A flask contains a mixture of compounds A and B. Both compounds decompose by first-order kinetics. The half-lives for A and B are 300 s and 180 s, respectively. If the concentrations of A and B are equal initially, the time required for the concentration of A to be four times that of B(in s) :
(Use ln 2 = 0.693)

Solution

Chemistry ma’am · 4 steps

B decays faster, so the ratio A/B grows. Count half-lives and find when it reaches 4.

1. Write both decays. For first order, [A] = [A]₀ e^(-k_A t) and [B] = [B]₀ e^(-k_B t), with k = 0.693/t_(1/2). Since [A]₀ = [B]₀, the starting amounts cancel in the ratio.

   [A]/[B] = e^((k_B - k_A)t) = 4 ⇒ (k_B - k_A) t = ln 4 = 2 ln 2

2. Find the rate difference. k_B - k_A = ln 2(1/180 - 1/300) = ln 2 × 120/54000 = (ln 2)/450 s⁻¹.
3. Solve for t. (ln 2)/450 t = 2 ln 2, so t = 900 s. Check: in 900 s, A goes through 3 half-lives (left with 1/8) and B through 5 (left with 1/32). The ratio is (1/8)/(1/32) = 4.
4. So. The concentration of A becomes four times that of B after 900 s, option (d).

The idea to keep. For two first-order decays starting equal, the ratio depends only on the difference of rate constants. When the time is a multiple of both half-lives, just count halvings.

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