The equilibrium constant K for the reaction $H_{2} + I_{2} ⇌ 2 H I$ at $700 K$ is $49$ . What is the equilibrium constant for the reaction $H I ⇌ \frac{1}{2} H_{2} + \frac{1}{2} I_{2}$ at the same temperature?

1.43

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0.143

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0.49

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0.02

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Solution

The correct option is B $0.143$
Given $H_{2} + I_{2} ⇌ 2 H I, k = 49$
$∴$ For the reaction:
$2 H I \to H_{2} + I_{2}, \frac{1}{k} = \frac{1}{49}$
$∴$ For the reaction : $H I \to \frac{1}{2} H_{2} + \frac{1}{2} I_{2}$
${(\frac{1}{k})}^{1 / 2} = {(\frac{1}{49})}^{1 / 2} = 0.143$

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