A reaction that is second-order in one reactant has a rate constant of 3.8 × 10^-2 L/(mol ∙ s).If the initial concentration of the reactant is 0.280 mol/L,how long will it take for the concentration to become 0.140 mol/L?

Question

Quizplus · Accepted Answer

For a second-order reaction, the rate law is given by $\frac{1}{[A]_t} - \frac{1}{[A]_0} = kt$, where $[A]_t$ is the concentration at time $t$, $[A]_0$ is the initial concentration, and $k$ is the rate constant. Plugging in the given values: $\frac{1}{0.140} - \frac{1}{0.280} = (3.8 \times 10^{-2})t$, solving for $t$ gives $t = 93$ seconds.

A Reaction That Is Second-Order in One Reactant Has a Rate