The rate constant for the forward reaction: A(g) ⇌ 2B(g) is 1.5 × 10^-3 s^-1 at 300 K. If 10^-5 mole — Chemical Equilibrium Chemistry Question

💡 Solution & Explanation

Reaction: $\text{A}(g) \rightleftharpoons 2\text{B}(g)$; $k_f = 1.5\times10^{-3}\ \text{s}^{-1}$ \textbf{Step 1 — Equilibrium concentrations} (V = 10 L): \[ [\text{A}] = \frac{10^{-5}\ \text{mol}}{10\ \text{L}} = 10^{-6}\ \text{M}, \qquad [\text{B}] = \frac{100\ \text{mol}}{10\ \text{L}} = 10\ \text{M} \] \textbf{Step 2 — Equilibrium condition:} At equilibrium, rate of forward reaction = rate of backward reaction: \[ k_f[\text{A}] = k_b[\text{B}]^2 \] \textbf{Step 3 — Solve for $k_b$:} \[ k_b = \frac{k_f[\text{A}]}{[\text{B}]^2} = \frac{1.5\times10^{-3} \times 10^{-6}}{(10)^2} = \frac{1.5\times10^{-9}}{100} = 1.5\times10^{-11}\ \text{M}^{-1}\text{s}^{-1} \] \textbf{Answer: C} — $k_b = 1.5\times10^{-11}\ \text{M}^{-1}\text{s}^{-1}$