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Chemistry Question on Equilibrium

The value of equilibrium constant of the reaction, HI(g)⇌12H2(g)+12I2(g)  is  8.0HI(g) \rightleftharpoons \frac{1}{2} H_2 (g) + \frac{1}{2} I_2 (g)\, \, is\, \, 8.0 . The equilibrium constant of the reaction, H2(g)+I2(g)⇌2HI(g)H_2 (g) + I_2 (g) \rightleftharpoons 2HI(g) Will be

A

116\frac{1}{16}

B

164\frac{1}{64}

C

16

D

18\frac{1}{8}

Answer

164\frac{1}{64}

Explanation

Solution

HI(g)⇌12H2(g)+12I2(g)HI(g) \rightleftharpoons \frac{1}{2} H_2 (g) + \frac{1}{2} I_2 (g)
K=[I2]1/2[H2]1/2[HI]   .......(i)K = \frac{[I_2]^{1/2} [H_2]^{1/2}}{[HI]}\, \, \, .......(i)
H2(g)+I2(g)⇌2HI(g)H_2 (g) + I_2 (g) \rightleftharpoons 2HI(g)
K′=[HI]2[H2][I2]   .........(ii)K' = \frac{[HI]^2}{[H_2][I_2]}\, \, \, .........(ii)
From Eqs. (i) and (ii)
K×K′=1K \times \sqrt{K'} = 1
K′=1K2=1(8)2=164K' = \frac{1}{K^2} = \frac{1}{(8)^2} = \frac{1}{64}