1. Given equilibrium

$N_2O_4(g) \rightleftharpoons 2NO_2(g), \qquad \Delta H^\circ = +58\,\text{kJ}$

Since $\Delta H^\circ$ is positive, the forward reaction is endothermic.

2. Case (a): Temperature is decreased

By Le Chatelier’s principle:

• Decreasing temperature favors the exothermic direction.
• Since the forward reaction is endothermic, the reverse reaction is exothermic.

So equilibrium shifts towards reactant ($N_2O_4$).

Thus,

$\text{(a) shift } \to \text{ reactant}$

3. Case (b): Pressure is increased by adding } N_2 \text{ at constant } T

Here, $N_2$ is an inert gas with respect to this equilibrium.

At constant temperature, when an inert gas is added:

• If volume is constant, partial pressures of reacting gases do not change.
• Even though total pressure increases, the equilibrium position depends on the partial pressures of $N_2O_4$ and $NO_2$, not on total pressure alone.

So there is no shift in equilibrium.

Thus,

$\text{(b) no change}$

4. Match with options

• (a) towards reactant
• (b) no change

This corresponds to Option B.

5. Comparison with stored answer

Stored correct answer: B

Our derived answer: B

So the derived answer agrees with the stored answer.