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For a reaction: $H_{2} (g) + I_{2} (g) ⇌ 2 HI (g)$ at certain temperature, the value of equilibrium constant is 50. If the volume of the vessel is reduced to half of its original volume, the value of new equilibrium constant will be

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Detailed Solution

$\begin{array}{lcccc} H_{2} & + & I_{2} & ⇌ & 2 HI \\ Initial & 1 & 1 & 0 \\ Final & \frac{1 - α}{v} & \frac{1 - α}{v} & \frac{2 α}{v} \\ Since Δ n = 0 & [2 - (1 + 1)] \end{array}$

$∴$ K does not depend on volume.
So on reducing volume to half K does not change.
$∴$ K = 50.

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