Question

In which of the following equilibrium, increase in pressure shifts the equilibrium in the forward direction?

• A. \( {H}_2(g) + {I}_2(g) \rightleftharpoons 2 {HI}(g) \)
• B. \( {PCl}_5(g) \rightleftharpoons {PCl}_3(g) + {Cl}_2(g) \)
• C. \( {N}_2(g) + {O}_2(g) \rightleftharpoons 2 {NO}(g) \)
• D. \( {CO}(g) + 3 {H}_2(g) \rightleftharpoons {CH}_4(g) + {H}_2{O}(g) \)
• E. \( {CO}(g) + {H}_2{O}(g) \rightleftharpoons {CO}_2(g) + {H}_2(g) \)

Hint:

In gas-phase equilibria, increasing pressure generally shifts the equilibrium towards the side with fewer moles of gas. Always count the moles of gases on each side of the reaction to predict the effect of pressure.

Answer 1

The Correct Option is D

According to Le Chatelier's Principle, if a change in conditions (such as pressure) is applied to a system at equilibrium, the system will adjust in such a way as to counteract the change.

When pressure is increased, the equilibrium will shift in the direction where the number of moles of gas is decreased, because the system tries to reduce the volume (and hence the pressure).

Now, let's analyze each of the reactions:

(A) \( \text{H}_2(g) + \text{I}_2(g) \rightleftharpoons 2 \text{HI}(g) \) has 2 moles of gas on the left and 2 moles of gas on the right. Hence, changing the pressure does not affect the equilibrium position.

(B) \( \text{PCl}_5(g) \rightleftharpoons \text{PCl}_3(g) + \text{Cl}_2(g) \) also has 1 mole of gas on both sides. Thus, the equilibrium is not affected by a change in pressure.

(C) \( \text{N}_2(g) + \text{O}_2(g) \rightleftharpoons 2 \text{NO}(g) \) involves 2 moles of gas on both sides, so increasing pressure does not shift the equilibrium.

(D) \( \text{CO}(g) + 3 \text{H}_2(g) \rightleftharpoons \text{CH}_4(g) + \text{H}_2\text{O}(g) \) has 4 moles of gas on the left side (1 mole of CO and 3 moles of \( \text{H}_2 \)) and 2 moles of gas on the right side (1 mole of \( \text{CH}_4 \) and 1 mole of \( \text{H}_2\text{O} \)). Increasing the pressure will shift the equilibrium towards the right, where fewer moles of gas are present.

(E) \( \text{CO}(g) + \text{H}_2\text{O}(g) \rightleftharpoons \text{CO}_2(g) + \text{H}_2(g) \) has 2 moles of gas on each side, so changing the pressure will not shift the equilibrium.

Thus, the correct answer is (D), as the increase in pressure shifts the equilibrium in the forward direction where fewer moles of gas are present.