PCl3(g) + Cl2(g) ⇄ PCl5(g) Kc = 6.5 At a certain point in time, a 1.00 L rigid reaction vessel conta — Equilibrium Chemistry Question

STEPS:

1. Write the expression for the reaction quotient ($Q_c$): For the given reversible gaseous reaction, $\text{PCl}_3(g) + \text{Cl}_2(g) \rightleftharpoons \text{PCl}_5(g)$, the reaction quotient is written as:
$$Q_c = \frac{[\text{PCl}_5]}{[\text{PCl}_3][\text{Cl}_2]}$$
2. Calculate the initial molar concentrations: The volume of the rigid reaction vessel is $1.00\text{ L}$. Using the molarity formula ($M = \text{moles}/\text{volume}$):
* $[\text{PCl}_3]_0 = \frac{1.5\text{ mol}}{1.00\text{ L}} = 1.5\text{ M}$
* $[\text{Cl}_2]_0 = \frac{1.0\text{ mol}}{1.00\text{ L}} = 1.0\text{ M}$
* $[\text{PCl}_5]_0 = \frac{2.5\text{ mol}}{1.00\text{ L}} = 2.5\text{ M}$
3. Calculate the value of $Q_c$: Substitute these initial concentrations into your reaction quotient expression:
$$Q_c = \frac{2.5}{(1.5)(1.0)} = \frac{2.5}{1.5} \approx \mathbf{1.7}$$
4. Compare $Q_c$ to the equilibrium constant ($K_c$): The equilibrium constant for the reaction at this temperature is $K_c = 6.5$. Comparing the calculated $Q_c$ to $K_c$:
$$1.7 < 6.5 \implies \mathbf{Q < K_c}$$
Because $Q < K_c$, the system is not yet at equilibrium. To establish equilibrium, the reaction must shift to the right (the forward direction) to consume reactants and produce more products, which increases the value of $Q_c$ until it equals $K_c$.
5. Analyze the effect of the forward shift on gas moles: Examine the stoichiometric coefficients of the gaseous species:
* Reactants: $1\text{ mol PCl}_3(g) + 1\text{ mol Cl}_2(g) = \mathbf{2\text{ moles of gas}}$
* Products: $\mathbf{1\text{ mol PCl}_5(g)}$
As the reaction shifts forward, every cycle of the reaction consumes 2 moles of gas to produce only 1 mole of gas. Consequently, the total number of moles of gas ($n$) in the vessel decreases.
6. Relate the change in gas moles to the measured pressure: According to the Ideal Gas Law ($PV = nRT$), pressure is directly proportional to the number of moles of gas when volume and temperature are constant:
$$P = \left(\frac{RT}{V}\right)n$$
Since the container is rigid (constant $V$) and the temperature ($T$) is constant, the decrease in the total moles of gas ($n$) must cause the measured pressure inside the vessel to decrease, making Option C the correct answer.

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WHY_OTHERS_WRONG:

• Option A is incorrect: Although this option correctly notes that $Q < K_c$, it makes a conceptual error by claiming the pressure will increase. Because a forward shift converts 2 moles of reactant gas into only 1 mole of product gas, the total number of gas molecules decreases, which *decreases* the pressure rather than increasing it.
• Option B is incorrect: This option relies on two false premises. First, the calculated value of $Q$ ($1.7$) is less than $K_c$ ($6.5$), meaning the claim that $Q > K_c$ is mathematically false. Second, if $Q > K_c$ were true, the reaction would shift to the left to produce *more* moles of gas, which would increase the pressure.
• Option D is incorrect: While this option correctly states that the pressure would decrease if the system shifted to the right, it attributes the change to $Q > K_c$. As calculated above, $Q$ is less than $K_c$.