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A mixture of 0.3 mol of $H_{2}$ and 0.3 mol $I_{2}$ is allowed to react in a 10 lit Evacuated flask at 500°C. The reaction is $H_{2} + I_{2} ⇌ 2 H I$ . The $K_{C}$ is found to be 64. The amount of un reacted $I_{2}$ at equilibrium is

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0.06 mol

0.03 mol

0.15 mol

0.2 mol

answer is B.

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Detailed Solution

$∴ 64 = {(\frac{2 x}{0.3 - x})}^{2} \Rightarrow x = 0.24$

Un reacted $I_{2} = 0.3 - 0.24 = 0.06$ .

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