2 XY(g) ⇄ X2(g) + Y2(g) Kp = 230 A certain gas, XY(g), decomposes as represented by the equation abo — Equilibrium Chemistry Question

STEPS:

1. Identify the equilibrium constant and initial partial pressures: The balanced gas-phase equilibrium reaction is:
$$2\ \text{XY}(g) \rightleftharpoons \text{X}_2(g) + \text{Y}_2(g) \quad K_p = 230$$
The initial partial pressures from the table are:
* $P_{\text{XY}} = 0.010\text{ atm}$
* $P_{\text{X}_2} = 0.20\text{ atm}$
* $P_{\text{Y}_2} = 2.0\text{ atm}$
2. Understand the concept of the reaction quotient ($Q_p$): The reaction quotient, $Q_p$, has the same mathematical expression as the equilibrium constant, $K_p$, but uses the *initial* partial pressures of the gases rather than their equilibrium values. Comparing the value of $Q_p$ to $K_p$ allows us to determine which way the system must shift to establish equilibrium:
* If $Q_p < K_p$, the reaction shifts forward (to the right) to make more products.
* If $Q_p = K_p$, the system is already at equilibrium.
* If $Q_p > K_p$, the reaction shifts in reverse (to the left) to make more reactants.
3. Calculate the value of $Q_p$: Write the pressure expression for this reaction and substitute the initial values:
$$Q_p = \frac{P_{\text{X}_2} \cdot P_{\text{Y}_2}}{P_{\text{XY}}^2}$$
$$Q_p = \frac{(0.20\text{ atm}) \times (2.0\text{ atm})}{(0.010\text{ atm})^2}$$
$$Q_p = \frac{0.40}{0.0001} = \mathbf{4000}$$
4. Compare $Q_p$ to $K_p$ and determine the shift:
* Our calculated $Q_p = 4000$.
* The given equilibrium constant $K_p = 230$.
* Since $Q_p > K_p$ ($4000 > 230$), the ratio of products to reactants is currently too high. To reach equilibrium, the system must reduce the partial pressures of the products and increase the partial pressure of the reactant.
5. Conclude the direction of the reaction: The reaction must proceed in the reverse direction to form more reactant, which directly corresponds to Option B.

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WHY_OTHERS_WRONG:

• Option A is incorrect: The reaction would only proceed forward to form more products if $Q_p < K_p$. Because our calculated $Q_p$ (4000) is far larger than $K_p$ (230), the forward reaction is unfavorable, and the net reaction must go backward to decrease the excess products.
• Option C is incorrect: The mixture is not at equilibrium because $Q_p \neq K_p$. For the system to be at equilibrium with no net change, the calculated reaction quotient would have to equal exactly 230.
• Option D is incorrect: While volume is important for some equilibrium calculations (especially when converting between $K_c$ and $K_p$ or dealing with a change in the total number of moles of gas), it is not needed here. Because this reaction has the same number of moles of gas on both sides ($2 \text{ mol reactant} \rightarrow 2 \text{ mol product}$), the volume terms cancel out completely, and we can determine the direction of the reaction using only the partial pressures.