MCQ
The equilibrium constant of the reaction ${H_2}_{(g)} + {I_2}_{(g)}$ $\rightleftharpoons$ $2HI_{(g)}$ 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$32$
- ✓$64$
- D$128$