4.4: Integrated Rate Laws

Introduction

In reaction kinetic investigations the concentration versus time is usually monitored. Differential rate laws only describe the slope (rate of change) of the concentrations, not the concentration versus time. To compare more directly to measured values it is useful to consider integrated rate laws, which describe the concentration of species versus time. There are some simple cases where the differential rate laws can be integrated to generate the integrated rate laws.

Combined with pseudo-order (isolation) methods these can often be used to directly fit the observed concentration versus time to verify the order with respect to a particular reagent and extract apparent rate constants that can be combined with the known excess concentrations to extract the rate constant. This can require many fewer experiments than using the initial rates method.

Zeroth order reactions

If the reaction follows a zeroth order rate law, it can be expressed in terms of the time-rate of change of [A] (which will be negative since A is a reactant):

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

In this case, it is straightforward to separate the variables. Placing time variables on the right and [A] on the left

\[ d[A] = - k \,dt \nonumber \]

In this form, it is easy to integrate. If the concentration of A is [A]₀ at time t = 0, and the concentration of A is [A] at some arbitrary time later, the form of the integral is

\[ \int _{[A]_o}^{[A]} d[A] = - k \int _{t_o}^{t}\,dt \nonumber \]

which yields

\[ [A] - [A]_o = -kt \nonumber \]

or

\[ [A] = [A]_o -kt \nonumber \]

This suggests that a plot of concentration as a function of time will produce a straight line, the slope of which is –k, and the intercept of which is [A]₀. If such a plot is linear, then the data are consistent with 0^th order kinetics. If they are not, other possibilities must be considered.

First order reactions

A first order rate law would take the form

\[ \dfrac{d[A]}{dt} = k[A] \nonumber \]

Again, separating the variables by placing all of the concentration terms on the left and all of the time terms on the right yields

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

This expression is also easily integrated as before

\[ \int_{[A]=0}^{[A]} \dfrac{d[A]}{[A]} =-k \int_{t=0}^{t=t}\,dt \nonumber \]

Noting that

\[ \dfrac{dx}{x} = d (\ln x) \nonumber \]

The form of the integrated rate law becomes

\[ \ln [A] - \ln [A]_o = kt \nonumber \]

or

\[ \ln [A] = \ln [A]_o - kt \label{In1} \]

This form implies that a plot of the natural logarithm of the concentration is a linear function of the time. And so a plot of ln[A] as a function of time should produce a linear plot, the slope of which is -k, and the intercept of which is ln[A]₀.

It should also be noted that the integrated rate law (Equation \ref{In1}) can be rewritten in exponential form by taking the exponetial of both sides:

\[ [A] = [A]_o e^{-kt} \label{expdecay} \]

Because of this functional form, 1^st order kinetics are sometimes referred to as exponential decay kinetics. Many processes, including radioactive decay of nuclides follow this type of rate law.

Note

With modern non-linear least squares fitting tools data is best fit directly to equation \(\ref{expdecay}\) rather than fitting ln[A] versus t with a line, which over emphasizes noise at later times.

If the goal is just to quickly determine if the order with respect to [A] is 1 (or close to it), then plotting ln[A] versus t to see if it looks linear is still reasonable.

 

Example \(\PageIndex{1}\):

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

Time (s)	0	10	20	50	100	150	200	250	300
[A] (M)	0.873	0.752	0.648	0.414	0.196	0.093	0.044	0.021	0.010

 

Solution

Plots of a direct fit to an exponential and the linearized version show how using the exponential is better:

Figure \(\PageIndex{1}\): Comparison of fitting data from a first order process to an exponential and fitting the data to a line after taking the natural log. Note that in the natural log case the y-axis is actually a plot of the log of just the numeric part of the concentration. This is indicated by labeling the axis with ln[A]-ln(M) = ln(X.XX•M) - ln(M) = ln(X.XX) + ln(M) - ln(M) = ln(X.XX). Thus the graph is "offset" by whatever the value of ln(M) is, where 'M' stands for molarity.

From the exponential plot k= (0.014928 ± 0.000004) s^-1 (the inverse of 66.99) and [A]_o = (0.8731 ± 0.0001) M. From the linear fit k =  (0.014908 ± 0.000009) s^-1, which is less certain than from the exponential fit. You can also see from the residuals in the linear fit that the disagreement between the fit and the experimental values is worse at long times than it is for the exponential fit.

Second order reactions

If the reaction follows a second order rate law, similar 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

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

The integration then becomes

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

And noting that

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

the result of integration Equation \ref{eq1} is

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

or

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

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

\[A + B \rightarrow P \nonumber \]

If this has a simple second order rate law that depends in first order on each of the reactants

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

and

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

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

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

Separation of concentration and time variables results in

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

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

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

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

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

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

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

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

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

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

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

\[ m = ([B]_0 – [A]_0)k. \nonumber \]

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

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

which integrates as shown before to

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

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.

Contributors and Attributions

• Patrick E. Fleming (Department of Chemistry and Biochemistry; California State University, East Bay)

• J. Gutow (UW Oshkosh)

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