Question

The equilibrium constant for the reaction $H_2+I_2\rightleftharpoons 2HI$ is 50. If the volume of the container is reduced to half of its original value, the value of the equilibrium constant will be:

• A. 25
• B. 50
• C. 75
• D. 100

Answer 1

The correct option is B 50

At equilibrium the concentration of $H_2$ $I_2$ and HI is ($1-x),(1-x),(2x)$ respectively.
$K_c=50$
$=\frac{{\left[HI\right]}^2}{\left[H_2\right]I_2}$

$K_c=\frac{4x^2}{(1-x^2)}=50$

$K_c$ value does not depend on initial concentration. When volume of reaction vessel is reduced to half, concentration is doubled. $K_c$ remains same.
It can also be seen in another way that since $\Delta n$ of reaction is zero there would be no impact of a change in concentration.