A flask is filled with equal moles of A and B. The half lives of A and B are 100 s and 50 s respectively and are independent of the initial concentration. The time required for the concentration of A to be four times that of B is ___________ s.
(Given : ln 2 = 0.693)

Solution

Chemistry ma’am · 4 steps

Half-life independent of concentration means first order, so write both decays in half-lives and compare.

1. Spot the order. A half-life that does not depend on the initial concentration belongs to a first-order reaction. So after time t, each substance is left with (1/2)^(t/t_(1/2)) of its starting amount.
2. Write both amounts. Both start at the same concentration C₀. Then [A] = C₀(1/2)^(t/100) and [B] = C₀(1/2)^(t/50).

   [A]/[B] = 2^(t/50)/2^(t/100) = 2^(t/100) = 4 = 2²

3. Check with ln 2. In rate-constant form, kA = 0.693/100 and kB = 0.693/50. Then ln 4 = (kB - kA)t, so 2 × 0.693 = 0.693/100 t, which gives t = 200 s.
4. So. t/100 = 2, so t = 200 s. By then A has gone through 2 half-lives and is at 1/4 of C₀. B has gone through 4 half-lives and is at 1/16 of C₀, which makes A four times B.

The common slip: 100. Letting only B decay to one-fourth and ignoring A's own decay gives 100 s, but A also falls to half in that time.

The idea to keep. When two first-order species decay together, their ratio changes as e^((kB-kA)t). Only the difference in rate constants matters, because both of them keep decaying.

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