In a first-order reaction, the concentration of the reactant decreases from 1.0 M to 0.25 M in 40 minutes. What will be the concentration of the reactant after a further 60 minutes?

Step-by-step solution

For a first-order reaction, the concentration drops from 1.0 M to 0.25 M (which is one-fourth of the initial concentration). This corresponds to 2 half-lives, since 1 2^2 = 1 4 . Thus, 2 t_ 1/2 = 40 minutes , which gives t_ 1/2 = 20 minutes . The question asks for the concentration after a further 60 minutes. The total time elapsed from the start is 40 + 60 = 100 minutes . Total number of half-lives, n = 100 20 = 5 . Final concentration = [ A ]₀ 2^n = 1.0 2^5 = 1.0 32 = 0.03125 M . Answer: 0.03125 M