1. Given equilibrium

The reaction is $2NO_2(g) \rightleftharpoons N_2O_4(g)$

It is given that formation of $N_2O_4$ is exothermic. Therefore, the reverse reaction $N_2O_4(g) \rightleftharpoons 2NO_2(g)$ is the decomposition of $N_2O_4$ and is endothermic.

We need the condition that increases decomposition of $N_2O_4$, i.e. shifts equilibrium towards $NO_2$.

2. Check each option using Le Chatelier's principle

Option A: Lowering the temperature

Since forward reaction $2NO_2 \to N_2O_4$ is exothermic, lowering temperature favors the exothermic direction, i.e. formation of $N_2O_4$.

So decomposition of $N_2O_4$ decreases.

Thus, A is incorrect.

Option B: Increasing the pressure

On the left side, there are $2$ moles of gas; on the right side, there is $1$ mole of gas.

Increasing pressure favors the side with fewer moles of gas, i.e. $N_2O_4$ formation.

So decomposition of $N_2O_4$ decreases.

Thus, B is incorrect.

Option C: Addition of an inert gas at constant volume

At constant volume, adding inert gas increases total pressure, but the partial pressures of reacting gases remain unchanged because $p_i = \frac{n_iRT}{V}$ for each reacting gas, and $n_i, T, V$ remain the same.

Hence the reaction quotient does not change, so equilibrium does not shift.

Thus decomposition is not increased.

So, C is incorrect.

Option D: Addition of an inert gas at constant pressure

At constant pressure, adding inert gas causes the volume to increase.

As volume increases, the partial pressures of all reacting gases decrease. The equilibrium then shifts toward the side with more moles of gas to oppose this change.

Here,

• Left side: $2$ moles of gas ($2NO_2$)
• Right side: $1$ mole of gas ($N_2O_4$)

So equilibrium shifts to the left, i.e. toward $NO_2$.

That means decomposition of $N_2O_4$ increases.

Thus, D is correct.

3. Final answer

The decomposition of $N_2O_4$ is increased by: $\boxed{\text{D: addition of an inert gas at constant pressure}}$