Two systems (g) ⇌ (g) + (g) and COCl2(g) ⇌ (g) + (g) are simultaneously in equilibrium in a vessel a — Chemical Equilibrium Chemistry Question

💡 Solution & Explanation

Step 1 - Understand Simultaneous Equilibria and the Common Species We are given two gas-phase systems established simultaneously in a single reaction vessel of constant volume: System 1: \[\ce{PCl5(g) <=> PCl3(g) + Cl2(g)} \quad \text{with equilibrium constant } K_{c1} = \frac{[\ce{PCl3}][\ce{Cl2}]}{[\ce{PCl5}]}\] System 2: \[\ce{COCl2(g) <=> CO(g) + Cl2(g)} \quad \text{with equilibrium constant } K_{c2} = \frac{[\ce{CO}][\ce{Cl2}]}{[\ce{COCl2}]}\] Because both reactions occur within the same container, they share a common volume and gaseous environment. Crucially, diatomic chlorine gas ($\ce{Cl2}$) is a product of both reactions and is the **common species** linking the two independent equilibrium systems. This means that any change affecting the concentration or partial pressure of $\ce{Cl2}$ in one system will immediately affect the equilibrium position of the other system. Step 2 - Analyze the Direct Effect of Adding \ce{CO(g)} When some carbon monoxide gas ($\ce{CO}$) is introduced into the reaction vessel at constant temperature and volume: * The concentration of $\ce{CO}$ in the vessel instantly increases, causing the reaction quotient of the second system ($Q_{c2}$) to exceed its equilibrium constant ($K_{c2}$): \[Q_{c2} = \frac{[\ce{CO}]_{\text{inst}}[\ce{Cl2}]}{[\ce{COCl2}]} > K_{c2}\] * According to **Le Chatelier's Principle**, the second system will respond to counteract this stress by shifting its equilibrium position in the **backward direction** ($\ce{CO(g) + Cl2(g) -> COCl2(g)}$) to consume the added reactant. * As the second equilibrium shifts backward, it consumes not only the added $\ce{CO}$ but also the common species $\ce{Cl2(g)}$. Consequently, the concentration and total moles of free $\ce{Cl2}$ in the vessel decrease. Step 3 - Analyze the Cascading Effect on the First Equilibrium The decrease in the concentration of the common product, $\ce{Cl2(g)}$, disturbs the first equilibrium system: * The instantaneous reaction quotient for the first system ($Q_{c1}$) becomes less than its equilibrium constant ($K_{c1}$): \[Q_{c1} = \frac{[\ce{PCl3}][\ce{Cl2}]_{\text{new}}}{[\ce{PCl5}]} < K_{c1}\] * According to **Le Chatelier's Principle**, the first system will respond by shifting in the **forward direction** ($\ce{PCl5(g) -> PCl3(g) + Cl2(g)}$) to produce more chlorine gas and restore equilibrium. * This forward shift consumes the reactant $\ce{PCl5(g)}$ and produces both $\ce{PCl3(g)}$ and $\ce{Cl2(g)}$. Step 4 - Evaluate the Options Systematically Let us evaluate each of the options based on our analysis of the new simultaneous equilibrium state: * **(A) \ce{PCl5} increase:** Incorrect. Because System 1 shifts in the forward direction, reactant $\ce{PCl5}$ is consumed, meaning its moles must **decrease**, not increase. * **(B) \ce{PCl3} remain unchanged:** Incorrect. The forward shift of System 1 produces more of the product $\ce{PCl3}$, which causes its total moles to **increase**. * **(C) \ce{PCl5} decrease:** Correct. The forward shift of System 1 directly consumes $\ce{PCl5}$, resulting in a **decrease** in its overall moles. * **(D) \ce{Cl2} increase:** Incorrect. Although the forward shift of System 1 produces some $\ce{Cl2}$, Le Chatelier's shift can only partially counteract the initial loss of chlorine. Therefore, at the new equilibrium, the net concentration and moles of $\ce{Cl2}$ will be **lower** than in the initial equilibrium state. \[\boxed{\text{C}}\]