The rate of First order reaction is 0.04 mol L⁻¹ s⁻¹ at 10 minutes and 0.03 mol L⁻¹ s⁻¹ at 20 minutes after initiation. Half life of the reaction is _______ minutes.
(Given log 2=0.3010, log 3=0.4771)

Solution

Chemistry ma’am · 3 steps

For a first-order reaction, rate tracks concentration, so the ratio of rates is the ratio of concentrations.

1. Rate tracks concentration. For first order, rate = k[A], so the rate is proportional to [A]. The ratio of the two rates is the ratio of the two concentrations: [A]₁₀/[A]₂₀ = 0.04/0.03 = 4/3.
2. Find k. The concentration falls by this factor in 20 - 10 = 10 min. So k = 2.303/10 log 4/3, and log 4/3 = 2 log 2 - log 3 = 0.6020 - 0.4771 = 0.1249.

   t_(1/2) = (2.303 log 2)/k = (10 × log 2)/(log 4/3) = (10 × 0.3010)/0.1249

3. So. t_(1/2) = 24.1 min, which rounds to 24.

The common slip: 48. Taking the interval as 20 min, the time since the start, instead of the 10 min gap between the two readings doubles the half-life to 48.

The idea to keep. In a first-order reaction, rates at two times behave exactly like concentrations at those times. Use the time gap between the two readings, not the time since the start.

Answer from the NTA final answer key. Solution by PrepWiser.