3 O2(g) <=> 2 O3(g) K_c = 1.8 x 10^-56 at 570 K For the system represented above, [O2] and [O3] init — Equilibrium Chemistry Question

STEPS:

1. Write the expression for the reaction quotient ($Q_c$):
Like the equilibrium constant expression, the reaction quotient $Q_c$ is calculated using the initial concentrations of the products divided by the reactants, each raised to the power of their stoichiometric coefficients from the balanced equation:
$$3\ \text{O}_2(g) \rightleftharpoons 2\ \text{O}_3(g)$$
$$Q_c = \frac{[\text{O}_3]^2}{[\text{O}_2]^3}$$
2. Substitute the given initial concentrations to calculate $Q_c$:
Using $[\text{O}_2] = 0.150\text{ M}$ and $[\text{O}_3] = 2.5\text{ M}$:
$$Q_c = \frac{(2.5)^2}{(0.150)^3}$$
$$Q_c = \frac{6.25}{0.003375} \approx \mathbf{1.9 \times 10^3}$$
3. Compare the value of $Q_c$ to the equilibrium constant ($K_c$):
Compare your calculated $Q_c$ with the given equilibrium constant $K_c = 1.8 \times 10^{-56}$:
$$1.9 \times 10^3 \gg 1.8 \times 10^{-56} \implies \mathbf{Q_c > K_c}$$
4. Predict the direction of the shift to establish equilibrium:
* When $Q_c > K_c$, the ratio of products to reactants in the current mixture is much larger than it should be at equilibrium.
* To decrease this ratio and reach equilibrium, the system must convert products back into reactants.
* This forces the reaction to shift to the left (the reverse direction).
5. Determine the effect on the amount of $\text{O}_3(g)$:
Because the system shifts to the left, the product $\text{O}_3(g)$ is consumed (its amount decreases) while the reactant $\text{O}_2(g)$ is produced (its amount increases). This confirms Option D is the correct answer.

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WHY_OTHERS_WRONG:

• Option A is incorrect: This option incorrectly claims that $Q < K_c$. Even without calculating the exact value of $Q_c$, a student can spot that because we have a very large concentration of product ($2.5\text{ M}$) compared to the extraordinarily tiny equilibrium constant ($10^{-56}$), $Q_c$ must be much larger than $K_c$. Additionally, if $Q < K_c$ were true, the system would shift right, which would *increase* the amount of $\text{O}_3(g)$, not decrease it as Option B suggests.
• Option B is incorrect: Although this option correctly predicts that the amount of $\text{O}_3(g)$ will decrease, the explanation given ($Q < K_c$) is mathematically incorrect. If $Q < K_c$, the system would shift to the right, which would increase the amount of product.
• Option C is incorrect: While this option correctly identifies that $Q_c > K_c$, it makes a conceptual error in predicting the direction of the shift. It claims the amount of $\text{O}_3(g)$ will increase, but a shift to the right would only move the system even further away from equilibrium since $Q_c$ is already too large.