are removed. Factors Affecting Stability: There are numerous factors which affect the stability of a colloidal system. (i) Brownian movement: As a result of this movement, the particles are in constant rapid motion, whereby aggregation of particles is prevented. So, the sol remains stable. But as soon as Brownian motion ceases, the sol becomes unstable. It is, therefore, expected that a sol will be more stable, the higher the dispersity and greater the charge on the particles. (ii) Addition of electrolytes : As described before, a sol becomes unstable by the addition of suitable quantity of electrolytes. It is now established that the presence of electrolytes has a powerful effect on the potential difference between the particles of the dispersed phase and the dispersion medium and that this potential difference is closely connected with the stability of sols. (iii) Effect of concentration of the sol: Ghosh and Dhar (1927) showed that a number of positively charged sols follow the rule that, greater the concentration of the sol greater the amount of electrolyte required to coagulate the sol. In other words, the greater the concentration of the sol, greater is its stability. (iv) Effect of dilution: Chaudhury (1928) showed that the dilution of a sol also affects the stability, which is decreased. The reason is that the decrease of the charge and total surface of sol particles with dilution decrease the stability of the sol. Mukherjee (1930) attempted to find out a relation between dilution and stability of a sol by measuring the migration velocity of the particles, but found the relation to be complicated. (v) Rate of addition of electrolyte : Dhar (1925) showed that the amount of electrolyte required to confer stability on a sol depends upon the ra1e at which it is added. This phenomenon is known as 'acclimatization of sols '. There are two types of acclimatizations: in some cases, less electrolyte is required, when it is added slowly or in small quantities at a time, and in other cases, the amount required is more. The former and latter are known as positive and negative acclimatizations, respectively. This phenomenon is probably due to charge occurring in the stabilisation by electrolytes. (vi) Temperature: It is seen that decrease of temperature confers more stability on the sols. Reid and Burton (1928) showed that heat alone is sufficient to cause coagulation. (vii) Mechanical agitation: It has been seen that mechanical agitation decreases the stability of sols. Freundlich and Loebmann (1922) showed that the sol of CuO was found to be coagulated by mechanical agitation. The rate of coagulation is proportional to the square of the rate of stirring. (viii) Ultra-violet light and X-ray radiations: Many lyophohc sols are coagulated by U-V light and X-rays or rays even from radiations of
190
PHYSICAL CHEMISTRY-I
radium. It is seen that the coagulation effect of the rays is independent of the sign of the colloid. Lal and Ganguly (1930) studied the coagulating influence of U-V light on sols of AgI, Au, Ag, V205' Th(OHh, As 2S3 etc.
14. Isoelectric Point Isoelectric point is that point at which the concentration of the positi ve ions in solution becomes equal to that of the negative ions. Similarly, in case of colloidal solutions, a sol may be positively charged in presence of an acid, due to the preferential adsorption of H+ ions. On the contrary, a sol may be negatively charged in presence of an alkali, due to the preferential adsorption of OH- ions. However, there must be an intermediate H+ ion concentration at which the colloidal particles are neither positively nor negatively charged. Hence, the isoelectric point in case of colloidal solution, is that point at which the colloidal solutions have no charge at all. IsoeleclI"ic point plays an important part in the stabilization of a sol. The stability of a sol is due to the presence of zeta potential (see subsequent pages) which is zero at the isoelectric point. At this point, the colloidal particles will not move towards any electrode under the influence of electric field. Hardy said that at isoelectric point, the colloidal particles are electrophoretically inert. Isoelectric point also occurs in protein sols. A protein solution is amphoteric. In acidic medium, a protein sol is positively charged, while in an alkaline medium, it is negatively charged. At a certain point or at a certain pH, the particles will have equal positive and negative charges. Different protein sols have different isoelectric points, e.g., isoelectric point for a gelatin sol occurs at pH == 4.7. Below pH 4.7, the colloidal particles of gelatin move towards the cathode, while at pH greater than 4.7 the particles move towards the anode.
15. Emulsions and Characteristics Emulsions are liquid-liquid colloidal systems. In other words, an emulsion may be defined as a dispersion of finely divided liquid droplets in another liquid. Generally, one of the two liquids is water and the other. which is immiscible is designated as oil. Either liquid can constitute the dispersed phase. . Types of Emulsions : There are two kinds of emulsions: (i) Water in oil type: In it water is the dispersed phase and oil acts as dispersion medium, e.g., butter etc. It is designated by WIO or w-in-o. (ii) Oil in water type: In it oil particles form the dispersed phase and water is the dispersion medium, e.g., milk, vanishing creams etc. It is designated by O/W or o-in-w. Tests for Types of Emulsion : The two types of emulsions can be
experimentally identified in a number of ways, which are as follows:
191
COLLODIAL STATE
(i) Conductivity method: If the conductivity of the emulsion is large, the emulsion is of O/W type, as these emulsions are more conducting. But if the conductivity is low, it is of W/O type, as they are less conducting. (ii) Filter paper method: In this method, a drop of emulsion is kept on a piece of filter paper. If the liquid spreads easily and readily, le~ving a spot at the centre, then the emulsion is of O/W type. In case the emulsion does not spread, the emulsion is of W/O type. Preparation of Emulsions: Emulsions can be prepared by shaking or stirring the two phases with the addition of a suitable emulsifier. The type of emulsion formed, depends on the angles of contact of the two liquids with the solid emulsifier.
=::=::=:=:=::=:=:Na+ =:=:=1-_-=:,"",_:=-_-=:,-'_==t--Oil droplet
~~~~!i~;~~~~~~~--=~~~:=::-~~~~~~~~ :::::.::::::: -=:=-------- :=:=:= ------- ------ B-::-- --
:~;:~~~~~~~~ ~~-~~~~~~~:~;~~-----t----:~-----
~~~~:~:~;:=~ -------------~:~~:~:~~~;-~~ --- --- - ----
~~~~~~+:~.:::~ ~::._:N~:~~~~ ---------------------------- ------------------------------------------------------------------------ --------------
Polar head Hydrocarbon tail
----------- -------------- - --------- - -------------
Water
Fig. 11
The liquid to be dispersed is added in small quantities to the dispersion medium, where it is spread into a thin, unstable film which spontaneously breaks up into droplets under the influence of surface tension. Many emulsifying machines are used, such as homogenizers, in reducing the size of globules. A condensation method is used t t epare concentrated O/W emulsions. It consists in allowing vapours of oil to pass through an aqueous solution containing an emulsifying agent. Emulsifiers or Emulsifying Agents : The two types of emulsions i.e., W/O or O/W type, do not remain stable and after sometime, the two layers separate. It means that an emulsion formed by merely shaking two liquids is unstable. But in order to get stable emulsions it is necessary to add a third substance known as emulsifier or emulsifying agent in suitable small
192
PHYSICAL CHEMISTRY-I
quantity. By adding an emulsifier, a stable emulsion of high concentration can be obtained. The emulsifiers are generally long chain compounds with polar groups. such as soaps of various kinds, long chain sulphonic acids and alkyl sulphates. The function of an emulsifier is to lower the interfacial tension between the dispersed phase and the dispersion medium so as to facilitate the mixing of the two liquids. In compounds like soap, the aliphatiC portion is soluble in oil. while the end group (sulphonic acid etc.) called a polar group (because its unsymmetrical grouping contributes a dipole moment to the compound) is soluble in water. The soap molecules get concentrated at the interface between water and oil in such a way that their polar end (-COONa) and hydrocarbon chain (R-) dip in water and oil, respectively as shown in figure 11. This allows the two liquids to come in close contact with each other. Soluble substances like iodine also act as emulsifiers, in the case of etherlwater emulsions. Some mixed stabilising agents also act as emulsifiers, e.g., ethyl alcohol and lycopodium powder, acetic acid and lamp black, clay and sodium oleate etc. Properties of Emulsions: Since an emulsion is a colloidal system of liquid dispersed in liquid, therefore, its properties are common to those of colloidal solutions. (a) Concentration and size of the particle : In an emulsion, the amount of solid substance dissolved is very small as compared to that of the dispersion medium. An emulsion contains droplets of diameter 0.008-0.05 mm. (b) Electrical conductivity: It has been found that emulsions of OIW type are characterised by higher electrical conductivity, while emulsions of WIO type have a lower or no electrical conductivity. (c) Electrophoresis: Like colloidal particles, the droplets of emulsion are also electrically charged. Hence, they migrate towards the oppositely charged electrode under the influence of electric current. (d) Dilution: On increasing the amount of dispersion medium, a separate layer is formed. (e) Brownian motion: Just like colloidal particles, droplets of emulsion are also in a state of constant rapid zig-zag motion. (f) Optical properties: Droplets of emulsion scatter light to different extent, depending upon their size. Reversal of Phase: The change of an emulsion of OIW type to W10 type or vice-versa is known as 'reversal of phase'. An OIW emulsion. e.g., olive oil in water, containing a sodium or potassium soap as emulsifying agent, may be converted into WIO type by the addition of salts of bivalent or trivalent cations.
COLLODIAL STATE
193
Breaking of Emulsions: Emulsions can be broken or demulsified to get the constituent liquids by heating, freezing, centrifuging or by addition of appreciable amounts of electrolytes. They are also broken by destroying the emulsifier. For example, an oil-water emulsion stabilised by soap is broken by the addition of a strong acid. The acid converts soap into insoluble free fatty acids. Uses of Emulsions : Emulsions find numerous applications in daily life, medicine, industry and cosmetics. They may be described as follow& . Daily articles of life: Milk is an emulsion of fat dispersed in water stabilised by casein and as all know is practically a complete food. Ice cream, is an emulsion, in which ice particles are dispersed in cream, stabilised by gelatin. Artificial beverages, coffee, fruit jellies are all emulsions in nature. The cleansing action of ordinary soap is due to a large extent on the production of O/W emulsion. A number of medicines and pharmaceutical preparations are emulsions in nature. It is assumed that in this form, they are more effective. Cod-liver oil, castor oil, petroleum oil are used as medicines which are all emulsions. Asphalt emulsified in water is used for building roads, without the necessity of melting the asphalt. Most of the cosmetics used are emulsions as they permit uniform spreading and promoting the penetration into the skin. Vanishing cream is an O/W type emulsion. Hair creams, cold creams are W/O type emulsions. A variety of emulsions of oils and fats are used in leather industry to make leather soft and pliable and also to make it waterproof. Emulsions are also used in oil and fat industry, paints and varnishes, plastic industry, adhesives, cellulose and paper industry etc.
16. Gels Graham applied the term gel to any coagulum from a sol, whereas according to Ostwald, a gel is a jelly like colloidal system in which a liquid is dispersed in a solid medium. For example, when a warm sol of gelatin is cooled, it sets to a semi-solid mass, which is gel. The process of conversion of a sol to a gel is known as gelation. Gelation may be brought about by either of the methods, (a) cooling the sol, (b) evaporating the sol, (c) addition of electrolytes.
Classification of Gels (i) On the basis of dispersion medium: Gels produced as a result of coagulation are known as coagels. When the dispersion medium in the gel is either water, alcohol or benzene, Qlen the gel formed is known as hydrogel, alcohol or benzogel, respectively. They may be cut, bent or broken as desired. If gels are subjected to prolonged pressure they begin to flow, assuming the shape of the containing vessel.
194
PHYSICAL CHEMISTRY-I
(ii) On the basis of chemical composition: Gels are called inorganic and organic according to their chemical composition. e.g .. gels containing inorganic solvents are known as inorganic gels. Similarly, organic gels are those gels in which organic solvents are used. (iii) On the basis of size of particles: Gels may be colloidal or coarse depending upon whether the size of particles is small or comparatively bigger. Common jellies, agar-agar solution are colloidal gels. (iv) On the basis of their properties: Such gels are classified on the basis of thei~ mechanical properties. Hence, there may be elastic gels, nonelastic gels, rigid gels etc. (q) Heat reversible or elastic gels: Such gels when partially dehydrated'change into a solid mass which, however, change back i~lto the original form on simple addition of water followed by slight warming. if necessary. Gelatin, agar-agar and starch form reversible gels. (b) Heat irreversible or non-elastic gels: Such gels when dehydrated become glassy or change into a powder which on addition of water and followed by warming do not change back into the original gel. Silica, alumina and ferric oxide gels form irreversible gels. Preparation of Gels : Gels may easily be prepared by any of the following methods : (i) By cooling of colloidal solutions: Certain substances form gels when their hot solutions are cooled, e.g., gelatin, agar-agar etc. The setting or gelation of such a solution is characteristic of: (a) temperature of gelation, (b) time of gelation, (c) viscosity of the medium, (d) minimum cOllcentration of the substance at which the gelation may occur. It has been found that substances like sulphates, tartarates, acetates, citrates promote the rate of gelation, whereas certain substances like chlorides, nitrates etc. retard the rate of gelation. The gelation of proteins is retarded by acid and alkalies. When pure alcohol is added to the aqueous solution of calcium acetate, the whole of the salt goes into alcohol, which then sets into a gel containing the liquid. (ii) By double decomposition: The gels of some sols are prepared by the process of double decomposition. On adding water to sodium silicate we get a gel of silicic acid. During the reaction, silicic acid is prepared in the free state which rapidly sets to a gel. (iii) By exchange of solvents : Sometimes gelation may occur due to the exchange of solvent in which the sol is insoluble. (iv) By chemical reactions : Gels can be prepared by this method from concentrated solutions, if one product of the reaction is insoluble and the particles have a tendency to form linear aggregates. On shaking concentrated solutions of barium thiocyanate and manganous sulphate, we get a gel of barium sulphate. Gel of aluminium hydroxide can be prepared by mixing concentrated solutions of aluminium salt and ammonium hydroxide.
COLLODIAL STATE
195
(v) By coagulation or by decrease of solubility: Many gels can be prepared by the coagulation of colloidal solutions. Sols of aluminium hydroxide or ferric hydroxide can be converted into gels when some coagulating agents are added to the colloidal solutions and provided the concentration of the sol is sufficiently high. Properties of Gels (i) Swelling of gels: Hydrophilic gels or elastic gels when placed in water absorb a definite amount of the liquid, whereby the volume increases. The process is known as swelling or imbibition. (ii) Optical properties: Gels show the phenomenon of double refraction. It may be present in the gel from the start or it may be produced by pressure or tension. e.g .• a dry gel of gum shows double refraction like that of the glass under tension and compression. (iii) Electrical properties: It has been seen that the electrical conductivity does not change during the transition from the sol to the gel state. (iv) Syneresis: Syneresis is phenomenon of the excluslOll of the liquid constituent of gels on standing, whereby the gel contracts or shrinks. This may be regarded as reverse of swelling. Gelatin and agar-agar shows syneresis at low concentration, while silicic acid shows it at high concentration. (v) Some of the gels, particularly gelatin and silica liquefy on shaking, changing into the corresponding sol. The solon standing reverts back to the gel. This phenomenon of reversible sol-gel transformation is generally refen'ed to as thixotropy. (vi) Colloidal system usually shows a slow spontaneous aggregation. This is known as ageing. In gels, ageing results in the gradual formation of a denser network of gelling agent. (vii) Gels also show the mechanical properties of rigidity, tensile strength and elasticity that are characteristic of solids. Structure of Gels : There is still a difference of opinion about the structure of gels. Various theories have. however, been proposed to explain the structure of gels. (i) Honey-comb theory: Butschli (1897-1900) concluded as a result of investigations of gels under a microscope that many gels have fine honeycomb structure. (ii) Martin-Fisher solvation theory: According to this theory, a gel may be regarded as a system of two components. one liquid solute and the other solid solvent. X-ray studies prove that the structure of gels is miscellar in nature. (iii) Zsigmondy's theory: Zsigmondy and his co-workers as a result of ultra-microscopical observations with gelatin gel could not confirm the honey-comb structure and suggested that the structure of the gel is much finer than that assumed by Butschli. Therefore. according to Zsigmondy and Bachmann. gels have granular fine structure.
196
PHYSICAL CHEMISTRY-I
(iv) Fibrillar theory : It may also be assumed that in gelatin, the particles arrange themselves into fibres, which are then inter-twined. These threads cause cohesion and great elasticity of gels. (v) von Wiemarn's theory: According to von Wiemarn's concept, gels may be classified according to the degree of dispersion of their primary structural elements. (vi) Thomas-Sibi-theory: Thomas and Sibi (1928-1930) conducted experiments and suggested view close to von Wiemarn's theory. They assumed that solon cooling forms cluster of needles which interpenetrate to give a firm mass. They further showed that units of structure may be intertwining hair instead of straight needles. Uses of Gels: Gels have found several applications. Silica gel is used in laboratory and in industry and is also used to support the platinum catalyst,
when this is used in the contact process of H2S04 manufacture. It is resistant to catalytic poisoning. Solidified alcohol (gel) is used as fuel in picnic stoves and is made from alcohol and calcium acetate.
17. Electrical Double Layer or Zeta Potential In order to explain the origin of charge on a colloidal particle, von Helmholtz postulated the existence of an electrical double layer of opposite charge at the surface of separation between a solid and liquid, i.e., at solid liquid interface. According to modern views, when a solid is in contact with a liquid, a double layer appears at the surface of separation. One part of the double layer is fixed on the surface of the solid. It is known as fixed part of the double layer and it consists of either negative or positive ions. The second part of the double layer consists of a mobile or diffuse layer of ions which extends into the liquid phase. This layer consists of ions of both the charges but the net charge is equal and opposite to that on the fixed part of the double layer. This arrangement is shown in figure (12). In figure (a), the fixed part of the double layer comprises of positive charges, while in figure (b), the fixed part of the double layer consists of negative charges. The presence of charges of opposite signs on the fixed and diffuse parts of the double layer produces a potential between the two layers. This potential is known as electrokinetic potential or zeta potential. It is represented by S(zeta). It is therefore the electromotive force which is developed between
the fixed layer and the dispersion medium. The ions which are preferentially adsorbed by the sol particles, are held in the fixed part of the double layer. It is these ions which give the characteristic charge to the sol particles. It is supposed that the charge on a colloidal particle is due to the preferential adsorption of either positive or negative ions on the particle surface. If the particles have a preference to adsorb negati ve ions they acquire negative charge or vice-versa. The negative charge on As 2S3 sol is due to the preferential adsorption of sulphide ions on the particle surface. The sulphide
1-97
COLLODIAL STATE
ions are given -by the ionisation of hydrogen sulphide, which is present in traces. Similarly, the negative charge on metal sol particles obtained by Bredig's method is due to the preferential adsorption of hydroxyl ions given by the traces of alkali present. The positive charge on sol of Fe(OH)3 prepared by the hydrolysis of ferric chloride is due to the preferential adsorption of ferric ions on the surface of the particles. The ferric ions are obtained by the dissociation of ferric chloride present in traces in the sol.
!·.:ii·················++:··:~:::~::~··;;:::=I~'i :.I..
+:+ · -- + + + ·-+ + + + :+ + + + I
+ + +:::: + + + +
--
::::
I
','
~
• •
I I I I
--.,.....,
Fixed ','
t
I I
,
+
--
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Difuse layer
layer
+
-........-. Fixed
\
+
+ :}
+
y
Difuse Layer
':-:
layer" .
;
: ·:·;:i:;:;:;: ~:~[; :;~: i:i;:~; :~:;i~:;:~: :;: :~:;:;: :~: :;: ;~; : ;~:~:;: :~';:;:~;~; : :' ~~~~: ~~. ·: :;: i~;~: :;·~;: ;i:~; : : : :·;: ·;:;: ·: : ;: : ;: ~; : :i·: : :; ~.:i:.:·: ~l:~: .:·: : ·: From the above, it is clear that the ion which is more nearly related chemically to the colloidal particle is preferentially adsorbed by it. So, in . AS ZS 3 sol, sulphide and not hydrogen ion is preferred and in Fe(OHh sol, ferric and not chloride ion is preferred as shown below : [As S ]Sz-12H+ and [Fe(OHh]Fe 3+ 13CIZ 3
Now consider stannic oxide sol. If a freshly prepared precipitate of stannic oxide is peptised by a small amount of HC!. the colloidal solution carries a positive charge. If the. precipitate is peptised by a small amount of NaOH, the colloidal solution carries a negative charge. In the first type of sol, a small amount of stannic chloride SnCI4 is formed and Sn 4+ ion is preferred over CC ion. So, the sol gets positively charged. In the second type of sol, a small amount of sodium stannate Na2Sn03 IS formed and now the SnO~- ion is preferred over Na+ ions. So, the sol gets negatively charged. The structure of the sol particles in the two cases can be represented as : Positive sol: [Sn02J Sn4+ 14CINegative sol: [SnOzJ SnO~-12Na+ The chloride and sodium ions form the diffuse part of the electrical double layer.
198
PHYSICAL CHEMISTRY-I
Another case is the formation of positively and negatively charged sols of silver IOdide. If a dilute solution of AgN0 3 is added to a slight exces~ of KI solution. a negative sol of AgI is obtained, due to the adsorption of iodide Ions. The structure of the particles is represented as rAgI] C I K-'- If a dilute solutIon of KI is added to a slight excess of AgN0 1 solutIon. a positive sol of AgI is obtamed, due to the adsorption of silver ions. The structure of the particle is shown as [AgIJAg + / NO). If on the contrary, equivalent amounts of AgN0 3 and KI are mixed, there is complete precipitation of silver iodide and no sol is formed. Problem 5 : (a) Discuss the origin of charge 011 the colloidal particle. Explain its significance also. (Meerut 2004, 2001) (b) Classify the following sols according to their charge :(Meerut 2002) (i) gold (ii) ferric hydroxide (iii) gelatin (iv) blood (v) sulphur (vi) AS1 S 3 (c) What will be the charge on the following: (i) AgJ in AgN03 (ii) AgJ ill KJ (iii) As 2S3 ill H 2S (iv) Fe(OHh ill FeCl 3
(1) Origin of Charge All the dispersed particles of a sol carry a positive or negative charge. In order to explain the origin of charge on a colloidal particle. following Vlew& have been gIven. (i) Charge is ionic ill nature: It has been shown that colloids, e.g., sodium palmitate and soap dissociate and produce ions. So, the charge is produced a~ a result of the formation of ions. But this view does not hold good for non-electrolytic colloids such as clay, smoke etc. as they also carry a charge. (ii) Charge is frictional in nature: In earlier days. it was regarded that the source of the charge on colloid particles is essentially physical and it was suggested that the charge resembled the frictional electricity brought about by the contact of such substances as glass and silk in this case, substance with hIgher dielectric constant will render the sol negatIvely charged. (iii) Association of electrolytes ; It has been observed that small quantities of electrolytes are associated with colloidal systems and that if they are removed by persistent dialysis or by other methods, the sols become unstable and the particles grow in size and are finally precipitated. It is probable that in many cases, traces of ions present in the sol are re~ponslble for the charge and the stability of colloidal particles. It is seen that stable sols of gold, silver etc. can be obtained by Bredig's arc method by passing an electric current between electrodes of the metals under water. Water contam~ a little KOH or any other alkali. One ion is retained by the collOIdal particles, giving the same charge to the colloidal particles and the other ion is retained by the dispersion medium.
199
COLLODIAL STATE
(iv) Formation of electrical double layer: See problem 4 (16). Significance of charge: The charge on a colloidal particle is of great importance as shown below. (i) In industrial applications of colloids: A number of processe& like electro-deposition of rubber, removal of smoke, punficatlOn of water. tanning of leather depend upon the presence of charge on a colloidal particle. (ii) In stability of sols: As all colloidal particles have the same charge. they are kept apart due to mutual repulsions and so they do not coalesce with one another to form bigger particles. If the charge is removed, they come closer together and get coagulated.
(2) Classification of Sols Positive sol: Fe(OHh. Negative sols: Gold, gelatin. blood, sulphur. AS 2S 1.
(3) Charge on Sols (i) Positive (ii) Negative (iii) Negative (iv) Positive. Problem 6: Explain thefollowingfacts:
(a)
A sulphur sol is coagulated by adding a little electrolyte, whereas gelatin sol is apparently unaffected.
(b)
What happens when a collOidal solution of gold influence of electric field?
(c)
What happens when an electrolyte is added to colloidal solution of gold?
(d)
What happens when a beam of light is passed through a colloidal solution of gold? A colloidal solution is stabilised by addition of gelatin. Presence of H 2S is essential ill AS2S3 sol though H 2S iOllises alld should precipitate the sol. Why ferric chloride or alum is used for stoppage of bleeding?
(e) (/) (g)
IS
(I
brought Hilder The
(a) A sulphur sol consists of negatively charged particles dispersed in water. When an electrolyte is added to it, the sol being lyophobic in nature I~ easily coagulated. On the other hand; gelatin sol is lyophilic in nature and gelatm particles are heavily hydrated and so water envelopes around them and prevents their coming in contact with electrolyte. Hence, it is not easily coagulated on addition of electrolyte. (b) When colloidal gold solution is brought under electric field, the gold particles move towards anode and lose their charge and get coagulated. This shows that gold particles carry a negative charge (electmpllOresis). (c) When an electrolyte is added to gold sol. the cation of the electrolyte neutralises the negative charge on the collOidal particles and the sol get!'> coagulated.
200
PHYSICAL CHEMISTRY-I
(d) When a beam of light is passed through a colloidal solution, the path of the light becomes visible when viewed through an ultramicroscope, due to scattering of light by colloidal particles (Tyndall effect). (e) A colloidal solution is stabilised by the addition of gelatin. This is due to the fact that gelatin being a lyophilic colloid forms a protective layer around the colloidal particles. This protective layer prevents the precipitating ions from reaching the sol particles and in this way they are prevented from being precipitated. (1) According to modem view, at the surface of separation of solid and liquid. an electrical double layer IS formed. One of the layer is called fixed layer which is towards the particle and the other layer called the mobile layer points towards the dispersion medium. The fixed layer contains the Ions of one charge, either positive or negative and those ions are preferentially adsorbed which are chemically alike. So. to stabilise AS ZS 3 sol, S2- ions given by H 2S from the fixed layer give negative charge to AS 2S3 particle, i.e., [AS ZS3]S2-12H+. Thus, because of the same charge, AS ZS3 particles do not come closer to get coagulated. So, unlike other electrolytes, HzS stabilises AS 2S3 sol. (g) Blood is a negatively charged sol in which albuminoid substance is dispersed in water. When FeCl 3 or alum is added, the trivalent ions, Fe3+ or A1 3+ coagulate the blood. Thus, bleeding is stopped.
Problem 7. Describe the applications of colloids in chemistry. The applications of colloids can broadly be divided into three types: (i) Natural applications, (ii) Technical applications, (iii) Analytical applications. [A] Natural Applications: (i) Rain: Cloud is a colloidal system in which water particles are distributed in air. When air which has become saturated with water vapours, reaches the cooler parts of the atmosphere. cloud is formed as a result of condensation. Further cooling and condensation form bigger drops of water which fall due to gravity in the form of rain. (ii) Blood: Blood is a collOIdal system, having albuminoid substance as the dispersed phase carrying a negative charge. The stoppage of bleeding by the application of alum or ferric chloride can be explained on the basis of coagulation, as in this case A1 3+ acts as active ion for the coagulation of negatively charged albuminoid particles. (iii) Bille colour of the sky: This is an application of Tyndall effect. The upper atmosphere contains colloidal dust or ice particles dispersed in air. As the sun rays enter the atmosphere [Fig. (13)], these strike the colloidal particles. They absorb sunlight and scatter light of blue colour (4600-51 00 A). The light that is incident at earth's surface is considerably reddened due to the removal of most of the blue light in the upper atmmiphere (figure 13).
201
COLLODIAL STATE
IY"'e.~u~::~:+!
.'. " . .'.
i;;;;;;~~. ;~ . '··;1 (iv) Articles ofdaily use: Milk which is a complete food is an emulsion of oil in water (OfW) type stabilised by casein. Butter is an emulsion of water dispersed in fat (O/W) type. Fruit juice is a colloidal system having juice dispersed in the solid tissue of the fruit. Ice cream is ice particles dispersed in cream. (v) Formation of deltas: The river water contains colloidal particles of sand and clay which carry negative charge. The sea water on the other
IY;':::::::::;;:~~:::::::+
• ~=~<~~ ~ .'.
~
~
~
e J: ~
R~~
j
4
: : :· i:~:~ ~ :~·: :;~;~; :;: ~:; :i~: ~ ~;~;:~:~;i~.~;~;·.·.~ ~.:·.·.~.~.~:;~i:;·:.;:i.:i~:i:~:; ;·i.~. ;~;t;i;·. ;:·:l:·~.: ~ ! hand, contains positive ions such as Na+, Mg2+, Ca:!+. As the river water meets sea water, these ions coagulate the sand or clay particles which are precipitate
\
202
PHYSICAL CHEMISTRY-I
[B] Technical Applications: These are numerous technical applications of colloids. but we shall explain only the important ones. (i) Rubber industry: Rubber is obtained from the sap of certain trees. The sap--a1so known as latex-is a colloidal system of colloidal particles dispersed in water. stabilised by protein contents. On coagulating the latex by various methods. collOIdal gel. known as rubber is obtamed. The rubber particles carry a negative charge. When rubber is to be deposited on any material. then two electrodes are made. The anode is made of that substance on which rubber has to be deposited. On passing an electric current. rubber thus deposits on the substance. (ii) Tanning ofleather : Both hide and leather are colloidal substances and possess gel structures. When hide is soaked in any tanning agent which consists of tannin (a sol containing negatively charged particles). mutual coagulation of the two takes place due to oppostie charges. This process is known as talllling and the hide becomes more and more leather like. imparting hardness to it. (iii) Soils and clays : The value of a soil. i.e .• its ability to retain moisture is dependent on the colloidal propertIes of humus-an important constituent of the soil. When soil is completely saturated with H+ ions, it is known as 'clayic acid' and the salts derived from this acid are known as 'clayates'. There are colloids. e.g., Fe(OHh, AI(OHh. amorphous silicates and bacteria which are present in soils. (iv) Purification of sewage and water: Sewage consists of impure particles dispersed in water and is thus colloidal in nature. carrying a negative charge. The dirt particles carrying a negative charge are coagulated on the oppositely charged positive electrode and are thus removed. The deposit is then used as manure. The method for sewage purification is known as 'actiWifed sludge process '. (v) Photography: Photographic plates are thin glass plates coated with a fine suspension of silver bromide and gelatin. Silver bromide dispersed in gelatin gel forms photographic emulsion. The AgBr sol may be produced by mixing gelatin solution containing dilute solution of AgN0 3 with dilute alkali bromide. whereby insoluble AgBr is formed in a fine suspension. stabilised by gelatin which is then painted on thin glass plates. (vi) Metallurgy of alloys: The technical property of an alloy depends upon the size of the subdivided particles in it. Thus. alloys which are brittle and non-elastic have a coarse structure. As the degree of dispersion is increased. the hardness of the alloy is also increased. (vii) Colours of various materials: The colour of various materials is due to colloid metal and colloid metallic compound. The red colour of ruby glass is due to colloidal gold. Yellow glass owes its colour to colloidal silver. Colloidal chromium imparts colour to artificial rubies.
!
203
COLLODIAL STATE
(viii) Dyeing: Cotton, silk and wool are all colloidal in nature having a gel structure. The dyestuffs include a number of substances having different propertIes; their degree of dispersion is colloidal.
preCip~~i:n~O:;ok:~~
a:;s7e: " .: " :; : -:." ' ':' ' .:.;' ' ': : ~.::-;.:"':':.~::""':"::-:"'~'.,"'-:''''''''~-'''''D:-'u-'s:-t~fr"":e"":~'"'·::··"":-;:=.i.:
in which carbon particles are dis- :::: persed in air carrying a negative ): charge. In large industrial cities '.
:yO~~:~h~:sa~~:;a~i~~:sc;::~
ir
i:l!
ent in it are very injurious to :::: human health. So, to remove car- ::: bon particles, use is made of :-: 'Cottrell electrical precipitator' ..:.::' [Fig. (15)] in which the smoke is '.' made to pass through a positively.;.; charged anode in the chimney, when carbon particles settle down. According to a modern :;::
f
~~~~~~ t~:te~m~:h i: l:det:~li~
gases
:!::
Point electrode Plate electrode Gases carrying dust or smoke
I
•
]~--
knob charged to a very high pos- :;:. p.r.~~#~t~Q itive potential (50,000 volts). The A~f~r::~~r . gases free from smoke, pass from the top of the chimney. (x) Medicines: Most of H..~:-~~~~~ the medicines which are used, are l·~··::;Lii:~=~=::::::;:::::::c;::"::~~2~i3 colloidal and are effective due to their easy assimilation and ab~orption by the human system. Colloidal antimollY is effective in curing kalazar. Colloidal sulphur is effective in curing skin diseases and in kilhng germs. Colloidal calciltm and gold are used as intramuscular injections to increase the vitality of human system in serious diseases, Colloidal silver or argyrol is used for curing gmnulatio~. Milk of li!agllesia, an emulsion. is used for stomach troubles, [C] Analytical Applications: (i) Qualitative and quantitative analysis: A knowledge of the behaviour of colloids and their mode of formation play an important role in analytical problems. In volumetric analysis, hydrophilic colloids alter the end point, e.g .. in a titration of HCI and NaOH, the amount of deviation in the end point is increased with increasing amounts of coJloids. In the volumetric estimation of silver by Mohr's method, the phenomenon of adsorption comes into being.
204
PHYSICAL CHEMISTRY-I
In gravimetric analysis, definite crystalline precipitates are desired and the procedure thus adopted aims to get sufficiently bigger particles, because extremely small particles may pass through the filter paper. To prevent adsorption of the undesirable molecules with the desired precipitate, it is necessary to precipitate the solution at the boiling point and fairly dilute solution should be used. (ii) Detection of natural from artificial honey: Ley's test for detecting the natural honey from artiticial honey consists in treating few drops of the honey with an ammoniacal silver salt solution. If the honey is natural or pure, then the metallic silver produced assumes a reddIsh yellow colour due to the traces of albumin or ethereal oils, which act as protective colloids and maintain the colloidal silver in a high degree of dispersion. A dark yellow or greenish precipitate is formed with artificial honey. (iii) Identification of traces of noble metals : Noble metals when present in the colloidal form produce very bright and intense colours. Cassius purple test indic:ates the presence of colloidal gold.
Problem 8: Give an elementary idea about sol-gel transformation. Mention the physical changes occurring therein. What is the effect of dissolved substance on sol-gel transformation? Sol-gel transformation is a phenomenon in which a gel is formed from a sol. Fernau and Pauli showed that a freshly prepared sol of CeOz containing about 10 g. per litre was transformed into a gel by coagulation with electrolyte. But if the sol of CeOz was kept for 200-300 days, then it lost its power and gave a precipitate instead of a gel, by the addition of electrolytes. But in the case of some lyophilic sols, such as gelatin in water, agar in water, the sol-gel transformation is well known and the transformation is reversible within certain limits. For example, if a gelatin sol (of not too low concentration) is prepared by heating gelatin with water upto 70°C, and if it is then cooled, then it is seen that the sol sets to a gel at low temperature. If gel is again warmed, it liquefies to a sol and so on. This process can be repeated as and when desired. It appears that an irreversible chemical reaction is continuously taking place in gelatin sol, which can be easily recognised. If a gelatin sol is kept at a high temperature at which it does not gelatinise, the viscosity decreases continuously and reaches a considerable lower end value than its original value. The sol thus formed loses the property of gelatinizing and the gelatin thus transformed is known as f3-glutin or galactose. In other words, the sol ~ gel transformation is practically actually reversible, as long as the temperature 65-70° is not exceeded. This sol-gel transformation proceeds perfectly continuously, because there are no singular points or rapid changes in direction in the temperaturetime curve.
205
COLLODIAL STATE
[I] Physical Changes Occuring in Sol-Gel Transformation (i) Thermal: The cooling curves of solutions of low molecular weight substances show usually pronounced breaks or arrest points when solute begins to separate. With sols the onset on gelatin produces a small break, showing that the sol-gel transformation is accompanied by the liberation of heat. Similar behaviour is seen with soap sol when it is changed into gel. (ii) Optical: Arsiz (1915) showed that there is a complete continuous change in the Tyndall cone ofIight during the sol-gel transformation. A gelatin sol in water or water containing glycerine, shows a weak Tyndall cone at 70°C. But as the sol is cooled, the Tyndall cone becomes stronger, and becomes weaker on warming again. It alters in its intensity during the sol-gel transformation taking place at constant temperature. In other words, the Tyndall light increases and decreases with the viscosity of the solution. (iii) Electrical conductivity: No change is observed in the electrical conductivity during the sol-gel transformation. If the sols contain electrolytes, a slight increase in conductivity is observed, which may be ascribed to secondary effects during the transformation. (iv) Viscosity: Viscosity is the property which changes remarkably during the sol-gel transformation. The relationships are clear and well defined in the case of glycerosols of gelatin as studied and examined by Arsiz. (v) Diffusion: It has been found that the rate of diffusion of substances having low molecular weights is similar to the rate of diffusion of pure solvent. (vi) Volume changes: During the sol-gel transformation a change in volume occurs. It was found that in some cases expansion of volume takes place, as in the case of methyl cellulose-water system, while in some cases contraction of volume takes pa1ce, e.g., gelatin-water system. In the gelation of 9% sol of ferric hydroxide, no change in volume was observed. The change in volume can be ascribed to either the process of crystallisation or dissolution. (vii) Other properties: As regards a number of other properties, sols and gels appear to be exactly alike, e.g., in refractive index and vapour pressure etc.
[II] Effect of Dissolved Substances on Sol formation
~
Gel Trans-
Dissolved substances have a marked effect on the sol-gel transformation. Paschels (1890-1900) examined the changes in the solidifying point of gelatin sols in presence of some substances. The effect of anions appears to be in the order of S04> tartarate > acetate> halogens> thiocyanates. The first three groups favour gelatinization and thereby raise the solidifying point and thus shortens the time of gelatinization. The rest of the groups lower the l>olidifying point and thereby lengthen the time of gelatinization. Non-electrolytes also influence the time of gelatinization. Lewites found that substances such as sugar, polyhydric alcohols etc. shorten the time
206
PHYSICAL CHEMISTRY-I
of gelatinization and substances like urea. urethane, thiourea etc .. lengthen the time of gelatinization.
Problem 9: mite a note on thixotropy. Or What is thixotropy ? (Meerut 2006) When suitable quantities of electrolytes are added to a concentrated sol like Fe(OHh. A1 20 3, V205' Zr02, Sn02 etc. a pasty gel is formed which has a remarkable property of being liquefied when shaken. only to set again to gel form. on standing. So. in many cases. gels can be liquefied either by shaking or any other mechanical action. but returns to the gel state more or less rapidly as soon as the disturbing action is stopped. Peterfi (1927) suggested the name of 'thixotropy' (change by touch) for this type of phenomenon. It is also known as 'isothermal reversible sol-gel transformation'. Systematic observations on thixotropy were made by Szegvari and Shalek (1923), when they studied sols of Fe:P3 and many dilute sols of high polymers. Goodeve and Whitfield (1837-39) studied the equilibrium between the spontaneous building of an internal structure and its breakdown. The apparent viscosity of a thixotropic system was measured at different rates of shear. If 11n 11r represent the apparent viscosity and extrapolated residual viscosity, respectively, then we have, 8 11n = 11r + S where 8 is the coefficient of thixotropy and S is the shear. If 11n and lIS are plotted as ordinate and abscissa. we get curves of different types. The intercept on the Y-axis will give the value of 11m i.e., resistance of flow at extremely high shear ratio. It has been observed that thixotropy of carbon black dispersion is reduced on adding 3% linoleic acid. These curves also show the effect of thixotropy in the absence and presence of linoleic acid, respectively.
[I] Theories of Thixotropy Several theories have been proposed to account for thixotropy which are given below: (i) Solvation or hydration theory: According to this theory, thick envelopes called lyospheres of oriented water molecules are formed round the colloidal particles and these eventually become so large that freedom of movement is lost and the system becomes a gel. It is also regarded that lyospheres are destroyed on shaking, but partially, so that the gel is then liquefied. It is seen that finely divided inorganic substances suspended in inorganic solvents, also exhibit thixotropy. This fact goes against this theory. (ii) Oriented coagulation theory : According to this theory, the particles of sol capable of forming a thixotropic system are believed to be
207
COLLODIAL STATE
anisotropic and anisometric or both, so that the electrical charge and water of hydration, if any, are unequally distributed. When the correct amount of electrolyte is added to reduce the zeta potential to a sufficient extent, the particles tend to coalesce, but owing to their shape they can form a stable gel only if correctly oriented. The extent of the surfaces of the particle in contact is probably limited, and as a consequence. any stress will tend to reduce this and destroy the gel structure. It is interesting to note in connection with this theory, that nearly all thixotropic sols show marked streaming double refraction, so that the particles are probably anisotropic or anisometric. (iii) Three dimensional theory: This theory is based on the hypothesis of three dimensional gel structure as already described before. It is supposed that when a gel is thixotropic, shaking is only sufficient to break the cross linking, thus re-forming the sol. The particles would then be linear, thus accounting for the double refraction of now suggested in the second theory.
[II] Influence of Foreign Substances on Thixotropy Thixotropic gelation may be considered as a form of nocculation and is sensitive to all kinds of additions. The time of solidification (1) is strongly dependent on the concentration (c) of the electrolyte, as seen in the empirical equation given by Schalek and Szegvari (1923), log T=A - B.c It has been shown that thixotropic gelation is a slow coagulation. Time of soliditication is also changed with a change in the pH of the solution as mentioned by Freundlich. It is seen that alcohol promotes the thixotropic gelation and after evaporating the alcohol. the sol returns to the gel state.
[III] Influence of Temperature on Thixotropy As seen from Schalek and Szegvari's experiments, the increase of temperature will decrease the time of solidification.
[IV] Applications of Thixotropy The phenomenon of thixotropy has found several applications in biological fields, technology and in articles of daily use. Protoplasm has thixotropic properties. Myosin sols are known to form strongly thixotropic gels, which might indicate that thixotropy has a significance for muscular action. Most of the quicksands are thixotropic in nature. The appl ication of thixotropy i!> much used in drilling muds used in drilling oil wells. Drilling muds always contain a certain amount of plastic clays responsible for its thixotropy. Paints, varnishes and printing inks are all thixotropic in nature. A paint having a greater time of solidification is considered to be its good property. A suspension of graphite in a mineral oil shows a thixotropic gelation.
208
PHYSICAL CHEMISTRY-I
MULTIPLE CHOICE QUESTIONS 1.
2. 3.
4.
5.
6. 7.
8.
9.
10.
11.
12.
13.
The following will have the maximum coagulating power: (i) Na+ (ii) Sn4+ (iii) Bi+ Milk is . (i) Gel (ii) Sol (iv) Emulsion (iii) Aerosol The size of colloid particles is : (i) > 0.1 Il (ii) 1 mll- 0.1 Il (iii) > I mil (iv) > 5000 mil A freshly prepared precipitate ofSn02 peptised by HCI will carry the follOWIng charge: (ii) Positive (i) Neutral (iv) Amphoteric (iii) Negative Every colloid system is : (ii) Heterogeneous (i) Homogeneous (iii) Contains one phase (iv) Homogeneous and heterogeneous. The following is a hydrophobic COllOId: (i) Gelatin (ii) Gum (iii) Starch (iv) Sulphur Gold sol prepared by different methods will have different colours. It is due to : (ii) Gold shows variable valency (i) Difference in size of colloid particles (iii) Different concentrations of gold (iv) Presence of impurities. Following is an emulsifier: (ii) Water (i) Oil (iv) Soap (iii) NaCI The sky appears blue. It is due to : (i) Reflection (Ii) AbsorptiQn (iii) Scattering (iv) Refraction Which of the following reaction will give colloidal solution: (i) Cu + HgCl 2 --t Hg + CuCl 2 (ii) Cu + CuCI 2 --t CU2CI2 (iii) 2Mg + CO 2 --t 2MgO + C (iv) 2HN0 3 + 3H2S --t 3S + 4H20 + 2NO Fat is: (i) Emulsion (ii) Gel (iii) Colloidal solution (iv) Solid sol Colloidal solutions cannot be purified by : (i) Dialysis (ii) Electrodialysis (iii) Ultrafiltration (iv) Electrophoresis The charge on AS2S 3 sol is due to the adsorption of: (i) H+ (ii) OW (iii)Sz(iv) O2
14. Surface tension of lyophilic sols is : Lower than water (ii) More than water (iii) Equal to water (iv) Sometimes lower and sometimes more than water On addition of 1.0 ml of 10% NaCI to 10 rnl gold sol in the presence of 0.0250 gm of starch, the coagulation is just stopped. The gold number of starch is :
(i)
15.
COLLODIAL STATE
16.
17.
18.
19.
20.
209
(i) 0.025 (ii) 2.5 (iii) 2.5 (iv) 25.0 Which of the following is a lyophilic colloid? (i) Milk (ii) Fog (iii) Blood (iv) Gelatm At critical micelle concentration, the surfactant moiecules undergo: (i) Association (iJ) Aggregation (iii) Micelle formation (iv) All the above Which of the following electrolytes is least effective in causmg coagulatIOn of Fe(OHh sol? (i) K3Fe(CN)6 (ii) K2S04 (iii) KCl (iv) K2Cr04 The ability of an ion to bring about coagulation of a given colloid depends on : (i) Its charge (il) Sign of the charge only (ii) Magnitude of charge (iv) Both charge and magnitude Which of the following statement is false? (i) Elastic gels undergo imbibition (ii) Non-elastic gels are reversible in nature (iii) Inorganic gels usually exude solvent on standing (iv) Both elastic and non-elastic gels show thixotropy.
Fill in the Blanks 1. 2. 3. 4. 5. 6. 7.
True solutions are ............... systems. In a colloidal system .......... phases are present. In aerosol ............ is the dispersed phase and ............. is the dispersIOn medium. Hair cream is an example of .............. system. The process of converting a fresh precipitate into a colloidal solution is known as .......... .. Tyndall effect is due to ............. of light. When placed in water ............... gels ............. This phenomenon is called ........ .
8. 9. 10.
Breaking of emulsions is also known as .......... . The' emulsifier in milk is ................. . Formation of deltas ~s an example of .................. .
True or False State whether the following statements are true (T) or false (F) ? 1.
2. 3. 4. 5. 6. 7. 8.
The lower the gold number, the better is the protective power of a protective colloid. Lyophobic sols are irreversible in nature. The process of passing a precipitate into colloidal solution on adding an electrolyte is called electrophoresis (Meerut 2(02) Colloidal system is biphasic in nature. Colloidal solutions exhibit Brownian motion. The soap may be considered as a surfactant. Blood carries a positive charge. Lyophilic colloids can act as protective colloids.
210 9. 10. 11. 12. 13. 14. 15. 16.
PHYSICAL CHEMISTRY-I
The role of an emulsifier is to increase the surface tension between the two phases. Elastic gels are reversible in nature. (Meerut 2(03) Gum arabic is a hydrophobic colloid. The colour of a colloidal solution depends on the shape and size of the particles. Avogadro's number can be determined with the help of Brownian motion. Out of NaCI, BaCI 2, AICI 3 solutions, the coagUlating power of NaCI is maximum for coagulating a As 2S3 sol. The difference in potential between the fixed and diffuse layers of an electrical double layer is called streaming potential. Boot polish is not a colloidal system. (Meerut 2(01)
ANSWERS Multiple Choice Questions 1. (b), 10. (d) 19. (d).
2. (d), 3. (b), 4. (b), 5. (b), 11. (a) 12. (d) 13. (c), 14. (a), 20. (b),
6. (d), 15. (d),
7. (a), 8. (d), 9. (c), 16. (d). 17. (d). 18. (c),
Fill in the Blanks 1. Homogeneous 4. emulsion 7. elastic, swell, imbibition 10. coagulation
2. two 5. peptization 8. demulsification
3. liquid, solid 6. Scattering 9. casein
True or False 1.
(T)
2.
(F)
3.
(F)
4.
7.
(F)
8.
(T)
9.
(F)
10. (T)
13.
(T)
14- (F)
15.
(F)
16. (F)
(T)
•
5.
(T)
6.
11.
(F)
12. (T)
(T)
DOD
CHEMICAL KINETICS &CATALYSIS 1. CHEMICAL KINETICS Problem 1: (a) Define and discuss the following terms: (i) Rate of chemical reaction (ii) Velocity coeffICient (iii) Molecularity of a reaction (iv) Order of reaction (Meerut 2007, 2006, 2005) Or, Write a note on order and molecularity of reactions. (b) What is the difference between molecularity and order of reaction? (Meerut 2002 2004. 2000)
(C) Explain why reactions of higher orders are rare?
(d) What are the factors which affect reaction rates?
(Meerut 2003)
(a) [I] Rate of Chemical Reaction It is defined as, 'the rate at which the concentration of a reactant changes with time', i.e., . _ Change in concentration R eactton rate T" I tme mterva The rate at which a reaction proceeds can be followed by measuring the concentration of either the reactant or product. If dx represents the amount of the reactant changed during a small interval of time dt, then the reaction rate is represented by dxl dt. If, on the other hand, de represents the concentration of the reactant left behind after a short interval of time dt, then the reaction rate is also represented by - del dt. The negative sign implies that the concentration of the reactant decreases with time or rate of reaction decreases with time. The unit of reaction rate is mol L -1 S-1. For a reaction, A + B ~ C + D, the reaction rate can be expressed as _ d [A] or _ d [B] or + d [C] or + _ d [D] ~
~
~
~
C,onsider the reaction, 2N20 S (g) ~ 4N02 (g) + O2 (g). The rates of reaction can be expressed as, d [0 2 ] d [N 20 S] d [N0 2] or +~ dt or+ dt All the rates can be equated as, (211)
212
PHYSICAL CHEMISTRY-I
1 d [N20 5 ] 1 d [N02 ] d [0 2] dt =+4" dt =+~
-"2
[II] Velocity Coefficient Consider the following reaction : A + B ~ Products If a and b are the initial concentrations of the reactants A and B and if x be the number of moles of each reactant undergoing reaction after time t, then the concentrations A and B after time t will be (a - x) and (b - x) respectively. According to the law of mass action, the reaction rate (dxldt) is given by dx oc (a - x)(b - x) or dx = k (a - x) (b - x) dt dt where, k is a constant known as velocity constant, rate constant, velocity coefficient or specific reaction rate. It is characteristic of the reaction. If (a -x) = 1 and (b -x) = 1, then k=:
= Rate of reaction
So, velocity coefficient is defined as, the rate of reaction under conditions when the molecular concentration of each reactant is unity.
[III] Molecularity of Reaction It is defined as, the total number of molecules of all the reactants taking part in a chemical reaction as represented by a simple equation. Examples. (i) Inversion of cane sugar (Molecularity = 2) C 12H 220 11 + H 20
H+ ions --7
C6H 120 6 + C 6H I2 0 6
(ii) Dissociation of ammonia (Molecularity
= 2)
2NH3 ~ N2 + 3H2 (iii) In general, for a reaction, nlA + 1I2B ~ 113C + 11l2D the molecularity will be nl + 112' So, reactions having molecularity 1,2, 3 etc., are known as unimolecular, bimolecular, trimolecular reactions, respectively.
[IV] Order of Reaction Reactions are generally classified on the basis of their order of reaction. It is defined as, 'The total number ofreacting molecules whose concentration changes during the chemical reaction. ' In other words, it is also defined as, 'The total number of reacting molecules whose concentration determines the rate of reaction '. (a) If A ~ Products and (dxldt) = k [A], then Order of reaction = 1
213
CHEMICAL KINETICS & CATALYSIS
(b) If A + B
~
Products and (dx/dt) = k [A][B], then Order of reaction = 2 (c) If nlA + n2B + n3C + ... ~ Products, and :
= k [A)"' [Btl [C]"3 ... , then
Order of reaction = 11\ + 112 + 1I3 + ... dx (a) If A + 2B ~ Products, and - = k then dt ' Order of reaction = 0 In general, order of reaction is also defined as, the sum of the powers to which the concentration (or pressure) terms of the reactants are raised in order to determine the rate of a reacton.
(b) Relation between Molecularity and Order of Reaction Till recently, the molecularity and order of reaction were regarded as synonymous. But, it has now been found that in certain cases the two are not identical. For example, during the inversion of cane sugar into glucose and fructose or the hydrolysis of ethyl acetate in presence of acid catalyst, apparently two molecules take part, making the reactions bimolecular. But actually, the concentration of only one of reactants changes, i.e., concentration of water does not appreciably change. Hence, the reactions though bimolecular are of the first order.
H+ ions
CH3COOC 2H 5 + H 20 - - 7 CH300H + C 2H 50H Thus, often molecularity of a reaction coincides with its order, but the two need not be always identical. The molecularity must be an integral value, but the order of reaction may be zero, whole number or even fractional. Whereas molecularity can be given on the basis of some proposed theoretical mechanism so as to satisfy the experimental findings, the order of reaction can be obtained from experimental results. For complex reactions, occurring in steps, the molecularity of each step will be different, while the order of reaction will be determined by the slowest step. Table-I. Difference between order of reaction and molecularity Molecularity
Order of reaction
l. It is equal to the number of molecules of J. It is equal to the sum of the powers of the
reactants which take part in a single step molar concentrations of the reactants mJhe chemical reaction. rate expression. 2. It is a theoretical concept which depends 2. It is an expermientally determined quantity on the rate determining step in the reaction which is obtained from the rate for the mechanism. overall reaction.
214
PHYSICAL CHEMISTRY-I
3. It is always a whole number.
3. It may be whole number. zero or fractIOnal value. 4. It is obtained from a single balanced 4. It cannot be obtained from a balanced chemical equation. chemical equation. 5. It does not reveal anything about 5. It reveals some basic facts about reaction mechanism. reaction mechanism.
(c) Most of the reactions are of the first and second orders. According to kinetic theory of gases, the reaction occurs due to collisions of molecules. For a second order reaction, two molecules have to collide which is reasonably possible. But the chances of three, four or even more molecules colliding simultaneously are very remote and so reactions of higher orders are very rare. There are a number of complex reactions involving a number of molecules. But in such cases, the reactions occur in several stages and slowest stage is the rate determining step and it gives the order of reaction. The following reaction, though involving ten molecules is of the second order. KCI0 3 + 3H2S04 + 6FeS04 ~ 3Fe2(S04h + KCI + 3H20
(d)
Factors which Affect the Rate of Reactions
The following factors influence the rate of reactions. (i) Effect ofconcentration: We know that the rate of a reaction falls with time, so it is clear that the rate of reaction is directly proportional to the concentration of reactants. (ii) Effect of temperature: The rate of reaction increases with rise in temperature. In most cases, a rise of lOoC in temperature doubles and in some cases even trebles the rate of a reaction. The ratio of the velocity constants of a reaction at two temperatures differing by lOoC is called temperature coefficient of the reaction. . . Rate constant at 35°C _ k35 Temperature coeffiCient = R 250C ate constant at k25 (iii) Effect of nature of reactants: Rates of reactions are effected considerably by the nature of reactants. For example, consider the two well known oxidation reactions taking place in aqueous solution. One is the oxidation of ferrous ion, Fe 2+ and the other of oxalate ion, C:p/-, by permanganate ion in acid medium. (a) 5Fe2+ (aq) + Mn04- (aq) + 8H+ (aq) ~ 5Fe3+ (aq) + Mn2+ (aq) + 4H 20 (b) 5C20/- (aq) + 2Mn04- (aq) + 16H+ (aq) ~ IOCOz (g)
+ 2Mn-'+ (aq) + 8H20 The first reaction is much faster than the second. As Mn04" ion is common in both the reactions, the difference clearly lies in the nature of ferrous and oxalate ions. Fe2+ ion is a simple ion, whereas C 20/- ion is a poly atomic ion and contains a number of covalent L:mds which have to be broken in the oxidation reaction.
215
CHEMICAL KINETICS & CATALYSIS
(iv) Effect of catalyst: A catalyst generally increases the rate of a reaction at a given temperature. A catalyst is generally specific in its action, i.e., it may affect the rate of one particular process only. In very few cases, a catalyst may decrease the rate of a reaction. (v) Effect of surface area of reactants : In heterogeneous reactions, particle size decreases and so the surfae area for the same mass increases. The smaller particles, therefore, react more rapidly than the larger particles. (vi) Effect of radiation: We know that the energy associated with each photon of radiation is given by hv, where h is Planck's constant and V is the frequency of the radiation concerned. The rate of certain reactions may be speeded up by the absorption of photons of certain radiations. Such reactions are called photochemical reactions.
Problem 2: What do you understand by a zero order reaction? Derive the rate expression for it. What are the characteristics of zero order reactions? Given an example of such a reaction. Or, Write a short note on zero order reaction. (Meerut 2003)
[I] Zero Order Reaction A reaction in which the concentration ofthe reactants does not change with time and the reaction rate remains constant throughout is said to be a zero order reaction.
[II]
Rate Expression for Zero Order Reaction Consider the following zero order reaction, A -----7 Products dx :. Reaction rate, dt = k (k = velocity constant) or
dx =k dt
On integration, we get or x == kt + 1 where, 1 == integration constant When t=O, thenx=O.
... (1)
1=0
:. From equation (1), x == kt
or
k=~ t
... (2)
Equation (2) is known as zero order rate expression.
[III]
Characteristic
(i) The dimension of velocity constant is concentration x time-lor mole liC! time-! or mole liC! s-t, if time is expressed in second.
[IV]
Example
(i) Several photochemical reactions and heterogeneous reactions (enzymic) are of zero order, e.g., the photochemical combination of Hz and Cl 2 in presence of sunlight to give HCl is of zero order.
216
PHYSICAL CHEMISTRY-I
Sunlight
H2 (g) + CI 2 (g) ----t 2HCI (g)
Problem 3: What is a half order reaction? Derive rate equation for it and discuss its characteristic.
[I] Half Order Reaction A reaction in which the reaction rate depends on half power of concentration of a reactant is known as half order reactioll, i.e .• A ~ Products . cbc ReactIOn rate. dt
..
= k [A] 112
[II] Rate Equation for Half Order Reaction Consider the reaction. ~Products
A a
(Initial
COliC.)
(COliC.
at time t)
(a-x)
dx_ k (a _ x )112 dt dx
---l/-~ =
or
(a - x)
~
k dt
On integrating,
dx 112 =fkdt f (a -x) or
1 112 --(a-x) ===kt+I
2
... (1)
where I = integration constant. When t = O. x
= O. so
-! a 112 = I
2 Putting the value of J in equation (1). we get.
- 2"1 (a or
x)
112
= kt -
k= -
;t
2"1 a 112 [(a - x) 112 - a 112]
... (2)
Equation (2) is the required rate equation for a half order reaction.
217
CHEMICAL KINETICS & CATALYSIS
[III] Characteristic of Half Order Reaction (1)
The
unit
of rate
constant
will
be
_._1_ x conc l12 time
or
(mol L-I)1I2 (sri or mol 112 L -112 S-I.
Problem 4: What is a first order reaction? Derive the rate expression for it and discuss its characteristics. Give some examples of first order reaction. (Meerut 2(02)
[I] First Order Reaction A reaction in which the reaction rate depends only on one concentration term is said to be a first order reaction. For example, if A ~ Products then,
[II)
reaction rate = ~ =k [A]
Rate Expression for First Order Reaction
Consider the following simplest first order reaction. A ~ Products Let a mole/litre be the initial concentration of the reactant A. Suppose x mole/litre of A decomposes in time t, leaving behind (a - x) mole/litre of it. For a first order reaction, the reaction rate at any time t is given by :
:
:::: k {A] = k (a - x)
where, k is the velocity constant or rate constant.
dx --=kdt ... (1) a-x In order to get the value of k, we integrate equation (1). So, ... (2) - loge (a - x) == kt + I where, I is the integration constant. When t =0, x :::: 0, therefore, from equation (2), I =-loge a Substituting the value of I in equation (2), - 10& (a - x) = kt - loge a 1 a k::::-Io&-or ... (3) t a-x 2.303 I a ... (4) k ::::-- oglO-I a-x (As loge x == 2.303 10gIO x) Equations (3) and (4) are known asfust order rate expressions. The constant k is called the first order rate constant. or
218
[III]
PHYSICAL CHEMISTRY-I
Characteristics of First Order Reaction
(i) The value of velocity constant is independent ofthe units in which concentration ofthe reactant is expressed. If unit of concentration is changed to say n times its original value, then equation (4) becomes 2.303 I n.a 2.303 I =- oglO 11 (ana- x) k =- - oglO t n.a - n.x t
= 2.303 10glO _a_ t a-x
This is the same as equation (4). (il) The dimension offirst order rate constant is reciprocal of time i.e., t-1• From equation (4), we have k = _1_ . concentration = _1_ time concentration time
If time is expressed in second or minute, k is expressed in second-lor minute -1, respectively. (iii) The time taken to complete a certain fraction of a reaction is independent of the initial concentration of the reactant. Suppose tl/2 is the time taken to reduce the concentration of reactant A to half, i.e., from a to al2, then x:::: al2. From equation (4), we get
2.303
k :::: - - 10glO
or
a
2.303
(/2) :::: - - 10glO 2
tl/2 a- a tl/2 _ 2.303 I 2 - 2.303 03010 _ 0.693 tl/2 k oglO - k x. - k
... (5)
Equation (5) is thus independent of a, i.e., the initial concentration of the reactant.
[IV]
Examples of First Order Reaction (i)
Decomposition of hydrogen peroxide in aqueous solution. I
H 20 Z - - ? H 20 + "2 O2
(ii) Hydrolysis of methyl acetate in presence of acid. CH3COOCH3 + H 20
Acid --?
CH3COOH + CH30H
(iii) Hydrolysis of cane sugar in presence of acid. C I2H 220 Il + H20
Acid --?
CJI120 6 + C6H 1Z0 6
Problem 5: (a) What are pseudQ-unimolecular reactions? (Meemt 2000) (b) How will you study the kinetics of the hydrolysis of methyl acetate? (c) The rate of first order reaction increases with the concentration of the reactant. Explain with reason. (Meenlt 2005)
219
CHEMICAL KINETICS & CATALYSIS
(a)
Pseudo-Unimolecular Reactions
Reactions which are not unimolecular, but obey the first order rate expression are known as pseudo-unimolecular reactions. For example. hydrolysis of methyl acetate. inversion of cane sugar etc. are pseudo-unimolecular reactions. In general. when the order of reaction is generally less than the molecularity of a reaction, it is said to be a pseudo order reaction. H+
CH3COOCH3 + H 20 ~ CH3COOH + CH30H The reaction is bimolecular but the rate of reaction is given by, dx -dt = k [CH3COOCH.31 The concentration of water does not change during the course of the reaction, as it is taken in excess. Similarly, for the reaction, C 12H 220 Il + H20 the rate of reaction is given by
H+ ~
C6 H 120 6 + C6H 120 6,
dx dt = k [C12H220111 because the value of [H20] remains constant as it is taken in excess.
(b)
Kinetics of Hydrolysis of Methyl Acetate (Experimental Determination)
The hydrolysis of methyl acetate is catalysed by hydrogen ions and occurs as follows :
It is a pseudo-unimolecular reaction, for although two molecules take part in the change, the concentration of water does not appreciably change because of its being present in large excess. As acetic acid is produced in the reaction, the reaction can be followed by withdrawing a definite quantity of the reaction mixture after definite intervals of time and titrating it against a standard alkali solution, say NaOH. The amount of alkali used is equivalent to the total amount of HCI present originally and the amount of acetic acid formed in the reaction. As the amount of HCl originally present is known or can be determined by titrating against the same alkali at the start of the reaction, the amount of acetic acid formed (x) after different intervals of time (t) can be determined. The amount of acetic acid formed at the end of the reaction is equivalent to initial concentration (a) of the esier. 5 ml of reaction mixture (10 ml CH3COOCH3 + 100 ml 0.5 N HCl) IS withdrawn into a conical flask at the start of the reaction and some pieces of ice
220
PHYSICAL CHEMISTRY-I
are added to freeze the equilibrium. It is titrated against N/1O NaOH, using phenolphthalein as an indicator. Similarly, 5 ml of the reaction mixture is withdrawn after say 5,10,20,30 minutes and also at the end of 24 hours, when the reaction is supposed to be complete. The solution is then titrated as usual. Suppose the volumes of standard NaOH solution required for the neutralisation of 5 ml of the reaction mixture, at the start of the reaction, after time t and at the end of the reaction, are VO,VI and V~, respectively. Then x, the amount of acetic acid formed after time t is proportional to (VI - Yo), while a, the initial concentration of the ester is proportional to (V~ - Yo). Thus, (a - x), the concentration of the ester left behind unreacted after time t is proportional to (V~ - yo) - (VI - yo) or (V~ - VI)' SubstitutIng the values of a and (a - x) in the first order rate equation, k - 2.303 10 _a_ -
g,oa-x
t
k - 2.303 log
we get,
V~ - Vo
t 10 V~ - VI The values of k at different intervals of time come out to be constant. This shows that the reaction is of the first order.
(C) For a first order reaction, the rate of reaction is given
by, dx
or
dt = k [Reactant]
dx dt
oc
[Reactant]
So, the rate of reaction will increase with an increase in the concentration of the reactant.
. Problem 6: (a) Show that the time required for the completion of a definite fraction (say one halj) of the reaction is independent of the initial concentration for a reaction of the fust order. (b) Prove that if the half life period of a first order reaction is 30 minutes, the reaction will be 75 percent complete in 60 minutes (Meerut 2004)
(8) Suppose
is the time taken to complete a definite fraction, say one half of a reactant from initial concentration a to a12, i.e., x = a12. For a first order reaction, 2.303 I a 2.303 I a k =- - oglo - - =- - oglO-t a-x tll2 a tl/2
a-2"
or
2.303
tll7
a
2.303
=- k - loglO (aI2) =-k-1oglO 2 2.303 x 0.3010 0.693 k =-k-
=
Thus, tll2 is independent of initial concentration, a.
221
CHEMICAL KINETICS & CATALYSIS
(b) 75 percent completion of a reaction involves two half life periods, so the time taken for the reaction to be 75 percent complete will be 2 x t1l2 =2 x 30 minutes =60 minutes.
Problem 7: What is a second order reaction? Derive the rate expression for a second order reaction. Discuss the characteristics of a second order reaction. Write some examples of second order reactions and study the kinetics of anyone reaction. (Meerut 2002, 2000) Or What is the unit of k in a second order reaction? (Meerut 2006)
[I] Second Order Reaction A reaction is said to be of the second order if its reaction rate depends on two concentration terms of reactants. For example, 2A ~ Products. The rate of reaction at any given time is given by dx dt
= k [A]2
[II]
Rate Expression for Second Order Reactions (a) When there is only one reactant or both reactants have equal concentrations : Consider the following reactions : 2A
~
Products.
or A + B ~ Products. Let a be the initial concentration of each of the two reactants and (a - x) be their concentration after any time t. Then, according to the law of mass action, the reaction rate is given by dx 2 - = k (a - x) (a - x) = k (a - x) dt
dx
or
(a _x)2
= k dt
where k is the velocity constant. In order to get the value of k, we integrate the above expression, i.e.,
f or
dx 2 =fkdt (a -x)
1 --=kt+1 (a -x)
(I = integration constant) ... (1)
When t = 0, x = 0, therefore, from equation (1),
*
It is assumed that the concentration of the substance which is taken in excess does not appreciably change during the course of the reaction.
222
PHYSICAL CHEMISTRY-I
1=1.. a
Substitutmg the value of 1 in equation (1). we get
-1-=kt+1.. a-x a 1 1 ----=kt a-x a
or or
k=1..[_l _1..] t a-x a
or
k=1...
x
... (2)
a (a - x)
t
Equation (2) is known as second order rate expression. (b) When the initial concentrations of both reactants are not equal : Let a and b be the initial concentrations of reactants A and B and let (a - x) and (b - x) be their respective concentrations after any time t. The reaction rate will thus be given by :
= k (a -x)(b -x)
_---=dx=-__ =k tit (a -x)(b - x)
or
_1_ [_1___1_]
or
(a - b) (b -x)
dx = k dt
.. ,(3)
(a -x)
In order to get the value of k, we integrate equation (3), i.e.,
_1_ f[-I- -_1_] f k (a - b)
(a -x)
dx -
(a~b)[f (b~X) -f (a~x)l= f
or or
(b -x)
1
(a _ b) [-loge (b - x)
+ lo~ (a - x)]
dt
kdt
= kt + 1
... (4)
(l = Integration constant)
When t = 0, x = 0, therefore, from (4), 1 (a-b) [-Iogeb + loge a] =1 or
1 a 1 = (a _ b) loge b
Substituting the value of 1 in equation (4), we get _1_ (a-x) _ _1_ ~ (a-b) loge (b-x) -kt+ (a-b) loge b
or
k=
1 10 b (a -.i2 t (a - b) ge a (b - x)
... (5)
223
CHEMICAL KINETICS & CATALYSIS
Equation (5) is known as second order rate expression.
[III]
Characteristics of Second Order Reactions
(i) The rate constant is not independent of the unit in which the concentration is expressed. Let the new unit be 11 times the first value. So, from equation (2), k,=lx nx _1 x .1 t na x n (a - x) t a (a - x) 11 The new value of k, i.e., k' is lin times the original value. (ii) Unit of rate constant. From equation (2), the rate constant will be expressed in terms of (litime) x (conc/conc 2) or (concr l (timer'. If concentration is expressed in mole L-I and time in second, then k will be expressed in (mol L-Ir l (srI or mol- I L S-I. (iii) The time taken to complete a certain fraction of the reaction is inversely proportional to the initial concentration of the reactant. Let tll2 be the time required for the completion of half the reaction, i.e., x = O.5a. Substituting this value in equation (2), we get k =_1_ . O.5a =_1_ . O.5a 1 tll2 a (a - O.5a) tl/2 a X O.5a a tl/2 1 1 or tll2 = -k or tll2 oc .a a (iv) If one of the reactants is present in excess, the second order reaction becomes of the first order. For different initial concentrations of the reactants, we have from equation (5),
k=
1 10 b (a - x) t (a - b) ge a (b - x) When one of the reactants, say A is present in very large excess, we can neglect x and b in comparison to a. The above equation then reduces to : 1 ba l b k=-log =-log
ta e a (b - x) ta e b ~x Since a remains constant throughout the change, we have 1 b k.a=-lo~-t b-x k,=l.log _b_ t e b-x This equation is identical with first order rate expression.
[IV]
Examples of Second Order Reactions (i) Hydrolysis of an ester by an alkali. CH3COOC 2H5 + NaOH (ii)
~
CH3COONa + C 2H50H
Thermal decomposition of acetaldehyde.
224
PHYSICAL CHEMISTRY-I
2CH3CHO
~
2CH4 + 2CO
(iii) Decomposition of ozone into oxygen. 203
[V]
~
302
Study of Kinetics of Hydrolysis of Ethyl Acetate (Experimental Determination)
The reaction between ethyl acetate and alkali also known as saponification of ester occurs as follows : CH3COOC 2H5 + NaOH ~ CH3COONa + C 2H50H Take 50 ml each of 0.01 M ester solution and 0.01 M NaOH solution in a flask. The reaction mixture is kept in a constant temperature bath. 5 ml of the reaction mixture is withdrawn at regular intervals, say at the start and after 5, 10, 20, 30, 40 minutes. Ice is added to the reaction mixture to freeze the equilibrium. The mixture is titrated against a standard acid say 0.05 M HCI. The reaction mixture is then kept for 24 hours when it is supposed to be complete and 5 ml of it is titrated as before. The volume of acid in each case corresponds to the amount of unchanged caustic soda or ethyl acetate, i.e., (a - x) at that time. The volume of acid required initially corresponds to the original concentration, a. Thus, x will be given by a - (a - x). Inserting the values of a, x and (a - x) in second order rate expression,
k=!.
x t a (a -x) we find that the values of k come out to be constant. This shows that the reaction is of the second order.
Problem 8 : Show that the time required for the completion of a definite fraction (say one half) of the reaction is inversely proportional to the initial concentration for a second order reaction. (Meerut 2(03) Suppose the initial concentration of the constant is a. Let tl/2 be the time required to complete a definite fraction, say one half of the reaction, then x = a12. For a second order reaction, k =! . x =_1_ . al2 t a (a - x) 11/2 a (a - al2) ~ __1_ k . a (aI2) - k . a 1 tl/2 ocor a i.e., the time is inversely proportional to the initial concentration of the reactant.
or
1
_!
112 -
Problem 9: What are third order reactions? Derive the rate expressions and d~cuss their characteristics. Give examples of third order reactions.
225
CHEMICAL KINETICS & CATALYSIS
[I] Third Order Reaction Reactions of third and higher orders are rare, but there are in fact reactions which are definitely of third and sometimes of higher order. This is due to the fact that the probability of three molecules coming to a single point simultaneously, i.e., probability of trimolecular collisions is much less as compared to unimolecular or bimolecular collisions. So, a third order reaction is one in which the reaction rate depends on three concentration terms of the reactant/so For a third order reaction : A + B + C - 7 Products The reaction rate is given by, dx dt = k (a - x)(b - x)(c - x)
[II] Rate Expressions for Third Order Reactions Case I. When a =b =c, we have dx 3 dt = k(a -x) dx
or
(a _x)3
kdt
On integration, we get
f(a
:t
X
)3
=f
k dt
1 ---=kt+1 2 (a _x)2 where 1 = integration constant. When t =0, x =0, we then have
... (1)
_1_=1 2a 2 Substituting the value of 1 in equation (1), we get 1
1
2(a -x)
2a
----=-2 = kt +-2
1 kt=---,. 2(a -xi - 2a2
=1. [a 2- (a - X)2] =! 2 or
a2 (a
_x)2
k =.l . x2(2a -x)2 2t a (a -x)
Case II. When a -:t b -:t c, we have
x (2a - x) 2 a2(a _x)2 ... (2)
226
PHYSICAL CHEMISTRY-I
dx
dt = k(a - x) (b - x) (c - x) dx
ill
=k&
(a - x) (b - x) (c - x)
Breaking into partial fractions, we get dx
~-~~-~~-~
+
dx
~-~~-~~-~
+
dx
~-~~-~~-~
=k~
On integrating, we have k =..!.l(b-c) t
log~+(c-a) IOg¥+(a-b) 'Og~l
1
(b - c)(c - a)(a - b)
... (3)
Case III. When two of the molecules are identical as in 2A + B -----7 Products the concentrations at any time are (a - 2x) and (b - x). The rate equation then becomes :
= k (a - 2x)2 (b - x)
On integrating, we get k= 1 (2b - a) t (2b - a)2 a (a - 2x)
[2x
+ In b (a -
2x)]
a (b - x)
[III] Characteristics of Third Order Rate Expressions (1) From equation (2), we have k = _._1_ . (concentration x concentration)
tIme (concentration)2 x (concentration)2 1 1 =-.-x 2 tIme (concentration) The unit of k will be (timer l (concentrationr 2• If concentration is expressed in mole/litre, then k will be expressed in (timer l (mole/litrer2.
(2) The time taken to complete a certain fraction of a reaction is inversely proportional to the square of the initial concentration of the reactants. The time (to.5) taken for the completion of half the reaction is given by x = a12. Therefore, k =-_,_ al: (2a - a12) a~(Il-aI2)
2t05
or i.e.,
1 2t
(,':~. 3a!2 '---2--
to.5 -= - . -
a-. a 14
3 1 . ---;;2k a-
=-
227
CHEMICAL KINETICS & CATALYSIS
[IV] Examples of Third Order Reactions The following reactions belong to third order reactions: (i) 2NO + O2 ~ 2N02 (ii) 2NO + Cl 2 ---7 2NOCI (iii) 2NO + Br2 ---7 2NOBr (iv). keaction between p-nitro benzoyl chloride and n-butyl alcohol. (v) Reaction between silver acetate and sodium formate. HCOONa + 2CH3COOAg ~ 2Ag + CO 2 + CH3COOH + CH3COONa
Problem 10: What are nth order reactions? Derive the rate equation and discuss the characteristic.
[I] Definition A reaction in which the reaction rate depends on nIh power of the concentration of the reactant is known as nth order reaction, e.g., nA ~ Products dx =k [At dt
[II] Rate Equation In general, a reaction of the nth order where all the initial concentrations are the same is represented as, nA ~ Products a
(Initial cone.) (Cone. at time t)
(a - x)
dx =k [At =k(a - xt dt On integrating, within the boundary conditions, t =0, x
fo
x
dx (a-xr
=0, we get
=fl kdt 0
[(n-I)(:_X)(n-I)I~: =k[t(~
or
_1_[ "I
or
(n-I)
_I_]-kt
(aJx)(n-l)- a(n-l) \"
k ~ I. [ 1( t(n-l) (a-x)n-
or
I) --( 1 I)] a n-
... (1)
This equation is applicable for all orders except when n = 1, because when n = 1, n - 1 = 0. 1 [
So, k
0
t
X0
1 (a _ x)' -
:0
j
..,
228
PHYSICAL CHEMISTRY-I
=-7 (~) =Indeterminate Therefore, the equation becomes indeterminate.
[III] Characteristic Halflife value. The time taken to complete half reaction is inversely proportional to (n - l)th power of initial concentration of the reactant. For half change, a
x ="2'
t= t1l2
Putting the values in equation (1), '112 =
k(n
~ 1
1)
i{" - "J
[(a _ [2(n
l
a }-
1]
-I)
= k(n-l) a(n-I) - a(n-I)
1]
1 [2(n -I)
= k(n-l) a(n-I) - a(n-I) =
1
[2(n-l) _ 1]
k(n-l)a(n-I) t1l2
or or
=
(2(n-I)-1) 1 k(n _ 1) . a(n -
1)
.•• (2)
1
t 112 = Constant X (i1=1) a
1
t 112 oc (i1=1)
a
Problem 11: Discuss the methods employed in determining the order of a reaction. (Meerut 2005, 2001) There are several methods to determine the order of a reaction.
[I]
Integration Method (Hit and Trial Method)
In this method, the initial concentrations of all reactants are determined. The concentration of the reacting substances is then determined by analysing the reaction mixture at different intervals of time. The different values of a and x are then substituted in rate expressions of the first, second and third order reactions. The order of reaction is given by that equation which gives a nearly constant value of k. This method, therefore, involves the trial of one equation after another till the correct one is found. This method is extensively used for simpler reactions.
229
CHEMICAL KINETICS & CATALYSIS
[II] Fractional Change Method or Half-change Method (Method of equi-fractional parts) We know that the time required to complete a certain fraction (say one half) of the reaction. is independent of initial concentration for a first order reaction, is inversely proportional to the initial concentration for second order reaction, is inversely proportional to the square of the initial concentration for a third order reaction. In general. the time required for the completion of the same fraction of the reaction is inversely proportional to the initial concentration raised to the power which is One less than the order of reaction, i.e., 1 toc-n-I a
where, n is the order of reaction. If tl and t2 be the times for the completion of the same fraction of reaction with different initial concentrations al and a2 and if n is the order of reaction. then
or or or For gaseous systems, if PI and pz be the pressures at times tl and t 2, then log (t1/t2) n = 1 + -"-'--=---=log (P2/PI) So, in this method, we start with two different concentrations of the reactants and note the time for the completion of any fraction, say, half change in each case. By substituting these values in the last equation, we can calculate the order of reactiOIf. n.
[III]
Ostwald's Isolation Method
This method is applicable particularly for those reactions where two or more reactants are involved. The experiments are carried out by taking all reactants except one in excess. The reactant which is not taken in excess is said to be isolated*. The order of reaction is then determined with respect to the isolated reactant. The order of the whole reaction will then be the sum of these individual orders of reaction.
*
It is assumed that the concentration of the substance which appreciably change during the course of the reaction.
IS
taken in excess does not
230
PHYSICAL CHEMISTRY-I
Consider a general reaction nlA + n2B + n3C
~
Products
Let the order of reaction be nl when A is isolated, i.e., Band Care taken in excess, n2 when B is isolated, i.e., A and C are taken in excess and n3 when C is isolated, i.e., A and B are taken in excess. The total order of the reaction will thus be given by nl + n2 + n3.
[IV] Graphical Method In this method, the values of x (amount decomposed) are plotted against t (time). The value-Of dxldt at any time is determined from the graph (Fig. 1) by measuring the angle e, since dxldt =tan 8. ::::::::··::t:·:·:·:·:::-::·::::·::::·::::·;':·:·:·:::::::::::::::·::::::::::::::::::\l!·~
( ....
:dx Slope = tan 9 = dx I dt
C ..... !S dt
:;.:
U1<-_~.,{?_i'._O"_~0_1. q;:_0~ ::::
I
'5 ...... ~
·o~
~
_ _ _--,
....
(a-x)_
: : :. ~========~: The various values of dxl dt (Y-axis) are plotted against the corresponding concentrations (a - x) on X-axis as shown in figure (2). If we get a straight line, the reaction is of the first order. In case, a straight line is obtained by plotting dxl dt and (a - x)2 or (a - x)3, the reaction is of second or third order, respectively. Problem 12: Di1icuss the energy of activation and temperature coeffi(Meerut 2000) cient. Or Di1icuss the effect of temperature on reaction velocity in the light of Arrhenius theory. How the energy of activation of a chemical reaction can be determined experimentally?
[I] Temperature Coefficient See problem 1 (d).
[II] Energy of Activation Arrhenius (1889) put forward the hypothesis that all the molecules of a reactant do oot take part in the chemical reaction. It is only a certain number of molecules known as active molecules that may take part in the reaction.
231
CHEMICAL KINETICS & CATALYSIS
The molecules not capable of taking part in the reaction are known as passive molecules. On raising the temperature, the equilibrium between active and passive molecules is supposed to shift over rapidly, so that not only we have an increased collision frequency, but still more, we have a largely increased number of active molecules ready to take part in the reaction. The equilibrium is represented as Mpassive ~ Mactive
K = [M active] a
[M passive)
Rate of the reaction oc [M active] = Ka [M pa"ive] According to this concept, passive or non-reactive molecules can be activated by absorption of energy. An expression connectmg the temperature with the equilibrium constant (K) of a reversible reaction was given by van't Hoff on thermodynamical grounds. The relation is given by dlogeK !ill dt - RT-
... ( 1)
where, aE is the heat of reaction or energy change. For a reversible reaction : A+B ~C+D k) [CUD] we have, K,. = k2 = [A][B] where k) and k2 are the velocity constants for forward and backward reactions, respectively. From van't Hoff isochore, we have dlog e K,. _ aE or dT - RT-
d loge (k)/k2)
aE
Rr
---_.
dT
It follows that
... (2)
and
... (3)
where E) - E2 = !ill and B is a constant. It has been found that B is independent of temperature and is equal to zero, so that d loge k) E) -RTdT d loge k2 E2 and dT -RT-
232
PHYSICAL CHEMISTRY-I
dlogek dT
In general,
EI
=
Rr
or d log e k
E = --. dT
Rr
... (4)
On integration, we get E
loge k = - RT + constant k = Ae-E1/
or more general,
... (5)
... (6)
where A = constant, known asfrequency or Arrhenius factor of the reaction or collision number. It is also known as pre-exponential factor. £ is a term which has the dimensions of energy. If R is measured in calories, then £ is measured in calories per mole. Equation (6) is the famous Arrhenius equation for the rate constant. On taking logarithms, equation (6) can be written as follows: £ loge k = loge A - RT Differentiating it with respect to temperature, we get £ dlog k=-~ dT e
Ri
... (7)
Integrating equation (7) between proper limits, we have k'-
f k
or
I
dlogek=
IT'T
I
£ -ry
dT
RJ'
k? £ ( 1 1) 10glO k~ = 2.303 R TI - Tz
... (8)
[III] Determination of Activation Energy If we measure the velocity constants of a reaction at two different temperatures, it is possible to calculate the value of activation energy (E) from equation (8). The value of E can also be evaluated graphically. If 10glO k is plotted against liT, we get a straight line, the slope of which is equal to - E/2.303 R or - £/4.576. Knowing the slope, E can be easily evaluated. The intercept on loglo k axis gives the value of log A. The Arrhenius equation holds equally good for homogeneous and heterogeneous reactions. Heterogeneous and catalytic reactions also give straight lines over a very wide range of temperatures. If there is a marked deviation from a straight line on plotting log k and liT, then it gives an indication that the observed reaction is a composite one made up of two or more concurrent reactions influenced by one of the reactions and it may predominate, so that the slope of the curve corresponds to the value of E proper to this reaction.
233
CHEMICAL KINETICS & CATALYSIS
For many reactions taking place at ordinary temperatures, the energy of acti vation is of the order of 20,000 cal/mole and under such circumstances the temperature coefficient (tc) is found to satisfy van't Hoff rule, e.g., at 300K, the value of tc is given by kT + 10 d loge k log t = log - - = e c e kT dt
10
ge
kT + 10 = 20,000 kT 2 X (300)2
X
X
E 10=-x 10
Rr
10 "" 1
kTk+TIO "" e "" 2.7
Problem 13 Discuss activation energy, potential energy barrier and Arrhenius law. We know that for a chemical reaction to occur, collisions between reactant molecules must occur. It has been postulated that collisions between those reactant molecules only which are associated with a certain minimum amount of energy can cause a chemical reaction. The minimum amount of energy which must be associated with molecules so that mutual collisions may result in chemical reaction is termed as threshold energy. Molecules possessing energy less than threshold energy will not react on collisions. Molecules possessing energy equal to or in excess of threshold energy constitute only a small fraction of the total number of molecules. This explains why a small fraction of the total number of collisions result in chemical reaction. The number of molecules possessing energy equal to or in excess of threshold energy increases appreciably even with a small rise in temperature. So, the rate of reaction increases considerably even with a small increase of temperature. We can thus conclude that most of he molecules have much less kinetic energy than the threshold energy. The excess energy that the reactant molecules having energy less than the threshold energy must acquire in order to react and give the final products is known as activation energy. Therefore, Activation energy = Threshold energy - Energy possessed by the molecules initially. In other words, passive or non-active molecules, i.e., molecules possessing energy less than threshold energy, can be activated by absorption of excess energy known as activation energy. As already stated, the reactant molecules have to acquire a minimum amount of energy before they can yield products on mutual collisions, i.e., there is an energy barrier placed between reactants and products (Fig. 3). This barrier has to be crossed before the reactants can yield products. It determines the magnitude of threshold energy which the reactant molecules must acquire before they can react to give products.
234
PHYSICAL CHEMISTRY-I
.,., ... ",. ',','.,.,.,',' ....'.'.'. """"""'."""';"""")11
..... -::::
Activation energy
~{ ....
:I
!
>.
Activation energy of reverse process
~.
w
E1 Reactants ' A ~-~-.------- - t_ - - - --
t- --__
AE = Heat of reaction ~-------------L----------~----------~--B
}~
Products
p~~g~~~.S reacti~n.
!li:·\::-:................................................................
.. Of
=. .. . . . . . . . .
.....
);l..;.
:.:::::::::::::::::::::::::::::::::::::::::::::::::::::\::::::::::;:::-:::::::.:.::: :/.:.: ::::. ~i~..3.. :::::::;::::::::::;:::::::::::::}}::::::::::;::::::::::::::::::::::::::::::::::::::\\:
Further, different reactions require dillerent amounts of activation energy. The concept of activation energy gives us an idea whether a given reaction is slow or fast at a given temperature. A reaction which has lower activation energy will proceed at a faster rate at a given temperature or vice-versa. The differences ill activation el!el~f?Y are mainly responsible for observed difference in the rates of reactio/l. The concept of activation energy as applied to chemical reaction can be explained by plotting energy against the progress of the reaction (A ~ B) as shown in figure (3). The point A represents the initial energy (E 1) of the reactants, AC represents the energy barrier which the reactants have to cross and B represents the products having energy, E 2 . At the beginning of rising portion of the curve AC, the molecules come very close to each other, but they cannot react as they possess energy less than the threshold energy. If, however, sufficient energy is given to them, they can go up and reach the summit C of the energy barrier. The molecules are then said to be in an activated state. In this state, the molecules are under conditions of acute strain. The bonds between atoms of the reacting molecules become very feeble. Now, the condition is such that the probability of formation of new bonds between atoms of the molecules of the reactant is fairly strong. This probability is shown by the curve CB moving down the barrier.
235
CHEMICAL KINETICS & CATALYSIS
In the above case, the reactants are shown to possess a total energy higher than the products. So, according to the law of conservation of energy. there will be a release of energy, i.e., heat will be evolved. The reaction is exothermic and the amount of heat evolved gives the heat of reaction. This release of energy is shown by llE. Evidently, llE = EI - E2, where EI is the normal energy of the reactants and E2 that of the products. In this case, llE is negative as energy is released and not absorbed. An exothermic reaction has thus a lower activation energy or a faster rate of reaction. Now, consider the reverse reaction (B ~ A). Since the energy (E2) of molecules of B is less than the energy (E)) of A, as shown in the figure, energy will be absorbed and not released, i.e., the reverse reaction will be endothermic. An endothermic reaction has always a greater activation energy and hence has a slow~r rate of reaction than the opposing reaction.
Problem 14. Discuss the collision theory for llnimolecillar reactions. (Meerut 2007, 2005)
It seems to be difficult as to how the collision theory could possibly be used to explain the mechanism of unimolecular reactions. In unimolecular processes only one molecule takes part in the reaction, then a question arises: How do molecules in unimolecular reactions attain their energy of activation? Lindemann (1922) suggested its answer by pointing out that the behaviour of unimolecular reactions can be explained on the basis of bimolecular collisions provided we postulate that a time lag exists between activation and reaction during which activated molecules may either react or be deactivated to ordinary molecules. Thus, the rate of reaction will not be proportional-to all the molecules activated, but only to those which remain active. Lindemann suggested that the above reaction takes place as follows : kj
First stage:
A+A
Second stage:
A
* k3
~
A*+A
Products
The first stage involves collision of reacting molecules forming a few activated molecules represented by A*, with a velocity cQnstant k b after which time-lag occurs. During this time-lag the activated molecules lose their excess energy and revert to the original state, with a velocity constant k 2• Alternatively, the activated molecules may decompose into products, with a velocity constant k 3• 2 Rate of activation =k l [AJ = k2 [A*] [A] Rate of deactivation
236
PHYSICAL CHEMISTRY-I
Rate of decomposition = k3 [A *] According to the stationary or steady state principle, whenever a short-lived reaction intermediate occurs in a system, its rate of formation can be taken as equal to its rate of disappearance. Applying this principle, we have kl [A]2 = k2 [A'] [A]
or
+ k3 [A']
= [A·] {k2 [A]
,
+ k3}
kl [A]2
[A ] = k2 [A]
+ k3
Since the rate of reaction is proportional to the concentration of actIvated molecules, we can, therefore, write that, Rate of reaction = - d ~~] = k3 [A'] or
d [A] klk3 [Af -~= k2 [A] +k3
(i) At ~igh pressure:
At sufficiently high pressure, the term k2 [A] is far greater than k3 , which can thus be neglected. Equation (1) then reduces to _ d [A] = klk3 [A] = k'[A] dt k2
Hence, the reaction is of the first order, as the rate of reaction is proportional to the concentration of only one molecule of the reactant. Thus, if concentration of A is high, the reaction should be of the first order. (ii) At low pressure: At low pressure, k3 > > k2 [A], therefore, k2 [A] can be neglected in comparison to k 3 • Therefore, equation (1) reduces to ... (2)
Hence, the reaction is of the second order. Thus, if the concentration of A is low, the reaction becomes of the second order. Problem 15. Discuss the mathematical treatment of transition state theory. Compare this theory with collision theory. According to transition state theory, the rate of a reaction is the number of activated complexes passing per second over the top of potential energy barrier. This rate is equal to the concentration of activated complex times the average velocity with which a complex moves across to the product side. The activated complex is not in a state of stable equilibrium, since it lies at a maximum potential energy.
237
CHEMICAL KINETICS & CATALYSIS
Postulates of Transition State Theory (i)
As the reacting molecules approach each other there is a continuous series of changes in bond distances. These changes are accompanied by energy changes.
Activated complex
---
1- - - - - - - - - -
---------
-I --~, =
EI E2
_ 0 0 eneraVl::::1
of reactants E2
= Activation
",",2rn'",':'"
of products
E
E' -------------------------
Products
Reaction path_
(ii)
The reactant molecules are changed into an energy rich intermediate called activated complex or transition state (fig. 4). (iii) The activated complex may be formed by some loose association or bonding of reactant molecules with necessary rearrangement of valence bonds and energy. If it is a unimolecular reaction, the reactant molecule may produce the activated complex by rearrangement of atoms and redistribution of energy. (iv) The activated complex, though unstable, has a transient existence. It is treated formally as a definite molecule with an independent entity. The activated complex remains in equilibrium with the reactants and its potential energy is maximum. Finally, the activated complex decomposes into products. (v) The activation energy of reaction in the light of this theory, is the additional energy which the reacting molecules must acquire to form the activated complex.
Thermodynamic or Mathematical Treatment of Transition State Theory Consider a bimolecular reaction between reactants A and B. According to transition state theory,
238
PHYSICAL CHEMISTRY-I
A + B ~ x* ~ Products Reactants
Acti vated complex
The equilibrium constant (K) for the formation of activated complex is,
_ [x*]
K -
[A] [B]
or
[X*] = K [A] [B]
... (1)
According to transition state theory, the rate of reaction is the number of activated complexes which pass over the potential energy barrier per unit time. This, in turn, is equal to the concentration of activated complex multiplied by the frequency at which the complex would decompose into products. Mathematically,
~ = [X*] x Rate (or frequency) of dissociation of activated complex ... (2) From equations (1) and (2), we get
~=K
[A] [B] x Rate (or frequency) of dissociation of activated complex.
The activated complex would decompose only if enough vibrational energy is supplied to the system, so that the atoms vibrate with certain critical frequency, leading to bond breaking. Therefore, Frequency of dissociation of activated complex = Evib1h ... (3) where, Evib = average vibrational energy at temperature Tand h =Planck's constant. RT (.: k= RIN) ... (4) But, EVib=kT=/i From equations (3) and (4) frequency of dissociation of activated complex =RTINh
~ = K* [AJ [B]. Z~
... (5)
For conversion of reactants into products, dx (k = rate constant) ... (6) -=k[AJ [BJ dt From equations (5) and (6), k [A] [B] = K [A] [B]. RTINh
k=K. RTINh ... (7) Equation (7) is the mathematical statement of transition state theory. According to thermodynamics, K can be correlated with ilG* through the following relation, ilG* = - RTln K where ilG* = (Free energy of activated complex) - (Free energy of reactants). ilG* is known as standard free energy change
239
CHEMICAL KINETICS & CATALYSIS
AG' == AH· - TAS' - RTln K' ~ AF - TAS~
or
u*
Inl\.
== - (AF - TAS'l
RT
' - (M!' - TtS')/RT K =- e From equations (7) and (8) we get. k RT - (!lH' - TtS' )IRT =- Nh . e
or
k == ~~.
... (8)
e-!lH'IRT. etS'IR
•.•
(9)
where, AFt = standard enthalpy change, i.e., standard heat of activation, As" =standard entropy change, i.e., standard entropy of activation. Equation (9) can be applied not only to bimolecular, but also to unimolecular and trimolecular processes. Morever, it can be applied to reactions in solution also.
Comparison of Transition State Theory With Collision Theory. (i) From transition state theory, i.e., according to equation (9) k == RT . e -W'IRT • etS'lR Nh
From collision theory,
k = PZe - EIRT
where P = probability factor or steric factor Z = collision number or collision frequency. Comparing both equations, we have
or
(:. E=AJt)
or
or The steric factor P is thus related to entropy of activation. (ii) In collision theory, no account is taken of the internal motions of the reactant molecules, whereas in transition state theory, account is taken of the internal degrees of freedom of reactant molecules and the changes these undergo on reaction. (iii) The concept of entropy of activation in transition state theory is very useful for qualitative purposes. Thus. for bimolecular processes, this is an advantage over the colliSIOn theory.
240
PHYSICAL CHEMISTRY-I
Important Formulae to Remember 1.
20
Unit of reaction rate = mol L -I Units of rate constant (i) Zero order reaction
S-I
= c~nc = (conc) (timer l = mol C time
l S-I
(If concentration is expressed in mol L-I and time in second) o=l (,0,0) FO,rst ord er reaction -0-
tIme
000) S econ d (III
. )-1 = s-I = ( time
o l Ofd er reaction =--2
(conc)
0
1
- 0 -
tIme
0 )-1 = (conc )-2 ( time
= (mol L-lr2 (sri = mol-2 L 2s-1
30
Rate expressions (i) Zero order reaction: k = :!. t
00 0 k =2.303 a (1 1) FOIrst order reaction: - IoglO-t a-x where, a = initial concentration of reactant, x = concentration of reactant undergoing reaction in time to (iii) Second order reaction (i)
40
50
k=lo t
x
[Whena=b]
a(a-x)
(ii) k = 20303 log b(a - x) [When a :;t: b] tea - b) 10 a(b -x) Order of reaction (n) can be calculated by half change method by the following formula: log t, - log t2 n = 1 + .,.---"--'---:-"log P2 -log PI where,PI =pressure of the system at time tl' P2 = pressure of the system at time t20 The activation energy (E) can be calculated from the following formula:
(1 1)
k2 E loglO kl = 2.303 R TI - T2
60
where, kl and k2 are the rate constants at absolute temperatures TI and T20 For a reaction, ilIA + n2B ~ n3C + f!4D, we can equate the reaction rates as : 1 ReactIOn rate =- o
III
d[A] d[B] - =- -I dt
112
dt
241
CHEMICAL KINETICS & CATALYSIS
=+ ~ d [C] =+ ~ d [D] n3
7.
dt
dt
n4
Rate expression for third order reaction (i) k = J... . x(2a - x) 2t a 2(a - x)2
k=1.[(a-b)
(ii)
[When a = b
= c]
log~+ (b-c) IOg~+ (c -a) IOg¥Jl (a - b)(b - c)(c - a)
t
[When a
* b * c)
NUMERICAL PROBLEMS Ex. 1: The decomposition of hydrogen peroxide was studied by titrating it at different intervals of time with potassium permanganate. Calculate the velocity constant for it from the following data, if the reaction is of the first order.
o
600
1200
22.8
13.8
8.2
t (sec) KMn04 (ml)
. IS: . k SoI. The fiIrst ord er rate expressIon
2.303 a =- Iog--
t a-x The amount of KMn04 is proportional to the amount of H 20:! present, so the volume ofKMn04 used at zero time corresponds to initial concentration (a) and the volume used after time t, corresponds to (a - x) at that time. Inserting these values in the above equation, we get
I 22.8 =0 000837 -1 k 600 = 2.303 600 og 13.8' s I 22.8 k 1200 = 2.303 1200 og 8.2
=0000852. s
1
The average value of velocity constant,
k = 0.000837 + 0.000852 = 0.000844 sec- I 2
Ex. 2: The optical rotation of sucrose in presence of dil. HCI at various intervals is given in the following table: t (min.) Rotation (degree)
0
10
20
40
100
00
32.4
28.8
25.5
19.6
6.7
-14.1
Show that reaction is of the first order. Sol. The inversion of sucrose will be a first order reaction if the above data conforms to the equation,
242
PHYSICAL CHEMISTRY-I
2.303 I a 2.303 I ro - r~ k = - - oglo--=-- o g - t a-x t rt-r~ where ro, rl and r~ are the optical rotations at the start of the reaction, after time t and at the completion of reaction, respectively. Inserting the values in the above equation, we get k
10
= 2.303 I 32.4 - (- 14.1) = 2.303 I 46.5 = 0008060 ·-1 10 og 28.8 _ (_ 14.1) 10 og 42.9' mm
I 32.4 - (- 14.1) = 2.303 I 46.5 = 0 008037 ·-1 k20 = 2.303 20 og 25.5 _ (_ 14.1) 20 og 39.6' mm 40
= 2.303 I 32.4 - (-14.1) = 2.303 I 46.5 = 0 008054 ·-1 40 og 19.6 _ (_ 14.1) 40 og 33.7' mm
100
= 2.303 I 32.4 - (- 14.1) = 2.303 I 46.5 = 0 008040 ·-1 100 og 6.7 _ (-14.1) 100 og 20.8' mm
k
k
The constancy in the value of k shows that the reaction is of the first order.
Ex. 3: Decomposition of diazobenzene chloride was foUowed at constant temperature by measuring the volume of nitrogen evolved at different times. The data is given below: t (min)
o o
1
20
so
70
10
2S
33
162
Calculate the specific reaction rate and order of reaction. Sol. Diazobenzene chloride deomposes as : CJIsN = NCI ~ CJIsCI + N2
Thus, the amount of N2 evolved will be a measure of diazobenzene chloride decomposed; i.e., x The total N2 evolved at infinite time will thus give the initial concentration i.e., a. Now, substituting the values in the first order rate expression, k - 2.303 10 _a_ t glo a-x
2.303 I 162 2.303 I 162 000322 ·-1 k20 = -W ogIO 162 _ 10 = -W ogIO 152 =. mm 2.303 1 162 2.303 I 162 000336 ---so ogIO 162 _ 25 = ---so ogIO l37 =. mm 2.303 I 162 2.303 I 162 000326 k 70 = -m ogiO 162 _ 33 = -m oglo 129 =. mm kso =
·-1
·-1
243
CHEMICAL KINETICS & CATALYSIS
The constancy in the values of k shows that the reaction is of the first order. The specific rate constant, k is thus given by : . -I 000328 .-1 k = 0.00322 + 0.00336 + 0.00326 mm =. min 3 Ex. 4: A first order reaction is 15% complete in 20 minutes. How long will it take to be 60% complete? Sol. For a first order reaction, we have 2.303 I a k = - - oglO-t a-x or
2.303 I al 2.303 I a2 - - oglO--=-- oglO-tl
al -XI
Here,
t2
... (i)
a2 -X2
XI
15 = 100 al = 0. 15a1 tl = 20
x2
=100 a2 =O. 6a l t2 =?
60
Inserting these values in equation (i), we get 2.303 I al 2.303 I a2 oglO al - 0. 15a l = ~ oglo a2 - 0.6a2
-W
1 al 1 a2 loglo -085 = . IOglO -04 20 . al t2 . a2
or or
_ 20 t2 -
log (1/0.4) _ 20
x log (1/0.15) t2 =
log (100/40)
x log (100/85) 112.76 minutes.
Ex. 5: Show that the case of unimolecular reaction, the time required for 99.9% of the reaction to take place is ten times that required for half of the reaction. (Meerut 2005) Sol. For unimolecular reaction, the rate expression may be written as k = 2.303 log _a_ fl 10 a-x For half the reaction, let the time taken be fl, then 2.303 a .. k == - - loglO 05 tl
or
fl
X
== 0.5a.
a- . a
0.693 ==-k-
Suppose the time taken for 99.9% of the reaction is x=0.999a. 2.303 a k == -t2- loglO a - 0999 . a
... (i) t2'
then ... (ii)
244
PHYSICAL CHEMISTRY-I
or
k
= 2.~03
log 103 or
t2
= 2.30: x 3 = 6.~09
Dividing (ii) by (i), we get
!1 = 6.909 + tl
6.693
k
k
= 10 or t2 = 10 t 1•
Ex. 6: For a reaction A + B ~ C + D, the initial rates for different initial concentrations of reactants have been found as follows: IniJial conc. (mollir 1)
Initial rate
[A]
[B)
I
2.0
2.0
2.0 x 10-3
II
4.0
2.0
4.0 x 10-3
III
6.0
2.0
6.0 x 10-3
IV
2.0
4.0
2.0 x 10-3
V
2.0
6.0
2.0 x 10-3
VI
2.8
8.0
2.0 x 10-3
Give: (a) Order of reaction with respect to A (b) Order of reaction with respect to B (c) Overall order of reaction (d) Rate law equation (e) Rate constant
(Meerut 2007) (Meerut 2007) (Meerut 2007)
Sol. (a) According to sets I, II, III, we can infer that order of reaction with respect to A {when [B] is kept constant} will be 1. (b) According to sets IV, V, VI, we can infer that order of reaction with respect to B {when [A] is kept constant} will be O. (c) Overall order of reaction = 1 + 0 = 1. (d) The rate law equation will be dx dt = k [AJ
... (.) 1
(e) From set I, we can substitute the values of dx/dt and [AJ in equation (i). Therefore,
20 X 10-3 = k x 2.0 or
3
k = 2.0 X 10- = 10-3 s-1 2.0 Ex. 7: 1 ml of methyl acetate was added to 20 ml of N/20 HCI at 2SOC. 2 ml of the reaction mixture was withdrawn at different times and titrated with a standard alkali. Time (millutes)
ml qf alkali used
.
-
245
CHEMICAL KINETICS & CATALYSIS
Show that the hydrolysis of methyl acetate is a pseudo u1limolecular reaction.
Sol. The reaction will be of the first order or pseudo unimolecular reaction if the data conforms to the first order rate expression k = 2.303 log _a_ = 2.303 log V= - Vo t a-x t V=-Vt where Vo, Vt and V= are the volumes of alkali used at the start of the reaction,
after time t and at the end of the reaction, respectively. Thus, k = 2.303 I 42.03 - 19.24 = 2.303 I 22.79 = 0 00327 ·-1 75 75 og 42.03 _ 24.20 75 og 17.83' mm
k 119
= 2.303 I 42.03 - 19.24 = 2.303 I 22.79 = 0 00327 " - I 119 og 42.03 _ 26.60 119 og 15.43" mm
2.303 I 42.03 - 19.24 2.303 I 22.79 000319 " - I k 183 = 183 og 42.03 _ 29.32 = 183 og 12.71 =. mm As the values of k are nearly constant, the reaction is pseudo unimolecular. Ex. 8: The rate of the reaction aA + bB ----7 cC + dD is given by the foUowing expression : -d [A] =k [A] [B]2 dt
where, k is the velocity constant. What is (;) order of reaction if A is present in excess (ii) order of reaction if B is present in excess (iii) total order of reaction (iv) molecularity of the reaction? Sol. (i) Order of reaction when A i£ in excess = 2 (ii) Order of reaction when B is in excess = 1 (iii) Total order of reaction = 2 + 1 = 3 (iv) Molecularity of the reaction = a + b. Ex. 9: The half life periods for the thermal decomposition of phosphine at three different pressures are given below:
-
Initial pressure (mm)
707
79
37.5
Half life (sec)
84
84
84
Calculate the order of reaction. (Meerut 2000) Sol. 1st Method: Since the half life periods are the same. irrespective of the initial concentrations or pressures, the reaction is of the first order. 2nd Method: We know that: log (t1/t2) n = 1 + -"'--'--'--=log (a2/al) log (tl/t2) or 11=1+ (":aocp) log (P2/PI) For first set:
246
PHYSICAL CHEMISTRY-I
_ log (84/84) .. n - 1 + log (79/707) 1 + 0 or n = 1 For second set: _ log (84/84) _ _ _ n - 1 + log (07.5/79) - 1 + 0 - lor n - 1 :. Order of reaction = 1. Ex. 10 : The timefor half change (t)for a gaseous reaction was measured for various initial pressures (p). The following data were obtained :
Calculate the order of reaction. Sol. 1st method: We know that
(Meerut 2006)
1
or p. tll2 = constant p (i) p x tll2 = 200 x 150 = 30000 mm min (ii) p x tll2 = 300 x 99·8 = 29940 mm min (iii) p x tll2 = 400 x 75.3 = 29120 mm min As the values of ptll2 are constant, the reaction is of the second order. (.: tll2 oc lip) 2nd Method: We knew that: log (t)/t2) n = 1 + ---':::....:.-'--..:::.... log (P2/P) For first set: _ 1 log (150/99.8) _ 1 0.1769 •. II + log (300/200) - + 0.1760 = 1 + 1.004 = 2.004 :::: 2 For second set: _ 1 log (99.8/75.3) _ 1 0.1223 /1+ log (400/300) - + 0.1249 tll2 oc -
= 1 + 0.979 = 1.979:::: 2 So, the order of reaction is 2. Ex. 11: Time for half change (t) of a gas undergoing thermal decomposition for various initial pressures was found as : PreHure (mm)
750
500
250
1¥me (mill utes)
105
235
950
Find the order of the reaction. Sol. The order of reaction is given by log (t)/t2) II = 1 + _=-,--=--..:=.o... log (P2/P)
247
CHEMICAL KINETICS & CATALYSIS
For first set:
..
_ 1 log (105/235) _ 1 0.3498 n - + log (500/750) - + 0.1760
= 1 + 1.98 =2.98 "" 3 For second set:
_ 1 log (235/950) _ 1 0.6066 n - + log (250/500) - + 0.3010·
= 1 + 2.01 :. Order of reaction = 3.
=3.01"" 3
Ex. 12: The following data were obtained for a gaseous reaction : Initial pressure (mm)
200
300
400
Half period (min)
150
99.8
75.3
Calculate the order of reaction. log (tJ/t2) Sol. n = 1 + ---'''---'-''---'''log (P2/PJ)
n Similarly.
=1 =1
n
log (150/99.8)
+ log (300/200)
=1
log (150/75.3) = 1
+ log (400/200)
0.1769
+ 0.1761
=1
0.2992 = 1
+ 0.3010
+ +
1 =2 1= 2
Order of reaction = 2. Ex. 13: Dinitropentaoxide decomposes as follows : - d [N20SJItit = kl [N20S]
d [N02J1tit = k2 [N20s1 d [02J1dt = kJ [N2 0 s1
What is the relation between k., k2 and k J ?
Sol. The decomposition of dinitropentaoxide is a first order reaction, so
According to stoichiometry, the rate pf the above reaction can be expressed by the following relations : d [N 20 51 I d [N0 21_1 d [0 21 1 dt dt 2 fit 2
or
.
1
kJ [N 20 51= 2" k2 [N 20 51= 2k3 [N 20 5]
248
PHYSICAL CHEMISTRY-I
or
kl
or
k2
=2= 2k3
,.. (i)
... (ii)
2kl = k2 = 4k3
So, relation between k 1, k2 and k3 can be expressed by either equation (i) or (ii). Ex. 14: The first order reaction has k = J.5 X 10-6 per second at 200C. If
the reaction is allowed to run for 10 hours, what percentage of the initial concentration would have changed into product? Sol.
For a first order reaction, k - 2.303 10 _a_ t glOa-x
or
-6 2.303 100 1.5 x 10 = 10 x 60 x 60 log (l00 -x)
100 log (100 - x) or
or or
100 100 - x
1.5 X 10-6 x 10 x 60 x 60 2.303
=
0 023447 .
= 1.055
100 = 105.5 - 1.055x 5.5 5 3 x = 1.055 = .21
:. 5.213% of the initial concentration had changed into product. Ex. 15: An acid solution of sucrose was hydrolysed to the extent of 57% after 66 min. Assuming the reaction to be of first order, calculate the time taken for 75% hydrolysis. k = 2.303 10gIO _a_ t a-x
For 57% hydrolysis, x
..
= 0.57 a,
2.303 a k = ~ 10gIO a _ 0.57a
2.303
= ~ 10gIO
100 43
For 75% hydrolysis, x = 0.75a,
k = 2.303 10 t
a = 2.303 10 0 100 glO a - 0.75a t ",10 25
2.303 I 100 _ 2.303 I 100 66 oglO 43 t og 25
or
100 log 25 t = --100 log-43
0.6020
x 66 = 0.3665 x 66
249
CHEMICAL KINETICS & CATALYSIS
or t = 108.4 min. Ex. 16: For a first order reaction, the rate constant is found to be 7.0 x 10-7 at 'PC and 9 x 10-4 at 5'PC. Calculate the energy of activation and its specific reaction rate at 12'PC.
[1 1]
, k2 E Sol. ,We know that, log k; = 2.303 R T\ - T2
T\=273+7=280K,k l =7xl0-7,
where, E=activation emvgy, 4 T2 = 273 + 57 = 330 K, k2 = 9 X 10- . So, from equation (8), we get 7
log (9 x 10-4) - log (7 x 10- ) = 2.303
!
1.987 [ 2!0 -
E
3~0 ]
[330-280]
= 2.303 x 1.987 280 x 330
-4 -7 Ex50 or log (9 x 10 ) -log (7 x 10 ) = 2.303 x 1.987 x 280 x 330
Ex50 4.9542 -7.8451 = 2.303 x 1.987 x 280 x 330
or
- 2.303 x 1.98750 x 330 x 3.1091 -~ 26•29 x 103 ca II moIe. or E Spccific reaction rate at 127°C, i.e., 400 K is given by 26290 [1 1] log k400 - log k280 = 2.303 x 1.987 280 - 400 26290 [400 - 280] 2.303 x 1.987 400 x 280 or
7
log k400 - log (7 x 10-)
26290 x 120 2.303 x 1.987 x 400 x 280 = 6.153.
log k400 = log (7 x 10-7) + 6.153
or
=-7.8451 x 6.153 =-
1.6921
k400 = 0.02032 min-to or Ex. 17: Bodenstein by studying the kinetics ofdecomposition ofa gaseous hydrogen iodide gave the values of specific reaction rates to be 3.517 x 19-7 and 3.954 x 10-2 at 556 K and 781 K. Cakulate the energy of activatfo'n of the reaction and frequency factor.
k2 E [ 1 1] (a) We know that log k\ = 2.303 R T\ - T2 2
3.954 X 10- _ E [_1___1_] g 3.517 x 10-7 - 2.303 x 1.987 556 781
10
250
PHYSICAL CHEMISTRY-I
E
=2.303 x 1.987 E=
[781 - 556J 557 x 781
3.954 x 10-2
log
7X
3.517 x 10-
2.303
X
1.987 X 556 x 781
225 = 44610 cal/mole.
where, A = frequency factor. Hence, 3.517 x 10-7 = A.e-44610/1.987 x 556 or
-7 44610 log 3.517 x 10 = log A - 1.987 x 556
- 6.4538 = log A - 40.2910 or log A = 33.8372 or or A = 6.874 X 1034• Ex. 18: In the saponification of ethyl acetate by sodium hydroxide using equal concentrations the progress of the reaction was foUowed by titrating 25 ml of the reaction mixture at regular intervals against standard acid. The foUowing data were obtained: Time (min)
0
15
25
35
Volume of acid used (ml)
16
6.13
4.32
3.41
Show that the reaction is of the second order. Sol. We have, a oc Vo and (a - x) oc VI' Therefore, x oc (Vo - VI)' For a second order reaction, when reactants are of equal concentrations, k =! t
.
x a (a - x)
=!. (Vo -
VI)
Vo . VI
t
Substituting the different values, we get k 15=15' 16x6.13 =~ .
1
00067
. -I -1 mIn cons
1 (16 - 4.32) k25 = 25' 16 x 4.32
00069
. -I -I mm cons
k
(16 - 6.13)
- 1
35 -
(16 - 3.41) 35' 16 x 3.41
=.
0.0068 min- 1 cons-I
The constancy in the values of k shows that the reaction is of the second order. Ex. 19: Decomposition of a gas is of second order when the initial concentration of the gas is 5 x 10-4 mole per litre. It is 40% decomposed in 50 minute~. What is the value of velocity constant? Sol. For a second order reaction,
k=!. t
x
a (a - x)
251
CHEMICAL KINETICS & CATALYSIS
Here, a =0.00005; x
=0.0005 x (40/100) =0.0002
1 0.0002 26 661't I -1 ·-1 . . k = 50 x 0.0005 (0.0005 _ 0.0002) = . I mo e mID .
Ex. 20: A second order reaction where a = b is 20% completed in 500 seconds. How long wiU it take for the reaction to go to 60% completion? Sol. For a second order reaction, where a = b,
k=.!..
x .. t a (a - x) When the reaction is 20% complete, x = 0.2a k=_l_. 0.2a 1. 0.2a _ __ .. 500 a (a - 0.2 a) 500 a X 0.8a 2000 a Suppose it takes t\ seonds for the reaction to go to 60% completion. N ow, or
k =1. . tl
0.6a _ 1 0.6a a (a - 0.6a) t 2 · a X OAa 1 .3 =-tl 2a 2000a t) =
=1. . l tJ
2a
3000 sec.
Ex.4: A second order reaction with two reactants is started with O.IM concentrations of each reactant. It is 20% completed in 500 seco~ds. How long will it take the reaction to go to 70% completion? For a second order reaction when the concentrations are equal, we have
k=.!.. t
x
a (a - x)
For 20% completion: a =0.1, x= 0.2 X 0.1 =0.02; t =500 k = _1_ X 0.02 =_1_ X 0.02 =_I_ .. 500 0.1 (0.1 - 0.02) 500 0.1 x 0.01 200 For 60% completion: a = 1; x =0.6 X 0.1 = 0.06; t = ? k=! X 0.:.::.06=--_ .. t 0.1 (0.1 - 0.06) 1 1 0.06 or -=-x 200 t 0.1 - 0.04 200 x 0.06 or t =Q.l X 0.04 =3000 seconds.
2. CATALYSIS Problem 1. Define catalyst and catalysis. Mention the types and classification of catalysis. Discuss the characteristics of catalytic reactions.
[I]
Catalyst and CatalysiS
Berzelius (1835) found that the speed of a number of reactions is increased due to the presence of a small quantity of a foreign substance. He also
252
PHYSICAL CHEMISTRY-I
found that these substances remain chemically unchanged at the end of the reaction. He termed these substances as catalysts and the phenomenon itself is known as catalysis. A familiar example is that of the decomposition of KCI03 . The decomposition of KCl0 3 is very slow even at high temperature, but a small quantity ofMn02 increases the rate of decomposition to a very great extent and Mn02 remains chemically unchanged at the end of the reaction. 2KCl03 + [Mn02] ~ 2KCl + 302 + [Mn02] But later on it was observed that there are certain substances which can retard the rate of a chemical reaction. Hence. Ostwald defined that. "A catalyst is a substance which influences the speed of a 'chemical reaction without itself undergoing any chemical change at the end of the reaction. " Catalysis is mainly divided into two types. viz .• homogeneous catalysis and heterogeneous catalysis.
[A] Homogeneous Catalysis When the catalyst is present in the same phase as that ofthe reactants, the phenomenon is known as homogeneous catalysis. (a) Examples of homogeneous catalysis in gas phase (i) Oxidation of sulphur dioxide (S02) to sulphur trioxide (S03) with nitric oxide (NO) as catalyst.
2S02 + O2 + [NO] Gas
Gas
Gas
~
2S03 + [NO] Gas
(ii) Decomposition of acetaldehyde (CH 3CHO) with iodine (1 2) as catalyst. Vapour
Gas
Vapour
Gas
(b) Examples of homogeneous catalysis in solutiop-pbase Many reactions in solutions are catalysed by acids (J-r) and bases (OH-). (i) Decomposition of hydrogen peroxide (H20 2) in the presence of iodide ion (r) as catalyst, I
2H20 2 ~ 2H20 +02 (ii) Hydrolysis of cane sugar in aqueous solution in the presence of mineral acid as catalyst.
Cane sugar
Glucose
Fructose
(iii) Hydrolysis of an ester in the presence of acid or alkali. +
H/OH
-
CH3COOC 2Hs + H20 --~ CH3COOH + C2H50H Ethyl acetate
Acetic acid
Ethanol
253
CHEMICAL KINETICS & CATALYSIS
[8] Heterogeneous Catalysis When the catalyst is in a different phase than that of reactants, the phenomenon is known as heterogeneous catalysis. Some examples of heterogeneous catalysis with reactants in the gas, liquid or the solid phase are given below. (a) Heterogeneous catalysis with gaseous reactants (contact catalysis) (i) Oxidation of ammonia to nitric oxide (NO) in the presence of a platinum gauze (a stage in the manufacture of nitric acid), 4NH3 + 502 + [Pt] ~ 4NO + 6H20 + [Pt] Gas
Gas
Solid
(ii) Combination of sulphur dioxide (S02) and oxygen in the presence of finely divided platinum or vanadium pentoxide, V 20 5, (contact process for sulphuric acid). 2S02 + O2 + [Pt] ~ 2S03 + [Pt] Gas
Gas
SolId
(iii) Hydrogenation reactions of unsaturated organic compounds are catalysed by finely divided nickel. H2C=CH2+ H2 + [Ni] ~ H3C-CH3 + [Ni] Ethene (gas)
Gas
Solid
Ethane
Vegetable oils are tri-esters of glycerol with higher unsaturated acid (oleic acid). When hydrogen is passed through the vegetable oils in presence of nickel, the carbon-carbon double bonds of the acid portions are hydrogenated to yield solid fats (vanaspati ghee). (iv) Combination of nitrogen and hydrogen to form ammonia in the presence of finely divided iron, (Haber's process for ammonia). N2 + 3H2 + [Fe] ~ 2NH3 + [Fe] Gas
Gas
Solid
(b) Heterogeneous catalysis with liquid reactants (i) The decomposition of aqueous solutions of hydrogen peroxide (H20 2) is catalysed by manganese dioxide (Mn02) or platinum in colloidal form. 2H20 2 + [Pt] ~ 2H20 + O2 + [Pt] Liquid
Solid
(ii) Benzene and ethanoyl chloride (CH3COCI) react in the presence
of anhydrous aluminium chloride to form phenyl methyl ketone (CJI5 COCH3)· C614 + CH3COCI + [AICI3] ~ C6H5COCH3 + HCI + [AICI3] Liquid
Liquid
Solid
(c) Heterogeneous catalysis with solid reactants The decomposition of potassium chlorate (KCI0 3) is catalysed by manganese dioxide (Mn02).
254
PHYSICAL CHEMISTRY-I
2KCI03 + [Mn02J ----) 2KCI + 302 + [Mn02J Solid
Solid
The above reaction is heterogeneous. though both reactants are in the same phase (solid). because the solid forms a new phase.
[II] Classification of Catalysis Catalytic reactions are of the following four types: (a) Positive catalysis (b) Negative catalysis (c) Auto-catalysis (d) Induced catalysis (a) Positive Catalysis: When the catalyst used accelerates the speed ofa chemical reaction, it is known as a positive catalyst and the phenomenon is known as positive catalysis. For example. the rate of decomposition of hydrogen peroxide increases in the presence of colloidal platinum as catalyst. Catalyst
Other reactions are : (ii)
2KCI03
Mn°2
--~)
2KCI + 302
PtorNO
(iv) 2S02 + O2 ) 2S03 (b) Negative CMaJysis: When the foreign substance retards the speed of a chemical reaction, it is known as a negative catalyst and the phenomenon is known as negative catalysis. The following are examples of this type. (i) Decomposition of hydrogen peroxide H,P04
2H20 2 ~ 2H20 + O 2 (ii) Oxidation of chloroform 4CHCl3 + 302
C,H,OH
) 4COCI 2 + 2Cl2 + 2H20 (iii) Tetraethyllead as antiknock When tetraethyllead. Pb(C2H5)4 is added to petrol. it retards the too rapid or explosive combustion of the fuel which is responsible for the working of the engine. Explanation of Negative Catalysis The mechanism of negative catalysis could be different for different reactions, e.g .• (1) By poisoning a catalyst. A negative catalyst may work by poisoning a catalyst which already happens to be present in the reaction mixture. -
255
CHEMICAL KINETICS & CATALYSIS
For example, the traces of alkali dissolved from the glass of the container, catalyse the decomposition of hydrogen peroxide (H20 2). However, the addition of an acid would destroy the alkali catalyst and thus prevents decomposition. (2) By breaking a chain reaction. In some cases, negative catalysts are believed to operate by breaking the chain of reactions. For example, the combination of H2 and C1 2, which is a chain reaction, is negatively catalysed by nitrogen trichloride (NCI 3). Cl2 ~ Cl'+Cl' Free radicals
H' + Cl2 ~ HCl + Cl' .NCI3 breaks the chain of reactions by absorbing the propagating species (CI') and the reaction stops. I
NCl3 + CI· ~ "2N2 + 2Cl2
(c) Auto-Catalysis: When one of the products formed in the reaction itself acts as a catalyst, the
substance is known as an autocatalyst and the phenomenon is known as auto-catalysis. In auto-catalysis the initial rate of the reaction rises as the catalytic product is formed, instead of decreasing steadily (See fig. 5). The curve plotted between reaction rate and time shows a maximum when the reaction is complete. (i) For example, hydrolysis of ethyl acetate by water is an auto-catalytic reaction, since acetic acid liberated in this reaction acts as a catalyst. CH3COOC2Hs + H20
tI --------------------Completion of reaction c
.Q
i Sigmoid curve
Time---+ Fig. S : Curve showing the rise of rate of reaction with time.
~
CH3COOH + C2HsOH Auto-catalyst
(ii) The oxidation of oxalic acid by acidic KMn04 is catalysed by the presence of Mn2+ ions formed in the solution. In the beginning, the colour of KMn04 disappears slowly, but as Mn2+ is formed in the solution, the colour discharges rapidly. So, Mn2+ ions acts as auto-catalyst.
256
PHYSICAL CHEMISTRY-I
2Mn04" + 5C20~- + 16W ~ 2Mn2+ Violet
+ lOC0 2 + 8H20
Colourless
(iii) The free arsenic produced by the decomposition of arsine (AsH3) auto-catalyses the reaction. 2AsH3 ~ 2As + 3H2 Catalyst
(d) Induced Catalysis: When one reaction influences the speed of other, which is not possible under ordinary conditions, the phenomenon is known as induced catalysis. For example, sodium sulphite solution is readily oxidised in air but sodium arsenite solution is not oxidised by passing a current of air through it. However, if air is passed through a mixture of sodium sulphite and sodium arsenite solution, the oxidation of both take place. Here the oxidation of sodium sulphite acts as a catalyst for the oxidation of sodium arsenite solution.
[III] Characteristics of Catalytic Reactions Although there are different types of catalytic reactions, the following features or characteristics are common to most of them. These features are often referred to as the criteria of catalysis. (1) A catalyst remains unchanged in mass and chemical composition at the end of the reaction. Qualitative and quantitative analysis show that a catalyst undergoes no change in mass or chemical composition. However, it may undergo a physical change. Thus, granular manganese dioxide (Mn02) used as a catalyst in the thermal decomposition of potassium chlorate is left as a fine powder at the end of the reaction. (2) A small quantity of catalyst is generally needed to produce almost unlimited reaction. Sometimes, a trace of a metal catalyst is required to affect very large amounts ~f reactants. For example, one ten-millionth of its mass of finely divided platinum is, however, needed to catalyse the decomposition of hydrogen peroxide. On the other hand, there are catalysts which need to be present in relatively large amount to be effective. Thus, in Friedel-Crafes reaction, Anhy. Alel3
C6H6 + C 2HsCi ) C6H5CzH5 + HCI Anhydrous aluminium chloride functions as a catalyst effectively when present to the extent of 30 percent of the mass of benzene. (3) A catalyst cannot, in general, initiate a reaction. In most cases, a catalyst speeds up a reaction already in progress and does not initiate (or start) the reaction. But there are certain reactions where the reactants do not combine for very long period (perhaps years). For example, a mixture of hydrogen and oxygen, which remains unchanged almost
257
CHEMICAL KINETICS & CATALYSIS
indefinitely at room temperature, can be brought to reaction by the catalyst platinum black in a few seconds. Room temp
) No reaction
Ptblack
2H2 + O2 ) 2H20 It is thus now considered that the catalyst can initiate a reaction. According to this view, the reacting molecules (in the absence of catalyst) do not posses minimum kinetic energy for successful collisions. The molecules rebound from collisions without reacting at all. (4) A catalyst is more effective when finely divided. _ In heterogeneous catalysis, the solid catalyst is more effective when in a state of fine sub-division than when used in bulk. So, a lump of platinum will have much less catalytic activity than colloidal or platinised asbestos. Finely divided nickel is a better catalyst than lumps of solid nickel, because former occupies greater surface area than the latter. (5) A catalyst is specific in its action. While a particular catalyst works for one reaction, it will not necessarily work for another reaction. Different catalysts, moreover, can bring about completely different reactions for the same substance. For example. formic acid gives carbon dioxide and hydrogen when passed over hot copper. eu HCOOH ~ CO 2 + H 2 , However, with hot aluminium oxide, formic acid gives carbon monoxide and water. Al,O.
HCOOH ~ CO + H2 0 (6) Change of temperature changes the rate of a catalytic reaction as it would do for the same reaction without a catalyst. We have already studied the effect of temperature change on reversible reactions under Le-Chatelier's principle. Some catalysts are, however, physically altered by a rise in temperature and hence their catalytic activity may be decreased. This is particularly true with colloidal solutions like that of platinum, since a rise in temperature may cause their coagulation. In such a case, the rate of reaction increases up to a certain point and then gradually decreases. The rate of reaction is maximum at a particular temperature
which is known as the optimum temperature. (7) A catalyst does not alter the final position of equilibrium, although it decreases the time required to establish it It means that in a reversible reaction the catalyst accelerates the forward and the reverse reactions equally. Thus, the ratio of the rates of two opposing reactions, i.e., the equilibrium constant, remains unchanged. The effect of a catalyst on the time required for equilibrium to be established for the reaction
258
PHYSICAL CHEMISTRY-I
A+B ~ C+D is shown in fig. (6). In the beginning, the concentrations of A and B are maximum and so the rate of forward reaction is maximum. As the time passes, the rate of the reaction decreases till the equilibrium is established. For the reverse Equilibrium reaction, the initial concentrations of C and D are zero and the rate of reaction is lowest. As the time passes, the rate of reaction increases till the equilibrium is estabTime--. lished. Similar curves of the rates of reactions with the catalyst show that the rates of the forward reac- :::::: time required for ih:e equilibrium :::::: tion and the reverse reaction are )t. . . . '.' . . . . . . .~. ~. ~~~~.~~~~:......................... mm changed equally but the equilibrium is established in a much shorter time. For example, in the Haber's process for ammonia,
+ 3H2
Fe ~
2NH3 the reaction is very slow. In the presence of the catalyst, the equilibrium is reached much earlier but the percentage yield remains unchanged. The iron catalyst decreases the time to attain equilibrium but cannot change the percentage yield. Energy considerations also show that the final state of equilibrium cannot be changed by the catalyst. Suppose the catalyst accelerates the forward reaction more than the reverse reaction. This will shift the equilibrium point, which cannot happen without the supply of energy to the system. But a catalyst unchanged in mass and~composition at the end of the reaction, cannot supply the required energy. Problem 2. Write notes on the following: (a) Catalytic promoters (b) Catalytic poisons N2
(A) Catalytic Promoters The activity of a catalyst can often be increased by the addition of a small quantity of a second material. This second substance is either not a catalyst itself for the reaction or it may be a feeble catalyst. A substance which, though itself not a catalyst, promotes the activity of a catalyst is called a promoter or an activator.
[I] Examples of Promoters (i) In some reactions, mixtures of catalysts are used to obtain the maximum catalytic efficiency. For example, in the synthesis of methanol
259
CHEMICAL KINETICS & CATALYSIS
(CH 30H) from carbon monoxide and hydrogen, a mixture of zinc oxide and chromium oxide is used as a catalyst.
(ii) Molybdenum (Mo) or aluminium oxide (Ali-03) promotes the activity of iron catalyst in the Haber's synthesis for the manufacture of ammonia. Fe
N2 + 3H2
~ 2NH3 +Mo
[II] Explanation of Promoter's Action The theory of promotion of a catalyst is not clearly understood. It may be due to: Peak
& 1'%
-~i-~i-~i-~i-~i\~ I I I I I
-Ni-Ni-Ni-Ni-Ni-
I
I
I
I
I
I
I
I
I
I
-Ni-Ni-Ni-Ni-Ni-Ni-Ni-Ni-Ni-Ni-
I I I I I
I
I
I
I
I
I
I
I
I
I
I
-Ni-Ni-Ni -Ni-Ni-Ni -Ni-Ni-Ni
.;,:
~
U
I
-Ni-Ni-Ni-Ni-Ni-
I I I I I
(1) Increase of peaks and cracks. The presence of the promoter increases the peaks and cracks on the catalyst surface. This increases the concentration of the reactant molecules and hence the rate of reaction. (2) Change of lattice spacing. The lattice spacing of the catalyst is changed thus enhancing the spaces between the catalyst particles. The adsorbed molecules of the reactant (say H 2) are further weakened and cleaved. This makes the reaction go faster. The phenomenon of promotion is a common feature of heterogeneous catalysis.
(8) Catalytic Poisons Very often a heterogeneous catalyst is rendered ineffective by the presence of small amounts of impurities in the reactants. A substance which destroys the activity of the catalyst to accelerate a reQction, is called a catalytic poison and the process is called catalytic poisoning.
260
PHYSICAL CHEMISTRY-I
IF''''''''·'''''''''''·''''''·'·'''·''·'·'·'·'·'''·'·'·'·'·''·"'·'·'·'·'·"·'·'·"~;?;S,:"·""""""·'·'·"·"'·'·""""·''''11
~ :.:.. .:.:.:.:.:.:.:.:. .:.:.:.:.:.:. :.-.:.:.:.:.:.:.:.: .
I"
Distance between /f catalyst particles ~
catalyst makes the reaction go faster?
':::'::-:::::": ;.:.: :::::
::::::::::::::-::::::::::::::.........................................................................................................................................................................::}}»}:.:;.:
(a) Examples of Catalytic Poisoning (1) The platinum catalyst used in the oxidation of hydrogen is poisoned by carbon monoxide. Pt
2H2 + O2 ----+) 2H20 Poisoned by CO
(2) The platinum catalyst used in the oxidation of sulphur dioxide (contact process) is poisoned by arsenic oxide (AsP3)' Pt
2S0 2 + O2 ----+) 2S03 Poisoned by AS20 3
(3) The iron catalyst used in the synthesis of ammonia (Haber's process) is poisoned by H 2S. N2 + 3H2
Fe
----+) 2NH3
(b) Explanation of Catalytic Poisoning (1) The poison is adsorbed on the catalyst surface in preference to the reactants: Even a monomolecular layer makes the surface unavailable for further adsorption of the reactants. The poisoning of iron catalyst by H2S comes in this class. (2) The catalyst may combine chemically with the impurity: The poisoning of iron catalyst by H2S comes in this class.
261
CHEMICAL KINETICS & CATALYSIS
o II c, , , ,
,
~~;i;l
o
o
, , ,, ,,
,c, , , ,,
II
II c
,
Fig.9: Poisoning of platinum catalyst ~~~ by carbon monoxide.
Fe + H2S
~
FeS+H2
Problem 3. Discuss the theories of catalysis and also mention the industrial applications of catalysts.
[I] Theories of Catalysis Many theories have been put forward to explain the catalytic activity of catalyst. A few important theories are given below : 1. Intermediate Compound Formation Theory According to this theory, a catalyst first combines with one of the reactants to form an intermediate compound of activity greater than that of the reactants. This intermediate compound then reacts with another reactant fufurm the product and so gives back the catalyst. If A and B are two reactants and C is a catalyst, then according to this theory. . A+C~
AC
AC+B
~
AB+C
A+B+[C]
~
AB+[C]
This theory can be fully explained by the following examples : (1) In the oxidation of S02 by air, NO which acts as a catalyst, first combines with oxygen to form N0 2 (intermediate compound) which oxidises S02 and gives back nitric oxide. 2NO + O2 ~ 2N0 2 [S02 + N0 2 ~ S03 + NO] x 2 2S0 2 + O2 + [2NO] ~ 2S03 + [2NO] (2) In the formation of ether from alcohol, H2S04 which is used as catalyst first forms an intermediate compound C2H sHS0 4 . C2HsOH + H2S04 ~ C2HsHS0 4 + HP C 2HsHS04 + C2HsOH
~
C2HsOC2HS + H2S04
262
PHYSICAL CHEMISTRY-I
(3) Fonnation of methyl benzene, C6H5CH3 by reaction between benzene (C6H6) and methyl chloride (CH3CI) using anhydrous aluminium chloride. AICI 3, as catalyst (Friedel-Craft's reaction). AICI)
C6H6 + CH3CI ~ C6H5CH3 + HCI Mechanism: CH3CI + AICI 3 ~ [CH3 [AICI4 f
t
Intermediate compound
C6~ + [CH3t [AICI 4f ~ C6HSCH3 + AICI 3 + HCI (4) Thermal decomposition of potassium chlorate (KCI0 3) in the presence of manganese dioxide (Mn02)' MnO,
2KCI03 ~ 2KCI + 302
Mechanism: 2KCI0 3 + 6MnOz
~
6Mn03
+ 2KCI
Intermediate
compound
6Mn03 ~ 6MnOz + 30 2 It may be noted that the actual isolation of intermediate compounds which would prove their existence is very difficult. As already stated, by their very nature they are upstable. In general, the intermediate compounds suggested as being formed are usually possible rather than proved.
2. Adsorption Theory or Modem Theory of Heterogeneous Catalysis Adsorption theory explains the mechanism of a reaction between two gases catalysed by a solid (heterogeneous or contact catalysis). According to this theory, the catalyst acts by adsorption of the reacting molecules on its surface. Generally speaking, four steps can be put forward for heterogeneous catalysis. For example, for the following reaction, Catalyst
A(g) + B(g) ~ C(g) + D(g) the steps are as follows : Step 1. Adsorption of reactant molecules. The reactant molecules A and B strike the catalyst surface. They are held up at the surface by weak vander Waals forces (physical adsorption) or by partial chemical bonds (cllemisorptioll). Step 2. Formation of activated complex. The particles of the reactants adjacent to one another JOIn to form an intermediate complex (A - B). The activated complex is unstable. It has only a fleeting existence. Step 3. Decomposition of activated complex. 'The activated complex breaks to form the products C and D. The separated particles of the products are held to the catalyst surface by partial chemical bonds.
263
CHEMICAL KINETICS & CATALYSIS
Step 4. Desorption of products. The particles of the products are desorbed or released from the surface. They are stable and can lead to an independent existence.
r··'·······'·N... . . ,. . . . ". . . . . . . . . . . . . '. . 1
:~; 3
''''~;;;''''''''''''''''''".S2,,''
2
Catalyst
"""""'''
"""
""""'''''
;;;~;
. . . . . . . . . ..
""""""""~"'''''''''''''' Catalyst
."
4 """'''''''''''''''''''''''''''''''''''''''''',.
!!;,t;i;;;i·;;~;:;?~~;.·~~~~~~?~~;:::~t··g::.····.·
The mechanism of contact catalysis may vary in details, depending on the nature of the reactants. Consider the example of hydrogenation of ethene in presence of nickel. In this casse, ethene adds hydrogen in the presence of nickel as a catalyst to yield ethane.
H _N_i.....:(c_a_ta....;ly'-st-'-)~)
H
Ethene gas
H
I I H-C--C-H I I H
Ethane (gas)
The catalyst functions according to the following steps. Step 1. Adsorption of hydrogen molecules. Hydrogen molecules are adsorbed on the nickel surface due to the residual valence bonds of the nickel atoms. Step 2. H-H bonds are broken. The H-H bond is smaller (0.74 A) than Ni-Ni bond. Therefore, the H-H bond of the adsorbed hydrogen molecule is stretched and weakened. The weakened bond breaks, separating the hydrogen atoms. The separated hydrogen atoms are held to the nickel surface by chemical bonds. Step 3. Formation of the activated complex. The chemisorbed hydrogen atoms then attach to ethene molecule by partial chemical bonds. Therefore, unstable activated complex is formed. Step 4. Decomposition of the activated complex and desorption of ethane molecule. The unstable activated complex decomposes to form ethane molecule. The freed catalyst surface is again available for further action.
264
PHYSICAL CHEMISTRY-I
il'::"""'''''''''~'''''''''''''''''''''''''''''''''''''''''M""""""""'·":'~
:1 H,,--
Step 1
/H
H,,--
j H/r:I"--H )~ :'.!
Step 3
Step
2;'--H
I
I
HH/C--C"--HH
@--@
I
Step 4
\
:::.:\::\U\ '~i~: .~~ ;.~~~~~ ~~.~~~ ~;~~~~~~~~~~. ~~. j/:::;:i{::::U/!::::·:. \u:=::nt} .......... ...~~~.e~e..o.~. ~i.~~.~~.~~r:r~c~'....................I)/\/(I(U:: The adsorption theory explains the catalytic activity as follows: (1) Metals in a state of fine sub-division or colloidal form are rich in free valence bonds and hence they are more efficient catalysts than the metal in lumps. (2) A promoter increases the valence bonds on the catalyst surface by changing the crystal lattice and thereby increasing the active centres. (3) Catalytic poisoning occurs because the so-called poison blocks the free valence bonds on its surface by preferential adsorption or by chemical combination.
[II] Industrial Applications of Catalysts Tht" presence of a catalyst is very useful in many industrially important reactions, which are either very slow or take place at a very high temperature. :.', .......:.<::.. ,,-:.:.:::- :-:-:: ',':::.: '';'' :;:>'
','
...
,',::-::-:,:::::::::,:':':-:::':::':,:',.;::.:::,:,:;:;::::';::'';;';''':-:-':;:::';''::::''\:{
."" .
265
CHEMICAL KINETICS & CATALYSIS
Hence. to decrease the cost of production it is essential to make use of a suitable catalyst. A few important examples of heterogeneous catalytic reaction of industrial applications are given as follows: Catalyst and other favourable conditions
Reaction 1.
Haber's process for the manufacture of Finely divided Fe + molybdenum (as ammonia promoter) 200 atm. pressure and N2 + 3H2 ~ 2NH3 temperature.400°-450°C.
2.
The manufacture of chlorine by Deacon's Cupric chloride + excess of air at a process temperature of 500°C. 4HCI + 02 .~ 2H20 + 2Cl2
3.
Ostwald's process for the manufacture of Platinised asbestos + excess of air (as HN0 3 promoter) and temp. 300°C. 4NH3 + 502 ~ 4NO + 6H20 2NO + 02 4N02 + 2H20 + 02
~
~
2N02 4HN03
4.
Manufacture of hydrogen by Bosch's Ferric oxide + Cr203 (as a promoter) at a process temp. of 400°-600°C. (CO + Hz) + H20 ~ C02 + H20
5.
Manufacture of methyl alcohol from ZnO + Cr203 (as a promoter). 200 atms. water gas pressure and temp of 450°C. CO+H2 +H2 ~ CH30H
Water gas
Water gas
6.
Chamber process for the manufacture of H2S04 ZS02 + 02 + [NO] ~ 2S03 + [NO] S03 + H20 ~ H2S04
7.
Acetic acid from acetaldehyde 2CH3CHO + 02 ~ 2CH3COOH
Nitric oxide
Problem 4. What is biochemical or enzYme catalysis? Discuss the characteristics and some examples of enzYme catalysis. Discuss the kinetics of enZJme catalysis. Or Explain enZJme catalysed reactions. (Meerut 2006, 2004)
Biochemical or Enzyme Catalysis Enzymes are complex nitrogeneous organic compounds. They are produced in the living cells of plants and animals. On dissolving in water they form the colloidal solution. hence they behave as very active catalysts in certain biochemical reactions and are known as biochemical catalysts and the phenomenon itself is known as biochemical catalysis.
[I] Characteristics of Enzyme Catalysts (1) Enzymes form a colloidal solution in water and hence they are very
active catalysts.
1
266
PHYSICAL CHEMISTRY-I
(2) Like inorganic catalysts they cannot disturb the final state of equilibrium of a reversible reaction. (3) They are highly specific in nature, i.e., one catalyst cannot catalyse more than one reaction. (4) They are highly specific to temperature. The optimum temperature of their activity is 35°C to 40°C. They are deactivated at 70°C. (5) Their activity is increased in the presence of certain substances, known as co-enzymes. (6) A small quantity of enzyme catalyst is sufficient for a large change. (7) They are destroyed by U.v. rays. (8) Their efficiency is decreased in presence of electrolytes.
[II] Examples of Enzyme Catalysis The following are some examples of biochemical or enzyme catalysis. (1) Manufacture of ethyl alcohol from cane sugar Glucose
Fructose
Zymase
) 2C 2H50H + 2C0 2 C6H120 6 (2) Manufacture of acetic acid from ethyl alcohol
C2H50H + O2
Mycoderma aceti
)
C
H3
CO H 0 0 + H2
(3) Conversion of starch into maltose 2(C~1005)n
+ nH20
Diastase
--~
nC 12H220 II
Starch
Maltose
C 12H220 U + H20
Maltase ---~)
2C6H l2 0 6
Maltose
Glucose
(4) In the estimation of urea Urease enzyme completely converts urea into ammonium carbonate. /,NH2
/'
O=C",-
+ 2H20
Urease
~
(NH4hC0 3
"'- NH2
(5) In digestive tract (a) In stomach, pepsin enzyme converts proteins into peptides, whereas in intestines, pancreas trypsin converts proteins into amino acids by hydrolysis. These amino acids are absorbed by blood and are used in the building of tissues. (b) The enzyme ptyalin present in human saliva converts starch into glucose. Ptyalin
) nC 6H l2 0 6 Gluco;e
J
267
CHEMICAL KINETICS & CATALYSIS
[III] Kinetics of Enzyme Catalysis or Michaelis-Menten Equation A reactant in an enzyme catalysed reaction is known as substrate. According to the mechanism of enzyme catalysis, the enzyme combines with the substrate to form a complex, as suggested by Henri (1903). He also suggested that this complex remains in equilibrium with the enzyme and the substrate. Later on in 1925, Briggs and Haldane showed that a steady state treatment could be easily applied to the kinetics of enzymes. Some photochemical reactions and some enzymic reactions are reactions of the zero order. With S representing substrate, E the enzyme, ES an enzyme-substrate complex and P the products, the mechanism of the enzyme catalysed reaction is presumed to be adequately represented by kj
k
E+S~ ES~E+P k2
where kl' k2' k3 are the rate constants for the respective reactions. The rate of formation of the complex ES is, evidently given by the following equation,
d[!S] = 0 = kl [E] [S] - k2 [ES] - k3 [ES] ... (1)
where [E], [S] and [ES] represent molar concentrations of the free enzyme, substrate and the complex, i.e., bound or reacted enzyme, respectively. Now [E] cannot be experimentally measured. The equilibruim between the free and bound enzyme is given by the enzyme conservation equation,
i.e., [E]o = [E] + [ES] where [E]o refers to the total enzyme concentration. So, [E] = [E]o - [ES] On substituting the value of [E] in equation (1), we get d
~S] = kl
{[E]o - [ES]} [S] - (k'l + k3) [ES]
=0
... (2)
As the reaction proceeds, the intermediate complex formed in accordance with the suggested mechanism, decomposes instantaneously according to the same mechanism. On applying the steady state principle, we have
d [ES] = 0 dt
268
PHYSICAL CHEMISTRY-I
At the stationary state, equation (2) may be written as, k\ {[E]o - [ES]} [S] = (k2 + k 3) [ES]
or
k\ [E]o [S] = {(k2 + k3) + k\ [S]} [ES]
[ES] =
=
k\ [E]o [S] (k2 + k3) + kI [S]
[E]o [S] k2 + k3 kI
+ [S]
The rate of formation of the product, P, i.e., r is represented by the equation, r = d [P] dt
=k3 [ES]
... (3)
Substituting the value of [ES] in equation (3), we get d [P] k3 [E]o [S] r=--= dt k2 + k3 k;-+[S] . (k2 + k3) The quantIty k is known as Michaelis constant and may be denoted by Km. Therefok, d [P] _ k3 [E]o [S] ... (4) dt - Km + [S]
Equation (4) is known as Michaelis-Menten equation. Further simplification of equation (4) can be made. If it is assumed that all the enzyme has reacted with the substrate at high concentrations the reaction will be going on at maximum rate. No free enzyme will remain so that [E]o = [ES]. So, from equation (3), we get rrnax = Vrnax = k3 [E]o where "'max reters to maximum rate, using the notation of enzymology. So, Michaelis-Menten equation can also be written as, Vmax [S] r= Km + [S] If r = Vma J2, i.e., if the rate of formation of product is equal to half of the maximum rate at which the reaction proceeds at high concentration of substrate, then Km = [S] Thus, Michaelis COli stant is equal to that concentration of substrate, S at which the rate of formation of the product is half the maximum rate obtained at a high cOllcentration of substrate.
CHEMICAL KINETICS & CATALYSIS
269
From equation (4), we can draw the following conclusions: (i) If [S] is very small as compared to Km, the factor Km/[S] will be very large as compared to unity and so the rate of formation of P, i.e., d [P]Idt will be directly proportional to [8]. In other words, the reaction will be of the first order with respect to S (Fig. 13). . 1 Maxlmum rate, 2 Vmax
Vmax [SJIKm (first order)
Substrate concentration (S) _
(ii)
If [S] is very large as compared to Km, the factor Km/[S] will be
negligibly small as compared to unity and so the rate of formation of P, i.e., d [P]/dt will be independent of the concentration [S]. In other words, the reaction will be of zero order with respect to S (Fig. 13). (iii) If [S] is very small or very large, the reaction remains of the first order with respect to the total concentration, [E]o of the enzyme.
Problems 5. mite a short note on acid-base catalysis.
(Meerut. 2007)
As a result of the work of Bronsted, Lowry and others, it has become evident in recent years that a variety of atomic, molecular and ionic species are capable of catalysing chemical reactions. For some processes, hydrogen ions appear to be effective, while other reactions are catalysed by hydroxyl ions, cations of weak bases, anions of weak acids, undissociated molecules of acids and bases etc. This general acid catalysis involves cases where all acids act as catalysts, while general base catalysis refers to processes catalysed by bases of all kinds. In some cases, both acids and bases are effective, while in others a particular species is effective. General acid-base catalysis is illustrated by mutarotation of glucose, which is catalysed by hydrogen, hydroxyl and complex ions, as well as by the acids and bases, though the most effective catalyst is the hydrogen ion.
270
PHYSICAL CHEMISTRY-I
Mechanism of acid-base catalysis: It is accepted that acid-base catalysis involves a reversible acid-base reaction between the substrate and catalyst. This is in agreement with the protonic concept of acids and bases, since acid catalysis depends on the tendency of the acid to lose a proton, while base catalysis depends upon the tendency of the base to gain a proton. The mechanism of reaction involving Wand OIr ion catalysis may be expressed as follows. by taking the example of hydrolysis of esters. (a) With W ions as catalyst ORO
II I
n+
R
0
I I+ HO '
R
I+ I
CH,-C-O + n ---> CH,-C-O -H -'---> CH,-C/\H
H
H
(b) With OS- ions as catalyst ORO
II I
R
0
I I I
R
I I I
CH3-C-O + OIr ~ CH3-C-O+ ~ CH3-C-O+-H H-O H-O CH3COOH + ROH The rate of reaction is given by dx -d = kH+ C H+ Cester + kOH- C OH Cester + kH,o CH,o Cester t where kH+ and koH- are the catalytic coefficients of hydrogen and hydroxyl ions.
~
MULTIPLE CHOICE QUESTIONS 1.
The dimension of first order rate constant is : (i) time-1 (ii) time (iii) time x conc
2.
(iv)time-l x conc- l
The dimension of second order rate constant is : (i) times x conc (ii) time- l x conc- l (iii) time x conc-1 (iv) time-1 x conc-2
271
CHEMICAL KINETICS & CATALYSIS
3.
4.
5.
hydrolysis of methyl acetate by acid is of : Zero (ii) First order Second order (iv) Third order hydrolysis of ethyl acetate by NaOH is of : Zero (ii) First order Second order (iv) Third order reaction rate of a reaction 2A ~ 3B is given by : (a) _ d [A] (b) _! d [A] ~ 2 ~
The (i) (iii) The (i) (iii) The
(c)
6.
+ ~ d ~~]
(d)
+ d ~]
For a reaction A + B ~ Products, the reaction rate is given by, : dx = k[A] [Br dt The order of reaction with respect to A and B are : (ii) 1, 1 (iii)O. 1 (i) 1,0
7.
8.
9.
(iv) 2, 0
The half life period of a first order reaction is 20 min. The time required for the concentration of the reactant to change from 0.4 M to O.I,M is : (ii) 40 min (iii) 60 min (iv) 80 min (i) 20 min The ratio of the time required for 75% of a first on':,,! rr,ction to complete to that required for 50% of the reaction is : (i) 4: 3 (ii) I : 2 (iii)2: I ' (iv)3: 2 The time required to decompose half of the reaction for a n'h order reaction is also doubled. The order of reaction is : (ii) 1 (iii) 2 (iv) 3 (i) 0
10. The minimum energy required for reactant molecules to enter into chemical
11.
12.
13.
14.
reaction is known as : (i) Kinetic energy (ii) Potential energy (iii) Activation energy (iv) Threshold energy A first order reaction is 75% completed in 32 minutes. For 50% completion, it will take: : (i) 4 min (ii) 8 min (iv) 32 min (iii) 16 min The elementary step of the reaction 2Na + Cl 2 ~ 2NaCI is found to follow third order reaction kinetics. The molecularity of the reaction is : (i) 0 (ii) 1 (iii) 2 (iv) 3 In a reaction A ~ B, the reaction rate is doubled on increasing the concentrations of reactants four times. The order of reaction is : (i) 0, (ii) 1 (iii) 2 (iv) 3 On plotting loglok against Iff, the slope of the straight line is given by::
.
(1)
Ea R
"
(n)
E (iii) 2.30; R
Ea -R
272 15.
16.
17.
18.
19.
20.
21.
PHYSICAL CHEMISTRY-I
If the plot of 10glO [A] against time is a straight line with a negative slope, the order of reaction is : (ii) 1 (iii) 2 (iv) 3 (i) 0 In the hydrogenation of oils the catalyst used is : (i) Iron (ii) Platinum (iii) Nickel (iv) Molybdenum The effect of a catalyst in a chemical reaction is to change the: (i) Activation energy (ii) Equilibrium concentration (iv) Final product (iii) Heat of reaction The catalyst used in the contact process of sulphuric acid manufacture is : (i) Oxides of nitrogen (ii) Nickel (iii) Vanadium pentoxide (iv) Manganese dioxide Which of the following is used as a contact catalyst? (i) Boron (ii) Germanium (iii) Nickel (iv) U~anium Which of the following statements is universally correct? (i) A catalyst remains unchanged at the end of the reaction (ii) A catalyst physically changes at the end of the reaction (iii) A catalyst takes part in the chemical reaction (iv) A catalyst can induce chemical reaction The catalyst used for the oxidation of ammonia to nitric acid is : (i) Cupric chloride (ii) Iron oxide (iii) Platinum (iv) Manganese dioxide
22. A substance that regards the rate of chemical reaction in the presence of a catalyst
23.
24.
25.
26.
27.
is called : (i) An inhibitor (ii) A positive catalyst (iii) An auto-catalyst (iv)A promoter A catalyst poison is essentially: (i) A homogeneous catalyst (ii) A heterogeneous catalyst (iii) An inhibitor (iv) An auto-catalyst Catalyst poisons (temporary poisoning) act by : (i) Chemically combining with the catalyst (ii) Getting adsorbed on the active centres on the catalyst surface (iii) Chemical combination with anyone of the reactants (iv) Coagulating the catalyst Which of the following types of the metal make the most effiCIent catalyst? (i) Transition metals (ii) Alkali metals (iii) Alkaline earth metals (iv) Coloured metals Which one of the following statements is incorrect? (i) Presence of a catalyst does not alter the equilibrium concentration in a reversible reaction (ii) Change of temperature alter the rate of catalysed reaction in the same proportion as of the reaction without catalyst (iii) Homogeneous catalysis depends upon the nature and extent of the surface (iv) Change of a catalyst may change the nature of the reaction Enzymes are: (i) Micro-organisms (ii) Proteins (iii) Inorganic compounds (iv) Moulds
CHEMICAL KINETICS & CATALYSIS
28.
'
..
1;
1
273
A pho~chemical reaction is : (i) Catalysed by light (ii) Initiated by light (iii) Accompanied with emission of light (iv) Accompained with absorption of light
Platinised asbestors is used as catalyst in the manufacture of H2S04, It is an example of : (i) Homogeneous catalyst (ii) Auto-catalyst (iii) Heterogeneous catalyst (iv) Induced catalyst 30. The enzyme ptyalin used for digestion of food is present in : (ii) Blood (i) Saliva (iii) Intestine (iv) Adrenal gland 31. A catalyst is a substance with : (i) Increase the equilibrium concentration of the product (ii) Change the equilibrium constant of the reaction . (iii) Shortens th time to each equilibrium (iv) Supplies energy of the reaction 32. An example of auto-catalytic reaction is : (i) The decomposition of nitroglycerine (ii) Thermal decomposition of KCl0 3• Mn02 mixture 29.
~
(iii) Break down of 14C6 33.
34.
3? 36.
37.
38.
(iv) Hydrogenation of vegetable oil using nickel catalyst The efficiency of an enzyme in catalysing a reaction is due to its capacity: (i) To form a strong enzyme-substrate complex (ii) To decrease the bond energies in the substrate molecule (iii) To change the shape of the substrate molecule (iv) To lower the activation energy of the reaction A catalyst: (i) Increases the energy change in a reaction (ll) Increases the free energy change in a reaction (iii) Neither decreases nor increases the free energy change in a reaction (iv) Can increase or decrease the free energy change in a reaction but it depends on the catalyst Starch is converted into disaccharide in the presence of : (i) Diastase (ii) Maltase (iii) Lactase (iv)Zymase Glucose or fructose is converted into C2H50H in the presence of : (i) Invertase (ii) Maltase (iii) Zymase (iv) Diastase A ~talyst increases the rate of a chemical reaction by : (i) Increasing the activation energy (ii) Decreasing the activation energy (iii) Reacting with reactants (iv) Reacting with products Which of the following statement is correct? (i) Enzymes are in colloidal state (ii) Enzymes are catalysts (iii) Enzymes catalyse any reaction (iv) Urease in an enzyme
274 39.
40.
41.
42.
43.
44.
45.
46.
47.
48.
49.
PHYSICAL CHEMISTRY-I
A catalyst is used: (i) To vaporise the compound (ii) To kill the enzyme (iv) To balance the reaction (iii) To alter the velocity of reaction Enzymes Il.re : (i) Substances made by chemists to activate washing powder (ii) Very active vegetable catalysts (iii) Catalysts found in organisms (iv) Synthetic catalysts Which of the following catalysts is used for preparing toluene by reacting benzene with CH 3Cl? (i) Ni (ii) Anhydrous AlCl 3 (iii) Pd (iv) Pt The rusting of iron is catalysed by which of the following? (i) Fe (ii) O2 (iii) Zn (iv) H+ In which of the commerical process, a catalyst is not used? (i) Haber's process (ii) Deacon's process (iii) Solvay process (iv) Lead chamber process Which of the following statements is correct for a catalyst? (ii) It alters the rate of the reaction (i) It supplies energy to the system (iii) It alters the equilibrium constant (iv) It is used up in the reaction Organic catalysts differ from inorganic catalysts in : (i) By acting at very high temperature (ii) By acting at low temperature (iii) Being used up (iv) Being proteneous in nature Which one of the following statements regarding catalyst is not true? (i) A catalyst can initiate a reaction (ii) A catalyst remains unchanged at the end of the reaction (iii) A catalyst does not alter the equilibrium in a reversible reaction (iv) Catalysts are sometimes very specific in terms of reactions Which statement is incorrect for heterogeneous catalysis? (i) Catalyst is absorbed on the surface (ii) Active centres are found on the surface of catalyst (iii) Catalyst increases the energy of activation (iv) None of these Which of the following is used as a contact catalyst? Oi) Nickel (i) Boron (iii) Germanium (iv) Uranium Which one of the following statements is incorrect in the case of heterogeneous catalyst? (i) The catalyst lowers the energy of activation (ii) The catalyst actually forms a compound with the reactant (iii) The surface of the catalyst plays a very important role (iv) There is no change in the energy of activation
Fill in the Blanks 1. 2.
A first order reaction is 15% complete in 20 min. It will take ............ min to be 60% complete. The rate ofreaction is nearly doubled on increasing the temperature by .......... .
275
CHEMICAL KINETICS & CATALYSIS
3.
Order of reaction for the hydrolysis of ethyl acetate by HCI is ............ .
4.
For a reaction A ~Products, the rate law is reaction is ...... ... .
~ = k [A]/2.
The order of
t
5.
The order of reaction of decomposition of H20 2 is ............... .
6.
14.
If the rate constant, k = .! . -x()' the order of reaction is ............. . t a a-x If the rate constant has tl-te UnIt sec-\ the order of reaction is ............. . The equation k = Ae- EIRT is known as ................... equation. In equation k = Ae- EIRT, A is known as ............... . The hydrolysis of ethyl acette in acidic medium is ............. order reaction. A substance which changes the rate of reaction is known as ............... . The substance which retards the reaction rate is known as ................. cataryst. The ................. of a catalyst increases when it is finely divided. The substance which increases the efficiency of a catalyst is known as ............ .
15.
In the conversion of urea into ammonium carbonate ................. acts as a catalyst.
16.
2502 + O2 ~ 2S0 3 is an example of .............. catalysis.
17.
CH 3COOCH 3 + H20~ CH 3COOH + C 2H50H is an example of .......... . catalysis.
7. 8. 9.
10. 11.
12. 13.
18. 19.
20.
He!
~ 2NH , Mo acts as ............. . In N2 + 3H2 ~ 3
A catalyst poison is essentially a ............... . The presence of a catalyst ............ the activation energy of the reaction.
True or False State whether the following statements are true (T) or false (F)?
.
,,
1. 2. 3. 4. 5.
6. 7. 8.
9.
10. 11.
12. 13.
14.
The value of temperature coefficient is nearly 10. All radioactive emanations are of first order. The inversion of cane sugar by HCI is of second order. The reaction rate is proportonal to the surface area of reactant. The half life period of a first order reaction is O.~93. Order of reaction and molecularity are similar. Order of reaction can even be 4 or more. Most of the reactions are of first and second order. hv
The reaction H2 + Cl 2 ~ 2HCI is of zero order. The rate of zero order reaction depends on the concentration of the reactant. A catalyst is a substance which can only increase the reaction rate. When one of the products formed in the reaction itself acts as a catalyst, the phenomenon is called auto-catalysis. A catalyst is specific in action. A catalyst can change the position of equilibrium.
276 15. 16. 17. 18. 19. 20. 21. 22.
PHYSICAL CHEMISTRY-I
A catalyst remains unchanged in mass and chemical composition at the end of a reaction. A large quantity of catalyst is required to bring about a reaction. The substance which increases the activity of a catalyst is called an activator. A promoter decreases the peaks and cracks on the catalyst surface. Enzyme ptyalin present in h)lman saliva changes starch into glucose. The presence of a catalyst increases the activation energy of the reaction. In homogeneous catalysis, the intermediate compound is formed at lower activation energy. A catalyst cannot be recovered unchanged chemically at the end of the reaction.
ANSWERS 1. (a) ,2. (b) 3. (b) 4. (c) 5. (b) 6. (a) 7. (b) 8. (c) 9. (a) 10. (d) 11. (c) 12. (d) llWK~~~~~nWm~~~·Wn~nWn~u~ 25. (a) 26. (c) 27. (b) 28. (b) 29. (c) 30. (a) 31. (c) 32. (a) 33. (d) 34. (d) 35. (a) 36. (c)
37. (b) 38. (c) 39. (c) 40. (c) 41. (b) 42. (d) 43. (c) 44. (b) 45. (d) 46. (a) 47. (c) 48 (d) 49. (d)
Fill in the Blanks 1. 5. 9. 13. 17.
112.8 min one Frequency factor efficiency autolhomogeneous
2. 6. 10. 14. 18.
W'C two First promoter promoter
3. 7. 11. 15. 19.
(F) one one Catalyst Urease inhibitor
2. 6. 10. 14. 18. 22.
(T), (F),
3. (F), 7. (F), 11. (F), 15. (T), 19. en,
4. 8. 12. 16. 20.
112 Arrhenius negative homogeneous decreases
True or False 1. 5. 9. 13. 17. 21.
(F), (T), (T), (T), (T), (T),
(F); (F), (F), (F)
4. 8. 12. 16. 20.
en, (T),
(T), (F), (F),
DOD
