The reaction: A(g) → 2B(g) + C(g) follows first-order kinetics. The reaction is started with pure 'A — Chemical Kinetics Chemistry Question
Question
The reaction: A(g) → 2B(g) + C(g) follows first-order kinetics. The reaction is started with pure 'A' in a rigid closed vessel maintained at constant temperature. After 10 s, a pin hole is developed in the vessel. If the molar ratio of gases 'A' and 'B' coming out initially is 1 : 2, the rate constant of reaction is (Molar masses of A, B and C are 16, 4 and 8 g/mol, respectively, ln 2 = 0.7, ln 3 = 1.1)
💡 Solution & Explanation
Effusion rate r ∝ P / sqrt(M). Ratio r_A / r_B = (P_A / P_B) * sqrt(M_B / M_A) = 1/2. Given M_A = 16, M_B = 4, sqrt(M_B/M_A) = 1/2. Thus, P_A / P_B = 1. From stoichiometry A → 2B + C, P_B = 2x and P_A = P0 - x. So P0 - x = 2x => P0 = 3x. P_A = 2/3 P0. k = (1/10) ln(P0 / (2/3 P0)) = 0.1 ln(1.5) = 0.1 * 0.4 = 0.04 s^-1. Therefore, correct answer is A.