There are two main types of simple second-order reactions. The simplest form is one in which the reaction rate is proportional to the square of the concentration of a single species involved in the reaction. That is, when the velocity equation is of the form:

$$\displaystyle v = k \cdot [A]^2$$

The other possibility is that the reaction rate is proportional to the product of the concentrations of two species. I mean:

$$\displaystyle v = k \cdot [A] \cdot [B]$$

so that it is a first-order reaction with respect to each of the two species. We will analyze the two cases.


  • An example of the first type of second order reaction is the thermal decomposition of gaseous acetaldehyde (ethanal) which, at temperatures of the order of 500 ºC, decomposes into methane and carbon monoxide:

    $$\displaystyle CH_{3}\!\!-\!CHO (g)\longrightarrow CH_4 (g) + CO (g)$$

    For this reaction, it is known that:

    $$\displaystyle v = k \cdot [CH_{3}\!\!-\!CHO]^2$$

    The integration of this type of second order reactions is as follows.

    Given the:

    $$\displaystyle v = - \frac{d[A]}{dt} = k \cdot [A]^2$$

    we separate variables:

    $$\displaystyle - \frac{d[A]}{[A]^2} = k \cdot dt$$

    and we integrate:

    $$\displaystyle - \int_{[A]_0}^{[A]} \frac{d[A]}{[A]^2} = k \cdot \int_0^t dt \Rightarrow - \left. \left ( -\ \frac{1}{[A]}\right )\right|_{[A]_0}^{[A]} = k \cdot \left. (t)\right|_0^t$$

    that is to say:

    $$\displaystyle -\ \left [{\left (-\ \frac{1}{[A]}\right )\ -\ \left (-\ \frac{1}{[A]_0}\right ) }\right ] = k \cdot (t - 0)$$

    so that we finally have:

    $$\displaystyle \frac{1}{[A]} - \frac{1}{[A]_0} = k \cdot t$$

    As can be seen, from the expression obtained, we see that the graphical representation of the inverse of the concentration of $A$ against time should give us a straight line with slope equal to $k$.

    If we clear the concentration of $A$ we have:

    $$\displaystyle [A] =\frac{1}{k \cdot t + \frac{1}{[A]_0}} = \frac{[A]_0}{[A]_0 \cdot k \cdot t+1}$$

    so that plotting the concentration of $A$ against time will give us the hyperbolic curve:

  • As an example of the other type of simple second-order reactions that we can have, the ones we have mentioned in second place, we can put the reaction in an aqueous medium between the persulfate and iodide ions:

    $$\displaystyle {S_{2}O_{8}}^{2-{}} \ (aq) + 2 \ I^{-{}} \ (aq) \longrightarrow 2 \ {SO_{4}}^ {2-{}} \ (aq) + I_2$$

    whose velocity is known to take the form:

    $$\displaystyle v = k \cdot [{S_{2}O_{8}}^{2-{}} \ (aq)] \cdot [I^{-{}} \ (aq)]$$

    To integrate the reaction rate equation, we are going to assume that we have a reaction of the type:

    $$\displaystyle a\ A + b\ B \longrightarrow{Products}$$

    such that:

    $$\displaystyle v = - \frac{1}{a} \cdot \frac{d[A]}{dt} = - \frac{1}{b} \cdot \frac{d[B]}{dt } = k \cdot [A] \cdot [B]$$

    This is not necessarily so in all cases. There are reactions of the type:

    $$\displaystyle A \longrightarrow B + other \ products$$

    which also answer to the same type of rate equation,

    $$\displaystyle v = k\cdot[A]\cdot[B]$$

    appearing the concentration of one of the products in it. These types of reactions in which the speed, in addition to being proportional with respect to the reactant, is also proportional with respect to some of the products, are called autocatalytic reactions .

    To integrate the rate equation, we are going to use a variant of $\xi$, the extension of the reaction.

    After the reaction has started, at a certain instant of time we will have that the concentration of $A$ (if we choose the other species, $B$, the reasoning is identical) that remains unreacted will be:

    $$\displaystyle [A] = [A]_0 - \alpha$$

    where, evidently, $\alpha$ is the concentration part of $A$ that has already reacted. Since the number of moles of $A$ that have reacted is $a\cdot\xi$, then we will have:

    $$\displaystyle [A] = [A]_0 - \alpha = [A]_0 - a\cdot\frac{\xi}{V}$$

    For convenience, we will define the intensive variable $x$ as the extent of reaction per unit volume:

    $$\displaystyle x = \frac{\xi}{V}$$

    With which:

    $$\displaystyle [A] = [A]_0 - a \cdot x$$

    For obvious reasons (for every $a$ moles of $A$, $b$ moles of $B$ will have reacted) we have that:

    $$[\displaystyle B] = [B]_0 - b \cdot x$$

    So the rate of the reaction will be:

    $$\displaystyle -\ \frac{1}{a}\cdot\frac{d[A]}{dt} = -\ \frac{1}{a}\cdot\frac{d}{dt}{\ left ( {[A]_0 - a\cdot x} \right )} = \frac{dx}{dt}$$

    Therefore, equating the two expressions of the speed, we have:

    $$\displaystyle \left. {\begin{matrix}v = \frac{dx}{dt}\\v = k\cdot [A]\cdot [B]\end{matrix}} \right\} \Longrightarrow{}\frac{dx} {dt} = k\cdot [A]\cdot [B] = k\cdot\left ({[A]_0 - a\cdot x}\right )\cdot\left ({[B]_0 - b\cdot x}\right )$$

    We separate variables:

    $$\displaystyle \frac{dx}{\left ({[A]_0 - a\cdot x}\right )\cdot\left ({[B]_0 - b\cdot x}\right )} = k\ cdot dt$$

    and we integrate:

    $$\displaystyle \int_0^x\frac{dx}{\left ({[A]_0 - a\cdot x}\right )\cdot\left ({[B]_0 - b\cdot x}\right ) } = k\cdot \int_0^tdt$$

    The integral of the first member is the integral of a rational function that can be easily decomposed into simple fractions. We will omit the detailed development of the calculus, which we can find in any elementary integral calculus manual (#1). The decomposition of the integral leads us to the following expression:

    $$\begin {eqnarray*} \displaystyle\int_0^x\frac{dx}{\left ({[A]_0 - a\cdot x}\right )\cdot\left ({[B]_0 - b\ cdot x}\right )} =\\ & = & \int_0^x {\left ({\frac{\frac{a}{a\cdot [B]_0 - b\cdot [A]_0}}{[ A]_0 - a\cdot x} + \frac{\frac{-\ b}{a\cdot [B]_0 - b\cdot [A]_0}}{[B]_0 - b\cdot x}} \right )\cdot dx}\\ & = & -\ \frac{1}{a\cdot [B]_0 - b\cdot [A]_0}\cdot \int_0^x\frac{-\ a\cdot dx}{[A]_0 - a\cdot x} +\\ & + & \frac{1}{a\cdot [B]_0 - b\cdot [A]_0}\cdot \int_0^x\frac{ -\ b\cdot dx}{[B]_0 - b\cdot x}\\ & = & \frac{1}{a\cdot [B]_0 - b\cdot [A]_0}\cdot \left [ {- \ln([A]_0 - a\cdot x) + \ln([B]_0 - b\cdot x)}\right ]_0^x\\ & = & \frac{1}{a\cdot [B]_0 - b\cdot [A]_0}\cdot \left [{\ln{\left ({\frac{[B]_0 - b\cdot x}{[A]_0 - a\cdot x} }\right )}}\right ]_0^x\\ & = & \frac{1}{a\cdot [B]_0 - b\cdot [A]_0}\cdot \left ( {\ln{\left ({\frac{[B]_0 - b\cdot x}{[A]_0 - a\cdot x}}\right )} - \ln{\left ({\frac{[B]_0}{[A ]_0}}\right )}} \right )\\ & = & \frac{1}{a\cdot [B]_0 - b\cdot [A]_0}\cdot \ln{\left ({\f rac{\frac{[B]_0 - b\cdot x}{[A]_0 - a\cdot x}}{\frac{[B]_0}{[A]_0}}}\right )}\\ & = & \frac{1}{a\cdot [B]_0 - b\cdot [A]_0}\cdot \ln{\left ({\frac{[A]_0\cdot ([B]_0 - b \cdot x)}{[B]_0\cdot ([A]_0 - a\cdot x)}}\right )}\\ \end{eqnarray*}$$

    so that, finally, the equation resulting from the integration is:

    $$\displaystyle\frac{1}{a\cdot [B]_0 - b\cdot [A]_0}\cdot \ln{\left ({\frac{[A]_0\cdot ([B]_0 - b\cdot x)}{[B]_0\cdot ([A]_0 - a\cdot x)}}\right )} = k\cdot t$$

    We regroup some terms and remove the logarithm:

    $$\begin{eqnarray*} \displaystyle\ln{\left ({\frac{[A]_0\cdot ([B]_0 - b\cdot x)}{[B]_0\cdot ([A]_0 - a\cdot x)}}\right )} = (a\cdot [B]_0 - b\cdot [A]_0)\cdot k\cdot t\\ & \Longrightarrow & \frac{[A]_0\ cdot ([B]_0 - b\cdot x)}{[B]_0\cdot ([A]_0 - a\cdot x)} = e^{(a\cdot [B]_0 - b\cdot [A ]_0)\cdot k\cdot t}\\ \end{eqnarray*}$$

    Now we recover the concentrations of $A$ and $B$; Given the:

    $$\displaystyle [A] = [A]_0 - a\cdot x \Leftrightarrow a\cdot x = [A]_0 - [A]$$

    $$\displaystyle [B] = [B]_0 - b\cdot x \Leftrightarrow b\cdot x = [B]_0 - [B]$$

    then, the expression we have obtained becomes:

    $$\displaystyle\frac{[A]_0\cdot ([B]_0 - [B]_0 + [B])}{[B]_0\cdot ([A]_0 - [A]_0 + [A] )} = e^{(a\cdot [B]_0 - b\cdot [A]_0)\cdot k\cdot t}$$

    that is to say:

    $$\displaystyle\frac{[A]_0\cdot [B]}{[B]_0\cdot [A]} = e^{(a\cdot [B]_0 - b\cdot [A]_0)\ cdot k\cdot t} \Rightarrow \frac{[A]}{[B]} = \frac{[A]_0}{[B]_0}\cdot e^{(b\cdot [A]_0 - a \cdot [B]_0)\cdot k\cdot t}$$

    If we use the logarithmic form of this expression:

    $$\displaystyle\ln{\frac{[A]}{[B]}} = \ln{\frac{[A]_0}{[B]_0}} + (b\cdot [A]_0 - a \cdot [B]_0)\cdot k\cdot t$$

    We clearly see that the graph of $\ln {\frac{[A]}{[B]}}$ against time is a straight line whose slope is given by:

    $$\displaystyle (b\cdot [A]_0 - a\cdot [B]_0)\cdot k$$

    which allows us to easily identify this type of reaction from a table of experimental data.

#1: There is a huge amount of literature on the method of decomposing rational functions into simple fractions. Any math textbook for advanced high school students includes it. Some suggestions could be: