20. For which of the equilibrium systems represented below will the amount of product(s) at equilibr — Equilibrium Chemistry Question

STEPS:

1. Understand the physical impact of a volume change on gas-phase systems:
When the volume of a reaction vessel containing a gas-phase equilibrium system is increased at a constant temperature, the concentration and partial pressure of all gaseous species decrease.
2. Apply Le Chatelier's principle:
According to Le Chatelier's principle, if a system at equilibrium is subjected to a stress (such as a decrease in total pressure due to an increased volume), the system will shift its equilibrium position in the direction that counteracts the stress. To counteract a decrease in pressure, the system shifts in the direction that produces more moles of gas to restore some of the gas particles and pressure.
3. Establish the requirement for increasing the amount of products:
For the amount of product(s) to increase at equilibrium, the system must shift to the right (favoring the forward reaction) when the volume is increased. Therefore, the product side of the chemical equation must contain more moles of gas than the reactant side ($\Delta n_{\text{gas}} > 0$).
4. Analyze the stoichiometry of Option A:
$$\text{PCl}_5(g) \rightleftharpoons \text{PCl}_3(g) + \text{Cl}_2(g)$$
* Reactant gas moles: $1\text{ mole of gas } (\text{PCl}_5)$
* Product gas moles: $1\text{ mole of } \text{PCl}_3 + 1\text{ mole of } \text{Cl}_2 = 2\text{ moles of gas}$
* Since the product side has more moles of gas ($2 > 1$), increasing the volume of the vessel shifts the equilibrium to the right. This increases the amount of products, identifying Option A as the correct answer.

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WHY_OTHERS_WRONG:

• Option B is incorrect: In the reaction $2\ \text{NO}(g) + \text{O}_2(g) \rightleftharpoons 2\ \text{NO}_2(g)$, there are $3\text{ moles of gas}$ on the reactant side and only $2\text{ moles of gas}$ on the product side. Increasing the volume will shift the equilibrium to the left (toward the reactants) to favor the side with more moles of gas, causing the amount of products at equilibrium to decrease.
• Option C is incorrect: In the reaction $\text{N}_2(g) + \text{O}_2(g) \rightleftharpoons 2\ \text{NO}(g)$, there are $2\text{ moles of gas}$ on the reactant side and $2\text{ moles of gas}$ on the product side. Because the number of moles of gas is equal on both sides of the equation, a change in the volume of the vessel has no effect on the position of the equilibrium.
• Option D is incorrect: In the reaction $2\ \text{CO}(g) \rightleftharpoons \text{C}(s) + \text{CO}_2(g)$, carbon is in the solid phase and is excluded from gaseous calculations. There are $2\text{ moles of gas}$ on the reactant side and only $1\text{ mole of gas}$ ($\text{CO}_2$) on the product side. Increasing the volume will shift the equilibrium to the left (toward the reactants), which decreases the amount of products at equilibrium.