A and B decompose via first order kinetics with half-lives 54.0 min and 18.0 min respectively. Starting from an equimolar non reactive mixture of A and B, the time taken for the concentration of A to become 16 times that of B is _________ min. (Round off to the Nearest Integer).

Solution

Chemistry ma’am · 4 steps

Both of them are decaying, so track the ratio A/B. It doubles every 27 minutes, not every 18.

1. Each one alone. For first-order decay, [A] = [A]₀ (1/2)^(t/54) and [B] = [B]₀ (1/2)^(t/18), with t in minutes. The mixture is equimolar, so [A]₀ = [B]₀.
2. Take the ratio. Divide the two expressions. The starting concentrations cancel, and the ratio grows as a power of 2.

   [A]/[B] = 2^(t/18-t/54) = 2^(2t/54) = 2^(t/27)

3. Set it to 16. 16 = 2⁴, so t/27 = 4. Check: at 108 min, A has gone through 2 half-lives and B through 6, so [A]/[B] = 2⁶⁻² = 16.
4. So. t = 4 × 27 = 108 min.

The common slip: 72. Counting four half-lives of B alone, while forgetting that A is decaying too, gives 72 min, not 108.

The idea to keep. When two first-order species decay together, their ratio changes by a factor of 2^(n_B - n_A), where n_A and n_B are the numbers of half-lives each has gone through. Count the half-lives of both.

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