Integrated rate laws allow reaction rate to be defined as a function of time instead of initial concentrations.
Some reactions occur very fast. It may be difficult to obtain experimental measurements for initial concentration. An alternative way to determine the rate of such reactions is with an integrated rate law, a rate law that defines concentration as a function of time.
The rate of a zeroorder reaction does not depend on the concentrations of
reactants. The change in the concentration of reactants over time is not a factor in the rate of reaction.
${[\rm{A}]}=kt+[\rm{A}]_{0}$
In this equation:

t is time.

k is the rate constant.
 [A] is the concentration after time t has passed.
 [A]_{0} is the initial concentration.
The rate of a firstorder reaction is dependent on the concentration of only one reactant. The rate expression for this reaction is
$r=k\lbrack\rm A\rbrack$. Rate can also be written in a differential form.
$\rm{rate}=\frac{[\Delta {\rm{A}}]}{\Delta t}=k\lbrack\rm A\rbrack$
Using integration, this expression can be mathematically transformed.
$\ln\lbrack {\rm{A}}\rbrack\ln\lbrack {\rm{A}}\rbrack_0=kt$
This example illustrates calculating the time for a reaction. A contaminant, A, has leaked into a lake. The contaminant has an initial concentration of
$2.50\times10^{6}\;\rm M$. The contaminant A decomposes in a firstorder reaction, with a
k of 2.30 year
^{–1} at the average temperature of the lake. The integrated rate law for a firstorder reaction can be used to calculate how long will it take the concentration to drop to
$1.50\times10^{7}\;\rm M$.
$\ln\lbrack {\rm{A}}\rbrack\ln\lbrack {\rm{A}}\rbrack_0=kt$
[A]
_{0}, [A], and
k are known, so
t can be calculated.
$\begin{aligned}t&=\frac{\ln\lbrack {\rm{A}}\rbrack_0\ln\lbrack {\rm{A}}\rbrack}{k}\\&=\frac{\ln(2.50\times10^{6})\ln(1.50\times10^{7})}{2.30\;\rm{year}^{1}}\\&=\frac{2.81}{2.30}\\&=1.22\;\rm{years}\end{aligned}$
The rate of a secondorder reaction can depend on the square of the concentration of one reactant,
$\rm{rate}=k\lbrack\rm A\rbrack^2$, or it can depend on the concentrations of two reactants,
$\rm{rate}=k\lbrack\rm A\rbrack\lbrack\rm B\rbrack$.
A rate dependent on the concentration of one reactant can be written as
$\rm{rate}=\frac{\Delta\lbrack\rm A\rbrack}{\Delta t}=k\lbrack\rm A\rbrack^2$
This can be integrated to obtain
$\frac1{\lbrack\rm A\rbrack}=kt+\frac1{\lbrack\rm A\rbrack_0}$
This equation is similar to the integrated rate law for a firstorder reaction. It has the same four variables. Note that this equation is in the form
$y=mx+b$. If
$\frac1{\lbrack\rm A\rbrack}$ is plotted against
t, the graph is a straight line.
Compare this with the integrated rate law for a firstorder reaction:
$\ln\lbrack{\rm A}\rbrack\ln\lbrack{\rm A}\rbrack_0=kt$. This equation can also be rearranged into the form
$y=mx+b$.
$\ln\lbrack{\rm A}\rbrack=kt+\ln\lbrack{\rm A}\rbrack_0$
In this case, graphing
$\ln\lbrack \rm{A}\rbrack$ versus time gives a straight line.
The
reaction halflife is the time it takes for a reactant to drop to half its starting concentration during a chemical reaction. In terms of concentrations, halflife can be described as
$\lbrack\rm A\rbrack=\frac12\lbrack\rm A\rbrack_0$. Substitute this into the integrated rate law for a firstorder reaction.
$\begin{aligned}{\ln\lbrack{\rm A}\rbrack\ln\lbrack{{\rm A}\rbrack_0}}&={kt}\\{\ln\frac12\lbrack{\rm A}\rbrack_0\ln\lbrack{{\rm A}\rbrack_0}}&=kt_{\rm{half}}\\{\ln\frac12}&=kt_{\rm{half}}\\t_{\rm{half}}&=\frac{0.693}k\end{aligned}$
Substituting
$\lbrack\rm A\rbrack=\frac12\lbrack\rm A\rbrack_0$ into the zeroorder and secondorder integrated rate laws can be used to find those halflife equations.
Note that this equation is independent of initial concentration. Halflife is independent of concentration.
The following is an example of a halflife calculation. The organic pesticide capsaicin has leaked into soil. Suppose the pesticide decomposes in a firstorder reaction with
$k\;=\;0.0935\;\rm{day}^{1}$ at the average temperature of the soil. The modified equation for halflife can be used to calculate the halflife of the pesticide.
$\begin{aligned}t_{\rm{half}}&=\frac{0.693}k\\&=\frac{0.693}{0.0935\;\rm{day}^{1}}\\&=7.41\;\rm{days}\end{aligned}$