Given is the following reaction at equilibrium ${\mathrm{SO}}_2\left(\mathrm{g}\right)+\frac{1}{2}{\mathrm{O}}_2\left(\mathrm{g}\right)\rightleftharpoons {\mathrm{SO}}_3\left(\mathrm{g}\right)$. The addition of inert gas at constant pressure would cause

It is the reaction ${\mathrm{SO}}_2\left(\mathrm{g}\right)+\frac{1}{2}{\mathrm{O}}_2\left(\mathrm{g}\right)\rightleftharpoons {\mathrm{SO}}_3\left(\mathrm{g}\right)$ When the number of moles is reduced from 1.5 to 1, the forward reaction takes place. Additionally, when the number of moles is increased from 1 to 1.5, the opposite reaction takes place.

The equilibrium will change when inert gas is supplied at constant pressure, moving in the direction of an increase in the number of moles of gases. As a result, the ${\mathrm{SO}}_3$ will dissociate more when inert gas is added at constant pressure.