The gas phase reaction 2NO$$_2$$(g) $$\rightarrow$$ N$$_2$$O$$_4$$(g) is an exothermic reaction. The decomposition of N$$_2$$O$$_4$$, in equilibrium mixture of NO$$_2$$(g) and N$$_2$$O$$_4$$(g) can be increased by:

We consider the equilibrium

$$2\,\mathrm{NO_2}(g)\; \rightleftharpoons\; \mathrm{N_2O_4}(g)\qquad (\Delta H < 0,\;\text{exothermic})$$

Let the initial equilibrium partial pressures be $$P_{\mathrm{NO_2}}=p$$ and $$P_{\mathrm{N_2O_4}}=q$$. The equilibrium constant expressed in terms of partial pressure is

$$K_p=\frac{P_{\mathrm{N_2O_4}}}{\left(P_{\mathrm{NO_2}}\right)^2}=\frac{q}{p^2} \qquad\qquad (1)$$

To increase the decomposition of $$\mathrm{N_2O_4}$$, we want to drive the equilibrium toward the left-hand side, i.e. toward the formation of more $$\mathrm{NO_2}$$. Le Châtelier’s principle tells us that the direction of the shift depends on temperature, pressure and other constraints.

Effect of adding an inert gas at constant total pressure (Option A)

When we keep the total pressure $$P_{\text{total}}$$ constant and inject an inert gas, the overall number of moles in the container increases. Because the pressure is fixed, the volume must expand according to

$$P_{\text{total}}V = n_{\text{total}}RT \; \Longrightarrow \; V \propto n_{\text{total}}$$

The larger volume lowers the partial pressures of all reacting species:

$$P_{\mathrm{NO_2}} \;=\; \frac{n_{\mathrm{NO_2}}RT}{V} \quad\text{and}\quad P_{\mathrm{N_2O_4}} \;=\; \frac{n_{\mathrm{N_2O_4}}RT}{V}$$

Because $$V$$ has increased, both $$P_{\mathrm{NO_2}}$$ and $$P_{\mathrm{N_2O_4}}$$ decrease. However, what matters for equilibrium is the ratio $$q/p^2$$ in equation (1). The reaction possesses $$\Delta n = 1 - 2 = -1$$ (one mole on the right, two moles on the left). Mathematically

$$Q_p=\frac{P_{\mathrm{N_2O_4}}}{\left(P_{\mathrm{NO_2}}\right)^2} =\frac{q'}{(p')^2} =\frac{\dfrac{q}{\alpha}}{\left(\dfrac{p}{\alpha}\right)^2} =\alpha\,\frac{q}{p^2}=\alpha\,K_p,$$

where $$\alpha>1$$ because the volume has increased. Thus $$Q_p > K_p$$ immediately after the addition. To restore $$Q_p = K_p$$, the system must decrease $$Q_p$$, and the only way to do that is to shift the equilibrium toward the left (increase $$p$$, decrease $$q$$)—that is, toward the formation of more $$\mathrm{NO_2}$$. Consequently, the decomposition of $$\mathrm{N_2O_4}$$ increases.

Why the other options do not work

• Lowering the temperature (Option B) favors the exothermic forward reaction, producing more $$\mathrm{N_2O_4}$$, which is the opposite of what we want.

• Increasing the pressure (Option C) by compressing the system favors the side with fewer gas moles, again producing more $$\mathrm{N_2O_4}$$.

• Adding an inert gas at constant volume (Option D) leaves every partial pressure unchanged because $$V$$ and the amounts of reactants remain fixed; hence no shift occurs.

Therefore, only the first choice accomplishes a greater decomposition of $$\mathrm{N_2O_4}$$.

Hence, the correct answer is Option A.