rate = –d[A]/dt = d[B]/(2dt) = k[A] [A]₀ = mM [1 mM . . . 2.5 mM] k = s-1 [0 s-1 . . . 4 s-1]

In this example we consider a chemical process $A\rightarrow B+B$. The rate of the reaction is proportional to the concentration of $A$ $$rate\propto[A]$$ or $$rate=k[A]$$ where $k$ is the rate constant. Following the definition of the rate of a chemical reaction we may write $$-\dfrac{d[A]}{dt}=k[A]$$ for this process. This equation states that the rate of change of $[A]$ with respect to time, $t$, can be calculated as $k[A]$. In the lab, however, concentration–time plots are far more helpful than rate equations. In order to be able to draw concentration–time curves we first need to find the expression linking time to the concentration of $A$. We also realize that it is sufficient to establish the $[A]\sim t$ relation because we will always be able to calculate the concentration of $B$ for any given moment as $[B]=2([A]_0-[A])$. To find the expression linking time to $[A]$ we use integration but first we need to separate variables. Rearrangement yields $$-\dfrac{d[A]}{[A]}=kdt.$$ Both sides can now be integrated between the limits below:

 at the start some time later time $0$ $t$ conc. of $A$ $[A]_0$ $[A]$
as $$\int^{[A]}_{[A]_0}-\dfrac{d[A]}{[A]}=\int^t_0kdt$$ Since $k$ does not depend on time it can be moved before the integral sign and the right hand side becomes $$\int^t_0 k\,dt=k\int^t_0 dt=kt.$$ The left hand side rearranges into the familiar $$-\int^{[A]}_{[A]_0}\dfrac{1}{[A]}d[A]$$ form which can be found by recalling that $\int \frac{1}{x}dx=ln|x|+const$. It follows – switching now to definite integration – that $$-\int^{[A]}_{[A]_0}\dfrac{1}{[A]}d[A]=-\Big[ln [A]\Big]^{[A]}_{[A]_0}=-\big(ln[A]-ln[A]_0\big)=-ln\dfrac{[A]}{[A]_0}.$$ Hence, the integrated rate equation is $$-ln\dfrac{[A]}{[A]_0}=kt.$$ Taking the natural log of both sides after moving the negative sign over yields $$\dfrac{[A]}{[A]_0}=e^{-kt},$$ hence the concentration of $A$ at any given time can be calculated as $[A]=[A]_0e^{-kt}$ (see Figure 1) and that is what we sought to find. This formula can also be used to calculate the rate constant, $k$. For that we use the integrated rate equation in the following form $$-\big(ln[A]-ln[A]_0\big)=kt$$ which we rearrange into the more useful \begin{matrix} ln[A]&=&-k&\times&t&+&ln[A]_0\\ y&=&slope&\times&x&+&intercept \end{matrix} linear form used in Figure 3. Once the concentration of $A$ is measured during the course of the reaction and plotted against time, $k$ can be calculated as $k=-slope$. With $k$ known there is no need for further concentration measurements, $[A]$ and $[B]$ can be directly calculated for any given time as long as we remember the initial concentration of $A$.
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