TOPIC : EQUILIBRIUM

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CHEMISTRY: CHEMICAL EQUILIBRIUM

COURSE : B.Sc I

PAPER : I

TOPIC : EQUILIBRIUM

prepared by : Dr. KAMAL KISHORE SINGH

Chief coordinator

Department of chemistry, Nalanda

Open university , patna , Bihar.

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CHEMICAL EQUILIBRIUM

Chemical reactions can be divided into two categories

Reversible reaction

Those reactions in which products can react to form reactants back are called reversible

reactions. They proceed in both the directions and do not reach completion e.g.

3 Fe(s) + 4 H2O(g) Fe3O4(s) + 4H2(g)

N2(g) + 3H2(g) 2 NH3(g)

2 SO2(g) + O2(g) 2 SO3

Irreversible reactions

Those reactions which proceed only in one direction and it is not easily possible to

convert the products into reactants are known as Irreversible reactions

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They reach completion e.g.

BaCl2 + Na2SO4 BaSO4 + 2NaCl

AgNO3 + NaCl AgCl + NaNO3

Equilibrium

The stage at which the rate of forward reaction becomes equal to the rate of backward

reaction is called equilibrium. It is the condition in which the properties of system do not

change with time.

Physical Equilibrium

When there is equilibrium between two physical states of same substances it is called

physical equilibrium e.g.

H2O(s) H2O (1)

NH4Cl(s) + aq NH4Cl(aq)

Chemical Equilibrium

When there is equilibrium between two processes involving change of chemical species,

it is called chemical equilibrium.

Characteristics of Chemical Equilibrium

1. When chemical equilibrium is attained, there is no further change in pressure,

concentration, density or colour at a fixed temperature.

2. It is dynamic in nature i.e. reaction does not stop after the equilibrium is reached rather

it appears to stop because there is no further change in concentration, pressure or density

because rate of forward reaction is equal to rate of backward reaction.

3. It can be approached from either direction e.g.

H2 (g) + I2 (g) 2 HI (g)

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2HI (g) H2 (g) + I2 (g)

4. A catalyst can hasten the approach of equilibrium but does not alter the equilibrium.

Forward reaction

The reaction which proceeds from reactants to products is called forward reaction. The

rate of forward reaction is denoted by rf.

Backward reaction

The reaction which proceeds from products to reactants is called backward reaction. The

rate of backward reaction is denoted by rb.

Active Mass

The concentration of a reactant in moles per litre or partial pressure of a reacting gas in

atmosphere is known as its active mass. It is expressed by enclosing the symbol or

formula of the substance in square brackets e.g. [A]. These days active mass is replaced

by molar concentration.

Law of mass action

The rate at which a substance reacts at any instant is proportional to its active mass

(molar concentration) and the rate of chemical reaction is proportional to product of

active mass (molar concentration) of reactants e.g.

A B + C

Rate [A]

or, Rate = k [A] where ‘k’ is rate constant or velocity constant

A + B C + D

Rate [A] [B]

Rate = k[A] [B]

2A + B C + D

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Rate [A]2 [B]

Rate = k [A]2 [B]

Thus, the concentration term is raised to the power equal to Stoichiometeric coefficient in

the chemical equation.

Equilibrium Constant

It is defined as ratio of product of molar concentration of products to the product of molar

concentration of reactants raised to the power equal to their Stoichiometric coefficient in

a chemical reaction at particular temperature. The equilibrium constant measures the

extent to which a reaction takes place before equilibrium is reached.

If equilibrium constant (K) is large, the equilibrium mixture will have high proportion of

products. If it is small, the equilibrium mixture will contain high proportion of reactants.

aA + bB cC + dD

a b c df bk [A] [B] r [C] [D]

a b c df f b br k [A] [B] r k [C] [D]

Where kf and kb are rate constants for forward and backward reactions respectively.

At equilibrium fr = br

a b c df bk [A] [B] k [C] [D]

c d

fC a b

b

k [C] [D]K

k [A] [B]

Where Kc is equilibrium constant in terms of molar concentration of reactants and

products.

In case of gaseous reactions, it is more convenient to use partial pressures instead of

molar concentrations. The equilibrium constant in terms of partial pressures of reactants

and products is called Kp.

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aA bB cC dD

c d

p a b

(pC) (pD)K

(pA) (pB) Reaction coefficient

p q

c m m

[C] [D]Q

[A] [B]

Note 1 : If Q < K it means equilibrium has not reached the and system is moving

towards right.

At equilibrium p q

c m m

(pC) (pD)Q

(pA) (pB)

Note 2 : Q > K equilibrium has not reached. The reaction must proceed backwards.

pv = nRT

n

p RT CRTv

where ‘c’ is molar concentration in mol L-1

c d c d c d

(c d) (a b)p ca b a b a b

{[C]RT} {[D]RT} [C] [D] (RT)K K (RT)

{[A]RT} {[B]RT} [A] [B] (RT)

np cK K (RT) where n = (c + d) − (a + b)

Where n is no. of moles of gaseous products − no. of moles of gaseous reactants.

n may be zero, positive or negative e.g.

If n = 0, Kp and Kc are equal and have no units. In other case i.e. for n = +ve or −ve,

Kp is expressed as (Unit of pressure n) e.g. (atm)n, Kc is expressed as (Unit of

concentration)n e.g. (mol L−1)n

Relationship between degree of dissociation and Vapour density

In all cases, where number of moles are being changed from reactants to products, so

Vapour density will also change. If we consider, say dissociation of PCl5, the initial and

equilibrium states are represented as follows

5 3 2PCl (g) PCl (g) Cl (g)

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Initial 1 0 0

Moles at Equilibrium 1 − x x x

Total number of moles = 1 − x + x + x = 1 + x

Here we start with 1 mole of PCl5 and ‘x’ is degree of dissociation. If 1 mole at

temperature T and pressure P occupies a volume ‘v’ litres, the volume of resulting

mixture will be (1 + x) litres, where (1 + x) represents the number of moles at

equilibrium.

M

DV

1

Dvolume

(since mass is constant)

If ‘D’ is density of gas before dissociation and ‘d’ is density of equilibrium mixture then

in case of dissociation of PCl5

1

DV

....(i)

1

d(1 x)V

....(ii)

From (i) and (ii) D (1 x)V

d V

D

1 xd

D D d

x 1d d

D d

xd

For a molecule of gas which on dissociation gives ‘n’ molecules of products:

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D d

xd(n 1)

This relationship can be used for calculation of degree of dissociation from “Vapor

Density Measurement”

Illustration 1

pK for the reaction 5 3 2PCl PCl Cl at 250°C is 0.82. Calculate the degree of

dissociation at given temperature under a total pressure of 5 atm. What will be the degree

of dissociation if the equilibrium pressure is 10 atm. at same temperature.

Solution

Let 1 mole of 5PCl be taken initially. If `x’ moles of 5PCl dissociate at

equilibrium, its degree of dissociation = x.

Moles 5PCl 3PCl 2Cl

Initially 1 0 0

At equilibrium 1 – x x X

Total moles 1 – x + x + x = 1 + x

P = 5 atm. pK 0.82

PCl PCl Cl5 3 2

1– x xP xPP , P , P

1 x 1 x 1 x

PCl Cl5 2

p

PCl5

(P )(P )K

(P )

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2

p 2

xK P 0.82

1– x

2

2

x (5)0.82

1– x

2 0.82x

5.82

x = 0.375 (or 37.5%)

Now the new pressure P = 10 atm.

Let y be the new degree fo dissociation. As the temperature is same (250°C), the

value of pK will remain same.

2 2

p 2 2

(y) P (y) 10K 0.82

1– (y) 1– (y)

0.82

y10.82

or y = 0.275 (or 27.5%)

Note : By increasing pressure, degree of dissociation has decreased, as

depicted in illustration 1

xP

i.e., the system shifts to reverse direction. We can also

compare the result by applying Le Chatelier’s principle.

Illustration 2

Vapour density of the equilibrium mixture of NO2 and N2O4 is found to be 40 for the

equilibrium

N2O4 2NO2

Calculate (i) abnormal molecular weight

ii) degree of dissociation

iii) percentage of NO2 in the mixture

Solution

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i) N2O4 2NO2

Observed value of vapour density (d) = 40

Hence, abnormal molecular weight = 40 × 2 = 80

ii) D × 2 = theoretical molecular weight = 92

92

D 462

D d 46 40

x 0.15d 40

iii) N2O4 2NO2

Initial mol 1 0

At equilibrium (1 – x) 2x

0.85 0.30

Total moles at equilibrium = (1 + x) = 1 + 0.15 = 1.15

Percentage of NO2 = 2x 0.30

100 1001 x 1.15

= 26.08%

Heterogeneous equilibrium

When reactants and products are in different physical states, it is called heterogeneous

equilibrium e.g.

CaCO3 (s) CaO(s) + CO2(g) Kp = PCO2

3Fe (s) + 4H2O(g) Fe3O4(s) + 4H2(g) 4

2P 4

2

(pH )K

(pH O)

NH4 HS(s) NH3(g) + H2S(g) Kp = (pH2S) (pNH3)

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The molar conc. of solid and liquid reactant or product in equilibrium involving gases is

taken as unity. Kp is written for heterogeneous reactions and analysis is done in terms of

partial pressure.

Le-Chartelier’s Principle

When a system in equilibrium is subjected to change in temperature, pressure or

concentration, the equilibrium shifts in that direction so as to counterbalance the effect of

the change.

This principle is extremely useful in predicting the effect of changes in temperature,

pressure or concentration upon a system at equilibrium.

Effect of temperature

In case of exothermic reaction, increase in temperature favours backward reaction where

as decrease in temperature favours forward reaction e.g.

N2(g) + 3H2(g) 2NH3 (g) + heat

Increase in temperature, generally favours an increase in rate but extent of increase will

be different for backward and forward reactions owing to the differences in their energies

of activation. So the numerical value of ‘k’ (equilibrium constant) will change with

change in temperature.

The variation is expressed by the relation:

2 2 1

1 1 2

K T TH2.303log

K R T T

Where K1 and K2 are equilibrium constant at temperature T1 and T2 respectively

H = Standard enthalpy of reaction

Note 1 : For endothermic reaction

H = +ve and if T2 > T1 then 2

1

Klog ve

K

log K2 > log K1 K2 > K1

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i.e. K increases with increase in temperature for endothermic reactions.

Note 2 : For exothermic reactions

H = − ve and T2 > T1 2

1

Klog ve

K

log K2 < log K1 K2 < K1

i.e. K decreases with increase in temperature for exothermic reactions.

Note 3 : The new equilibrium state has a new value of equilibrium constant ‘k’ on

changing the temperature

Effect of pressure

If there is an overall increase in number of moles of products from reactants, low pressure

is favourable for forward reactions e.g. in dissociation of PCl5(g) low pressure is

favourable

PCl5 PCl3 (g) + Cl2 (g) n = 1

If there is decrease in number of moles from reactants to products, high pressure is

favourable for forward reaction e.g. high pressure is favourable for manufacture of NH3.

N2 (g) + 3H2(g) 2 NH3(g) n = −2

If the number of moles of reactants and products are equal is no effect of pressure e.g.

N2(g) + O2 (g) 2NO (g) n = 0

Effect of Concentration

If at equilibrium some quantity of reactant or product is added, the equilibrium will shift

in the direction in which added substance is consumed e.g. in N2(g) + 3H2 2NH3(g).

If at equilibrium, we increase the concentration of reactant (N2 or H2 or both), the rate of

forward reaction will increase. But if we increase the concentration of NH3, the reaction

proceeds towards backward direction. If we increase the conc. of NH3, the reaction

proceeds towards backward direction. If we decrease the conc. of NH3, rate of forward

reaction will increase. Therefore, product should be removed as soon as it is formed.

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Effect of Catalyst

Catalyst increases the rate of forward as well as backward reaction equally but

equilibrium is reached faster in presence of catalyst. It does not effect value of ‘k’.

Effect of addition of inert gas

At constant volume, the various concentration terms are not changed. Hence the

equilibrium will not be affected. However at constant pressure, the addition of inert gas

will increase the volume of the system and thus decrease the concentration and partial

pressures of the species involved. Hence the equilibrium will shift in the direction in

which concentration of species increase i.e. degree of dissociation should increase i.e.

formation of products should be favoured if n = +ve. If n = −ve, addition of inert gas

will favour backward reaction.

Note 1 : If K1 is equilibrium constant for

A + B C + D then 1

[C] [D]K

[A] [B]

K2 for C + D A + B is equal to 1

1

K

2

1

[A][B] 1K

[C] [D] K

K3 for nA + nB nC + nD is , n1(K ) , ‘n’ can be in fraction also.

n n

n3 13 n

[C] [D]K (K )

[A] [B]

K4 for nC + nD nA + nB is n

1

1

(K )

n n

4 n n n1

[A] [B] 1K

[C] [D] (K )

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Note 2 : If K1 be equilibrium constant for P Q and K1 = Q

[P]

If K2 be equilibrium constant for R S then K2 =

S

R

K for P + R Q + S in equal K1 × K2

1 2

[Q] [S]K K K

[P] [R]

Note 3 : If K1, be equilibrium constant or A B 1

[B]K

[A]

K2 be equilibrium constant for C D 2

[D]K

[C]

‘K’ for A + D B + C

1

2

K[B] [C]K

[A] [D] K

Illustration 3

Consider the reaction 2 2 2 2SO Cl (g) SO (g) Cl (g) ; at 37 C the value of

equilibrium constant for the reaction is 0.0032. It was observed that concentration of the

three species is 0.050 mol/lit. each at a certain instant. Discuss what will happen in the

reaction vessel ?

Solution

In this question, concentration of three species i.e. 2 2SO Cl , 2SO and 2Cl (g) each is

given, but is is not mentioned that whether the system is at equilibrium or not. So first

check it. Find reaction coefficient for given equation.

2 2

2 2

[SO ][Cl ] (0.05)(0.05)Q 0.05

[SO Cl ] (0.05)

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Q K , so system is not at equilibrium state.

As Q > K, the concentrations must adjust till Q = K for equilibrium. This can

happen only if reaction shifts backwards, and products recombine to give back reactants.

Hence in the reaction vessel, the system will move backward so that it can achieve

equilibrium state.

Relationship between KP and KC

Consider a gaseous reaction

1 2 1 2m A m B n C n D

n n1 2c D

p m m1 2A B

P PK

P P

…(i)

n n1 2

c m m1 2

[C] [D]K

[A] [B] …(ii)

Applying ideal gas equation

PV = nRT

n

P RTV

or P = CRT

Where C is molarity

A A B BP C RT ; P C RT

C C D DP C RT ; P C RT

Substittuing this in equation (i)

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n n1 2C D

m m1 2A B

C RT C RT

C RT C RT

=

n n1 2n n1 2C D

m m m m1 2 1 2A B

RTC C

C C RT

= (n n ) (m m )1 2 1 2cK (RT)

np cK K (RT)

n = number of moles of products in gaseous state – number of moles of reactants in

gaseous state.

If n 0

p cK K

Gaseous equilibria are of two types

Type I : In which n 0

p cK K

e.g. 2(g) 2(g) (g)H I 2HI

2(g) 2(g) (g)N O 2NO

Type II : In which n 0

So p cK K

e.g. 5(g) 3(g) 2(g)PCl PCl Cl ; n>0

2(g) 2(g) 3(g)N 3H 2NH ; n 0

Illustration 4

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For a homogenous gaseous reaction X 2Y Z, at 473 K, the value of cK 0.35

concentration units. When 2 moles of Y are mixed with 1 mole of X, at what pressure

60% of X is converted to Z?

Solution

Since pressure is to be calculated, so first find pK using the relation between cK and

pK ,

c n 1– 3 –2K 0.35, R 0.0821, T 473,

n 2 –4p cK K (RT) 0.35 (0.0821 473) 2.32 10

The expression for pK is : Zp 2

X Y

pK

p (p )

Moles X Y Z

Initially 1 2 0

At equilibrium 1 – x 2 – 2x X

total moles Tn 3 – 2x

Let P = equilibrium pressure

X Y

1– x 2 – 2xP P, P P,

3 – 2x 3 – 2x z

xP P

3 – 2x

2

p 2 2 2

xP

x(3 – 2x)3 – 2xKP (1– x)(2 – 2x)1– x 2 – 2x

P P3 – 2x 3 – 2x

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x = 0.6 (given)

2

–4p 2 2 2

0.6(3 –1.2)K 2.32 10

P (1– 0.6) –1.2)

2 2 2P (1.8 10 )

P = 180 atm

HOMOGENOUS GASEOUS REACTIONS

Here, we will discuss some important reversible reactions and the calculations for Kc and

Kp.

Illustration 5

The value of cK for the reaction : 2 2H (g) I (g) 2HI(g) is 45.9 at 773 K. If one

mole of 2H , two mole of 2I and three moles of HI are taken in a 1.0 L flask, find the

concentrations of 2I and HI at equilibrium at 773 K.

Solution

Reaction quotient = 2 2

2 2

[HI] 3 9

[H ][I ] 1 2 2

Note : When n 0 , not only p cK K , but volume terms always cancels in the

expression of K.

Q K (=45.9). Hence the reaction proceeds to forward direction to achieve

equilibrium. Let x mole of 2H and 2I combine to produce 2x mole of HI.

Moles 2H 2I 2HI

Initially 1 2 3

At equilibrium 1 – x 2 – x 3 + 2x

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2

C

2 2

[HI]K 45.9

[H ][I ]

concentration of species at equilibrium are :

2 2[H ] (1– x) /1, [I ] (2 – x) /1,[AB] (3 2x) /1

2

2

C

3 2x

(3 2x)1K 45.9

1– x 2 – x (1– x)(2 – x)

1 1

x = 0.68

2[I ] 2 – x 2 – 0.68 1.32M

(Note that volume = 1.0 L)

[HI] = 3 + 2x = 3 + 1.36 = 4.36 M

Illustration 6

0.15 mol of CO taken in a 2.5 L flask is maintained at 750 K along with a catalyst so that

the following reaction can take place; CO(g) + 2 3H (g) CH OH(g) . Hydrogen is

introduced until the total pressure of the system is 8.5 atm. At equilibrium 0.08 mol of

methanol is formed. calculate:

i) pK and cK

ii) The final pressure if the same amount of CO and 2H as before are used but no catalyst

so that the reaction does not take place.

Solution

2 3CO(g) 2H (g) CH OH(g)

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3C 2

2

[CH OH]K

[CO][H ]

Let y moles of 2H were initially.

Moles CO(g) 22H (g) 3CH OH

Initially 0.15 y 0

At equilibrium 0.15 – x y – 2x X

x = 0.08 (given)

moles of CO = 0.15 – 0.08 = 0.07,

moles of 2H y – 0.16, moles of 3CH OH 0.08

total moles = Tn 0.07 (y – 0.16) 0.08

Equilibrium pressure (P) = 8.5 atm

Volume of vessel (V) = 2.5 L, T = 750 K

Using the Gas equation, PV = nRT, we have :

T

PV 8.5 2.5n 0.345

RT 0.0821 750

0.345 = 0.07 + (y – 0.16) + 0.08

y = 0.355 moles

moles of 2H at equilibrium = y – 0.16 = 0.355 – 0.16 = 0.195

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Now find cK as follows :

3C 2 2

2

0.08[CH OH] 2.5K

[CO][H ] 0.07 0.195

2.5 2.5

= 187.84 concentration units .

Now find pK using the relation : np cK K (RT)

n 1–3 –2

–2pK 187.84 (0.0821 750) = 0.0495 atm units.

ii) when no reaction takes place, then the total pressure is simply due to 2H and CO

present initially.

T 2n n(CO) n(H ) 0.15 0.355 0.505

nRT 0.505 0.0821 750

P 12.438 atmV 2.5

FORMATION OF NH3

N2(g) + 3H2(g) 2NH3(g)

Suppose a moles of N2 are heated with b moles of H2 in a V litre container. Let x moles

of N2 be reacted at equilibrium.

N2(g) + 3H2(g) 2NH3(g)

Initial moles conc. a

V

b

V

2x

V

Conc. at eq. a x

V

b 3x

V

2x

V

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2

2 23

c 3 32 2

4x

[NH ] VK[N ][H ] a x b 3x

V V

If a = 1 and b = 3 then

2 2 2 2

c 3 4

4x V 4x VK

(1 x)(3 3x) 27(1 x)

If x 1

2 2

c

4x VK constant

27

2

2

1x

V or

1x

V but

1P

V So, x P

Thus fraction of N2 converted into NH3 is proportional to total pressure.

Illustration 7

Ammonia under a pressure of 15 atm. At 27°C is heated to 347°C is a closed vessel in the

presence of catalyst. Under these conditions, NH3 partially decomposes to H2 and N2. The

vessel is such that the volume remains effectively constant, whereas the pressure

increases to 50 atm. Calculate the % age of NH3 actually decomposed.

Solution

Ammonia decomposes to 2N and 2H as follows :

3 2 22NH N 3H

3 3NH NH

at 27°C at 347°C

(15 atm.) (P = ?) V remains constant

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First, let us find initial pressure of 3NH at 347°C.

P T (V is constant)

1 2 1 22

1 2 1

P P P T 15 620p 31atm.

T T T 300

Partial pressure 2NH3 3H2 N2

Initial 31 0 0

At equilibrium 31 – x 3x

2

x/2

Now final equilibrium pressure = 50 atm.

3 x

50 31– x x2 2

x = 19 atm.

% 3NH decomposed 19

100 61.2%31

Alternative method:

3 2 22NH 3H N

Let x be the degree of dissociation

Moles 32NH 23H 2N

At equilibrium 1 – x 3/2x x/2

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total moles = 1 + x

initial moles initial pressure

final moles final pressure

1 31

1 x 50

19x

31 % dissociation

19100 61.2%

31

Illustration 8

Solid NH4HS(s) (Ammonium hydrogen sulphide) dissociates to give NH3(g) and H2S(g) and

is allowed to attain equilibrium at 100°C. If the value of Kp for its dissociation is found to

be 0.34. find the total pressure at equilibrium and partial pressure of each component.

Solution

NH4HS(s) NH3(g) + H2S(g); since NH4HS is a solid, hence, NH HS4a 1 and its

undissociated amount will not effect the total pressure (due to gaseous NH3 and H2S

only). Let ‘x’ be its moles decomposed at equilibrium and P be the equilibrium pressure.

Moles NH4HS NH3 H2S

Initial 1 0 0

at equilibrium 1 – x x X

Total moles at equilibrium = moles of NH3 and H2S = 2x

P = ? Kp = 0.34

H S NH2 3

x P x PP P & P P

2x 2 2x 2

(for equimolar proportions, partial pressures are equal)

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P H S NH NH HS2 3 4K P P (a 1)

P P

0.342 2

2P

0.34 P 4 0.344

= 1.17

NH3

P 1.17P 0.585 atm

2 2 H S2

P 1.17P 0.585 atm

2 2

HETROGENEOUS SIMULTANEOUS EQUILIBRIUM

Illustration 9

Two solid compounds A and C dissociates into gaseous product at temperature T as

follows:

i) A(s) B(g) + D(g)

ii) C(s) E(g) + D(g)

At 20C pressure over excess solid A is 50 atm and that over excess solid C is 68 atm.

Find the total pressure of the gases over the solid mixture.

Solution

When A(s) dissociates alone in a vessel, the pressure over excess solid A is 50 atm. The

product gases B and D are present in the molar ratio of 1:1, this means that the pressure

of each gas will be half of total pressure.

P1K = 25 25 = 625 atm2

P2K = 34 34 = 1156 atm2

But when A(s) and C(s) are placed together in the container, the degree of dissociation of

each will be suppressed by the presence of other. (Le-Chatelier Principle)

A(s) B(g) + D(g)

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x x (x < 25)

C(s) E(g) + D(g)

y y (y < 34)

we have to deal with two equilibriums simultaneously. Write separate dissociation for

each equilibrium and then make the concentration or pressure of common species as

same.

A(s) B(g) + D(g)

X (x + y)

C(s) E(g) + D(g)

Y (x + y)

625 = x (x +y) …(1)

1156 = y(x +y) …(2)

From eq. (1) and eq. (2) we get

x = 14.8 atm, y = 27.37 atm.

Total pressure of the gases over the solid mixture=2(x +y)=84.34 atm.

EFFLORESCENT AND DELIQUESCENT

The moisture content of a gas in often expressed in terms of the dew point which is the

temperature to which a gas must be cooled before it becomes saturated with water

vapours. At this temperature, water or ice (depending on the temperature) will be

deposited on a solid surface. Dew point of H2O is 43°C at which vapour pressure of ice

formed is 0.07 Torr.

CaCl2·2H2O(s) CaCl2(s) + 2H2O(g)

CaCl2 (s) has its desiccating properties due to the formation of CaCl2·2H2O(s)

Page | 27


Kp =

2

2H O2

0.07p atm

760

A substance is the most effective drying agent, if it has the lowest vapour pressure of

water in equilibrium with it.

A hydrated salt will lose water (a property called efflorescent) below the vapor pressure

of water at that temperature.

Na2SO4· 10H2O(s) Na2SO4 (s) + 10H2O(g)

At 0°C, Kp = 4.08 × 10-25, Kp = 10H O2

p

H O2p = (4.08 × 10-25)1/10

= 3.64 × 10-3

Hence, NaSO4 · 10H2O will effloresce at a vapour pressure below 3.64 × 10-3 atm at 0°C,

or above relative humidity.

Relative humidity = 2

2

partial pressure of H O100

vapour pressure of H O

An anhydrous salt will absorbs water below relative humidity.

Illustration 10

Equilibrium constants are given (in atm) for the following reactions at 0°C

a) SrCl26H2O(s) SrCl22H2O(s) + 4H2O(g), Kp = 6.89 × 10–12

b) Na2HPO412H2O(s) Na2HPO47H2O(s) + 5H2O(g), Kp = 5.25 × 10–13

c) Na2SO410H2O(s) Na2SO4(s) + 10H2O(g) Kp = 4.08 × 10–25

The vapour pressure of water at 0°C is 4.58 torr.

Page | 28


i) Calculate the pressure of water vapour in equilibrium at 0°C with each of (a), (b) and

(c).

ii) Which is the most effective drying agent at 0°C?

iii) At what relative humidities with Na2SO410H2O be efflorescent when exposed to air

at 0°C?

iv) At what relative humidities will Na2SO4 be deliquescent (i.e,. absorb moisture) when

exposed to air at 0°C?

Solution

From the values of Kp, partial pressure of H2O in each case is calculated. Smaller the

vapour pressure, better the drying the agent.

i) For a) 4

12 3p H O H O2 2

K P 6.89 10 , P 1.62 10 atm

b) 5

13 3p H O H O2 2

K P 5.25 10 , P 3.53 10 atm

c) 10

25 3p H O H O2 2

K P 4.08 10 , P 3.64 10 atm

ii) SrCl22H2O is the most effective drying agent since it has the lowest vapour pressure

of water in equilibrium with it.

iii) The hydrate, Na2SO410H2O will lose water (efflorescence) below 3.64 × 10–3 atm

(2.77 torr) and relative humidity (R.H.) at 0°C will be

2.77

(R.H.) 100 60.5%4.58

iv) The dehydrated salt will absorb water above 60.5% R.H.

Illustration 11

For the reaction : 2 2 2CO (g) H (g) CO(g) H O(g), K is 0.63 at 700°C and 1.66 at

1000°C.

Page | 29


a) What is the average H for the temperature range considered ?

b) What is the value of K at 800°C ?

Solution

a) 1T 700 273 973 K

2T 1000 273 1273 K

1 2K 0.63, K 1.66

using the Vant’ Hoff equation :

2 2 1

1 1 2

K T – THlog

K 2.303R T T

1.66 H 1273 – 973

log0.63 2.303 2 1273 973

3H 8.0 10 cal 8.0 kcal

Note : The units of R and H must be same.

b) Let 2K be the equilibrium constant at 2T 1073 K

1T 973 K and then 1K 0.63

3

2K 8.0 10 1073 – 973log

0.63 2.303(1.99) 1073 973

2K 0.93

Le Chatelier’s Principle and Physical Equilibrium

Le Chatelier’s principle, as already stated, is applicable to all types of equilibria

involving not only chemical but physical changes as well. A few examples of its

application to physical equilbria are discussed below.

1. Vapour pressure of a liquid: Consider the equilibrium

Page | 30


Liquid Vapour

It is well known that the change of a liquid into its vapour is accompanied by absorption

of heat whereas the conversion of vapour into liquid state is accompanied by evolution of

heat. According to Le Chatelier’s principle, therefore, addition of heat to such a system

will shift the equilibrium towards the right. On raising the temperature of the system,

liquid will evaporate. This will raise the vapour pressure of the system. Thus, the vapour

pressure of a liquid increases with rise in temperature.

2. Effect of pressure on the boiling point of a liquid: The conversion of liquid into

vapour, as represented by the above equilibrium, is accompanied by increase of pressure

(vapour pressure). Therefore, if pressure on the system is increased, some of the vapours

will change into liquid so as to lower the pressure. Thus, the application of pressure on

the system tends to condense the vapour into liquid state at a given temperature. In order

to counteract it, a higher temperature is needed. This explains the rise of boiling point of

a liquid on the application of pressure.

3. Effect of pressure on the freezing point of a liquid (or melting point of a solid): At

the melting point, solid and liquid are in equilibrium:

Solid Liquid

Now, when a solid melts, there is usually a change, either increase or decrease, of

volume. For example, when ice melts, there is decrease in volume, or at constant volume,

there is decrease in pressure. Thus, increase of pressure on ice water system at a

constant temperature will cause the equilibrium to shift towards the right, i.e., it will

cause the ice to melt. Hence, in order to retain ice in equilibrium with water at the higher

pressure it will be necessary to lower the temperature. Thus, the application of pressure

will lower the melting point of ice.

When sulphur melts, there is increase in volume or at constant volume, there is increase

in pressure. From similar considerations, it follows that if the pressure on the system,

sulphur (solid) sulphur (liquid) is increased, the melting point is raised.

4. Effect of temperature on solubility: In most cases, when a solute passes into

solution, heat is absorbed, i.e., cooling results. Therefore according to Le Chatelier’s

principle, when heat is applied to a saturated solution in contact with solute, the change

will take place in that direction which absorbed heat (i.e., which tends to produce

cooling). Therefore, some more of the solute will dissolve. In other words, the solubility

of the substance increases with rise in temperature.

Page | 31


Dissociation of a few salts (e.g., calcium salts of organic acids) is accompanied by

evolution of heat. In such cases, evidently, the solubility decreases with rise in

temperature.

Illustration 12

2H and 2I are mixed at 400°C in a 1.0 L container and when equilibrium is established,

the following concentration are present : [HI] = 0.49 M, 2[H ] 0.08 M and 2[I ] 0.06

M. If now an additional 0.3 mol of HI are added, what are the new equilibrium

concentrations, when the new equilibrium 2 2H I 2HI is re-established ?

Solution

First determine the equilibrium constant CK for 2 2H I 2HI .

2 2

C

2 2

[HI] (0.49)K 50

[H ][I ] (0.08)(0.06)

When 0.3 mole of HI are added, equilibrium is disturbed. At instant,

[HI] = 0.49 + 0.3 = 0.79 M

CQ K since 2

C

(0.79)K 130

(0.08)(0.06)

backward reaction dominate and the equilibrium shifts to the left.

Let 2x = concentration of HI consumed (while going left) then concentration of each of

2 2H & I formed = x

[HI] = 0.79 – 2x, 2[H ] 0.08 x , 2[I ] 0.06 x and CK 50

2

C

(0.79 – 2x)K 50

(0.08 x)(0.06 x)

246x 10.2x – 0.35 0

Page | 32


x = 0.033 or –0.25 (neglecting the –ve value)

Finally, the equilibrium concentrations are :

[HI] = 0.79 – 2x = 0.79 – 0.033 2 = 0.724 M

2[H ] 0.08 x 0.08 0.033 0.113 M

2[I ] 0.06 x 0.06 0.033 0.093

RELATION BETWEEN STANDARD FREE ENERGY CHANGE (G0) AND

EQUILIBRIUM CONSTANT

The Gibbs free energy function is a true measure of chemical affinity under conditions of

constant temperature and pressure. The free energy change in a chemical reaction can be

defined as

G = G(products) – G(reactants)

When G = 0, there is no net work obtainable. The system is in a state of equilibrium.

When G is positive, net work must be put into the system to effect the reaction,

otherwise it cannot take place. When G is negative, the reaction can proceed

spontaneously with accomplishment of the net work. The larger the amount of this work

that can be accomplished, the farther away is the reaction from equilibrium. For this

reason –G has often been called the driving force of the reaction. From the statement of

the equilibrium law, it is evident that the driving force depends on the concentration of

the reactants and products. It also depends upon the temperature and pressure which

determine the molar free energies of the reactants and products.

The reaction conducted at constant temperature (i.e., in a thermostat)

–G = – H + TS

The driving force is made up of two parts, –H term and TS term. The –H term is the

heat of reaction at constant pressure and TS is heat involved when the process is carried

out reversibly. The difference is the amount of heat of reaction which can be converted

into net work (–G), i.e., total heat minus unavailable heat.

Page | 33


If the reaction is carried out at constant volume, the decrease in Helmholtz function –G

= –E + TS would be the proper measure of affinity of the reactant or the driving force

of the reaction.

Now we can see why Berthellot and Thompson were wrong in assuming that driving

force of the reaction was the heat of reaction. They neglected the TS term. The reasons

for the apparent validity of their principle was that for many reactions, H term far

outweighs the TS term. This is especially true at low temperature, since at higher

temperature, TS term increases.

The fact that driving force for a reaction is large (G is large negative quantity) does not

mean that the reaction will necessarily occur under any given conditions.

For example, the reaction

2 2 2

1H O H O; G 228.6kJ

2

does not occur at the laboratory temperature. The reaction mixture may be kept for years

without any detectable formation of water. Here H factor favours, but S factor

disfavours the reaction.

Similarly, the reaction

2Mg + O2 → 2MgO; G = – 570.6kJ

is not favoured. However, the thermite reaction

2Al + 3

2O2 → Al2O3

with large value of – G proceeds favourably.

Standard Free Energy and Equilibrium Constant

The change in free energy for a reaction taking place between gaseous reactants and

products represented by the general equation.

aA bB cC dD

Page | 34


According to Van’t Hoff reaction isotherm

c d0 C D

a bA B

p pG G RTln

p p

= G0 + RTlnQp

the condition for a system to be at equilibrium is that

G = 0

Thus at equilibrium

c d0 0 0C D

pa bA B

p p0 G RTln G RTln K

p p

Whence G0 = – RTlnK0p

Hence 0

0p

Gln K

RT

Note

1. In the reaction, where all gaseous reactants and products; K represents Kp

2. In the reaction, where all solution reactants and products; K represents Kc

3. A mixture of solution and gaseous reactants; Kx represents the thermodynamic

equilibrium constant and we do not make the distinction between Kp and Kc.

we may conclude that for standard reactions, i.e., at 1 M or 1 atm.

When G0 = –ve or K 1: forward reaction is feasible

G0 = +ve or K 1: reverse reaction is feasible

G0 = 0 or K = 1: reaction is at equilibrium (very rare)

Illustration 13

Page | 35


NO and Br2 at initial pressures of 98.4 and 41.3 torr respectively were allowed to react at

300K. At equilibrium the total pressure was 110.5 torr. Calculate the value of equilibrium

constant , Kp and the standard free energy change at 300K for the reaction

(g) 2(g) (g)2NO Br 2NOBr

Solution

From the initial and final pressures of the reaction species, partial pressures of each are

determined, hence Kp and G0 are calculated

2NO()g) + Br2(g) 2NOBr(g)

Initial 98.4 41.3 0

Used up 2x x ––

Produced –– –– 2x

Equilibrium (98.4 – 2x) (41.3 – x) 2x

Total pressures at equilibrium = NO Br NOBr2P P P

= (98.4 2x) (41.3 x) 2x 139.7 x 110.5

This gives, x = 29.2

Hence, NOP (98.4 2 29.2) 40 Torr

Br2P (41.3 29.2) 12.1 Torr

NOBrP 2 29.2 58.4 Torr

2 2

-1 -1NOBrp 2 2

NO Br2

P (58.4)K 0.1762 Torr 133.88 atm

P P (40) (12.1)

0G 2.303RT log K 2.303 8.314 300 log133.88

Page | 36


= 12216.26J = 12.22 kJ mol–1

SOLVED EXAMPLES

Example 1

The equilibrium constant for the reaction 2 2 32SO O 2SO at 1000K is 3.5. What

would partial pressure of oxygen gas have to be, to give equal moles of 2SO and 3SO ?

(a) 0.29 atm (b) 3.5 atm

(c) 0.53 atm (d) 1.87 atm.

Solution

2SO3

p 2OSO O 22 2

(p ) 1K

p(p ) p

Here, partial pressure of 3SO and 2SO same because moles of 2SO and 3SO are equal

at equilibrium.

O2p

1p 0.285

K

Ans. (a)

Example 2

Page | 37


5PCl is 50% dissociated into 3PCl and 2Cl at 1 atmosphere. It will be 40% dissociated

at :

(a) 1.75 atm (b) 1.84 atm

(c) 2.00 atm (d) 1.25 atm.

Solution

5 3 2PCl PCl Cl

Let x = degree fo dissociation of 5PCl at pressure P

y = degree of dissociation of 5PCl at pressure p

2 2

p 2 2

px p yK

(1– x ) (1– y )

given, x = 0.50, p = 1 atm

y = 0.40, p ?

1 0.25 p 0.16

0.75 0.84

p 1.75atm

Ans. (a)

Example 3

For the reaction

2 4(s) 3(g) 2(g)NH COONH 2NH CO

The equilibrium constant –5 3pK 2.9 10 atm . The total pressure of gases at equilibrium

when 1.0 mole of reactant was heated will be :

Page | 38


(a) 0.0194 atm (b) 0.0388 atm

(c) 0.0580 atm (d) 0.0667 atm.

Solution

2 4(s) 3(g) 2(g)NH COONH 2NH CO

Initially 1 0 0

At equi. (1–x) 2x x

Total moles of gaseous substances at equilibrium = 3x

Equilibrium pressure = P

NH CO3 2

2 pp p, p

3 3

2p NH CO3 2

K (p ) p

–5 342.9 10 p

27

p = 0.0580 atm

Ans. (c)

Example 4

One mole of 2 4(g)N O at 300 K is kept in a closed container under one atmosphere. It is

heated to 600 K when 20% of 2 4(g)N O decomposes to 2(g)NO . The resultant pressure is

(a) 1.2 atm(b) 2.4 atm

(c) 2.0 atm(d) 1.0 atm.

Solution

Page | 39


2 4 2N O 2NO

Initially 1 0

At equi. (1–x) 2x

Total mole at equilibrium = 1 + x = 1 .2 x 0.2

1 2 2

1 1 2 2

P P P1or

n T n T 1 300 1.2 600

or 2P 2.4atm

Ans. (b)

Example 5

For the reversible reaction, 2(g) 2(g) 3(g)N 3H 2NH at 500°C, the value of pK is

1.44 –510 when partial pressure is measured in atmospheres. The corresponding value

of cK with concentration in mole –1litre , is

(a) 1.44 –5 –210 / (0.082 500) (b)

–5 –21.44 10 / (8.314 773)

(c) 1.44 –5 210 / (0.082 773) (d)

–5 –21.44 10 / (0.082 773) .

Solution

– ng

c pK K (RT)

Since gn –2

–5 2cK 1.44 10 (0.082 773)

Ans. (d)

Example 6

Page | 40


One mole of N2O(g) is kept in a closed container along with gold catalyst at 450 K under

one atmosphere. It is heated to 900 K when it dissociates to N2(g) and O2(g) giving an

equilibrium pressure of 2.4 atm. The degree of dissociation of N2O(g) is

(a) 20% (b) 40%

(c) 50% (d) 60%.

Solution

122 (g) 2(g) 2(g)N O N O

0P (1– ) 0P 0P2

When temperature of 2N O is doubled, its pressure also doubles.

Pressure of 2N O at 900K, 0P 2atm

Total pressure at equilibrium = 2.4 atm

0P 1 2.42

0.4

Degree of dissociation of 2 (g)N O 40%

Ans. (b)

Example 7

For the reaction : (g) 2(g) 2(g)2HI H I ; the degree of dissociation ( ) of HI(g) is

related to equilibrium constant pK by the expression

(a) p1 2 K

2

(b)

p1 2K

2

Page | 41


(c) p

p

2K

1 2K (d)

p

p

2K

1 2 K.

Solution

(g) 2(g) 2(g)2HI H I

1– 2

2

2

p 2 2

P2

K(1– ) P

p2 K1–

p

p

2

1 2 K

Ans. (d)

Example 8

For which of the following reactions, the degree of dissociation cannot be calculated from

the vapour density data

I (g) 2(g) 2(g)2HI H I II 3(g) 2(g) 2(g)2NH N 3H

III (g) 2(g) 2(g)2NO N O IV 5(g) 3(g) 2(g)PCl PCl Cl

(a) I and III (b) II and IV

(c) I and II (d) III and IV.

Solution

Page | 42


The degree of dissociation cannot be calculated from the vapour density data if the

number of moles remains unchanged before and after reaching equilibrium.

Ans. (a)

Example 9

At a certain temperature the following equilibrium is established.

(g) 2(g) 2(g) (g)CO NO CO NO

One mole of each of the four gases is mixed in one litre container and the reaction is

allowed to reach equilibrium state. When excess of baryta water is added to the

equilibrium mixture, the weight of white precipitate obtained is 236.4 g. The equilibrium

constant cK of the reaction is

(a) 1.2 (b) 2.25

(c) 2.1 (d) 3.6.

Solution


Initially 1 1 1 1

At equi. 1–x 1–x 1 + x 1 + x

2 2 3 2CO Ba(OH) BaCO H O

Moles of 3

236.4BaCO 1.2

197

Moles of 2CO at equilibrium = 1.2

or, 1 + x = 1.2; x = 0.2

Page | 43


2 2

c

1 x 1.2K 2.25

1– x 0.8

Ans. (b)

Example 10

In a reversible reaction, study of its mechanism says that both the forward and backward

reactions follow first order kinetics. If the half life of forward reaction 1/2 f(t ) is 400 sec.

and that of backward reaction 1/2 b(t ) is 250 sec. The equilibrium constant of the reaction

is


(a) 1.6 (b) 0.433

(c) 0.625 (d) 1.109.

Solution

–1f

0.693k sec

400 ; –1

b

0.693k sec

250

f

b

k 250K 0.625

k 400

Ans. (c)

Example 11

At 27°C NO and 2Cl gases are introduced in a 10 litre flask such that their initial partial

pressures are 20 and 16 atm respectively. The flask already contains 24 g of magnesium.

After some time, the amount of magnesium left was 0.2 moles due to the establishment of

following two equilibrium

(g) 2(g) (g)2NO Cl 2NOCl

Page | 44


–12(g) (s) 2(s) pCl Mg MgCl ; K 0.2 atm

The final pressure of NOCl would be

(a) 7.84 atm (b) 18.06 atm

(c) 129.6 atm (d) 64.8 atm.

Solution

p

Cl2

1K 0.2

P Cl2

P at equilibrium = 5 atm

(g) 2(g) (g)2NO Cl 2NOCl

Initially 20 16 0

At equi. 20–2x 16–x–y 2x

2(g) (s) 2(s)Cl Mg MgCl

Initially 16 1 0

At equi. 16–x–y 1– y y

= 0.2

y= 0.8

(since y is the moles while y is pressure in atm)

0.8 0.082 300

y 1.9710

We know that,

16 – x – y 5

Page | 45


or, x = 9.03; NOClP 18.06 atm

Ans. (b)

Example 12

The 3CaCO is heated in a closed vessel of volume 1 litre at 600 K to form CaO and

2CO . The minimum weight of 3CaCO required to establish the equilibrium

3(s) (s) 2(g)CaCO CaO CO is p(K 2.25 atm)

(a) 2g (b) 4.57 g

(c) 10g (d) 100 g.

Solution

p CO2K P 2.25 atm

Number of moles of 2

2.25 1CO

0.0821 600

The minimum moles of 3CaCO required = 0.0457

The minimum weight 3CaCO required = 0.0457 100 = 4.57 g

Ans. (b)

Example 13

An acid reacts with glycerine to form complex and equilibrium is established. If the heat

of reaction at constant volume for above reaction is 1200 cal more than at constant

pressure and the temperature is 300 K, then which of the following expression is true ?

(a) p cK K (b) c pK K

(c) p cK K (d) none of these.

Solution

Page | 46


Given : E – H 1200 cal

H E nRT

n –2

np cK K (RT)

p –3

c

K1.648 10

K

p cK K

Ans. (a)

Example 14

When pure NH3 is maintained at 480°C and a pressure of 1 atm, it dissociates to give a

gaseous mixture containing 20% NH3 by volume. The degree of dissociation of NH3 is

(a) 2/3 (b) 3/2

(c) 5/2 (d) 2/5.

Solution

3(g) 2(g) 2(g)2NH N 3H

Initial a 0 0

At equi. a(1– ) a /2 3a /2

% of 3NH by volume = a(1– ) 1–

100 100a 3a 1

a(1– )2 2

By condition,

Page | 47


1–

10 201

2

3

Ans. (a)

Example 15

Steam at pressure of 1 atm is passed through a furnace at 2000 K wherein the reaction –51

22 (g) 2(g) 2(g) pH O H O ; K 6.4 10 occurs. The percentage of oxygen in the

exit steam would be

(a) 0.32 (b) 0.08

(c) 0.04 (d) 0.16

Solution

122 (g) 2(g) 2(g)H O H O

Initially 2 atm 0 0

At equi. 2 – x x x/2

1/2

–5x(x / 2)6.4 10

2 – x

3/2

–5x6.4 10

2(2 – x)

Dissociation of 2H O is so small that we can assume H O2P 2 atm

i.e. 2 – x 2

3/2 –5x 2 2 6.4 10 x or

3/2 3/2 –6x (2 16) 10

Page | 48


x = 2 –416 10 0.0032

% of 2O in the exit steam 0.0032 / 2

100 0.08%2

Ans. (b)

Example 16

Two moles of an equimolar mixture of two alcohols, 1R — OH and 2R — OH are

esterified with one mole of acetic aicd. If 80% of the acid is consumed and the quantities

of ester formed under equilibrium are in the ratio of 3 : 2, the value of the equilibrium

constant for the esterification of 1R — OH with acetic acid is

(a) 3.3 (b) 3.7

(c) 3.5 (d) 3.9 .

Solution

1 3 2 3 2 1 2R OH CH CO H CH CO R H O

(1– x) (1– x – y) x (x + y)

2 3 2 3 2 2 2R OH CH CO H CH CO R H O

(1– y) (1– x – y) y (x + y)

Since, the amount of acid consumed is 80% in complete esterification process, therefore

x y 0.8

Also, x 3

y 2

x = 0.48 and y = 0.32

Page | 49


Thus, equilibrium constant for first esterification process x(x y)

(1– x)(1– x – y)

C1

0.48 0.8K

0.52 0.2

C1K 3.7

Ans. (b)

Example 17

Solubility of a substance which dissolves with a decrease in volume and absorption of

heat will be favoured by

(a) high P and high T (b) low P and low T

(c) high P and low T (d) low P and high T.

Solution

The substance dissolves with a decrease in volume and absorption of heat. Therefore

according to Le Chalelier’s principle the process of dissolution will be favoured by high

pressure and high temperature.

Ans. (a)

Example 18

X2 + X– X3– (x = iodine)

This reaction is set up in aqueous medium. We start with 1 mol of X2 and 0.5 mol of X–

in 1L flask. After equilibrium is reached, excess of AgNO3 gave 0.25 mol of yellow ppt.

equilibrium constant is

(a) 1.33 (b) 2.66

(c) 2.00 (d) 3.00

Solution

Page | 50


X2 + X– 3X

1 0.5 0

(1 – x) (0.5 – x) x

(0.5 – x) = unreacted X–

X– Ag + 0.25

x = 0.25

3c

2

[X ] 0.25K 1.33

0.75 0.25[X ][X ]

Ans. (a)

Example 19

At temperature T, a compound AB2(g) dissociates according to the reaction 2AB2(g)

2AB(g) + B2(g) with degree of dissociation , which is small compared with

unity. The expression for Kp, in terms of and the total pressure, PT is

(a) 3

TP

2

(b)

2TP

3

(c) 3

TP

3

(d)

2TP

2

.

Solution

For the given equilibrium, the equilibrium concentrations are

2AB2(g) 2AB(g) + B2(g)

Equilib. conc. c(1 – ) c c

2

Page | 51


22

TB AB2

P 22AB2

c(c ) P(P )(P ) 2K

(P ) [c(1– )] [c(1 )]2

; 3

TP

2

PK

2(1– ) 12

Since, is small compared to unity, so 1 – 1 and 1 + 12

.

3

TP

PK

2

Ans. (a)

Example 20

2.0 mol of PCl5 were introduced in a vessel of 5.0 L capacity at a particular temperature.

At equilibrium, PCl5 was found to be 35% dissociated into PCl3 and Cl2. The value of Kc

for the reaction is

(a) 1.89 (b) 0.377

(c) 0.75 (d) 0.075.

Solution

Moles of PCl5 dissociated = 2 35

100

= 0.7

Moles of PCl5 left undissociated = 2 – 0.7 = 1.3 mol

[PCl5] = 1.3

5M, [PCl3] =

0.7

5M, [Cl2] =

0.7

5M

3 2

5

0.7 0.7

[PCl ][Cl ] 5 5K 0.75

1.3[PCl ]

5

Ans. (c)

Page | 52


SOLVED SUBJECTIVE EXAMPLES

Example 1

The equilibrium constant for the reaction H2 + I2 2HI; is found to be 64 at 450°C. If

6 mole of hydrogen are mixed with 3 mole of iodine in a litre vessel at this temperature;

what will be the concentration of each of the three components, when equilibrium is

attained ?

It the volume of reaction vessel is reduced to half; then what will be the effect on

equilibrium ?

Solution

The given equilibrium reaction is :

H2 + I2 2HI

At t = 0 6 3 0

At equilibrium 6 – x 3 – x 2x

concentration at

equilibriu

m

6 x

1

3 x

1

2x

1

2

C

(2x)K

(6 x)(3 x)

2

2

4x64

x 18 9x

Page | 53


2 216(x 9x 18) x

On solving x = 2.84

[HI] = 2x 2 2.84

5.681 1

[H2] = 6 x 6 2.84

3.161 1

[I2] = 3 x 3 2.84

0.161 1

In the present reaction n = 0 hence, volume change will not affect the equilibrium.

Example 2

5 gm of PCl5 were completely vaporized at 250°C in a vessel of 1.9 litre capacity. The

mixture at equilibrium exerted a pressure of one atmosphere. Calculate the degree of

dissociation, KC and Kp for this reaction.

Solution

No. of moles of PCl5

= Weight

Molecularweight =

50.024

208.5

The equilibrium for dissociation of PCl5 maybe represented as

PCl5 PCl3 + Cl2

At t = 0 0.024 0 0

At equilibrium (0.024 - x) X x

Page | 54


Total moles of gas components

= 0.024 – x + x + x = (0.024 + x)

We know, PV = nRT

1 1.9 = (0.024 + x) 0.0821 × 523

(0.024 + ) = 1.9

0.0821 523

x = 0.0202

Degree of dissociation = 0.0202

0.024 = 0.843

[PCl5] = 0.024 x 0.024 0.0202 0.0038

1.9 1.9 1.9

[PCl3] = [Cl2] = x 0.0202

V 1.9

Kc =

2

43 2

5

0.0202

[PCl ][Cl ] 4.08 101.9

0.0038[PCl ] 1.9 0.0038

1.9

= 0.0565 mole/litre

Kp = Kc (RT) n

Where Kc = 0.0565

n = 2 – 1 = 1

Kp = 0.0565 × 0.0821 × 523 = 2.43 atm

Example 3

16 moles of H2 and 4 moles of N2 are sealed in a one litre vessel. The vessel is heated at a

constant temperature until the equilibrium is established, when it is found that the

pressure in the vessel has fallen to 9/10 of its original value. Calculate KC for the reaction

Page | 55


N2(g) + 3H2(g) 2NH3(g)

Solution

The given equilibrium is

N2(g) + 3H2(g) 2NH3(g)

At t = 0 4 16 0

At equilibrium 4 – x 16 – 3x 2x

Total gaseous moles at equilibrium

= 4 – x + 16 – 3x + 2x = (20 – 2x)

Since, pressure has fallen to 9/10 of its original value, hence no. of mole will also fall up

to the same extent.

(20 – 2x) = 9

20 1810

x = 1

[N2] = 4 x 4 1

31 1

mole/litre

[H2] = 16 3x

1

= 13 mle/litre

[NH3] = 2x

1 = 2 mole/litre

KC = 2 2

3

3 32 2

[NH ] 2

[N ][H ] (3)(13) = 6.07 10–4

Page | 56


Example 5

The value of K for the reaction

O3(g) + OH(g) H(g) + 2O2(g)

Changed from 0.096 at 298K to 1.4 at 373K. Above what temperature will the reaction

become thermodynamically spontaneous in the forward direction assuming that H0 and

S0 values for the reaction do not change with change in temperature? Given that, S2980

=10.296JK–1.

Solution

We have,

0

2 2 1

1 1 2

K (T T )Hlog

K 2.303R T T

01.4 H 373 298

log0.096 2.303 8.314 373 298

H0= 33025J

Now the temperature above which the forward reaction will be spontaneous is actually

the temperature at which the reaction attains equilibrium, that is, when

K = 1 or log K = 0

G0 = – 2.303 RT log K = – 2.303 RT log 1.0 = 0

From thermodynamics, we get

G0 = H0 – TS0

0 = 33025 – T × 10.296

or T = 320.75 K

Example 6

Page | 57


A mixture of SO3, SO2 and O2 gases is maintained in a 10.0 litre flask at a temperature at

which equilibrium constant Kc for the reaction

2SO2(g) + O2(g) 2SO3(g) is 100.

i) If the number of mole of SO2 and SO3 in the flask are equal; how many mole of O2 are

present ?

ii) If the number of mole of SO3 in the flask is twice the number of mole of SO2,how

many mole of O2 are present ?

Solution

The given equilibrium is

2SO2(g) + O2(g) 2SO3(g)

2

3C 2

2 2

[SO ]K

[SO ] [O ] … (i)

i) If [SO3] = [SO2]

Then C

2

1K

[O ]

2

C

1 1[O ] 0.01

K 100 mole/litre

Total moles of O2 present = 0.01 10 = 0.1 mole

Volume of vessel is 10 litre

ii) When [SO3] = 2[SO2]

Then 2

3C 2

2 2

[SO ]K

[SO ] [O ] from Eq. (i)

2

4100

[O ]

Page | 58


[O2] = 4/100 = 0.04 mole/litre

Total moles of O2 in vessel at equilibrium = 0.04 10 = 0.4 mole

Example 7

A saturated solution of iodine in water contains 0.33g of I2 per litre of solution. More

than this can dissolve in KI solution because of the following equilibrium.

I2(aq) + I– I3–

A 0.1 M KI solution actually dissolves 12.5g of I2/litre, most of which is converted to I3–.

Assuming that the concentration of I2 in all saturated solutions is the same, calculate the

equilibrium constant for the above reaction.

Solution

I2 + I– I3–

Initially 12.5

M254

0.1M 0

At equb. 0.33

M254

12.17

0.1 M254

12.17M

254

Thus,

3

2

12.17[I ] 254K 708

0.33 13.23[I ][I ]

254 254

Example 8

For the reaction Ag(CN)2– Ag+ + 2CN–, the KC at 25°C is 4 10–19. Calculate

[Ag+] in solution which was originally 0.1 M in KCN and 0.03 M in AgNO3.

Solution

Page | 59


2KCN + AgNO3 Ag(CN 2) + KNO3 + K+

0.1 0.03 0 0 0

(0.1 – 0.06) 0 0.03 0.03 0.03

[Ag(CN 2) ] = 0.03 M

Now use Ag(CN 2) Ag+ + 2CN–

0.03 0 0.04

(0.03 – a) a 0.04 + a = 0.04

Since CK is too small and dissociation of Ag(CN 2) is very less and thus,

0.04 + a 0.04 and 0.03 – a 0.03

[Ag(CN 2) ] = 0.03; [Ag+] = a; [CN–] = 0.04

Now 2 2

C

2

[Ag ][CN ] a (0.04)K

0.03[Ag(CN) ]

a = 7.5 10–18

Example 9

When baking soda is heated in a sealed tube, following equilibrium exists:

2NaHCO3(s) Na2CO3(s) + CO2(g) + H2O(g)

If the equilibrium pressure is 1.04 atm at 398 K, calculate the equilibrium constant for the

reaction at 398 K.

Solution

Since, there are only two gaseous species in the above equilibrium

Page | 60


CO H O2 2

1p p

2 total pressure

1

1.04 0.52atm2

2p CO H O2 2

K p p [0.52]

Example 10

For the reaction

NH3(g) 1

2N2(g) +

3

2H2(g)

Show that degree of dissociation of NH3 is given as :

1/2

p

3 3 P1

4 K

where ‘P’ is equilibrium pressure. If Kp of the above reaction is 78.1 atm at 400°C,

calculate KC.

Solution

NH3(g) 1

2N2(g) +

3

2H2(g). Total moles

Initial moles t = 0 1 0 0 1

Moles equilibrium 1 – /2 3 /2 1 +

Partial pressure 1

p1

p2(1 )

3p

2(1 )

1/2 3/2

2 2p

3

[N ] [H ]K

[NH ]

Page | 61


1/2 3/23

p p2(1 ) 2(1 )

1p

1

2

2

p 27

4(1 )

Solving for ` ’ we get :

1/2

p

3 3 p1

4 K

We know,

np CK K (RT)

78.1 = CK (0.0821 673)1

KC = 1.413 moles litre–1

Example 11

Kp for the reaction N2O4 (g) 2NO2 (g) is 0.66 at 46C. Calculate the percent

dissociation of N2O4 at 46C and a total pressure of 0.5 atm. Also calculate the partial

pressure of N2O4 and NO2 at equilibrium.

Solution

This Example can be solved by two methods.

Method 1: Let the number of moles of N2O4 initially be 1 and is the degree of

dissociation of N2O4.

N2O4 2NO2

Initial moles 1 0

Moles at equilibrium 1– 2

Total moles at equilibrium = 1– + 2 = 1+

Page | 62


N O T2 4

1p P

1

NO T2

2p P

1

Kp =

2 2 2NO2 T

2N O2 4

p 4 P 4 0.5

p 1 1 1

= 0.5 i.e., 50% dissociation

Hence, partial pressure of N2O4 = 0.167 atm.

and partial pressure of NO2 = 0.333 atm.

Method 2 : Let the partial pressure of NO2 at equilibrium be p atm, than the partial

pressure of N2O4 at equilibrium will be (0.5–p)atm.

Kp =

2p0.66

0.5 p

p2 + 0.66 p–0.33 = 0

On solving p = 0.333 atm.

NO2p = 0.333 atm and N O2 4

p = 0.167 atm.

Example 12

At 1000 K, the pressure of iodine gas is found to be 0.112 atm due to partial dissociation

of I2(g) into I. Had there been no dissociation, the pressure would have been 0.074 atm.

Calculate the value of Kp for the reaction: I2(g) 2I(g).

Solution

Analysing in terms of pressure directly

Page | 63


Partial pressure I2 2I

Initial 0.074 0

At equilibrium 0.074- p 2p

total pressure at equilibrium

= (0.074 – p) + 2p = 0.112 (given) p = 0.038 atm

Kp =

2 2

I

I2

P 2p

P 0.074 p

Substituting value of p Kp = 0.16 atm

Example 13

Determine Kc for the reaction ½N2(g) + ½O2(g) + ½Br2(g) NOBr(g) from the

following information (at 298oK)

Kc = 2.4 1030 for 2NO(g) N2(g) + O2(g) ; '''CK = 1.4 for NO(g) + ½Br2(g) NOBr(g)

Solution

N2(g) + O2(g) 2NO (g)

Kc/ =

30c

1 1

K 2.4 10

i) ½N2(g) + ½O2(g ) NO(g)

Kc// =

/ 30 15c

1K 10 0.6455 10

2.4

ii) NO(g) + ½Br2(g) NOBr(g) '''CK = 1.4.

(i) + (ii) gives the net reaction:

Page | 64


½N2(g) + ½O2(g) + ½Br2(g) NOBr(g)

K =

1 1 1 1 1 1

2 2 2 2 2 22 2 2 2 2 2

NO NOBr NOBr

N O NO Br N O Br

= ''CK '''

CK = 0.6455 10-15 1.4 = 9.037 10-16

Example 14

In a mixture of N2 and H2 initially in a mole ratio of 1:3 at 30 atm and 300oC , the

percentage of ammonia by volume under the equilibrium is 17.8. Calculate the

equilibrium constant (KP) of the mixture, for the reaction

N2(g) + 3H2(g) 2NH3(g)

Solution

Let the initial moles of N2 and H2 be 1 and 3 respectively (this assumption is valid as KP

will not depend on the exact no. of moles of N2 and H2. One can even start with x and

3x).

Alternatively

N2(g) + 3H2(g) 2NH3

Initial 1 3 0

At eqb 1x 33x 2x

Since % by volume of a gas is same as % by mole,

2x

4 2x=0.178

4 0.178

x 0.302(2 2 0.178)

Mole fraction of H2 at equilibrium = 3 3x

0.61654 2x

Mole fraction of N2 at equilibrium = 1 0.6165 0.178 = 0.2055

KP =

2

NH T3

3

N T H T2 2

X P

X P X P

2

3

0.178 30

(0.2055 30) (0.6165 30)

= 7.31 10-4 atm–2

Example 15

Page | 65


The density of an equilibrium mixture of N2O4 and NO2 at 1 atm. and 348 K is 1.84 g

dm-3. Calculate the equilibrium constant of the reaction

N2O4(g) 2NO2(g).

Solution

Let us assume that we start with C moles of N2O4(g) initially.

N2O4(g) 2NO2(g)

Initial 0 0

At equilibrium C(1) 2C

Where is the degree of dissociation of N2O4(g)

Since =Total moles at equilibrium Vapour density initial

Total moles initially Vapour density at equilibrium

Initial vapour density = 92

462

C(1 ) 46

C d

Since vapour density and actual density are related by the equation,

V.D. = RT

2P

=

1.84 0.082 348

2

= 26.25

1 + = 46

1.75226.25

= 0.752

Kp =

2 2

T

T

2 C 2 0.75P 1

C(1 ) 1.752

C(1 ) 0.248P 1

C(1 ) 1.752

= 5.2 atm.

Example 16

In an evacuated vessel of capacity 110 litres, 4 moles of Argon and 5 moles of PCl5 were

introduced and equilibrated at a temperature of 250oC. At equilibrium, the total pressure

of the mixture was found to be 4.678 atm. Calculate the degree of dissociation, of PCl5

and KP for the reaction PCl5 PCl3 + Cl2 at this temperature.

Solution

Page | 66


PCl5(g) PCl3(g) + Cl2(g)

Initial moles 5 0 0

At equilibrium 5x x x

Total moles = 5 + x + 4 (including moles of argon) = 9 + x

Since total moles = PV 4.67 110

RT 0.082 523

= 12

x = 3

=3

0.65 ; KP =

23

4.6712

24.67

12

= 1.75

Example 17

256 g of HI was heated in a sealed bulb at 444°C till the equilibrium was attained. The

acid was found to be 22% dissociated at equilibrium. Calculate equilibrium constants for

synthesis and dissociation of HI?

Solution

2HI (g) H2 (g) + I2 (g)

Initial moles 256

128 0 0

Moles at equilibrium 2–2

(given = 0.22)

Equilibrium constant for dissociation

KC=

2

22 2

=

2

0.22 0.22

2 0.44

= 0.0199

Equilibrium constant for synthesis = C

1 1

K 0.0199 = 50.25

(H2 + I2 2HI )

Page | 67


[ Note : The equilibrium reaction considered for dissociation is 2HI H2 + I2 and not

HI ½ H2 + ½ I2 because for synthesis, the reaction is with 1 mole each of H2 and I2

i.e., H2 + I2 2HI ]

Example 18

Variation of equilibrium constant K with temperature T is given by Van’t Hoff equation

log K = log A –0H

2.303RT

log K

T–1

O

P

A graph between log K and T–1 was a straight line as shown and having OP = 10 and tan

= 0.5. Calculate

i) equilibrium constant at 298 K, and

ii) and equilibrium constant at 798 K, assuming H to be independent of temperature.

Solution

To calculate equilibrium constant, we need to know A and H, which are calculated as –

The given equation represents a straight line of slope = oH

2.303 R

= –tan = –0.5

H = 2.303 8.314 0.5 = 9.574 J/mol

Intercept = log A = OP = 10

log K = log A–oH 9.574

102.303 RT 2.303 8.314 298

K = 9.96 109

Page | 68


Now, to calculate equilibrium constant at some other temperature, we will use the

expression

2

1

Klog

K=

o

1 2

H 1 1

2.303 R T T

log 2

9

K

9.96 10 =

9.574

2.303 8.314

500

798 298

K2 (equilibrium constant at 798 = 9.98 109

Example 19

The equilibrium constant KP of the reaction

2SO2(g) + O2(g) 2SO3(g)

is 900 atm–1 at 800C. A mixture containing SO3 and O2 having initial partial pressures of

1 atm and 2 atm respectively is heated at constant volume to equilibrate. Calculate the

partial pressure of each gas at 800C.

Solution

It can be seen that as SO2 is not present initially, so equilibrium cannot be established in

the forward direction. Therefore it is established from reverse direction. Let n be the

increase in partial pressure of O2. Then at equilibrium the partial pressures of SO2, O2 at

SO3 are (0+2n), (2+n) and (1–2n) in atm respectively.

2SO2(g) + O2(g) 2SO3(g)

(0+2n) (2+n) (1–2n)

Also,

22SO3

P 2 2SO O2 2

p 1 2nK

p p 0 2n 2 n

As n is small (because equilibrium constant for the reverse reaction is very small i.e.,

1/900), it can be neglected in comparison to 2 and also

1–2n can be taken approximately to 1.

Page | 69


900 = 2

1

4n 2

Solving for n, we get n = 0.0118

Hence, O2p = 2+n = 2.0118 atm

SO2p = 2n = 0.0236 atm

SO3p = 1–2n = 1–0.0236 = 0.9764 atm

Example 20

In an experiment 5 moles of HI were enclosed in a 5 litre container. At 717 K equilibrium

constant for the gaseous reaction 2HI (g) H2 (g) + I2 (g) is 0.025. Calculate the

equilibrium concentrations of HI, H2 and I2. What is the fraction of HI that decomposes.

Solution

Let 2n be the number of moles of HI which is decomposed, the number of moles of H2

and I2 produced will be n moles each. Then molar concentrations of various species at

equilibrium are –

[HI] = 5 2n

5

mol/l, [H2] =

n

5 mol/l, [I2] =

n

5 mol/l

Also, KC =

2 2

2

H I

HI =

2

n n

5 5

5 2n

5

0.025 =

2

2

n

5 2n

Solving for n, we get n = 0.6

[HI] = 5 2 0.6

5

=

3.8

5= 0.76 mol/l

Page | 70


[H2] = 0.6

5 = 0.12 mol /l

[I2] = 0.6

5 = 0.12 mol /l

Fraction of HI decomposed = 2 0.6

5

= 0.24 or 24%

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TOPIC : EQUILIBRIUM