Question

The value of equilibrium constant of the reaction, $HI\left(g\right)\rightleftharpoons 1/2H_2\left(g\right)+1/2I_2\left(g\right)$ is $8.0$. The equilibrium constant of the reaction, $H_2\left(g\right)+I_2\left(g\right)\rightleftharpoons 2HI\left(g\right)$ will be:

• A. $16$
• B. $1/8$
• C. $1/16$
• D. $1/64$

Answer 1

The correct option is D $1/64$

The give reaction is :-

$HI\left(g\right)\rightleftharpoons \frac{1}{2}{H}_2\left(g\right)+\frac{1}{2}{I}_2\left(g\right)$

$\Rightarrow K_C=\frac{{\left[H_2\right]}^{1/2}{\left[I_2\right]}^{1/2}}{\left[HI\right]}=8\to \left(I\right)$

Now,

$H_2\left(g\right)+I_2\left(g\right)\rightleftharpoons 2HI\left(g\right)$

$\Rightarrow {K_C}^{{}^{\prime }}=\frac{{\left[HI\right]}^2}{\left[H_2\right]\left[I_2\right]}$

$={\left(\frac{1}{K_C}\right)}^2={\left(\frac{1}{8}\right)}^2=\frac{1}{64}$