In a first-order reaction, the concentration of the reactant decreases from 0.64 M to 0.04 M in 72 minutes . The half-life of this reaction is:

Step-by-step solution

For a first-order reaction, the concentration after n half-lives is given by: C_t = C₀ 2^n Given C₀ = 0.64 M and C_t = 0.04 M : 0.04 = 0.64 2^n 2^n = 0.64 0.04 = 16 2^n = 2^4 n = 4 This means 4 half-lives have passed in 72 minutes . 4 t_ 1/2 = 72 minutes t_ 1/2 = 72 4 = 18 minutes Answer: 18 minutes