Animated Solution for Chemistry - Chemical Equilibrium: Consider the following reaction : $N_2{O}_4(g)\rightleftharpoons 2N{O}_2(g);\Delta H^{\circ }=+58\text{ kJ}$ 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 $N_2$ at constant $T$.

Let's dive into the fascinating world of chemical equilibrium and Le Chatelier's principle. We are given the following reversible reaction:

Notice the enthalpy change provided: $\Delta H^{\circ }=+58\text{ kJ}$. The positive sign is our first major clue. It tells us that the forward reaction is endothermic—it absorbs heat from its surroundings to proceed. Conversely, the backward reaction must be exothermic, releasing heat.

The Temperature Tango

According to Le Chatelier's principle, a system at equilibrium acts like a stubborn child; it will always try to oppose any change you impose on it.

If we decrease the temperature of the system, we are essentially removing heat. How does the system respond? It tries to produce more heat to counteract our cooling effect. To generate heat, the equilibrium must shift in the exothermic direction.

Since the forward reaction is endothermic, the backward reaction is the exothermic one. Therefore, decreasing the temperature forces the equilibrium to shift backward, towards the reactant, $N_2{O}_4$.

The Inert Gas Illusion

Now, let's look at the second scenario. We are adding nitrogen gas ($N_2$) at a constant temperature. The problem states that the "pressure is increased by adding $N_2$ at constant $T$." This phrasing is critical. It implies that the gas is being pumped into a rigid, closed container. If the container could expand to keep the pressure constant, the pressure wouldn't increase! Thus, we are adding an inert gas at constant volume.

Nitrogen is an inert gas in this context because it does not react with either $N_2{O}_4$ or $N{O}_2$. When we add an inert gas at constant volume, the total pressure of the container undoubtedly increases. However, the equilibrium position depends on the partial pressures (or active masses) of the reacting gases.

The partial pressure of a gas is determined by its number of moles and the volume of the container. Since neither the moles of $N_2{O}_4$ and $N{O}_2$ nor the volume of the container have changed, their partial pressures remain completely unaffected. Because the active masses are unchanged, the reaction quotient $Q$ remains equal to the equilibrium constant $K_p$.

Result: There is absolutely no shift in the equilibrium.

Combining our findings from both cases: - Decreasing the temperature shifts the equilibrium towards the reactant. - Adding $N_2$ at constant volume results in no change.

This perfectly matches option (b). Always remember to read the conditions carefully—adding an inert gas at constant pressure would have yielded a completely different result!