X2(g) + Y2(g) ⇌ 2 XY(g) Kc = 3.0 A mixture of X2(g) , Y2(g) , and XY(g) is placed in a previously ev — Equilibrium Chemistry Question

STEPS:

1. Identify the reaction quotient ($Q_c$) expression for the system:
The reversible gas-phase reaction is:
$$\text{X}_2(g) + \text{Y}_2(g) \rightleftharpoons 2\ \text{XY}(g) \quad \text{with} \quad K_c = 3.0 \quad$$
The reaction quotient ($Q_c$) measures the relative amounts of products and reactants present in a reaction mixture at any given point in time. It is written using the same mathematical expression as the equilibrium constant $K_c$:
$$Q_c = \frac{[\text{XY}]^2}{[\text{X}_2][\text{Y}_2]} \quad$$

2. Understand how the comparison between $Q_c$ and $K_c$ dictates the system's shift:
To reach equilibrium, the reaction mixture's current ratio ($Q_c$) must adjust until it equals the constant ratio ($K_c$):
* If $Q_c < K_c$: The concentration of products is lower than what is required for equilibrium. The reaction shifts to the right (forward direction), consuming reactants to form more products.
* If $Q_c = K_c$: The system is already at chemical equilibrium; no shift occurs.
* If $Q_c > K_c$: The concentration of products is higher than what is required for equilibrium. The reaction shifts to the left (reverse direction) to consume product and form more reactants.

3. Calculate $Q_c$ for the initial concentrations in Option A:
Substitute the starting concentrations into the reaction quotient formula:
$$Q_c = \frac{(0.20\text{ M})^2}{(0.40\text{ M})(0.40\text{ M})} = \frac{0.040}{0.16} = \mathbf{0.25} \quad$$
Compare the calculated $Q_c$ to the given $K_c$:
$$Q_c\ (0.25) < K_c\ (3.0) \quad$$
Because $Q_c$ is less than $K_c$, the system must shift to the right to produce more product as it moves toward equilibrium, establishing Option A as the correct choice.

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WHY_OTHERS_WRONG:

* Option B is incorrect: Substituting the values into the quotient expression yields:
$$Q_c = \frac{(0.90\text{ M})^2}{(0.30\text{ M})(0.30\text{ M})} = \frac{0.81}{0.09} = \mathbf{9.0} \quad$$
Because $Q_c\ (9.0) > K_c\ (3.0)$, the system will shift to the left, consuming products and producing more reactants as it approaches equilibrium.
* Option C is incorrect: Substituting the values into the quotient expression yields:
$$Q_c = \frac{(0.30\text{ M})^2}{(0.15\text{ M})(0.15\text{ M})} = \frac{0.090}{0.0225} = \mathbf{4.0} \quad$$
Because $Q_c\ (4.0) > K_c\ (3.0)$, the system will shift to the left, leading to the consumption of products to form more reactants.
* Option D is incorrect: Substituting the values into the quotient expression yields:
$$Q_c = \frac{(0.20\text{ M})^2}{(0.10\text{ M})(0.10\text{ M})} = \frac{0.040}{0.010} = \mathbf{4.0} \quad$$
Because $Q_c\ (4.0) > K_c\ (3.0)$, this system will also shift to the left, which results in the formation of more reactants rather than products.