Question

At 450 K, K_p = 2.0 × 10¹⁰/bar for the given reaction at equilibrium.

2SO₂(g) + O₂(g) ⇌ 2SO₃(g) + 189.4 kJ

(a) What is K_c at this temperature?

(b) What is the value of K_c for the reserve reaction at the same temperature?

(c) What would be the effect on equilibrium if

(i) more SO₂ is added

(ii) pressure is increased

(iii) the temperature is increased?

Answer 1

(a) 2SO₂(g) + O₂ (g) ⇌ 2SO₃(g)

For the given reaction

∆n_g = n_p - n_r = 2 – 3 = –1

K_p = K_c (RT)^∆n

K_p = 2 × 10¹⁰ bar^–1

K_c = \(\frac{K_p}{(RT)^{\Delta n}}\)

= ( 2 × 10¹⁰ bar^–1 ) (0.0831 L bar K^–1 mol¹× 450 K)

= 7.48 × 10¹¹ L mol^–1

(b) For reserve reaction,

K_c = \(\frac{1}{K_c(forward\,reaction)}\)

= \(\frac{1}{748\times10^{11}}\)

= 0.134 x 10¹¹ L mol^-1

(c) (i) If more SO₂ is added, rate of forward reaction increases and more SO₃ will be formed.

(ii) If more pressure is increased, the reaction will shift in forward direction i.e., towards lesser number of moles.

(iii) Increase in temperature will favour backward reaction.