The half-life of a first-order reaction is 10 min. In what time, the rate of reaction will decrease from 6.0 × 10^21 molecules mL^-1 s^-1 to 4.5 × 10^25 molecules litre^-1 min^-1? (N_A = 6.0 × 10^23)

Question

Prashant Kotian · Accepted Answer

Correct answer: C. The rate of a first-order reaction is ∝ the concentration of the reactant: R = K[A]. Thus, R1/R2 = [A]1/[A]2. Converting R1 and R2 to consistent units: R1 = 6.0 × 10^21 molecules mL^-1 s^-1 = 10^-2 mol mL^-1 s^-1 = 10 mol L^-1 s^-1. R2 = 4.5 × 10^25 molecules L^-1 min^-1 = 75 mol L^-1 min^-1 = 1.25 mol L^-1 s^-1. Therefore, R1/R2 = 10/1.25 = 8 = 2^3. This indicates that 3 half-lives have elapsed. Total time = 3 × t_1/2 = 3 × 10 min = 30 min.

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