Question

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

• A. $0.04$
• B. $0.4$
• C. $25$
• D. $2.5$

Answer 1

The correct option is A $0.04$

The given reaction is :-

$\frac{1}{2}{H}_2+\frac{1}{2}{I}_2\rightleftharpoons HI$

$K_C=\frac{\left[HI\right]}{{\left[H_2\right]}^{1/2}{\left[I_2\right]}^{1/2}}=5\to \left(I\right)$

Now, the reaction is :-

$2HI\rightleftharpoons H_2+I_2$

$K_C=\frac{\left[H_2\right]\left[I_2\right]}{{\left[HI\right]}^2}\to \left(II\right)$

By using $\left(I\right)$ in $\left(II\right)$

$\Rightarrow K_C={\left(\frac{1}{K_C}\right)}^2={\left(\frac{1}{5}\right)}^2=\frac{1}{25}=0.04$