Ethylene is produced as: C4H8 → 2C2H4. The rate constant is 2.5 × 10^-5 s^-1. In what time (in hours — Chemical Kinetics Chemistry Question

💡 Solution & Explanation

Let initial C4H8 be a. At time t, remaining C4H8 is a-x, and C2H4 formed is 2x. Ratio = 2x / (a-x) = 1, so 2x = a-x => x = a/3. Remaining C4H8 = a - a/3 = 2a/3. Since it's a first-order reaction (implied by units of k): t = (1/k) ln(a / (2a/3)) = (1/k) ln(1.5) = (ln 3 - ln 2) / k = (0.48 - 0.30) / (2.5 × 10^-5) = 0.18 / (2.5 × 10^-5) = 7200 s. In hours, 7200 / 3600 = 2 hours. Therefore, correct answer is 2.