Question

The equilibrium constant for the reaction $H_2\left(g\right)+I_2\rightleftharpoons 2HI\left(g\right)$ is 64. If the volume of the container is reduced to one-fourth of its original volume, the value of the equilibrium constant will be:

• A. 16
• B. 32
• C. 64
• D. 128

Answer 1

The correct option is C 64

When a volume of a container is decreased to ${1/4}^{th}$ of the original volume, then the pressure of each of reactant and product species will vary accordingly.

Since here no. of moles on both sides are same, the pressure will affect both sides equally.

So, the volume of the equilibrium constant do not changes; it remains the same.