$H_{2} (g) + I_{2} (g) ⇌ 2 H I (g)$

The equilibrium constant for the given reaction is $64$ . If the volume of the container is reduced to half of the original volume, the value of the equilibrium constant will be:

16

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32

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64

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128

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Solution

The correct option is D $64$
$H_{2} (g) + I_{2} (g) ⟶ 2 H I (g)$

for this reaction, $Δ n_{g} = 0$

$∴$ The reaction and its equilibrium constant is not affected by a change in volume. Moreover, the equilibrium constant depends only on temperature.

Hence, the correct option is $C$

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