Question

The value of equilibrium constant of the reaction, $HI(g)\rightleftharpoons \frac{1}{2}{H}_2(g)+\frac{1}{2}{I}_2(g)$ is $8.0$. The equilibrium constant of the reaction, $H_2(g)+I_2(g)\rightleftharpoons 2HI(g)$ Will be

• A. $\frac{1}{16}$
• B. $\frac{1}{64}$
• C. $16$
• D. $\frac{1}{8}$

Answer 1

Answer: (B)

For reaction,
$H{I}_{(g)}\rightleftharpoons \frac{1}{2}{H}_{2(g)}+\frac{1}{2}{I}_{2(g)};K_c=8.0$
$K_c=\frac{[HI]}{{\left[H_2\right]}^{1/2}{\left[H_2\right]}^{1/2}}$
$\Rightarrow \frac{[HI]}{{\left[H_2\right]}^{1/2}{\left[H_2\right]}^{1/2}}=8\to \text{ (i) }$
Now, for the rection,
$H_{2(g)}+I_{2(g)}\rightleftharpoons 2H{I}_{(g)};K_c^{\prime }=?$
$K_c^{\prime }=\frac{\left[H_2\right]\left[I_2\right]}{[HI{]}^2}$
$K_c^{\prime }=\frac{1}{{\left(\frac{[HI]}{{\left[H_2\right]}^{1/2}{\left[H_2\right]}^{1/2}}\right)}^2}$
From $e^n$ (i), we have
$K_c^{\prime }=\frac{1}{(8{)}^2}=\frac{1}{64}$