Chemical Equilibrium

Irreversible and Reversible Reactions

Those reactions which proceeds in forward direction and reaches

almost to completion are called irreversible reactions. For example

2

3(s) MnO (s) 2(g)

2KClO _  2KCl_ 3O

(Thermal decomposition)

NaOH(aq)HCl(aq) NaCl(aq) H O2

(Strong acid–strong base

neutralisation reaction)

2AgNO3BaCl2  2AgCl  Ba(NO )3 2

(Precipitation reaction)

Whereas, those reactions which proceed in forward and backward

directions both and never reaches completion are called reversible

reactions. These reactions can be initiated in any direction. For

example,

5(g) 3(g) 2(g)

PCl ˆ ˆ †_{‡ ˆ ˆ} PCl Cl (Thermal dissociation)

3 _{strong base} 3 2

Weak acid

CH COOH NaOH _{‡ ˆˆ}ˆ ˆ† CH COONa H O (Neutralisation reactions)

2(g) 2(g) 3(g)

N 3H _{‡ ˆˆ}ˆ ˆ† 2NH

_{(synthesis reaction)}

But when Fe

(s)

is heated and water vapour is passed over it in open

vessel, it is converted to Fe

3

O

4(s)

along with the evolution of hydrogen

gas.

3Fe(s) 4H O2 (g) Fe O34(s) 4H2(g)

and when Fe

3

O

4

is reduced with hydrogen gas, it gives Fe

(s)

and H

2

O

3 4(s) 2(g) (s) 2 (g)

But, if the reaction is carried out in a closed container, this reaction

becomes reversible.

3Fe(s)4H O2 (g)ˆ ˆ†‡ ˆˆ Fe O3 4(s)4H2(g)

State of Equilibrium

It has generally been observed that many changes (physical and

chemical) do not proceed to completion when they are carried out in

a closed container. Consider for example vapourisation of water,

Water ˆ ˆ†_{‡ ˆˆ} Vapour

At any temperature, vapourisation of water takes place, initially the

concentration of water is much greater than the concentration of

vapour, but with the progress of time, concentration of vapour

increases whereas that of water remains constant and after a

certain interval of time, there is no change in concentration of

vapour, this state is known as state of physical equilibrium.

In a similar way, this has also been found for chemical reactions, for

example, when PCl

5(g)

is heated in a closed container, its dissociation

starts with the formation of PCl

3(g)

and Cl

2(g)

. Initially, only PCl

5(g)

was

taken, but with the progress of reaction, PCl

3(g)

and Cl

2(g)

are formed

due to dissociation of PCl

5(g)

. After a certain interval of time, the

concentration of PCl

5(g)

, PCl

3(g)

and Cl

2(g)

each becomes constant. It

does not mean that at this point of time, dissociation of PCl

5(g)

and

its formation from PCl

3(g)

and Cl

2(g)

has been stopped. Actually the

rate of dissociation of PCl

5

and the rate of formation of PCl

5(g)

becomes equal. This state is called the state of chemical

equilibrium. So, the state of chemical equilibrium is dynamic.

5(g) 3(g) 2(g)

PCl ˆ ˆ†‡ ˆˆ PCl +Cl

This can be shown graphically.

So, “state of chemical equilibrium can be defined as the state when

the rate of forward reaction becomes equal to rate of reverse

reaction and the concentration of all the species becomes constant”.

Law of Mass Action

Guldberg and Waage in 1807 gave this law and according to this

law, “At constant temperature, the rate at which a substance reacts

is directly proportional to its active mass and the rate at which a

chemical reaction proceeds is directly proportional to the product of

active masses of the reacting species”.

The term active mass of a reacting species is the effective

concentration or its activity (a) which is related to the molar

concentration (C) as

a = f



C

where f = activity coefficient. f



1 but f increases with dilution and

as V →



i.e. C → 0, f → 1 i.e., a → C. Thus at very low

concentration the active mass is essentially the same as the molar

concentration. It is generally expressed by enclosing the formula of

the reacting species in a square bracket.

To illustrate the law of mass action, consider the following general

reaction at constant temperature,

(g) (g) (g) (g)

aA bB _{‡ ˆ ˆ}ˆ ˆ † cC dD

Applying law of mass action,

Rate of forward reaction,

R

f



[A]

a

[B]

b

or R

f

= k

f

[A]

a

[B]

b

…(i)

Where k

f

= rate constant of forward reaction

Similarly,

Rate of reverse reaction

R

r



[C]

c

[D]

d

or R

r

= k

r

[C]

c

[D]

d

…(ii)

where k

r

= rate constant of reverse reaction.

At equilibrium,

Rate of forward reaction = Rate of reverse reaction

i.e., R

f

= R

r

So, from equations (i) and (ii), we get

k

f

[A]

a

[B]

b

= k

r

[C]

c

[D]

d

or,

f c d a b r k [C] [D] k [A] [B]

or

f c d c a b r k [C] [D] K k  [A] [B]

Where, K

c

is the equilibrium constant in terms of molar

concentration.

Equilibrium Constant (K

p

) in terms of Partial Pressures:

Consider the same general reaction taking place at constant

temperature,

From law of mass action,

c d c a b [C] [D] K [A] [B] 

_…(1)

From ideal gas equation

PV = nRT

or,

P n RT V 

At constant temperature,

n P V 

Thus, we can say in a mixture of gases, Partial pressure of any

component (say A)

P

A



[A]

Similarly, P

B



[B]

P

C



[C]

P

D



[D]

So, equation (1) can be rewritten as

c d C D p a b A B P P K P P   

…(2)

Relationship Between K

p

and K

c

From the above, for the same general reaction at constant

temperature.

c d c a b [C] [D] K [A] [B] 

_…(1)

and,

Cc Dd p a b A B P P K P P   

…(2)

From the ideal gas equation,

PV = nRT

n P RT V 

So, P

B

= [B]RT ; P

D

= [D]RT

Similarly,

PA [A]RT; PC [C]RT

Substituting the values of P

A

, P

B

, P

C

and P

D

in equation (2), we get

c c d d p a a b b [C] (RT) [D] (RT) K [A] (RT) [B] (RT)   

or

p ca db (c d)(a b) [C] [D] (RT) K [A] [B] (RT)    

or

n p c K K (RT) 

Where,



n = (c + d) – (a + b)

i.e.,



n = sum of no. of moles of gaseous products – sum of no. of

moles of gaseous reactants.

If



n



0, K

p



K

c

If



n = 0, K

p

= K

c

and if



n



0, K

p



K

c

Illustration 1: Calculate the Kc and Kp for the following

reactions and also deduce the relationship

between Kc and Kp

i) 2SO2(g) + O2(g)

ˆ ˆ †_{‡ ˆ ˆ}

2SO3(g)

ii)

1N_2(g)+ 3H_2(g)ˆ ˆ †_{‡ ˆ ˆ} NH_3(g)

2 2

Solution: i) 2SO2(g)

+ O

2(g) ˆ ˆ †‡ ˆ ˆ

2SO

3(g)

Applying law of mass action,

3 2 c 2 2 2 [SO ] K [SO ] [O ] 

_…(i)

3 2 2 2 SO p 2 SO O P K P P  

…(ii)

p  c



n = 2 – (2+1) = – 1



c p K K RT 

ii)

2(g) 2(g) 3(g) 1 3 N H NH 2 2 ˆ ˆ †‡ ˆ ˆ

3 c 1/ 2 3/ 2 2 2 [NH ] K [N ] [H ] 

_…(i)

3 2 2 NH p 1/ 2 3/ 2 N H P K P P  

…(ii)

Now,

n p c K K (RT)

n 1 3 1 1 2 2     _  _    

c p K K RT 

Illustration 3: At 400K for the gaseous reaction

2A + 3B

ˆ ˆ †_{‡ ˆ ˆ}

3C + 4D

the value of Kp is 0.05. Calculate the value of Kc (R

= 0.082 dm

3

_{atm K}

–1

_mol

–1

₎

Solution: From the given data

For the reaction 2A + 3B

_{‡ ˆ ˆ}ˆ ˆ †

3C + 4D



n = (3 + 4) – (2 +3) = 2

Since R is given in dm

3

_{atm. K}

–1

_mol

–1

_{, we shall take}

Substituting these values in the equation

p c n CRT K K , we get P     _{ }   2 c 1 0.082 400 0.05 K 1      _{ }  

c 0.05 K 0.082 0.082 400 400    

K

c

= 4.648 × 10

–5

Type of Chemical reactions

Irreversible reactions: The reactions which goes to completion and the products fail to recombine to give back reactants, are called irreversible reactions.

Reversible reactions: The reactions which never go to completion and that can occur in either direction i.e. both possibilities are there in reversible reaction (products can give reactants and reactants can give products).

In chemical equilibrium, we will analyse reversible chemical reaction, consider the following reactions.A chemical reaction is never complete. This is due to the reversible nature of it. Reversible means reaction taking place both in the forward and backward directions. The reaction between reactants yielding product(s) is the forward reaction while the reaction between products giving back reactants is the backward reaction. As the reaction progresses, a state of equilibrium is reached when the rate of forward reaction becomes equal to that of the backward reaction. After this there is no progress in the reaction although reactions are taking place in both the directions. This is called the state of chemical equilibrium. It is due to the attainment of a chemical equilibrium that chemical reaction does not go to completion. However, in some chemical reaction equilibrium is attained when the reaction is almost complete. Such reversible reactions are called irreversible reactions. The reversibility of a reaction is denoted by the symbol “_{‡ ˆ ˆ}ˆ ˆ† _{” as shown below:}

i) N₂3H₂ˆ ˆ †_{‡ ˆ ˆ} 2NH₃

ii) H₂ I₂ˆ ˆ †_{‡ ˆ ˆ} 2HI

A reversible reaction may be made irreversible if one of the products is allowed to escape away from the reaction mixture. For example the reaction

3(s) (s) 2(g) CaCO ˆ ˆ†_{‡ ˆ ˆ} CaO CO

will be reversible in a closed container but it would become irreversible in an open container due to escaping of CO2(g). That is CaCO3 CaO(s) + CO2(g) (open container)

Each reversible reaction consists of two reactions: one is forward reaction i.e. reactants combine to give products and other is backward reaction i.e. product reaction recombine to give back reactants.

proceed with the same speed, the reaction is said to be in equilibrium.

Law of Mass Action: In 1863 two Chemistry Guldberg and Waage stated that, at constant temperature, the rate at which a substance reacts is directly proportional to it’s active mass and rate of a chemical reaction is directly proportional to the product of active masses (raised to power their stoichiometric coefficients) of reactants.

Suppose,

A  Products R [A]

or A B Products

R [A][B]

Now if we consider the reversible reaction.

m A m B₁  ₂ ˆ ˆ †_{‡ ˆ ˆ} n C n D₁  ₂ Forward reaction: m A m B1  2 n C n D1 2 Rate of forward reaction m1 m2

f

(r ) [A] [B]

[ ]: denotes activity or active mass m1 m2

f f

r k [A] [B]

where kf is know as rate constant or velocity constant or specific reaction rate for forward reaction.

Backward reaction: n C n D₁  ₂ m A m B_{1 2}

or n1 n2

b b

r K [C] [D]

where Kb is rate constant for backward reaction. At equilibrium rf rb m1 m2 n1 n2 f b k [A] [B] K [C] [D] 1₁ 2₂ n n f eq m m b K [C] [D] K K  [A] [B]

Keq is known as equilibrium constant.

Initially A and B are mixed and they react. When they react, the rate of forward reaction decreases since concentration of A and B decrease. As C and D are formed, they react to form A and B. The rate of backward reaction increases with time because there is an increase in the concentration of C and D. When the two rates are equal, we say that system is in equilibrium.

At equilibrium state

i) Rate of forward reaction = Rrate of backward reaction.

ii) Equilibrium is dynamic in nature i.e. reaction does not stops at equilibrium but two reactions have equal rate.

iii) Once equilibrium is achieved the relative concentrations of reactants and products will remain constant.

iv) External change in pressure, temperature and concentration at equilibrium causes disturbance in equilibrium state. Again a new equilibrium will be reached after some adjustment (in detail this is to be discussed in Le-Chatlier’s principle).

Note: i) Activity or active mass is denoted by “a” or [ ].

ii) Activity of dilute aqueous solutions is expressed in concentration isn mole/litre i.e. molartiy. Actually for solutions.

a C f 

C = Concentration in mole/litre f = activity coefficient

for dilute solutions f = 1 so a = C

iv) Activity of pure solids and pure liquids is taken to be unity.

v) When activities are expressed in concentration (mole/litre) then equilibrium constant Keq is denoted as Kc.

vi) For gaseous reactions, the activities are expressed in terms of partial pressure (total pressure × mole fraction) the equilibrium constant Keq is denoted as Kp. Suppose a reaction N2(g) 3H2(g)‡ ˆ ˆˆ ˆ † 2NH3(g)

 

 

3 2 2 2 NH p 3 N H P K P P  

vii) Only a change in temperature can change the value of equilibrium constant Kp or Kc. There is no effect of change in pressure or concentrations on the value of equilibrium constant (exact variation of equilibrium constant with temperature will be discussed latter).

viii) A very large value of Kc or Kp signifies that reaction is going in forward direction i.e. the equilibrium largely lies towards forward direction or we can say reaction almost goes to completion.

ix) The equilibrium constant of a reaction and it’s reverse reaction are reciprocal of each other.

If A B ˆ ˆ †_{‡ ˆ ˆ} C D (Equilibrium constant = K) Then C D _{‡ ˆ ˆ}ˆ ˆ† A B (Equilibrium constant K 1

K

x) If a chemical equation is multiplied by a certain factor, it’s equilibrium constant must be raised to a power equal to that factor in order to obtain the equilibrium constant for new reaction.

If 2 2

1

NO O NO

2

 _{‡ ˆ ˆ}ˆ ˆ † _{Equilibrium constant = K}

Then 2NO O ₂ˆ ˆ †_{‡ ˆ ˆ} 2NO₂ Equilibrium constant _{K K}2

ix) If a reaction is obtained by adding two reactions then it’s equilibrium constant will be equal to the product of the equilibrium constants of two reactions. If K1 and K2 are step-wise equilibrium constant for

Aˆ ˆ †_{‡ ˆ ˆ} B …(i) B_{‡ ˆ ˆ}ˆ ˆ † C …(ii)

Then for Aˆ ˆ †_{‡ ˆ ˆ} C K K 1K2

x) If a reaction is obtained by substracting two reactions then it’s equilibrium constant will be equal to the ratio of equilibrium constants of two reactions. If K1 be the equilibrium constant for Aˆ ˆ †‡ ˆ ˆ B

K2 be the equilibrium constant for Bˆ ˆ †‡ ˆ ˆ C Then equilibrium constant for reaction

1 2 K A C B D i.e. A+D B C is K  ˆ ˆ †_{‡ ˆ ˆ}  ˆ ˆ †_{‡ ˆ ˆ} 

Reaction Quotient (Q)

At any point in a reversible reaction, ratio of the concentration terms in the same form as in the expression of equilibrium constant can be formulated. This ratio is referred to as reaction quotient (Q). Consider a general reaction.

m A m B₁  _{2 ‡ ˆ ˆ}ˆ ˆ † n C n D₁  ₂ 1₁ 2₂ n n m m [C] [D] Q [A] [B]  (in terms of concentration) or

   

   

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

 (in terms of partial pressure)

Comparing the value of Q with the standard value of Keq at a given temperature, one can comment on the status of reversible system. If

eq

Q K _{It means equilibrium has not yet reached. The system is moving in forward} direction increasing product concentration i.e. at this instant forward reaction is dominant over backward reaction.

eq

Q K

It means equilibrium is established and concentrations of reactants and products will remain constant.

eq

Q K _{It means equilibrium has not yet established. The reaction must proceed in} backward direction for the attainment of equilibrium.

Illustration 1: Kc for equilibrium A(g)‡ ˆˆˆ ˆ† B(g)C(g) is 0.45 at 200°C. One litre of a container holds 0.2 mole of A, 0.3 mole of B and 0.3 mole of C at equilibrium. Calculate the new equilibrium concentrations of A, B and C if the volume of container at same temperature is: a) Halved; (b) Doubled

Solution: A(g) B(g) + C(g) t = teq 0.2 M 0.3 M 0.3M

0.4 M 0.6 M 0.6 M Volume is reduced to half 0.4 x 0.6-x 0.6-x t = tnew eq.

C

0.6 0.6 K 0.4

Q   , So system is not at equilibrium and shifts in

backward direction. KC = 0.45 = (0.6 )2 0.4 x x   on solving, x = 0.12  [B] = [C] = 0.48 M [A] = 0.52 M

t = teq 0.2 M 0.3 M 0.3M 0.1 M 0.15 M 0.15M Not at equal -x +x +x_______ (0.1 – x) (0.15 + x) (0.15 + x) KC = 0.45 = (0.15 )2 0.1 x x   On solving, x = 0.03M  [A] = 0.07 M [B] = [C] = 0.18M DRILL EXERCISE – I

1. Three moles of N2 react with 2 moles of O2 in a two litre container to form one mole of

NO. What will be the number of mole of N2 per litre at equilibrium.

(A) 0.75 (B) 1.50

(C) 1.25 (D) None of these

1. 0.75

2. For the reaction A + 2B ˆ ˆ†_{‡ ˆˆ} 2C + D, initial concentration of A is a and that of B is 1.5 times that of A. concentration of A and D are same at equilibrium. What should be the concentration of B at equilibrium.

(A) a

4 (B) a 2

(C) 3a

4 (D)All of the above

2. a

2

3. Under similar conditions KP1 3 2 2 2NH ˆ ˆ ˆ†_{‡ ˆ ˆˆ} N 3H KP2 3 2 2 1 3 NH N H 2  2 ˆ ˆ ˆ† ‡ ˆ ˆˆ

Which of the following is true for KP1?

(A) KP2 (B) 2 P K 2 (C)(K )P₂ 2 (D) 1 2 P (K ) 3. (K )P₂ 2

Illustration 1: Consider the reaction SO Cl (g)2 2 ‡ ˆˆˆ ˆ† SO (g) Cl (g)2  2 ; 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/lt each at a certain instant. Discuss what will happen in the reaction vessel? Solution: In this question, concentration of three species i.e. SO Cl2 2, SO2 and Cl (g)2 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)   

 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 K

p

and K

c

Consider a gaseous reaction

m A m B₁  _{2 ‡ ˆ ˆ}ˆ ˆ † n C n D₁  ₂

   

 

 

1 2 1 2 n n c D p m n A B P P K P P    …(i)

1₁ 2₂ n n c m m [C] [D] K [A] [B]  …(ii)

Applying ideal gas equation PV = nRT P n RT V  or P = CRT Where C is molarity PAC RT ; PA B C RTB PCC RT ; PC DC RTD Substittuing this in equation (i)



 











1 2 1 2 n n C D m m A B C RT C RT C RT C RT   =

 

 

1 2 1 2 1 2 1 2 n n n n C D m m m m A B RT C C C C _RT    _  = (n n ) (m m )1 2 1 2 c K (RT)    n p c K K (RT) n

 = [number of moles of gaseous products in gaseous state – number of moles of gaseous

reactants in gaseous state..

If Case – I  n 0

Kp Kc

e.g. H2(g) I2(g)‡ ˆ ˆˆ ˆ † 2HI(g)

N2(g)O2(g)‡ ˆ ˆˆ ˆ † 2NO(g)

Gaseous equilibria are of two types Type I: In which  n 0 to Kp Kc e.g. H2(g) I2(g)‡ ˆ ˆˆ ˆ † 2HI(g) N2(g)O2(g)‡ ˆ ˆˆ ˆ † 2NO(g) Type II In which  n 0 Case - II n  0 So Kp Kc

Kp = Kc(RT)1 Kp Kc Case – III n  0 N2(g)3H2(g)‡ ˆˆˆ ˆ† 2NH3(g)      n 2 (3 1) 2 Kp = Kc(RT)–2  Kp Kc N2(g) 3H2(g)‡ ˆ ˆˆ ˆ † 2NH3(g); n 0 

Illustration 2: fFor a homogenous gaseous reaction X 2Y ˆ ˆ†_{‡ ˆˆ} Z, at 473 K, the value

of Kc 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 Kp using the relation between c K and Kp, Kc 0.35, R 0.0821, T 473,   n 1 – 3 –2 Kp K (RT)c n 0.35 (0.0821 473)2 2.32 10–4       

The expression for Kp is :

Z p 2 X Y p K p (p )  Moles X Y Z Initially 1 1 0 At equilibrium 1 – x 2 – 2x X  total moles nT 3 – 2x Let P = equilibrium pressure

 X Y 1– x 2 – 2x P P, P P, 3 – 2x 3 – 2x   z x P P 3 – 2x  2 p 2 2 2 X P _{x(3 – 2x)} 3 – 2X K P (1– x)(2 – 2x) 1– x 2 – 2x P P 3 – 2x 3 – 2x         x = 0.6 (given) 2 –4 p 2 2 2 0.6(3 –1.2) K 2.32 10 P (1– 0.6) –1.2)     2 2 2 P (1.8 10 )  P = 180 atm DRILL EXERCISE – II

1. A mixture of hydrogen and iodine (molecular ratio is 2 : 1) is reacted to form HI. The total moles at equilibrium will be:—

(A)3 – x (B) 3 (C) 2 – 2x (D)3 – 4x

1. 3

2. In the reaction, A + B _{‡ ˆˆ}ˆ ˆ† C + D in a one-litre container, concentration of B at fixed temperature was n mole and initial concentration of A was 3n mole. If the concentration of C at equilibrium is equivalent to that of B, the concentration of D will be

(A) n 4 (B) n (C) n 3 (D) n 2 2. n 4

3. Which of the following should be added in the equilibrium reaction 2SO2 + O2 ‡ ˆˆˆ ˆ†

2SO3 for generation of heat, if the forward reaction is exothermic and the backward

reaction is endothermic?

(A)SO3 (B) SO2 and O2

(C) SO2, O2 and SO3 in equal amount (D)None of these 3. SO2 and O2

i) The vapour density of a mixture consisting of NO2 and N2O4 is 38.3 at 26.7°C. Calculate the

number of moles of NO2 in 100 gm mixture.

ii) Establish a relationship between Kc and Kp for the following reactions.

a) N2(g) + O2(g) ‡ ˆ ˆˆ ˆ † 2NO(g)

b) (NH ) CO_{4 2 3(s)}ˆ ˆ †_{‡ ˆ ˆ} 2NH_3(g)CO_2(g) H O_{2 (g)}

Application of Law of Mass Action

1. Synthesis of Hydrogen Iodide: Suppose ‘a’ moles of H2 and ‘b’ moles of I2 are

heated at 444°C in a closed container of volume ‘V’ litre and at equilibrium, 2x moles of HI are formed.

H2(g) + I2(g) ˆ ˆ †‡ ˆ ˆ 2HI(g)

Initial concentration (mol L–1₎ a

V b V 0 Equilibrium concentration(mol L–1₎ a x V  b x V  2x V 2 c 2 2 [HI] K [H ][I ]  …(i)

Substituting the equilibrium concentrations of H2, I2 and HI in equation (i), we get 2 2 c 2x 4x V K a x b x (a x)(b x) V V                      …(ii) 2 2 2 HI p H I P K P P   …(iii)

PHI = mole fraction of HI × Total pressure PHI = 2x P a b  Similarly, 2 H (a x) P P (a b)     and I2 b x P P (a b)    

Substituting these values in equation (iii) we get

2 2 p 2x P a b K a x b x P P a b a b    _         _{ }  _  _   _       p 2 4x K (a x)(b x)    …(iv)

So, one can see from equations (iii) and (iv), that Kp = Kc

This is so, because n= 0 for the synthesis of HI from H2 and I2.

2. Thermal Dissociation of Phosphorus Pentachloride

PCl5(g) dissociates thermally according to the reaction,

PCl5(g) ‡ ˆ ˆˆ ˆ † PCl3(g) + Cl2(g)

Let us consider that 1 mole of PCl5 has been taken in a container of volume V litre and at

equilibrium x moles of PCl5(g) dissociates. Thus

PCl5(g) ˆ ˆ †‡ ˆ ˆ PCl3(g) + Cl2(g)

Initial concentration (mol L–1_{)1 0 0}

Equilibrium concentration(mol L–1₎ 1 x V  x V x V

According to law of mass action, at constant temperature, Kc = 3 2 5 [PCl ][Cl ] [PCl ] …(1) and Kp = 3 2 5 PCl Cl PCl P P P  …(2)

Substituing the values of equilibrium concentration, in equation (1), we have Kc = x x V V 1 x V   or 2 c x K V(1 x)  

Mole fraction of PCl3 = Mole fraction of Cl2 = x

1 x

and mole fraction of PCl5 =

1 x 1 x

 

Suppose total pressure at equilibrium is P, then we have from equation (2),

2 p x x P 1 x 1 x K 1 x P 1 x     _  or 2 p 2 x P K (1 x )   x  1  p 2 p K 1 K x P x x P P      

Similarly, we can apply law of mass action on any reaction at equilibrium. BRAIN TEASER 2:

Derive an expression for Kc and Kp for the reaction

N2(g) + 3H2(g) ˆ ˆ †‡ ˆ ˆ 2NH3(g)

Assuming that in a container of volume V, initially 1 mole of N2 and 3 moles of H2 were taken and at equilibrium 2x moles of NH3 is formed.

Illustration 3: For the chemical equilibrium 2NO(g) + O2(g) ‡ ˆˆˆ ˆ† 2NO2(g)

Kp = 1.3 × 10–2 at 1000K. What is the value of Kc at 1000K? Solution: 1.3  10-2 _{= K}

C (0.821  1000) -1

DRILL EXERCISE – III

1. What concentration of CO2 be in equilibrium with 2.5  10–2 mol litre–1 of CO at 100°C

for the reaction

FeO(s) + CO(g) _{‡ ˆˆ}ˆ ˆ† Fe(s) + CO2(g); KC = 5.0 1. 12.5  10 _{mol litre}–2 –1

2. The equilibrium constant KC for A(g) ‡ ˆˆˆ ˆ† B(g) is 1.1. Which gas has a molar

concentration greater than 1?

2. If [B] = 1; [A] = 0.91

3. 0.1 mole of N2O4(g) was sealed in a tube under atmospheric conditions at 25°C.

Calculate the number of mole of NO2(g) present, if the equilibrium N2O4(g) ˆ ˆ†‡ ˆˆ

2NO2(g), (Kp = 0.14) is reached after some time. 3. NO2 = 0.017  2 = 0.034 mole

The degree of dissociation at a certain or given temperature of PCl5 at 2 atm is found to be 0.4. At what pressure, the degree of dissociation of PCl5 will be 0.6 at the same temperature? Also calculate the equilibrium constant for the reverse reaction.

Solution: The dissociation of PCl5 takes place according to the equation,

PCl5 ‡ ˆ ˆˆ ˆ † PCl3(g) + Cl2(g)

Let 1 mole of PCl5 is taken in a closed container. Then

PCl5(g) ˆ ˆ †‡ ˆ ˆ PCl3(g) + Cl2(g)

Mole(s) before dissociation 1 0 0 Mole(s) at equilibrium 1 –   

As degree of dissociation() is given at 2 atm 1 –  = 1.0 – 0.4 = 0.6

 Total moles at equilibrium = 1 –  +  +  = 1 – 0.4 + 0.4 + 0.4 = 1.4 Now, 3 2 5 PCl Cl p PCl P P K P   ₌ 0.4 0.4 2 2 1.4 1.4 0.6 2 1.4     p 0.4 0.4 4 1.4 0.32 K 0.6 2 1.4 1.4 0.84        

Now, as temperature remains the same, Kp will also remain the same. So for

 = 0.6

PCl5(g) ˆ ˆ †‡ ˆ ˆ PCl3(g) + Cl2(g)

Mole before dissociation 1 0 0 Moles at equilibrium 1 – 0.6 0.6 0.6

Again, 2 p 0.6 0.6 P 1.6 1.6 K 0.4 P 1.6     p 0.6 0.6 1.6 K P 0.4 1.6 1.6      32 9 84 16 P P = 32 4 128 9 21 189  = 0.677 atm BRAIN TEASER 3:

i) Calculate the percentage dissociation of H2S(g), if 0.1 mole of H2S is kept in 0.4 litre

vessel at 1000K. For the reaction, 2H2S(g) ˆ ˆ †‡ ˆ ˆ 2H2(g) + S2(g), the value of Kc = 1.6 × 10–6

ii) A sample of HI was found to be 22% dissociated when equilibrium was reached. What will be the degree of dissociation if hydrogen is added in the proportion of 1 mole for every mole of HI originally present, the temperature and volume of the system being kept constant?

Equilibrium Constant for Heterogeneous Equilibria

The equilibrium which involves reactants and products in different physical states. The law of mass action can also be applied on heterogeneous equilibria as it was applied for homogeneous equilbria (involving reactants and products in same physical states).

i) Thermal Dissociation of Solid Ammonium Chloride: The thermal dissociation of NH4Cl(s) takes place in a closed container according to the equation:

NH4Cl(s) ˆ ˆ †‡ ˆ ˆ NH3(g) + HCl(g)

NH4Cl(s) ‡ ˆ ˆˆ ˆ † NH3(g) + HCl(g)

Initial moles 1 0 0

Moles at equilibrium 1 – x x x

Applying law of mass action, c 3 4

[NH ][HCl] K

[NH Cl]



As NH4Cl is a pure solid, so there is no appreciable change in its concentration. Thus,

Kc = [NH3] [HCl] 2 c 2 x x x K V V V    and Kp PNH3PHCl

ii) Thermal Dissociation of Ag2CO3: Ag2CO3(s) dissociates thermally according to

the equation.

Ag2CO3(s) ‡ ˆ ˆˆ ˆ † Ag2O(s) + CO2(g)

Applying law of mass action, at constant temperature, we get,

2 2 c 2 3 [Ag O][CO ] K [Ag CO ] 

Now, let us consider that 1 mol of Ag2CO3(s) is heated in a closed container of volume V

and x mol of Ag2CO3(s) dissociates at equilibrium, then

Ag2CO3(s) ‡ ˆ ˆˆ ˆ † Ag2O(s) + CO2(g)

Initial moles 1 0 0

Equilibrium moles 1 – x x x

Now, as Ag2CO3 and Ag2O are solids, so their concentration can be assumed to be

constants.tThus c 2 x K [CO ] V   2 P CO K P

The Le–Chatelier’s Principle

This principle, which is based on the fundamentals of a stable equilibrium, states that

“When a chemical reaction at equilibrium is subjected to any stress, then the equilibrium shifts in that direction in which the effect of the stress is reduced”.

Confused with “stress”. Well by stress here what we mean is any change of reaction conditions e.g. in temperature, pressure, concentration etc.

This statement will be explained by the following example. Let us consider the reaction: 2NH3 (g)

endo exo

ˆ ˆ ˆ †ˆ

‡ ˆ ˆ ˆˆ N2 (g) + 3H2 (g)

Let the moles of N2, H2 and NH3 at equilibrium be a, b and c moles respectively. Since the

reaction is at equilibrium,

 

 

2 3 2 3 N H 2 NH P P P  = Kp =









2 3



2 3 N T H T 2 NH T X .P X .P X .P Where,

X terms denote respective mole fractions and PT is the total pressure of the system.

 3 T T p 2 T a b P P a b c a b c _K c P a b c  _   _  _{ }   _{ }      _  _   _{ }    Here, a a b c  = mole fraction of N2 b a b c  = mole fraction of H2 c a b c  = mole fraction of NH3



 





2 3 T 2 2 P ab c  a b c  = Kp Since PT =





a b c RT V  

( assuming all gases to be ideal)

 ab₂3 RT 2

V c

 

_ _ = KP …(1)

Now, let us examine the effect of change in certain parameters such as number of moles, pressure, temperature etc.

If we increase a or b, the left hand side expression becomes QP ( as it is disturbed from

equilibrium) and we can see that QP > KP

The reaction therefore moves backward to make QP = KP.

If we increase c, QP < KP and the reaction has to move forward to revert

back to equilibrium.

If we increase the volume of the container (which amounts to decreasing the pressure), QP < KP

and the reaction moves forward to attain equilibrium.

If we increase the pressure of the reaction, then equilibrium shifts towards backward direction since in reactant side we have got 2 moles and on product side we have got 4 moles. So pressure is reduced in backward direction.

If temperature is increased, the equilibrium will shift in forward direction since the forward reaction is endothermic and temperature is reduced in this direction.

However from the expression if we increase the temperature of the reaction, the left hand side increases (QP) and therefore does it mean that the reaction goes backward (since QP > KP)?.

Does this also mean that if the number of moles of reactant and product gases are equal, no change in the reaction is observed on the changing temperature (as T would not exist on the left hand side)?. The answer to these questions is No. This is because KP also changes with

temperature. Therefore, we need to know the effect of temperature on both QP and KP to decide

the course of the reaction.

Effect of Addition of Inert Gases to A a Reaction Aat Equilibrium 1. Addition at constant pressure

Let us take a general reaction aA + bB cC + dD

We know, c d C D T T p a b A B T T n n P P n n K n n P P n n                                     









Where,

nC nD, nA, nB denotes the no. of moles of respective components and PT is the total

pressure and n = total no. of moles of reactants and products. Now, rearranging, n c d c D T P a b A B n n P K n n n          _



_ Where n = (c + d) – (a + b)

Now, n can be equal to = 0,  0 or  0 Let us take each case separately.

a) n = 0 : No effect b) n = ‘+ve’ :

Addition of inert gas increases the n i.e. PT

n       



 is decreased and so is n T P n        



 . So

products have to increase and reactants have to decrease to maintain constancy of Kp.

So the equilibrium moves forward.

In this case PT n       



 decreases but n T P n       





 increases. So products have to decrease

and reactants have to increase to maintain constancy of Kp. So the equilibrium moves

backward.

2. Addition at Constant Volume: Since at constant volume, the pressure increases with addition of inert gas and at the same time n also increases, they almost counter balance each other. So

n T P n       





 can be safely approximated as constant. Thus addition of

inert gas has no effect at constant volume. .

Dependence of Kp or Kc on Temperature

Now we will derive the dependence of KP on temperature. Starting with Arrhenius equation of rate constant

af

E / RT

f f

k A e … (i)

Where, kf = rate constant for forward reaction, Af = Arrhenius constant of forward reaction, Eaf = Energy of activation of forward reaction

E / RTar

r r

k A e …(ii) Dividing (i) by (ii) we get,

  ar af E E f f RT r r k A e k A   We know that f r k K k  (equilibrium constant )  K = f f EarRTEaf r r k A e k A   At temperature T1

  ar af 1 1 E E RT f T r A K e A   …(iii) At temperature T2   ar af 2 2 E E RT f T r A K e A   … (iv)

Dividing (iv) by (iii) we get

ar af 2 2 1 1 E E 1 1 T R T T T K e K            _{  }    log 2 r f 1 T a a T 2 1 K E E _{1 1} K 2.303 R T T     _  _  

The enthalpy of a reaction is defined in terms of activation energies as Eaf Ear = H  2 1 T T 2 1 K H 1 1 log K 2.303 R T T     _  _   log T2 T 1 2 K ΔH 1 1 = -K 2.303 R T T       …(v) f a E a H Reaction co-ordinate r a E E E ne rg y

For an exothermic reaction,



H would be negative. If we

increase the temperature of the system

(T

2

>T

1

), the right hand side of the equation (V) becomes

negative.



K < KT2 T1

, that is, the equilibrium constant at the higher

temperature would be less than that at the lower

temperature.

Now let us analyse our question. Will the reaction go forward or

backward?

Before answering this, we must first encounter another

problem. If temperature is increased, the new K

P

would

either increase or decrease or may remain same. Let us

assume it increases.

Now, Q

P

can also increase, decrease or remain unchanged. If K

P

increases and Q

P

decreases, then

QPT2 KPT2

, therefore the

reaction moves forward. If K

P

increase and Q

P

remains

same, then

QPT2

 

QPT1 KPT2

. Again, the reaction moves forward.

What, if K

P

increase and Q

P

also increases?

Will

QPT2 KPT2

or

QPT2 KPT2

or

QPT2 KPT2

? This can be answered by

simply looking at the dependence of Q

P

and K

P

on

temperature. You can see from the equation that K

P

depends on temperature exponentially. While Q’s

dependence on T would be either to the power g,l,t……..

Therefore the variation in K

P

due to T would be more than

in Q

P

due to T.



K

P

would still be greater than Q

P

and the reaction moves

Therefore, to see the temperature effect, we need to look at K

P

only. If it increases, the reaction moves forward, if it

decreases, reaction moves backward and if it remains

fixed, then, no change at all.Variation of Equilibrium

Constant with Temperature

As we have discussed that only a change in temperature can change the value of the equilibrium constant.

According to Arrhenius equation _{K A.e} –Ea/ RT

K = rate constant A = frequency factor Ea  Energy of activation so for forward reaction – E / RTaf

f f

K A .e

similarly for backward reaction – E / RTab b b K A .e ab af E – E RT f f eq b b K A K e K A        

Let K1 be the equilibrium constant at temperature T1 and K2 be the equilibrium constant at temperature T2 ab af E – E f RT 1 b A K e A  ab af 2 2 E – E 1 _– 1 R T T f 2 b A K e A        2 ab af 1 2 1 E – E K 1 1 ln – K R T T    _{ }   but  H E – Eaf ab 2 1 2 1 K – H 1 1 ln – K R T T     _{ }   2 1 1 2 K – H 1 1 ln – K R T T     _{ }   2 2 1 1 1 2 K H T – T log K 2.303 R T T     _{ }  

Notes: i) If T2 T1 and reaction is endothermic ( H 0)  then right hand side is +ve so K2 K1 i.e. for endothermic reactions increase in temperature increases the value of equilibrium constant and decrease in temperature will decrease the value of equilibrium constant.

ii) For exothermic reactions ( H 0)  and increase in temperature decreases the value of equilibrium constant and decrease in temperature increases the value of equilibrium constant. Illustration 4: Under what conditions will the following reactions go in the forward

direction ?

1. N2(g)+ 3H2(g) 2NH3(g) + 23 k cal. 2. N2(g) + O2(g) 2NO(g) - 43.2 k cal. 3. C(s) + H2O(g) CO2(g) + H2(g) + X k cal. 4. N2O4(g) 2NO2(g) - 14 k cal.

Solution: 1. Low T, High P, excess of N2 and H2.

2. High T, any P, excess of N2 and O2

3. Low T, Low P, excess of C and H2O

4. High T, Low P, excess of N2O4.

DRILL EXERCISE – IV

1. In the reaction, C(s) + CO2(g) ‡ ˆˆˆ ˆ† 2CO(g), the equilibrium pressure is 12 atm. If

150% of CO2 reacts, calculate Kp. 1. 16 atm.

2. At 298 K, the equilibrium between N2O4 and NO2 may be represented by the following

equation N2O4(g) ‡ ˆˆˆ ˆ† 2NO2(g). If the total pressure of the equilibrium mixture is P

and the degree of dissociation of N2O4(g) at 298 K is x, which one of the following is

the pressure of N2O4(g) under this condition:

(A) (1 x) P (1 x)    (B) 2x P (1 x)   (C) 2x P (1 x)   (D) 2 P 3  2. (1 x) P (1 x)   

3. One mole of SO3 was placed in one litre vessel at a certain temperature. The following

equilibrium was established 2SO3 ˆ ˆ†‡ ˆˆ 2SO2 + O2. At equilibrium 0.6 moles of SO2

were formed. The equilibrium constant of the reaction will be: (A) 0.36 (B) 0.45

(C) 0.54 (D) 0.675

3. 0.675

4. 2 moles of PCl5 were heated in a closed vessel of 2 litre capacity. At equilibrium, 40%

of PCl5 dissociated into PCl3 and Cl2. The value of equilibrium constant is:

(A) 0.267 (B) 0.53 (C) 2.63 (D) 5.3

4. 0.267

5. For reaction: PCl3(g) + Cl2(g) ˆ ˆ†‡ ˆˆ PCl5(g)

the value of KC at 250°C is 26. The value of Kp at this temperature will be:

(A) 0.61 (B) 0.57 (C) 0.83 (D) 0.3

Illustration 4: Under what conditions will the following reactions go in the forward direction ? 1. N2(g)+ 3H2(g) 2NH3(g) + 23 k cal.

2. N2(g) + O2(g) 2NO(g) - 43.2 k cal. 3. C(s) + H2O(g) CO2(g) + H2(g) + X k cal. 4. N2O4(g) 2NO2(g) - 14 k cal.

Solution: 1. Low T, High P, excess of N2 and H2.

2. High T, any P, excess of N2 and O2

3. Low T, Low P, excess of C and H2O

4. High T, Low P, excess of N2O4.

BRAIN TEASER 4:

Under what conditions will the following reactions go in the forward direction? 1. 2SO2(g) + O2(g) 2SO3(g) + 45 k cal.

2. 2NO(g) + O2(g) 2NO2(g) + 27.8 k cal. 3. PCl5(g) PCl3(g) + Cl2(g)- X k cal.

Le-Chatelier’s Principle and Physical Equilibria: 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

Liquid _{‡ ˆ ˆ}ˆ ˆ † 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:

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 ˆ ˆ †_{‡ ˆ ˆ} 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) ˆ ˆ †_{‡ ˆ ˆ} 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.

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.

Free Energy and Chemical Equilibrium

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

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 + TS

The driving force is made up of two parts, –H term and TS term. The –H term is the heat of reaction at constant pressure and TS 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.

If the reaction is carried out at constant volume, the decrease in Helmholtz function –G = –E + TS would be the proper measure of affinity of the reactant or the driving force of the reaction. Now we can see why Berthollet and Thompson were wrong in assuming that driving force of the reaction was the heat of reaction. They neglected the TS term. The reasons for the apparent validity of their principle was that for many reactions, H term far outweighs the TS term. This is especially true at low temperature, since at higher temperature, TS 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

1

H O H O; H 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

is not favoured. However, the thermite reaction 2Al(s) + 3

2O2(g)   → Al2O3(s)

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 _{‡ ˆ ˆ}ˆ ˆ † cC dD

According to Van’t Hoff reaction isotherm

c d 0 C D a b A B p p G G RT ln p p       = G0 + RTlnQp

the condition for a system to be at equilibrium is that

G = 0 Thus at equilibrium c d 0 C D 0 0 p a b A B p p 0 G RT ln G RT ln K p p       _{ }       Whence G0 _{= – RTlnK}0 p Hence 0 0 p G ln K RT  

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

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 45: Kc for the reaction N2O4 ˆ ˆ †‡ ˆ ˆ 2NO2 in chloroform at 291 K is 1.14.

Calculate the free energy change of the reaction when the concentration of the two gases are 0.5 mol dm–3 _{each at the same temperature. (R = 0.082}

lit atm K–1 _mol–1_.)

Solution: From the given data

T = 291 K; R = 0.082 lit atm K–1 _mol–1

Kc = 1.14; CNO2 CN O2 4= 0.5 mol dm

–3

The reaction quotient Qc for the reaction N O_{2 4}‡ ˆ ˆˆ ˆ † 2NO ,₂

2 2 c 2 4 [NO ] 0.5 0.5 Q 0.5 [N O ] 0.5    

Since Qp = Qc(RT)n and n = 2 – 1 = 1 in this case

Qc = 0.5 (0.082 × 291) = 11.93

Kp = Kc(RT)n = 1.14 (0.082 × 291) = 27.1

Substituting these values in the equation

G = G0 _{+ RTlnQ}

p = – RT ln Kp + RT ln Qp, we get

= – 2.303 RT (log Kp – log Qp)

G = – (0.082 × 291 × 2.303) [log 27.2 – log 11.93]

= – 54.95 (1.4346 – 1.0766) = – 19.67 lit atm

Illustration 6: Ammonium hydrogen sulphide dissociates according to the equation NH4HS(s) NH3(g) + H2S(g). If the observed pressure of the mixture is 1.12 atmosphere at 106°C, what is the equilibrium constant Kp of the reaction?

Solution: NH4HS(s) NH3(g) + H2S(g) t = 0 - 0 0

t = teq x atm x atm PT = 2x = 1.12

x = 0.56 KP = x2 _{= 0.3136 amt}2

Illustration 56: Calculate the pressure of CO2 gas at 700K in the heterogeneous equilibrium reaction CaCO3(s) ‡ ˆ ˆˆ ˆ † CaO(s) + CO2(g)if G0 for this reaction is 130.2 kJ mol–1_.

Also, G0  RT ln Kp  0 3 1 p 1 1 G 130.2 10 Jmol ln K RT (8.314JK mol )(700K)         2 10 CO p p K 1.94 10  atm

Illustration 67: For the equilibrium NiO(s) + CO(g) ‡ ˆ ˆˆ ˆ † Ni(s) + CO2(g), G0 (J mol–1)

= – 20,700 – 11.97 T. Calculate the temperature at which the product gases at equilibrium at 1 atm will contain 400 ppm (parts per million) of carbon monoxide.

Solution: For the given reaction Kp pCO2/ pCO

Since pCO pCO2, hence p 6 CO 1 1 K 2,500 p 400 10     G0  RT ln Kp  p 0 G 20,700 11.97T ln K RT RT    

The equation when solved for T using R = 8.314 K–1 _mol–1_{, gives T = 399K.} Illustration 8: Equilibrium constants are given (in atm) for the following reactions at 0°C a) SrCl26H2O(s) ‡ ˆˆˆ ˆ† SrCl22H2O(s) + 4H2O(g), Kp = 6.89 × 10–12 b) Na2HPO412H2O(s) ‡ ˆˆˆ ˆ† Na2HPO47H2O(s) + 5H2O(g), Kp = 5.25 × 10– 13 c) Na SO10HO _{‡ ˆˆ}ˆ ˆ† Na SO + 10H O K = 4.08 × 10–25

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

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, Na2SO210H2O 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 : (a) KP = PH O4₂ = 6.89  10-12  PH O2 = 1.62  10 -3 atm KP = H O2 5 P = 5.25  10 -13 _ 2 H O P = 3.5  10-3 _atm KP = H O2 10 P = 4.08  10 -25 _ 2 H O P = 3.6  10-3 _atm

(b) SrCl2.2H2O, Since it has lowest vapour pressure water in equilibrium

(c) The hydrate will loss water (efforesce) below. 2.77 torr i.e., 2.77  torr 100% 60.5%

4.58torr   relative humidity (d) The dehydrated salt will absorb H2O above 60.5% relative humidity.

BRAIN TEASER 5:

i) Calculate the partial pressure of HCl gas above solid a sample of NH4Cl(s) as a result of its

decomposition according to the reaction: NH4Cl(s) ˆ ˆ †‡ ˆ ˆ NH3(g) + HCl(g) Kp =?

ii) Calculate the equilibrium constant of a reaction at 300 K if G0 _{at this temperature for} the reaction is 29.29 kJ mol–1_.

Illustration 78: For the formation of ammonia the equilibrium constant data at 673K and 773K respectively are 1.64 × 10–4 _{and 1.44 × 10}–5 _{respectively. Calculate heat of the reaction.} Given R = 8.314 JK–1 _mol–1_.

Solution: Substituting the values in the equation 2 1 p _{2 1} p 2 1 K H T T ln , we get K R T T      _{ }   5 4 1.44 10 H 773 673 2.303log 8.314 773 673 1.64 10    _      _     2.303log(0.0878) H(100) 8.314 773 673     100H = 2.303 (–1.0565) × 673 × 773 × 8.314 whence, H 673 773 2.303 1.0565 8.314 100        = – 105216J = – 105.216kJ BRAIN TEASER 6:

The equilibrium constant KP for the reaction N2(g) + 3H2(g) 2NH3(g) is 1.6  10-4 atm at 400o_{C. What will be the equilibrium constant at 500}o_{C if heat of the reaction in this} temperature range is 25.14 k cal?

Relation between Vapour Density and Degree of Dissociation

In the following reversible chemical equation A _{‡ ˆ ˆ}ˆ ˆ † yB

Initial mol 1 0

At equilibrium (1 –x) yx x = degree of dissociation Number of moles of A and B at equilibrium = 1 – x + yx = 1 + x(y –1)

If initial volume of 1 mole of A is V, then volume of equilibrium mixture of A and B is = [1 + x(y – 1)]V

Molar density before dissociation, D molecular weight m volume V

 

Molar density after dissociation d m [1 x(y 1)]V     D [1 x(y 1)]1 d     x D d d(y 1)   

Note: y is the number of moles of products from one mole of reactant. D

d is also called Van’t

Thus for the equilibria I: PCl5(g)ˆ ˆ †‡ ˆ ˆ PCl3(g)Cl2(g), y = 2 II: N O2 4(g)‡ ˆ ˆˆ ˆ † 2NO2(g), y = 2 III: 2 2 4 1 2NO N O , y = 2 ˆ ˆ† ‡ ˆ ˆ  x D d d   2(d D) x d  

Also D × 2 = Molecular weight (theoretical value)

d × 2 = Molecular weight (abnormal value) of the mixture

Illustration 89: Vapour density of the equilibrium mixture of NO2 and N2O4 is found to be

40 for the equilibrium N2O4 ˆ ˆ †‡ ˆ ˆ 2NO2

Calculate (i) abnormal molecular weight ii) degree of dissociation

iii) percentage of NO2 in the mixture

Solution: i) N2O4 ˆ ˆ †‡ ˆ ˆ 2NO2

Observed value of vapour density (d) = 40 Hence, abnormal molecular weight = 40 × 2 = 80 ii) D × 2 = theoretical molecular weight = 2

 D 92 46 2    x D d 46 40 0.15 d 40      iii) N2O4 ˆ ˆ †‡ ˆ ˆ 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

100 100

1 x  1.15 =

DRILL EXERCISE –V

1. An equilibrium mixture for the reaction: 2H2S(g) ˆ ˆ†‡ ˆˆ 2H2(g) + S2(g) has 0.5 mole

H2S, 0.1 mole of H2 and 0.4 mole S2 in a one litre vessel. The equilibrium constant of

this reaction is given by:

(A) 0.004 mole litre–1 _{(B) 0.08 mole litre}–1

(C) 0.016 mole litre–1 _{(D) 0.160 mole litre}–1 1. 0.016 mole litre–1

2. Mole percentage of CO in vapour state is 50 at equilibrium at constant temperature and pressure for the reaction, C(Graphite) + CO2(g) ‡ ˆˆˆ ˆ† 2CO(g). What will be degree of

dissociation.

(A) 0.33 (B) 0.25 (C) 0.033 (D) 0.50

2. 0.33

3. The volume occupied by 3.6 g of PCl5 is 1 litre on complete vaporization at 1

atmospheric pressure and 200°C. Partial dissociation of PCl5 gives PCl3 and Cl2. What

should be the percentage of dissociation of PCl5 at the same temperature

(A) 4.9% (B) 0.49% (C) 49% (D) 0.049%

3. 49%

4. If x, y and z are the initial moles of PCl5, PCl3 and Cl2 for PCl5 ‡ ˆˆˆ ˆ† PCl3 + Cl2 and a

is amount of dissociation and P is total pressure, the mole fraction of PCl3 will be

(A) y ax x y z ax     (B) x az x y z ax     (C) x y ax x y z ax      (D) z ax x y z ax    

4. y ax

x y z ax

   

Answer to Drill Exercises

Drill Exercise – I 1. 0.75 2. a 2 3. (K )P₂ 2 Drill Exercise – II 1. 3 2. n 4 3. SO2 and O2

Drill Exercise – III 1. 12.5  10 _{mol litre}–2 –1

2. If [B] = 1; [A] = 0.91

3. NO2 = 0.017  2 = 0.034 mole

Drill Exercise – IV 1. 16 atm.

2. (1 x) P (1 x)    3. 0.675 4. 0.267 5. 0.61

Drill Exercise – V 1. 0.016 mole litre –1

2. 0.33 3. 49% 4. y ax x y z ax    

Solved Problems Objective

BRAIN TEASER 7:

N2O4(g) ‡ ˆ ˆˆ ˆ † 2NO2(g). In this reaction, NO2 is 20% of the total volume at equilibrium. Calculate

a) Vapour density

b) abnormal molecular weight c) percentage dissociation of N2O4

Brain Teaser 1: i) 0.43 ii) a) Kp = Kc b) Kp = Kc (RT)4 Brain Teaser 2: 2 2 c 4 4x V K 27(1 x)   and 2 2 p 4 2 16x (2 x) K 27(1 x) P    Brain Teaser 3: i) 2%; (ii) 0.037

Brain Teaser 4: 1. Low T, High P, excess of SO2 and O2.

2. Low T, High P, excess of NO and O2

3. High T, Low P, excess of PCl5

Brain Teaser 5: (i) 1.02 × 10–8 _{atm; (ii) K}0

p = 7.96 × 10–6

Brain Teaser 6: KP2  4.835