For a reaction A → 2 B+C the half lives are 100 s and 50 s when the concentration of reactant A is 0.5 and 1.0 mol L⁻¹ respectively. The order of the reaction is ______________ . (Nearest Integer)

Solution

Chemistry ma’am · 4 steps

Doubling [A] halved the half-life, so t_(1/2) ∝ 1/[A]: the mark of a second-order reaction.

1. Half-life rule. For a reaction of order n, the half-life depends on the starting concentration a as t_(1/2) ∝ a¹⁻ⁿ. Take the ratio of the two half-lives and the constant cancels.

   (t_(1/2))₁/(t_(1/2))₂ = (a₂/a₁)ⁿ⁻¹

2. Put in numbers. 100/50 = (1.0/0.5)ⁿ⁻¹, so 2 = 2ⁿ⁻¹.
3. Compare powers. n - 1 = 1, so n = 2. This fits second order, where t_(1/2) = 1/ka: the concentration doubled and the half-life halved.
4. So. The order of the reaction is 2.

The common slip: 0. Using t_(1/2) ∝ aⁿ⁻¹ instead of a¹⁻ⁿ gives 2 = (1/2)ⁿ⁻¹, so n = 0, not 2.

The idea to keep. Remember t_(1/2) ∝ a¹⁻ⁿ. A constant half-life means first order, t_(1/2) ∝ a means zero order, and t_(1/2) ∝ 1/a means second order.

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