The rate constant for a first-order reaction is 2.31 10⁻⁴ s ⁻¹ . If the initial concentration of the reactant is 0.80 M , what will be its concentration after 150 minutes?

Step-by-step solution

For a first-order reaction, the half-life is given by: t_ 1/2 = 0.693 k Substituting the given rate constant: t_ 1/2 = 0.693 2.31 10⁻⁴ = 3000 s Converting the half-life into minutes: t_ 1/2 = 3000 60 = 50 minutes The total time given is t = 150 minutes . The number of half-lives n is: n = t t_ 1/2 = 150 50 = 3 The concentration of the reactant after n half-lives is: [A]_t = [A]₀ 2^n Substituting the initial concentration [A]₀ = 0.80 M and n = 3 : [A]_t = 0.80 2^3 = 0.80 8 = 0.10 M Answer: 0.10 M