Reaction A: 4 HCl(g) + O2(g) ⇄ 2 Cl2(g) + 2 H2O(g) Reaction B: N2O4(g) ⇄ 2 NO2(g) Reaction C: H2(g) — Equilibrium Chemistry Question

STEPS:

1. Identify the core chemistry concept (Le Chatelier's Principle): Le Chatelier’s Principle states that if a system at equilibrium is subjected to a stress or disturbance, the system will shift its equilibrium position in a direction that counteracts or minimizes that stress.
2. Understand the effect of reducing the container volume: Reducing the volume of a sealed container holding gaseous reactants and products increases the concentration and the partial pressures of all gaseous species present. This increases the frequency of collisions and raises the overall pressure inside the container.
3. Determine how the system counteracts a volume reduction: To minimize the stress of increased pressure, the equilibrium will shift in the direction that decreases the total number of gas molecules in the container. Therefore:
* A decrease in volume (increase in pressure) shifts the equilibrium toward the side of the reaction with fewer moles of gas.
* An increase in volume (decrease in pressure) shifts the equilibrium toward the side with more moles of gas.
4. Analyze the stoichiometry of each reaction to count the gaseous moles on both sides:
* Reaction A:
$$4\ \text{HCl}(g) + \text{O}_2(g) \rightleftharpoons 2\ \text{Cl}_2(g) + 2\ \text{H}_2\text{O}(g)$$
* Reactant gas moles: $4 + 1 = \mathbf{5\text{ moles}}$
* Product gas moles: $2 + 2 = \mathbf{4\text{ moles}}$
* *Since the product side has fewer moles of gas ($4 < 5$), reducing the volume will shift this equilibrium to the right (forward direction), increasing the amount of products.*
* Reaction B:
$$\text{N}_2\text{O}_4(g) \rightleftharpoons 2\ \text{NO}_2(g)$$
* Reactant gas moles: $\mathbf{1\text{ mole}}$
* Product gas moles: $\mathbf{2\text{ moles}}$
* *Since the reactant side has fewer moles of gas ($1 < 2$), reducing the volume will shift this equilibrium to the left (reverse direction), decreasing the amount of products.*
* Reaction C:
$$\text{H}_2(g) + \text{I}_2(g) \rightleftharpoons 2\ \text{HI}(g)$$
* Reactant gas moles: $1 + 1 = \mathbf{2\text{ moles}}$
* Product gas moles: $\mathbf{2\text{ moles}}$
* *Since both sides have an equal number of gas moles ($2 = 2$), a change in volume has no effect on the equilibrium position or the amount of products.*
* Reaction D:
$$2\ \text{NH}_3(g) \rightleftharpoons \text{N}_2(g) + 3\ \text{H}_2(g)$$
* Reactant gas moles: $\mathbf{2\text{ moles}}$
* Product gas moles: $1 + 3 = \mathbf{4\text{ moles}}$
* *Since the reactant side has fewer moles of gas ($2 < 4$), reducing the volume will shift this equilibrium to the left (reverse direction), decreasing the amount of products.*
5. Select the correct reaction: Reaction A is the only system where a volume decrease shifts the equilibrium to the right to increase the amount of products, confirming Option A is the correct answer.

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WHY_OTHERS_WRONG:

• Option B is incorrect: In Reaction B, the product side has more moles of gas than the reactant side ($2 > 1$). Reducing the volume shifts the equilibrium to the left to favor the reactants, which decreases the amount of product ($\text{NO}_2$).
• Option C is incorrect: In Reaction C, there are exactly 2 moles of gas on both sides of the equation. Because the ratio of gas moles is $1:1$, changing the volume of the container does not shift the equilibrium in either direction, leaving the overall amount of product ($\text{HI}$) unchanged.
• Option D is incorrect: In Reaction D, the product side has more moles of gas than the reactant side ($4 > 2$). Reducing the volume shifts the equilibrium to the left to favor the reactant, which decreases the amount of products ($\text{N}_2$ and $\text{H}_2$).