Ncert Exercises & Solutions

Write a relation between $∆ G$ and Q and define the meaning of each term and answer the following.

1. Why a reaction proceeds forward when Q

2. Explain the effect of increase in pressure in terms of reaction quotient Q. For the reaction, $CO (g) + 3 H_{2} (g) ⇌ {CH}_{4} (g) + H_{2} O (g)$

The relation between

∆ G

and Q is

∆ G = ∆ G^{⊝} + RTlnQ

∆ G = change in free energy as the reaction proceeds .

∆ G^{⊝} = standard free energy

Q = reaction quotient

R = gas constant

T = absolute temperature in K

(a)

∆ G^{⊝} = - RT \ln K

∴ ∆ G = - RT \ln K + RT \ln Q

∆ G = RT \ln \frac{Q}{K}

If Q < K,

∆ G

will be negative and the reaction proceeds in the forward reaction.

If Q = K,

∆ G = 0

reaction is in equilibrium and there is no net reaction.

(b) $CO (g) + 3 H_{2} (g) ⇌ {CH}_{4} (g) + H_{2} O (g)$

K_{c} = \frac{[{CH}_{4}] [H_{2} O]}{[CO] {[H_{2}]}^{3}}

On increasing pressure, volume decreases. If we doubled the pressure, volume will be halved but the molar concentrations will be doubled. Then,

Q_{c} = \frac{2 [{CH}_{4}] \cdot 2 [H_{2} O]}{2 [CO] {2 [H_{2}]}^{3}} = \frac{1}{4} \frac{[{CH}_{4}] [H_{2} O]}{[CO] {[H_{2}]}^{3}} = \frac{1}{4} K_{c}

Therefore, Q_c is less than K_c, so Q_c will tend to increase to re-establish equilibrium and the reaction will go in forward direction.