Chapter 13, Problem 88E

The f̀rst-order integrated rate law for a reaction A $\to$ products is derived from the rate law using calculus as follows

Rate = k $\left[\text{A}\right]$ (first-order rate law)

Rate = $\frac{d\left[\text{A}\right]}{dt}$

$\frac{d\left[\text{A}\right]}{dt}$ =- k $\left[\text{A}\right]$

The above equation is a first-order, separable differential equation that can be solved by separating the variables and integrating

$\begin{array}{l}\frac{d\left[\text{A}\right]}{dt}=kdt\\ {\displaystyle {\int }_{{\left[\text{A}\right]}_0}^{\left[\text{A}\right]}\frac{d\left[\text{A}\right]}{\left[\text{A}\right]}}={\displaystyle \underset{0}{\overset{t}{\int }}kdt}\end{array}$

In the integral, [A]₀ is simply the initial concentration of A. You can then evaluate the integral

$\begin{array}{l}\left[\text{In}\left[\text{A}\right]\right.\left.\right]\frac{\left[\text{A}\right]}{{\left[\text{A}\right]}_0}=k{\left(t\right)}_0^t\\ \text{In}\left[\text{A}\right]-\text{In}{\left[\text{A}\right]}_0=-kt\\ \text{In}\left[\text{A}\right]=kt+\text{In}{\left[\text{A}\right]}_0\left(\text{integrted}\text{rate}\text{low}\right)\end{array}$

1. Use a procedure similar to the one just shown to derive an integrated rate law for a reaction A $\to$ products, which is one-half order in the concentration of A (that is, Rate = k[A]1/2).
2. Use the result from part a to derive an expression for the half-life of a one-half-order reaction.