In an experiment carried out at $1377 K$ , $HI$ was found to be 25% dissociated. The $K_{c}$ for the dissociation $2 HI (g) ⇌ H_{2} (g) + I_{2} (g)$ is:

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$\frac{9}{4}$

$9$

$\frac{1}{9}$

$\frac{1}{36}$

answer is D.

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Detailed Solution

Here, for the reaction,

$\begin{array}{l} \begin{array}{ccc} 2 HI & ⇌ & H_{2} + I_{2} \\ 1 - 0.25 & \frac{0.25}{2} & \frac{0.25}{2} \end{array} \\ K_{C} = \frac{{(\frac{0.25}{2})}^{2}}{(0.75)^{2}} = \frac{1}{36} \end{array}$

Hence, the correct option is d.

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