If 75% of a first order reaction was completed in 90 minutes, 60% of the same reaction would be comp — Chemical Kinetics Chemistry Question
Question
If 75% of a first order reaction was completed in 90 minutes, 60% of the same reaction would be completed in approximately (in minutes) _____. (take: log 2 = 0.30; log 2.5 = 0.40)
💡 Solution & Explanation
**Step 1: Find the rate constant (k) using first-order kinetics** For a first-order reaction: t = (2.303/k) × log([A]₀/[A]ₜ) When 75% is completed, 25% remains: [A]ₜ/[A]₀ = 0.25 90 = (2.303/k) × log(1/0.25) 90 = (2.303/k) × log(4) 90 = (2.303/k) × 2log(2) 90 = (2.303/k) × 2(0.30) 90 = (2.303/k) × 0.60 k = (2.303 × 0.60)/90 = 0.01535 min⁻¹ **Step 2: Calculate time for 60% completion** When 60% is completed, 40% remains: [A]ₜ/[A]₀ = 0.40 t = (2.303/k) × log(1/0.40) t = (2.303/0.01535) × log(2.5) t = 150 × 0.40 t = 60 minutes Therefore, the answer is 60.00.