Chemical EquilibriumMediumBloom L2
Question
The equilibrium $\text{SO}_2\text{Cl}_2(g) \rightleftharpoons \text{SO}_2(g) + \text{Cl}_2(g)$ is attained at $25^\circ\text{C}$ in a closed container, and an inert gas (helium) is introduced at constant volume. Which of the following statements is correct?
Options
A.Concentrations of $\text{SO}_2$, $\text{Cl}_2$, and $\text{SO}_2\text{Cl}_2$ change
B.More $\text{Cl}_2$ is formed
C.Concentration of $\text{SO}_2\text{Cl}_2$ is reduced
D.All of the above statements are incorrect
Solution
**Given:** Equilibrium $\text{SO}_2\text{Cl}_2(g) \rightleftharpoons \text{SO}_2(g) + \text{Cl}_2(g)$ at $25^\circ\text{C}$; an inert gas (He) is added at constant volume.
**Key Principle:** When an inert gas is added at **constant volume**, the volume of the container does not change. Therefore, the concentrations (moles per unit volume) of all reacting species — $\text{SO}_2\text{Cl}_2$, $\text{SO}_2$, and $\text{Cl}_2$ — remain unchanged.
**Analysis of each option:**
$$K_c = \frac{[\text{SO}_2][\text{Cl}_2]}{[\text{SO}_2\text{Cl}_2]}$$
Since volume is constant, adding He does not alter any concentration, so $Q_c = K_c$ and equilibrium is undisturbed.
- **Option A:** Incorrect — concentrations do not change at constant volume.
- **Option B:** Incorrect — no additional $\text{Cl}_2$ is formed; equilibrium does not shift.
- **Option C:** Incorrect — concentration of $\text{SO}_2\text{Cl}_2$ does not decrease.
**Answer: (D)**
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