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Q. At $500\,K$, equilibrium constant, $K_c$ for the following reaction is $5$.
$\frac{1}{2} H_{2\left(g\right)}+\frac{1}{2}I_{2\left(g\right)}\rightleftharpoons HI_{(g)}$
What would be the equilibrium constant $K_c$ for the reaction :
$2HI_{(g)} \rightleftharpoons H_{2(g)} +I_{2(g)}$ ?

Equilibrium

Solution:

$\frac{1}{2} H_{2\left(g\right)}+\frac{1}{2}I_{2\left(g\right)}\rightleftharpoons HI_{(g)}$ ; $K_c=5 \ldots(i)$
Multiply eqn $(i)$ by $2$,
$H_{2(g)} +I_{2(g)} \rightleftharpoons 2HI_{(g)}$; $K_c=(5)^2 \ldots(ii)$
Now, reverse the reaction
$2HI_{(g)} \rightleftharpoons H_{2(g)} +I_{2(g)}$; $K_{c}=\frac{1}{\left(5\right)^{2}}$
$\therefore K_{c}=\frac{1}{25}=0.04$