rate = –d[A]/dt = d[B]/dt = k

[A]₀ =

 [1 mM . . . 5 mM]

k =

 [0 mM s^{1} . . . 10 mM s^{1}]

In this example we consider a chemical process $A\rightarrow B$ where the rate of the reaction is
not proportional to the concentration of $A$. $$rate\require{cancel}\cancel{\propto}[A]$$ This peculiar behaviour can occur if the reaction requires a catalist. When all catalytic active sites are engaged the concentration of $A$ has no effect on the rate which typically depends on the number of active sites and the transport rates between the bulk and the active sites. All being constant to a good approximation (at least for large enough concentratins of $A$), the reaction rate can be mathematically captured as $$rate=k$$ where $k$ denotes the rate constant. Following the definition of the rate of a chemical reaction we may write
$$\dfrac{d[A]}{dt}=k$$ for this process. This equation states that the rate of change of $[A]$ with respect to time, $t$, remains the same during the course of the reaction and is equal to the constant value of $k$. 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 enough to establish the $[A]\sim t$ relation because we will always be able to calculate the concentration of $B$ for any given instant as $[B]=[A]_0[A]$. To find the expression linking time to $[A]$ we use integration but first we need to separate variables. Rearrangement yields
$$d[A]=kdt.$$
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
$$\int^{[A]}_{[A]_0}d[A]=\int^t_0kdt$$
Since $k$ does not depend on time it can be moved before the integral sign and the right hand side becomes
$$\int^t_0 k\,dt=k\int^t_0 dt=k\,\Big[t\Big]^{t}_{0}=k\,(t0)=kt.$$
It follows – as we have seen above – that
$$\int^{[A]}_{[A]_0}d[A]=\Big[[A]\Big]^{[A]}_{[A]_0}=\big([A][A]_0\big)=[A]_0[A].$$
Therefore, the integrated rate equation is
$$[A]_0[A]=kt$$
meaning that the concentration of $A$ at any given time can be calculated as $[A]=[A]_0kt$ (see Figure 1) which is what we set out to find. This formula can also be used to obtain the rate constant, $k$, as
\begin{matrix}
[A]&=&k&\times&t&+&[A]_0\\
y&=&slope&\times&x&+&intercept
\end{matrix}
linear form used in Figure 3. Once $[A]$ 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 $A$.