An optically active substance, A, decomposes into optically active substances 'B' and 'C' as: A → 2B — Chemical Kinetics Chemistry Question
Question
An optically active substance, A, decomposes into optically active substances 'B' and 'C' as: A → 2B + C (k = 0.001 min^-1). The specific rotations of A, B and C are +40°, +10° and -30° per mole, respectively. If initially A and C were present in 4:3 mole ratio, the time (in min), after which the sample becomes optically inactive, is (ln 2 = 0.7, ln 5 = 1.6, ln 7 = 2.0, ln 13 = 2.5)
💡 Solution & Explanation
Let initially n_A = 4, n_C = 3, n_B = 0. At time t: n_A = 4 - x, n_B = 2x, n_C = 3 + x. Optical rotation = 40(4 - x) + 10(2x) - 30(3 + x) = 160 - 40x + 20x - 90 - 30x = 70 - 50x. Setting rotation to 0 gives x = 1.4. Remaining A = 4 - 1.4 = 2.6. t = (1/k) ln(A0/A) = 1000 * ln(4/2.6) = 1000 * ln(20/13). Using given logs: ln(20/13) = 2ln 2 + ln 5 - ln 13 = 1.4 + 1.6 - 2.5 = 0.5. t = 1000 * 0.5 = 500 min. Therefore, correct answer is 0500.