If the reaction follows a second order rate law, the some methodology can be employed. The rate can be written as

\[ -\dfrac{d[A]}{dt} = k [A]^2 \label{eq1A}\]

The separation of concentration and time terms (this time keeping the negative sign on the left for convenience) yields

\[ -\dfrac{d[A]}{[A]^2} = k \,dt \]

The integration then becomes

\[ - \int_{[A]_o}^{[A]} \dfrac{d[A]}{[A]^2} = \int_{t=0}^{t}k \,dt \label{eq1}\]

And noting that

\[ - \dfrac{dx}{x^2} = d \left(\dfrac{1}{x} \right)\]

the result of integration Equation \ref{eq1} is

\[ \dfrac{1}{[A]} -\dfrac{1}{[A]_o} = kt\]

or

\[ \dfrac{1}{[A]} = \dfrac{1}{[A]_o} + kt\]

And so a plot of \(1/[A]\) as a function of time should produce a linear plot, the slope of which is \(k\), and the intercept of which is \(1/[A]_0\).

Other 2^{nd} order rate laws are a little bit trickier to integrate, as the integration depends on the actual stoichiometry of the reaction being investigated. For example, for a reaction of the type

\[A + B \rightarrow P\]

That has rate laws given by

\[ -\dfrac{d[A]}{dt} = k [A][B] \]

and

\[ -\dfrac{d[B]}{dt} = k [A][B] \]

the integration will depend on the decrease of [A] and [B] (which will be related by the stoichiometry) which can be expressed in terms the concentration of the product [P].

\[[A] = [A]_o – [P] \label{eqr1}\]

and

\[[B] = [B]_o – [P]\label{eqr2}\]

The concentration dependence on \(A\) and \(B\) can then be eliminated if the rate law is expressed in terms of the production of the product.

\[ \dfrac{d[P]}{dt} = k [A][B] \label{rate2} \]

Substituting the relationships for \([A]\) and \([B]\) (Equations \ref{eqr1} and \ref{eqr2}) into the rate law expression (Equation \ref{rate2}) yields

\[ \dfrac{d[P]}{dt} = k ( [A]_o – [P]) ([B] = [B]_o – [P]) \label{rate3} \]

Separation of concentration and time variables results in

\[\dfrac{d[P]}{( [A]_o – [P]) ([B] = [B]_o – [P])} = k\,dt\]

Noting that at time \(t = 0\), \([P] = 0\), the integrated form of the rate law can be generated by solving the integral

\[\int_{[A]_o}^{[A]} \dfrac{d[P]}{( [A]_o – [P]) ([B]_o – [P])} = \int_{t=0}^{t} k\,dt\]

Consulting a table of integrals reveals that for \(a \neq b\)[1],

\[ \int \dfrac{dx}{(a-x)(b-x)} = \dfrac{1}{b-a} \ln \left(\dfrac{b-x}{a-x} \right)\]

Applying the definite integral (as long as \([A]_0 \neq [B]_0\)) results in

\[ \left. \dfrac{1}{[B]_0-[A]_0} \ln \left( \dfrac{[B]_0-[P]}{[A]_0-[P]} \right) \right |_0^{[A]} = \left. k\, t \right|_0^t\]

\[ \dfrac{1}{[B]_0-[A]_0} \ln \left( \dfrac{[B]_0-[P]}{[A]_0-[P]} \right) -\dfrac{1}{[B]_0-[A]_0} \ln \left( \dfrac{[B]_0}{[A]_0} \right) =k\, t \label{finalint}\]

Substituting Equations \ref{eqr1} and \ref{eqr2} into Equation \ref{finalint} and simplifying (combining the natural logarithm terms) yields

\[\dfrac{1}{[B]_0-[A]_0} \ln \left( \dfrac{[B][A]_o}{[A][B]_o} \right) = kt\]

For this rate law, a plot of \(\ln([B]/[A])\) as a function of time will produce a straight line, the slope of which is

\[ m = ([B]_0 – [A]_0)k.\]

In the limit at \([A]_0 = [B]_0\), then \([A] = [B]\) at all times, due to the stoichiometry of the reaction. As such, the rate law becomes

\[ \text{rate} = k [A]^2\]

and integrate direct like in Equation \ref{eq1A} and the integrated rate law is (as before)

\[ \dfrac{1}{[A]} = \dfrac{1}{[A]_o} + kt\]

Example \(\PageIndex{2}\): Confirming Second Order Kinetics

Consider the following kinetic data. Use a graph to demonstrate that the data are consistent with second order kinetics. Also, if the data are second order, determine the value of the rate constant for the reaction.

**time (s)** |
0 |
10 |
30 |
60 |
100 |
150 |
200 |

**[A] (M)** |
0.238 |
0.161 |
0.098 |
0.062 |
0.041 |
0.029 |
0.023 |

**Solution**:

The plot looks as follows:

From this plot, it can be seen that the rate constant is 0.2658 M^{-1} s^{-1}. The concentration at time \(t = 0\) can also be inferred from the intercept.

[1] This integral form can be generated by using the method of partial fractions. See (House, 2007) for a full derivation.