© All Rights reserved @ LearnWithDash
Step-by-Step Explanation
Step 1: Identify the Relevant Reaction
The equilibrium reaction under consideration is:
$$2\,\text{NO}_2(\text{g}) \rightleftharpoons \text{N}_2\text{O}_4(\text{g}).$$
In the forward direction ($2\,\text{NO}_2 \to \text{N}_2\text{O}_4$), the reaction is exothermic. Conversely, in the reverse direction ($\text{N}_2\text{O}_4 \to 2\,\text{NO}_2$), the reaction is endothermic.
Step 2: Recall Le Chatelier’s Principle
Le Chatelier’s Principle states that if a dynamic equilibrium is disturbed by a change in conditions (e.g., pressure, temperature, volume), the system responds by shifting its equilibrium position to counteract the change.
Step 3: Analyze the Addition of Inert Gas
When an inert gas is added to the system, its effect on the equilibrium depends on whether the process occurs at constant volume or at constant pressure:
Constant Volume: Adding an inert gas at constant volume does not change the partial pressures of the reactants or products. Therefore, there is typically no shift in the equilibrium position.
Constant Pressure: Adding an inert gas at constant pressure causes an increase in the total volume. Since the pressure must remain the same, the equilibrium shifts in the direction that increases the number of gas moles, in order to partially offset the change.
Step 4: Determine the Effect on the Given Equilibrium
In this reaction,
$$\text{N}_2\text{O}_4(\text{g}) \rightleftharpoons 2\,\text{NO}_2(\text{g}),$$
the forward reaction produces more moles of gas (2 moles of NO2 for every 1 mole of N2O4). When an inert gas is added at constant pressure, the increase in total volume will favor the side of the reaction with a greater number of moles of gas. Hence, the equilibrium shifts towards the decomposition of N2O4 (the side with more moles).
Step 5: Conclusion
Therefore, “addition of an inert gas at constant pressure” increases the decomposition of dinitrogen tetraoxide, moving the equilibrium towards the side with more gas moles, namely 2NO2. This explains why the correct answer is the addition of an inert gas at constant pressure.