What is the equilibrium constant for the reaction at 27°C? — Chemical Equilibrium Chemistry Question

💡 Solution & Explanation

Given ΔH = E_a(fwd) - E_a(bwd) = -24.942 kJ/mol. Also E_a(fwd) / E_a(bwd) = 2/3, so E_a(fwd) = (2/3)E_a(bwd). Substituting this: (2/3)E_a(bwd) - E_a(bwd) = -24.942 => (-1/3)E_a(bwd) = -24.942 => E_a(bwd) = 74.826 kJ/mol. Then E_a(fwd) = 49.884 kJ/mol. The equilibrium constant K = k_f / k_b = (A e^(-E_af/RT)) / (A e^(-E_ab/RT)) = e^((E_ab - E_af) / RT) = e^(-ΔH / RT). Using ΔH = -24942 J/mol, R = 8.314 J/(mol·K), and T = 300 K: K = e^(24942 / (8.314 × 300)) = e^(24942 / 2494.2) = e¹⁰. Therefore, correct answer is C.