D32.1 Integrated Rate Law
Thus far, we have used various aspects of rate laws to explore details of reaction kinetics, e.g. how activation energy affects the rate of a reaction (rate constant), how to alter reactant concentrations to experimentally determine the rate law of a reaction, and how to use experimentally-determined rate laws to verify the validity of a proposed mechanism.
Sometimes, though, we are interested in practical aspects of a chemical reaction, such as figuring out how long it would take for a certain quantity of a compound to react. For such applications, we can make use of an integrated rate law, which is another method for determining the rate law and rate constant from experimental data (i.e., a different approach from the method of initial rates). An integrated rate law relates the concentration of a reactant or product to the elapsed time of the reaction. Thus, it can also be used to determine the concentration of a reactant or a product present after a certain period of time, or to estimate the time required for a reaction to proceed to a certain extent.
For a given rate equation, calculus can be used to derive an appropriate integrated rate law. If you are unfamiliar with calculus and integration, don’t worry—you can still use the integrated rate law for a reaction without deriving it yourself.
For a generic reaction:
The rate of the reaction can be expressed as:
![\text{rate} = - \dfrac{\Delta[\text{A}]}{{\Delta}t} = k[\text{A}]^m](https://wisc.pb.unizin.org/app/uploads/quicklatex/quicklatex.com-993c56d50eafb8fa31f7b92f75e5a516_l3.png "Rendered by QuickLaTeX.com")
In calculus, the definition of a derivative is:
![\dfrac{d[\text{A}]}{dt} = \lim_{{\Delta}t \rightarrow 0} \dfrac{\Delta [\text{A}]}{{\Delta}t}](https://wisc.pb.unizin.org/app/uploads/quicklatex/quicklatex.com-496cf409e6f49fc103799cc258fccf13_l3.png "Rendered by QuickLaTeX.com")
Therefore, the instantaneous rate for this reaction can also be written as:
![\text{rate} = -\dfrac{d[\text{A}]}{dt} = k[\text{A}]^m](https://wisc.pb.unizin.org/app/uploads/quicklatex/quicklatex.com-df9f1226067e9b7d5d003f1a61f9bfd3_l3.png "Rendered by QuickLaTeX.com")
Rearranging the equation gives:
![-\dfrac{d[\text{A}]}{[\text{A}]^m} = k\;dt](https://wisc.pb.unizin.org/app/uploads/quicklatex/quicklatex.com-3d20cee23f557e0f549c7bed5eda4fd2_l3.png "Rendered by QuickLaTeX.com")
Integrating both sides of this equation gives:
![\displaystyle{\int^{[\text{A}]_t}_{[\text{A}]_0} -\dfrac{d[\text{A}]}{[\text{A}]^m} = \int^t_0 k\;dt}](https://wisc.pb.unizin.org/app/uploads/quicklatex/quicklatex.com-d14dc79394ed7fa5fbb0f846353a01f4_l3.png "Rendered by QuickLaTeX.com")
Assuming that the reaction starts at t = 0, the right side integration simply becomes kt. Then multiplying both sides by -1 gives:
![\displaystyle{\int^{[\text{A}]_t}_{[\text{A}]_0} \dfrac{d[\text{A}]}{[\text{A}]^m} = -kt}](https://wisc.pb.unizin.org/app/uploads/quicklatex/quicklatex.com-223fa25d0949cbede50ff4a6d6282ad9_l3.png "Rendered by QuickLaTeX.com")
[A]0 is the initial reactant concentration and [A]t is the reactant concentration at time t. The solution to the left side integration has a different mathematical form depending on the order of the reaction with respect to [A] (that is, it depends on the value of m). We will explore cases where m = 1, 2, and 0 in the next three sections.
Please use this form to report any inconsistencies, errors, or other things you would like to change about this page. We appreciate your comments. 🙂
