In a first order decomposition reaction, the time taken for the decomposition of reactant to one fourth and one eighth of its initial concentration are t₁ and t₂ (s), respectively. The ratio t₁/t₂ will be:

Solution

Chemistry ma’am · 3 steps

One fourth is two half-lives and one eighth is three, so the ratio of times is 2 to 3.

1. Fixed half-life. For a first order reaction, t_(1/2) = 0.693/k does not depend on concentration. So every half-life takes the same time, whatever amount is left.
2. Count the halvings. Going to one fourth means halving twice: 1 → 1/2 → 1/4, so t₁ = 2 t_(1/2). Going to one eighth means halving three times, so t₂ = 3 t_(1/2).

   t₁/t₂ = (1/k ln 4)/(1/k ln 8) = (2 ln 2)/(3 ln 2) = 2/3

3. So. The integrated rate law t = 1/k ln[A]₀/[A] gives the same result: t₁/t₂ = 2/3, which is option (d).

The idea to keep. In first order kinetics, time is proportional to the number of half-lives. If the fraction left is (1/2)ⁿ, the time taken is n × t_(1/2).

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