Consider the following reaction:
$$\text{N}_2\text{O}_4(g) = 2\text{NO}_2(g);\;\Delta H^0 = +58\,\text{k}$$
For each of the following cases (a, b), the direction in which the equilibrium shifts is:
(a) Temperature is decreased.
(b) Pressure is increased by adding $$\text{N}_2$$ at constant T.

To determine how the equilibrium position shifts in the reaction

$$N_2O_4(g) \rightleftharpoons 2NO_2(g), \qquad \Delta H^\circ = +58\ \text{kJ mol}^{-1},$$

we apply Le Chatelier's Principle, which states that when a system at equilibrium is subjected to a change in concentration, temperature, or pressure, the equilibrium shifts in a direction that opposes the imposed change.

Since $$\Delta H^\circ$$ is positive, the forward reaction is endothermic and may be represented as

$$\text{Heat} + N_2O_4(g) \rightleftharpoons 2NO_2(g).$$

When the temperature is decreased, the system tends to produce heat to oppose the change. Therefore, the equilibrium shifts in the reverse (exothermic) direction, favouring the formation of $$N_2O_4$$. Hence, the equilibrium shifts towards the reactant side.

When the pressure is increased by adding $$N_2$$ (an inert gas) at constant temperature, the partial pressures of $$N_2O_4$$ and $$NO_2$$ remain unchanged. Since the equilibrium expression depends only on the reacting species, the equilibrium position is not affected, and no shift occurs.

Hence,

• Case (a): Equilibrium shifts towards the reactant.
• Case (b): No change in the equilibrium position.

Therefore, the correct answer is (a) towards reactant, (b) no change.