Rate of disappearance of reactant A in A ⇌ B at two temperatures: -d[A]/dt = (2.0×10^-3)[A] - (5.0×1 — Chemical Equilibrium Chemistry Question

💡 Solution & Explanation

Step 1 - Relate the rate equation to forward and backward rate constants For the reversible reaction \ce{A(g) <=> B(g)}, the net rate is: \[-\frac{d[\ce{A}]}{dt} = k_f [\ce{A}] - k_b [\ce{B}]\] At equilibrium, the net rate is zero, so: \[K_c = \frac{[\ce{B}]}{[\ce{A}]} = \frac{k_f}{k_b}\] Step 2 - Determine equilibrium constants at both temperatures At \(T_1 = 27^\circ\text{C} = 300\text{ K}\): \[k_{f1} = 2.0 \times 10^{-3}, \quad k_{b1} = 5.0 \times 10^{-4}\] \[K_1 = \frac{2.0 \times 10^{-3}}{5.0 \times 10^{-4}} = 4\] At \(T_2 = 127^\circ\text{C} = 400\text{ K}\): \[k_{f2} = 8.0 \times 10^{-2}, \quad k_{b2} = 4.0 \times 10^{-3}\] \[K_2 = \frac{8.0 \times 10^{-2}}{4.0 \times 10^{-3}} = 20\] Step 3 - Apply the Van't Hoff equation The integrated Van't Hoff equation: \[\log_{10}\left(\frac{K_2}{K_1}\right) = \frac{\Delta H^\circ}{2.303 \times R} \left(\frac{T_2 - T_1}{T_1 T_2}\right)\] Step 4 - Substitute and solve \[\log\left(\frac{20}{4}\right) = \frac{\Delta H^\circ}{2.303 \times 8.314} \cdot \frac{100}{300 \times 400}\] \[\log(5) = \frac{\Delta H^\circ \times 100}{2.303 \times 8.314 \times 300 \times 400}\] \[\Delta H^\circ = \left(\frac{2.303 \times 8.314 \times 300 \times 400}{100}\right) \times \log(5) \text{ J/mol}\] \[\boxed{\Delta H^\circ = \left(\frac{2.303 \times 8.314 \times 300 \times 400}{100}\right) \times \log(5) \text{ J/mol}}\] Step 5 - Explain each option * **Option (A)**: Incorrect. Uses log(50) — wrong ratio; K2/K1 = 20/4 = 5, not 50. * **Option (B)**: Correct. K1 = 4, K2 = 20, ratio = 5, Van't Hoff gives this expression exactly. * **Option (C)**: Incorrect. Uses log(2) — not derived from any correct ratio of the given constants. * **Option (D)**: Incorrect. Uses log(20) — represents K2 alone, not the ratio K2/K1.