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If the equilibrium constant of the reaction $2 HI ⇌ H_{2} + I_{2}$ is 0.25, then the equilibrium constant of the reaction $H_{2} + I_{2} ⇌ 2 HI$ would be

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Detailed Solution

$K_{1} for reaction 2 HI ⇌ H_{2} + I_{2} is 0 .25 K_{2} for reaction$
$H_{2} + I_{2} ⇌ 2 HI will be K_{2} = \frac{1}{K_{1}} = \frac{1}{0 .25} = 4$
Because II^nd reaction is reverse of I^st.

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