Half life of zero order reaction A → product is 1 hour, when initial concentration of reactant is 2.0 mol L ⁻¹. The time required to decrease concentration of A from 0.50 to 0.25 mol L⁻¹ is :

Solution

Chemistry ma’am · 3 steps

In a zero order reaction, concentration falls at a steady rate, so find k first and then divide.

1. Zero order law. For zero order, [A] = [A]₀ - kt. The concentration drops by the same amount every hour, so the half-life is t_(1/2) = [A]₀/2k. It depends on the starting concentration.
2. Find k. With [A]₀ = 2.0 mol L⁻¹ and t_(1/2) = 1 h, k = 2.0/(2 × 1) = 1.0 mol L⁻¹ h⁻¹.

   t = ([A]ₛₜₐᵣₜ - [A]_(end))/k = (0.50 - 0.25)/1.0 h = 0.25 h

3. So. 0.25 h = 0.25 × 60 = 15 min, which is option (b).

The idea to keep. Zero order means the concentration drops by the same amount each hour, so the half-life shrinks as concentration falls. Get k from the given data, then use time = drop in concentration ÷ k.

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