The following reaction has an equilibrium constant Kc equal to 3.07 × 10^-4 at 24°C: 2NOBr(g) ⇌ 2(g) — Chemical Equilibrium Chemistry Question

💡 Solution & Explanation

Step 1 - Relate Reaction Rates to the Reaction Quotient (\(Q_c\)) and Equilibrium Constant (\(K_c\)) For any general reversible reaction: \[\ce{aA(g) + bB(g) <=> cC(g) + dD(g)}\] The reaction quotient (\(Q_c\)) is a dynamic quantity calculated using the instantaneous molar concentrations of the reactants and products at any given moment: \[Q_c = \frac{[\ce{C}]^c [\ce{D}]^d}{[\ce{A}]^a [\ce{B}]^b}\] The relationship between the rate of the forward reaction (\(r_{\text{f}}\)) and the rate of the backward reaction (\(r_{\text{b}}\)) is determined by comparing the reaction quotient (\(Q_c\)) with the equilibrium constant (\(K_c\)): * **At Equilibrium (\(r_{\text{f}} = r_{\text{b}}\)):** The system is in dynamic balance, which occurs when \(Q_c = K_c\). * **Backward Direction Favored (\(r_{\text{f}} < r_{\text{b}}\)):** When \(Q_c > K_c\), there is an excess of products relative to reactants. The reaction shifts in the backward direction to establish equilibrium. * **Forward Direction Favored (\(r_{\text{f}} > r_{\text{b}}\)):** When \(Q_c < K_c\), there is an excess of reactants relative to products. The reaction shifts in the forward direction to establish equilibrium. Thus, for the rate of the forward reaction to be greater than that of the backward reaction (\(r_{\text{f}} > r_{\text{b}}\)), the required condition is: \[Q_c < K_c\] Step 2 - Express the Reaction Quotient for the Given System The given gas-phase decomposition reaction of nitrosyl bromide (\(\ce{NOBr}\)) is: \[\ce{2NOBr(g) <=> 2NO(g) + Br2(g)}\] The equilibrium constant for this reaction at \(24^\circ\text{C}\) is: \[K_c = 3.07 \times 10^{-4}\text{ M}\] The reaction quotient (\(Q_c\)) is expressed as: \[Q_c = \frac{[\ce{NO}]^2 [\ce{Br2}]}{[\ce{NOBr}]^2}\] Step 3 - Evaluate the Reaction Quotient (\(Q_c\)) for Each Option * **Option (A):** \[[\ce{NOBr}] = 0.06\text{ M}, \quad [\ce{NO}] = 0.015\text{ M}, \quad [\ce{Br2}] = 0.01\text{ M}\] \[Q_c = \frac{(0.015\text{ M})^2 \times 0.01\text{ M}}{(0.06\text{ M})^2} = \left(\frac{0.015}{0.06}\right)^2 \times 0.01 = \left(\frac{1}{4}\right)^2 \times 10^{-2} = 6.25 \times 10^{-4}\text{ M}\] Since \(Q_c = 6.25 \times 10^{-4}\text{ M} > K_c = 3.07 \times 10^{-4}\text{ M}\), the backward rate is greater. Incorrect. * **Option (B):** \[[\ce{NOBr}] = 0.15\text{ M}, \quad [\ce{NO}] = 0.015\text{ M}, \quad [\ce{Br2}] = 0.014\text{ M}\] \[Q_c = \frac{(0.015\text{ M})^2 \times 0.014\text{ M}}{(0.15\text{ M})^2} = \left(\frac{0.015}{0.15}\right)^2 \times 0.014 = (0.1)^2 \times 0.014 = 1.4 \times 10^{-3}\text{ M}\] Since \(Q_c = 1.4 \times 10^{-3}\text{ M} > K_c = 3.07 \times 10^{-4}\text{ M}\), the backward rate is greater. Incorrect. * **Option (C):** \[[\ce{NOBr}] = 0.18\text{ M}, \quad [\ce{NO}] = 0.012\text{ M}, \quad [\ce{Br2}] = 0.02\text{ M}\] \[Q_c = \frac{(0.012\text{ M})^2 \times 0.02\text{ M}}{(0.18\text{ M})^2} = \left(\frac{1}{15}\right)^2 \times 0.02 = \frac{1}{225} \times 0.02 \approx 8.89 \times 10^{-5}\text{ M}\] Since \(Q_c \approx 8.89 \times 10^{-5}\text{ M} < K_c = 3.07 \times 10^{-4}\text{ M}\), the forward rate is greater. Correct. * **Option (D):** \[[\ce{NOBr}] = 0.045\text{ M}, \quad [\ce{NO}] = 0.0105\text{ M}, \quad [\ce{Br2}] = 0.01\text{ M}\] \[Q_c = \frac{(0.0105\text{ M})^2 \times 0.01\text{ M}}{(0.045\text{ M})^2} = \left(\frac{7}{30}\right)^2 \times 10^{-2} \approx 5.44 \times 10^{-4}\text{ M}\] Since \(Q_c \approx 5.44 \times 10^{-4}\text{ M} > K_c = 3.07 \times 10^{-4}\text{ M}\), the backward rate is greater. Incorrect. Step 4 - Conclusion Only Option (C) satisfies the necessary condition of \(Q_c < K_c\). \[\boxed{\text{C}}\]