Reaction: A(g) + B(g) ⇌ C(g) + D(g), occurs in a single step. The rate constant of forward reaction — Chemical Equilibrium Chemistry Question

💡 Solution & Explanation

Step 1 - Express equilibrium concentrations using an ICE table For the single-step reversible reaction: \[\ce{A(g) + B(g) <=> C(g) + D(g)}\] Let initial concentrations \([A]_0 = [B]_0 = a\text{ M}\), and let \(x\text{ M}\) react: \[\begin{array}{lcccc} \text{Species} & \ce{A} & \ce{B} & \ce{C} & \ce{D} \ \hline \text{Initial (M)} & a & a & 0 & 0 \ \text{Change (M)} & -x & -x & +x & +x \ \text{Equilibrium (M)} & a - x & a - x & x & x \ \hline \end{array}\] Step 2 - Use the equilibrium condition \([A] = 2[C]\) to find \(x\) \[a - x = 2x \implies a = 3x \implies x = \frac{a}{3}\] Equilibrium concentrations: \[[A] = [B] = \frac{2a}{3}\text{ M}, \quad [C] = [D] = \frac{a}{3}\text{ M}\] Step 3 - Calculate \(K_c\) \[K_c = \frac{[C][D]}{[A][B]} = \frac{\left(\frac{a}{3}\right)^2}{\left(\frac{2a}{3}\right)^2} = \frac{a^2/9}{4a^2/9} = \frac{1}{4} = 0.25\] Step 4 - Find the backward rate constant using \(K_c = k_f / k_b\) For a single-step elementary reaction: \[K_c = \frac{k_f}{k_b} \implies k_b = \frac{k_f}{K_c} = \frac{2.0 \times 10^{-3}}{0.25} = \boxed{8.0 \times 10^{-3}\text{ mol}^{-1}\text{ L s}^{-1}}\] Step 5 - Evaluate all options - **Option (A) 5.0 × 10^-4**: Incorrect. This uses \(k_b = k_f \times K_c\) (inverted relation). - **Option (B) 8.0 × 10^-3**: Correct. \(k_b = k_f / K_c = 2.0 \times 10^{-3} / 0.25 = 8.0 \times 10^{-3}\). - **Option (C) 1.25 × 10^2**: Incorrect. Arithmetic error. - **Option (D) 2.0 × 10^3**: Incorrect. Powers of ten error.