Question

**Given:** $\text{PCl}_5(g) \rightleftharpoons \text{PCl}_3(g) + \text{Cl}_2(g)$ $$\Delta n_g = (1 + 1) - 1 = +1$$ Since $\Delta n_g > 0$, the forward reaction produces more moles of gas. Applying **Le Chatelier's Principle**: **(A) Introducing $\text{Cl}_2$ at constant volume:** Increases $[\text{Cl}_2]$, a product $\Rightarrow$ equilibrium shifts **backward**. ✗ **(B) Introducing an inert gas at constant volume:** Volume is unchanged, so the partial pressures (and concentrations) of $\text{PCl}_5$, $\text{PCl}_3$, and $\text{Cl}_2$ remain **unchanged**. $Q_c = K_c$ still holds $\Rightarrow$ **no shift** in equilibrium. ✗ **(C) Introducing an inert gas at constant pressure:** To maintain constant pressure, the volume **increases**, diluting all reacting species. Decrease in partial pressures of reactants and products shifts equilibrium toward more moles of gas (forward direction, since $\Delta n_g = +1$). ✓ **(D) Increasing the volume of the container:** Decrease in pressure shifts equilibrium toward the side with more moles of gas, i.e., the **forward direction**. ✓ **Answer: (C) and (D)**