Questions 22-25 refer to the following information. N2O4(g) ⇄ 2 NO2(g) Kp = 3.0 at 70°C colorless br — Equilibrium Chemistry Question

This question addresses a classic and frequently tested concept in chemical equilibrium: the effect of adding an inert (unreactive) gas to a gaseous equilibrium system at constant volume.

STEPS:

1. Understand the definition of partial pressure:
According to the Ideal Gas Law, the partial pressure of any individual gas ($i$) in a mixture is determined by the formula:
$$P_i = \frac{n_i RT}{V}$$
where $n_i$ is the number of moles of that specific gas, $R$ is the gas constant, $T$ is the absolute temperature, and $V$ is the volume of the container.
2. Analyze the conditions of the addition:
* Argon ($\text{Ar}$) is an inert noble gas that does not react with either $\text{NO}_2$ or $\text{N}_2\text{O}_4$.
* The argon gas is added at constant volume (the glass tube is rigid and its size does not change).
* The temperature ($T$) of the system is held constant at $70^\circ\text{C}$.
3. Determine the effect on the reacting species' partial pressures:
* Because the container volume ($V$) and temperature ($T$) are constant, and the number of moles ($n$) of $\text{NO}_2$ and $\text{N}_2\text{O}_4$ inside the tube has not changed, their individual partial pressures ($P_{\text{NO}_2}$ and $P_{\text{N}_2\text{O}_4}$) remain completely unchanged.
* *(Note: While the total pressure in the tube increases due to the addition of argon atoms, the partial pressures of the reacting gases do not change).*
4. Evaluate the reaction quotient ($Q_p$):
* The reaction quotient is defined as:
$$Q_p = \frac{(P_{\text{NO}_2})^2}{P_{\text{N}_2\text{O}_4}}$$
* Since the partial pressures of both $\text{NO}_2$ and $\text{N}_2\text{O}_4$ remain the same, $Q_p$ remains exactly equal to the equilibrium constant $K_p$ ($3.0$).
* Because $Q_p = K_p$, the system remains at equilibrium and no shift occurs.
5. Conclude:
* Since there is no shift in equilibrium, the partial pressures of the reacting species do not change, which identifies Option D as the correct answer.

*

WHY_OTHERS_WRONG:

• Options A and B are incorrect: These options predict a shift in the equilibrium position. A student might make this error if they incorrectly apply Le Chatelier's principle, thinking: *"An increase in total pressure always shifts the reaction toward the side with fewer moles of gas (the reactant side, $\text{N}_2\text{O}_4$)"*. However, a pressure-induced shift only occurs if the pressure change is caused by a change in volume. If volume is constant, the collision frequency between the reacting molecules is unaffected by the presence of the inert argon atoms, so no shift occurs.
• Option C is incorrect: This option suggests that the partial pressures of both reacting species will decrease. This violates Dalton's Law of Partial Pressures and the conservation of matter; adding a gas to a closed, rigid container cannot reduce the individual pressures or concentrations of the gases already present.