In a reaction A+B → C, initial concentrations of A and B are related as [A]₀=8[B]₀. The half lives of A and B are 10 min and 40 min , respectively. If they start to disappear at the same time, both following first order kinetics, after how much time will the concentration of both the reactants be same?

Solution

Chemistry ma’am · 3 steps

Write each concentration as start times a power of one half, set them equal, and solve for t.

1. First-order decay. For first order, every half-life halves the amount. After time t, [A] = [A]₀(1/2)^(t/10) and [B] = [B]₀(1/2)^(t/40), with t in minutes.

   8[B]₀(1/2)^(t/10) = [B]₀(1/2)^(t/40)

2. Solve for t. Cancel [B]₀ and rearrange: 8 = 2^(t/10 - t/40) = 2^(3t/40). Since 8 = 2³, you get 3t/40 = 3, so t = 40 min.

   Time	[A] in units of [B]₀	[B] in units of [B]₀
   20 min	8 × 1/4 = 2	(1/2)^(0.5) ≈ 0.71
   40 min	8 × 1/16 = 0.5	1/2 = 0.5
   60 min	8 × 1/64 = 0.125	(1/2)^(1.5) ≈ 0.35
   80 min	8 × 1/256 ≈ 0.031	1/4 = 0.25

3. So. At 40 min, A has gone through 4 half-lives and B through 1, and both stand at 0.5[B]₀. The answer is (b), 40 min.

The idea to keep. In first order, count half-lives: amount left is start × (1/2)ⁿ, where n = t/t_(1/2). To make two such amounts equal, the faster one must lose exactly its head start.

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