Physical Chemistry, Volume 1

P~9a~~-------------------------- PHYSICAL CHEMISTRY Dr. J.N. Gurtu Dr. H.C. Khera M.Sc., Ph.D. M.Sc., Ph.D. Forme...

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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: [email protected] Regd. Office: New Market, Begum Bridge, Meerut-250 001

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


[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


[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].


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) 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


(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 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


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


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


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


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


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


= 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


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


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


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


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


= 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


= 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


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


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'


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


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


33

(b) Trigonometric Series 3

5

' x x (1) smx=x-'3!+5"!......

00,

(-oo<x-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


(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


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


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


,.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.;.: 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)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


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


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


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


(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

123


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,

124


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


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


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


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


(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


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


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


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


(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


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


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


(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


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
(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


[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


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


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


(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


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


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


[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.

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


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


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


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


(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


(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

,

+

--

--

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


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


(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;':::::::::;;:~~:::::::+

• ~=~
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


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


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


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.


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


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


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


(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


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


[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]


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


(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


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


(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


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


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


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


[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


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


[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


=-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


[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


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.

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


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


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


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


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


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


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


=+ ~ 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


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


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


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


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


_ 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


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


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


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


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


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


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


[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


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


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


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


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


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


(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


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


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


(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


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


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


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


(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


[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


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.


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


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


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.


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


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.


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


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.


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

Physical Chemistry, Volume 1

Physical Chemistry, Volume 1

Physical Chemistry