SATHEE: Equilibrium Question 581

Equilibrium Question 581

Question: If for $H_{2} (g) + \frac{1}{2} S_{2 (S)}$ ⇌ $H_{2} S (g)$ and $H_{2 (g)} + B r_{2 (g)}$ ⇌ $2 H B r (g)$ The equilibrium constants are K1 and K2 respectively, the reaction $B r_{2} (g) + H_{2} S (g)$ ⇌ $2 H B r (g) + \frac{1}{2} S_{2 (S)}$ would have equilibrium constant [MP PMT 2001]

Options:

A) $K_{1} \times K_{2}$

B) $K_{1} / K_{2}$

C) $K_{2} / K_{1}$

D) $K_{2}^{2} / K_{1}$

Show Answer

Answer:

Correct Answer: C

Solution:

$K_{1} = \frac{[H_{2} S]}{[H_{2}] {[S_{2}]}^{1 / 2;}}$ ; $K_{2} = \frac{{[H B r]}^{2}}{[H_{2}] [B r_{2}]}$

$K_{3} = \frac{{[H B r]}^{2} \times {[S_{2}]}^{1 / 2;}}{[B r_{2}], \times [H_{2} S]}$ ; $\frac{K_{2}}{K_{1}} = K_{3}$

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