The volume of a closed reaction vessel in which the following equilibrium reaction occurs is halved:

$ 2SO_{2}(g) + O_{2}(g) \rightleftharpoons 2SO_{3}(g)$ As a result

A) The rates of forward and backward reactions will remain the same.

B) The equilibrium will not shift.

C) The equilibrium will shift to the right.

D) The rate of forward reaction will become double that of reverse reaction and the equilibrium will shift to the right


Option D


$K_{c} = \frac{\left[(SO_3)^2\right]}{\left[(SO_2)^2\right]\left[O_{2}\right]}$

If the volume is reduced to half, Kc will decrease to half. Thus, to maintain the equilibrium, the reaction should shift in the forward direction, i.e. towards the right. Also, to attain equilibrium back, the rate of forward direction will become double the rate of backward direction.