Dihydrogen gas is obtained from natural gas by partial oxidation with steam as per following endothermic reaction: CH 4 (g) + H 2 O (g) ⇌ CO (g) + 3H 2 (g) A. Write as expression for K p for the above reaction. B. How will the values of K p and composition of equilibrium mixture be affected by (i) increasing the pressure (ii) increasing the temperature (iii) using a catalyst?

Question

Philoid · Accepted Answer

a) For the given reaction,

CH₄ (g) + H₂O (g) ⇌ CO (g) + 3H₂ (g)

By using the formula,

K_p =

Where P_x and P_Y are the partial pressures of x and y(products)

P_A and P_B are the partial pressures of a and b (reactants)

Hence,

K_p =

b) (i) According to Le Chatelier’s principle, an increase in pressure applied to the system at equilibrium favours the reaction in the direction that takes place with a decrease in a total number of moles. Hence for the reaction,

CH₄ (g) + H₂O (g) ⇌ CO (g) + 3H₂ (g)

Number of moles of products (n_p)= 4

Number of moles of reactants (n_r) = 2

∴ n_p > n_r

According to Le Chatelier’s principle, the equilibrium will shift in the backward direction and value of K_p decreases.

(ii) If the temperature is raised, i.e., heat is applied to the system, then according to Le Chatelier’s principle, the equilibrium will shift to the side that absorbs heat, i.e., in the backward reaction. An endothermic reaction is favoured by low temperature.

Hence for the endothermic reaction,

CH₄ (g) + H₂O (g) ⇌ CO (g) + 3H₂ (g)

The equilibrium shifts in the forward direction and K_p increases.

(iii) By using a catalyst, equilibrium composition will not be disturbed but equilibrium will be attained quickly and value of K_p remains unaffected.

Note: Role of the catalyst is that it increases the rate of both forward and backward reactions equally; the equilibrium will be attained in less time, i.e., the same amount of product will be formed in less time. Catalyst does not effect equilibrium constant and heat of reaction.