# At 450 K, Kp = 2.0 × 10^10/bar for the given reaction at equilibrium. 2SO2(g) + O2(g) ⇌ 2SO3(g) + 189.4 kJ

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

2SO2(g) + O2(g) ⇌ 2SO3(g) + 189.4 kJ

(a) What is Kc at this temperature?

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

(c) What would be the effect on equilibrium if

(ii) pressure is increased

(iii) the temperature is increased?

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(a) 2SO2(g) + O2 (g) ⇌ 2SO3(g)

For the given reaction

∆ng = np - nr = 2 – 3 = –1

Kp = Kc (RT)∆n

Kp = 2 × 1010 bar–1

Kc$\frac{K_p}{(RT)^{\Delta n}}$

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

= 7.48 × 1011 L mol–1

(b) For reserve reaction,

Kc$\frac{1}{K_c(forward\,reaction)}$

$\frac{1}{748\times10^{11}}$

= 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.