Question

The volume of a closed reaction vessel in which the equilibrium -

$2S{O}_2\left(g\right)+O_2\left(g\right)\rightleftharpoons 2S{O}_3\left(g\right)$ sets is halved, Now -

• 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 more than that of reverse reaction and the equilibrium will shift

Answer 1

The correct option is D

The rate of forward reaction will become more than that of reverse reaction and the equilibrium will shift

In the reaction

$2S{O}_2\left(g\right)+O_2\left(g\right)\rightleftharpoons 2S{O}_3\left(g\right)$

In this reaction three moles (or volumes) of reactants are converted into two moles (or volumes) of products i.e. there is a decrease in volume and so if the volume of the reaction vessel is halved the equilibrium will be shifted to the right i.e. more product will be formed and the rate of forward reaction will increase i.e. double that of reverse reaction.