Which of the following changes would drive the position of equilibrium to the right side of $4 H C l (g) + O_{2} (g) \overset{}{⇌} 2 H_{2} O (g) + 2 C l_{2} (g), Δ H^{o} = 30 K J / m o l o f H C l$

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Addition of $C l_{2} (g)$

Reducing the volume of container at constant temperature

Decreasing temperature

Addition of inert gas $(H e)$ at constant pressure

answer is B.

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Detailed Solution

A) Chlorine is a product, adding chlorine will shift the reaction in backward direction
B) Oxygen will drive the reaction; increasing partial pressure of oxygen will drive the reaction in forward direction.
C) Since the reaction is endothermic, decreasing the temperature will drive reaction in backward direction.
D) Here $Δ n (g) < 0,$ increasing total pressure drives reaction in forward direction

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