rate = –d[A]/dt = d[B]/dt = k[A][B]Notice the initial rate acceleration. [A]₀ = mM [1 mM . . . 4 mM] [B]₀ = mM [0.1 mM . . . 1 mM] k = (Ms)-1 [0 (Ms)-1 . . . 2 (Ms)-1]

In this example we consider a reaction $A\rightarrow B$, but now we are accounting for the effect of autocatalysis. By autocatalysis we mean that the reaction is catalysed by its product $B$, a phenomenon that occur in some chemical and biochemical reactions. We can think of it as feedback where product molecules increase the rate of their own production. The simplest way of including autocatalysis into a rate equation is by multiplying the right hand side of the equation with $B$, which gives $$rate=k[A][B].$$ Now, as the concentration of $B$ increases (i.e. $[B]$ becomes larger and larger) during the reaction so does the rate, and that is precisely what feedback is. Using the definition of the rate of a chemical reaction we may write $$-\dfrac{d[A]}{dt}=k[A][B].$$ The extent of the reaction can be captured by introducing an auxiliary variable $x$ which is zero at the start of the reaction when the concentrations of $A$ and $B$ are $[A]_0$ and $[B]_0$, respectively. Subsequently, as $A$ is gradually transformed into $B$ in the reaction the value of $x$ continuously increases.

 at the start some time later time $0$ $t$ conc. of $A$ $[A]_0$ $[A]=[A]_0-x$ conc. of $B$ $[B]_0$ $[B]=[B]_0+x$

During the reaction the rate of decrease of $[A]$ is equal to the rate of increase of $[B]$, which in turn is equal to the growth rate of $x$ $$-\dfrac{d[A]}{dt}=\dfrac{d[B]}{dt}=\dfrac{dx}{dt},$$ thus we can rewrite the rate equation as $$\dfrac{dx}{dt}=k([A]_0-x)([B]_0+x).$$ Mathematically speaking, this is a differential equation stating that the rate of change of $x$ at a point in time, $t$, is equal to the expression on the right hand side. We can think of this equation as a precise statement of saying that the speed of reaction at a certain instant in time depends on the rate constant, $k$, the initial concentrations, $[A]_0$ and $[B]_0$, and also on how far the reaction has progressed. The latter information is stored in $x$, the value of which clearly changes as time goes. To be able to tell what the concentrations are at any given point in time we need to find the functions linking $[A]$ and $[B]$ to $t$, and that is exactly what we can do by integration. First, gathering terms containing $x$ gives $$\dfrac{1}{([A]_0-x)([B]_0+x)}dx=k\,dt$$ which now can be integrated between the limits $0$ and $t$ for time, and $0$ and $x$ for the extent of reaction. $$\int^x_0\dfrac{1}{([A]_0-x)([B]_0+x)}dx=\int^t_0 k\,dt.$$ 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 expression for $x$ looks a bit too complicated to be integrated in one step, so we break it up using the method of partial fraction decomposition. In doing so we suppose that the integrand can be transformed into a sum of two fractions with some terms ($u$ and $v$) in the numerators hoping that the two fractions are going to be easy to deal with separately. $$\dfrac{1}{([A]_0-x)([B]_0+x)}=\dfrac{u}{([A]_0-x)}+\dfrac{v}{([B]_0+x)}$$ To find $u$ and $v$ we multiply over as $$\dfrac{u}{([A]_0-x)}+\dfrac{v}{([B]_0+x)}=\dfrac{u([B]_0+x)+v([A]_0-x)}{([A]_0-x)([B]_0+x)}$$ and realize that we need to find $u$ and $v$ such that $$u([B]_0+x)+v([A]_0-x)=1$$ to be able to break up the integrand. For this, we expand and then regroup the expression on the left and write $$u[B]_0+v[A]_0+(u-v)x=1.$$ Since there is no $x$ on the right hand side of this equation there should be no term containing $x$ on the left either, which can be achieved if $u=v$ making $(u-v)x=0$. With the $(u-v)x$ term eliminated, $u[B]_0+v[A]_0$ must be equal to $1$, but since we have set $u=v$, it follows that $u([B]_0+[A]_0)=1$ which in turn means that $$u=\dfrac{1}{[A]_0+[B]_0}=v$$ With $u$ and $v$ determined, we now recast the integrand as $$\dfrac{1}{([A]_0-x)([B]_0+x)}=\dfrac{1}{[A]_0+[B]_0}\left(\dfrac{1}{([A]_0-x)}+\dfrac{1}{([B]_0+x)}\right),$$ so we can write $$\dfrac{1}{[A]_0+[B]_0}\int^x_0\left(\dfrac{1}{([A]_0-x)}+\dfrac{1}{([B]_0+x)}\right)dx$$ which separates into $$\dfrac{1}{[A]_0+[B]_0}\left(\int^x_0\dfrac{1}{([A]_0-x)}dx+\int^x_0\dfrac{1}{([B]_0+x)}dx\right)$$ We now use the method of integration by substitution to find $$\int\dfrac{1}{[B]_0+x}dx.$$ (The other integral can be found similary. For details see A + B → C.) Recall that the integral of $\frac{1}{g}$ is $ln |g|$ and that $dg=(\frac{dg}{dx})dx$. We make the substitution $g=[B]_0+x$, thus $dg=1dx$. Therefore, the integral becomes $$\int\dfrac{1}{g}dg=ln|g|+const=ln\left|[B]_0+x\right|+const.$$
Using the corresponding definite integrals we write $$\dfrac{1}{([A]_0+[B]_0)}\Big(\big[-ln([A]_0-x)\big]^x_0+\big[\,ln([B]_0+x)\big]^x_0\Big)$$ which expands into $$\dfrac{1}{([A]_0+[B]_0)}\Big(-ln([A]_0-x)+ln\,[A]_0+ln([B]_0+x)-ln\,[B]_0\Big).$$ Rearrangement yields $$\dfrac{1}{([A]_0+[B]_0)}ln\left(\dfrac{([B]_0+x)[A]_0}{([A]_0-x)[B]_0}\right).$$ Having integrated both sides of the original differential equation, we find – after rearrangement – that the concentrations of $A$ and $B$ can be expressed as a function of time, $t$, as \begin{matrix} ln\left(\dfrac{[B][A]_0}{[A][B]_0}\right)&=&([A]_0+[B]_0)k&\times&t\\ y&=&slope&\times&x \end{matrix} which gives a linear plot (see Figure 3).

By measuring the concentrations of $A$ and $B$ multiple times and then plotting $ln\left(\frac{[B][A]_0}{[A][B]_0}\right)$ versus $time$ one can conveniently calculate the rate constant of the reaction as $k=slope/([A]_0+[B]_0)$. Once $k$ is determined we can compute the concentration–time curves for any initial $[A]_0$ and $[B]_0$ values (as long as $[B]_0\neq0$); so there will be no need for any more concentration measurements. To achieve that we continue by taking the natural log of the equation for the linear plot $$\dfrac{([B]_0+x)[A]_0}{([A]_0-x)[B]_0}=e^{([A]_0+[B]_0)kt}.$$ Rearranging for $x$ gives $$x=\dfrac{[A]_0q-[B]_0}{1+q}$$ where $$q=\dfrac{[B]_0}{[A]_0}e^{([A]_0+[B]_0)kt}.$$ The latter two expressions can now be used to produce Figures 1 and 2.