The half-life of a reactant is the time required for its concentration to decrease to one-half of its initial value. Because the way in which half-life changes with concentration depends on reaction order, half-life measurements can be used to distinguish among zeroth-, first-, and second-order kinetics.
For a reactant \(A\),
\[ [A]_{t_{1/2}}=\frac{[A]_0}{2} \]where \([A]_0\) is the concentration at the beginning of the half-life interval and \(t_{1/2}\) is the time required for that concentration to fall by one-half.
The half-life equations are obtained by substituting
\[ [A]=\frac{[A]_0}{2} \]into the appropriate integrated rate law.
For a zeroth-order reaction,
\[ [A]=[A]_0-kt \]At the half-life,
\[ \frac{[A]_0}{2}=[A]_0-kt_{1/2} \]Solving for the half-life gives
\[ t_{1/2}=\frac{[A]_0}{2k} \]The half-life is directly proportional to the starting concentration. As the reaction proceeds and the concentration becomes smaller, each successive half-life becomes shorter.
For a first-order reaction,
\[ \ln[A]=\ln[A]_0-kt \]At the half-life,
\[ \ln\left(\frac{[A]_0}{2}\right)=\ln[A]_0-kt_{1/2} \]which simplifies to
\[ t_{1/2}=\frac{\ln 2}{k} = \frac{0.693}{k} \]The initial concentration does not appear in this expression. Therefore, a first-order reaction has a constant half-life: each successive halving of the concentration requires the same amount of time.
For a reaction that is second order in \(A\),
\[ \frac{1}{[A]}=\frac{1}{[A]_0}+kt \]At the half-life,
\[ \frac{1}{[A]_0/2} = \frac{1}{[A]_0} + kt_{1/2} \]Solving gives
\[ t_{1/2}=\frac{1}{k[A]_0} \]The half-life is inversely proportional to the starting concentration. As the reaction proceeds and the concentration decreases, each successive half-life becomes longer.
| Reaction order | Half-life equation | Behavior as concentration decreases |
|---|---|---|
| Zeroth order | \(\displaystyle t_{1/2}=\frac{[A]_0}{2k}\) | Successive half-lives become shorter |
| First order | \(\displaystyle t_{1/2}=\frac{\ln 2}{k}\) | Successive half-lives remain constant |
| Second order | \(\displaystyle t_{1/2}=\frac{1}{k[A]_0}\) | Successive half-lives become longer |
To apply the method experimentally, identify several intervals during which the reactant concentration falls by one-half. The pattern of those intervals reveals the reaction order.
For example:
Once the reaction order has been identified, the corresponding half-life equation can be rearranged to calculate the rate constant.
\[ \begin{aligned} \text{Zeroth order:}\qquad &k=\frac{[A]_0}{2t_{1/2}} \\[6pt] \text{First order:}\qquad &k=\frac{\ln 2}{t_{1/2}} \\[6pt] \text{Second order:}\qquad &k=\frac{1}{t_{1/2}[A]_0} \end{aligned} \]When using these equations, \([A]_0\) means the concentration at the beginning of the particular half-life interval, not necessarily the concentration at the beginning of the entire experiment.
Big picture: The method of half-lives identifies reaction order by examining how the time required for a concentration to fall by one-half changes as the reaction proceeds. A decreasing half-life indicates zeroth-order kinetics, a constant half-life indicates first-order kinetics, and an increasing half-life indicates second-order kinetics.
The concentration of a reactant \(A\) was measured as a function of time:
| Time (s) | \([A]\) (M) |
|---|---|
| 0 | 0.800 |
| 25 | 0.400 |
| 75 | 0.200 |
| 175 | 0.100 |
Use the half-life behavior to determine the reaction order and calculate the rate constant.
The first halving of the concentration is
\[ 0.800\ \mathrm{M}\rightarrow0.400\ \mathrm{M} \]which occurs from \(t=0\) to \(t=25\ \mathrm{s}\). Therefore,
\[ t_{1/2}=25\ \mathrm{s} \]for an interval beginning at \([A]_0=0.800\ \mathrm{M}\).
The second halving is
\[ 0.400\ \mathrm{M}\rightarrow0.200\ \mathrm{M} \]which occurs from \(t=25\ \mathrm{s}\) to \(t=75\ \mathrm{s}\):
\[ t_{1/2}=75-25=50\ \mathrm{s} \]The third halving is
\[ 0.200\ \mathrm{M}\rightarrow0.100\ \mathrm{M} \]which occurs from \(t=75\ \mathrm{s}\) to \(t=175\ \mathrm{s}\):
\[ t_{1/2}=175-75=100\ \mathrm{s} \]| Starting concentration (M) | Ending concentration (M) | Half-life (s) |
|---|---|---|
| 0.800 | 0.400 | 25 |
| 0.400 | 0.200 | 50 |
| 0.200 | 0.100 | 100 |
Each time the starting concentration is reduced by one-half, the half-life doubles:
\[ 25\ \mathrm{s}\rightarrow50\ \mathrm{s}\rightarrow100\ \mathrm{s} \]This behavior is characteristic of a reaction that is second order in \(A\), because
\[ t_{1/2}=\frac{1}{k[A]_0} \]Thus,
\[ \boxed{\text{Rate}=k[A]^2} \]Rearrange the second-order half-life equation:
\[ k=\frac{1}{t_{1/2}[A]_0} \]Using the first half-life interval,
\[ [A]_0=0.800\ \mathrm{M} \]and
\[ t_{1/2}=25\ \mathrm{s} \]Therefore,
\[ k = \frac{1} {(25\ \mathrm{s})(0.800\ \mathrm{M})} \] \[ k=0.0500\ \mathrm{M^{-1}\,s^{-1}} \]Therefore,
\[ \boxed{k=5.00\times10^{-2}\ \mathrm{M^{-1}\,s^{-1}}} \]For the second interval, the starting concentration is \(0.400\ \mathrm{M}\), not \(0.800\ \mathrm{M}\):
\[ k = \frac{1} {(50\ \mathrm{s})(0.400\ \mathrm{M})} \] \[ k=0.0500\ \mathrm{M^{-1}\,s^{-1}} \]The same value is obtained, confirming that the data are consistent with second-order kinetics.
Physical interpretation: For a second-order reaction, the half-life becomes longer as the reactant concentration decreases. Halving the starting concentration doubles the time required for the next halving.
The concentration of a reactant \(A\) was measured as a function of time:
| Time (s) | \([A]\) (M) |
|---|---|
| 0 | 0.800 |
| 40 | 0.400 |
| 60 | 0.200 |
| 70 | 0.100 |
Use the half-life behavior to determine the reaction order and calculate the rate constant.
The first halving of the concentration is
\[ 0.800\ \mathrm{M}\rightarrow0.400\ \mathrm{M} \]which occurs from \(t=0\) to \(t=40\ \mathrm{s}\). Therefore,
\[ t_{1/2}=40\ \mathrm{s} \]for an interval beginning at \([A]_0=0.800\ \mathrm{M}\).
The second halving is
\[ 0.400\ \mathrm{M}\rightarrow0.200\ \mathrm{M} \]which occurs from \(t=40\ \mathrm{s}\) to \(t=60\ \mathrm{s}\):
\[ t_{1/2}=60-40=20\ \mathrm{s} \]The third halving is
\[ 0.200\ \mathrm{M}\rightarrow0.100\ \mathrm{M} \]which occurs from \(t=60\ \mathrm{s}\) to \(t=70\ \mathrm{s}\):
\[ t_{1/2}=70-60=10\ \mathrm{s} \]| Starting concentration (M) | Ending concentration (M) | Half-life (s) |
|---|---|---|
| 0.800 | 0.400 | 40 |
| 0.400 | 0.200 | 20 |
| 0.200 | 0.100 | 10 |
Each time the starting concentration is reduced by one-half, the half-life is also reduced by one-half:
\[ 40\ \mathrm{s}\rightarrow20\ \mathrm{s}\rightarrow10\ \mathrm{s} \]This behavior is characteristic of a zeroth-order reaction, because
\[ t_{1/2}=\frac{[A]_0}{2k} \]The half-life is directly proportional to the concentration at the beginning of the interval. Therefore,
\[ \boxed{\text{Rate}=k} \]Rearrange the zeroth-order half-life equation:
\[ k=\frac{[A]_0}{2t_{1/2}} \]Using the first half-life interval,
\[ [A]_0=0.800\ \mathrm{M} \]and
\[ t_{1/2}=40\ \mathrm{s} \]Therefore,
\[ k = \frac{0.800\ \mathrm{M}} {2(40\ \mathrm{s})} \] \[ k=0.0100\ \mathrm{M\,s^{-1}} \]Therefore,
\[ \boxed{k=1.00\times10^{-2}\ \mathrm{M\,s^{-1}}} \]For the second interval, the starting concentration is \(0.400\ \mathrm{M}\) and the half-life is \(20\ \mathrm{s}\):
\[ k = \frac{0.400\ \mathrm{M}} {2(20\ \mathrm{s})} \] \[ k=0.0100\ \mathrm{M\,s^{-1}} \]The same value is obtained, confirming that the data are consistent with zeroth-order kinetics.
Physical interpretation: In a zeroth-order reaction, the reactant concentration decreases by the same amount during each equal time interval. Because less reactant remains after each halving, the next halving requires less time.
Examine the successive half-lives in the concentration data. First determine whether the reaction is zeroth, first, or second order. Once the order is correct, calculate the rate constant.
1. What is the order of the reaction?
Big picture: The method of half-lives determines reaction order by examining how the time required for each successive halving changes as the reactant concentration decreases. After the order is identified, the corresponding half-life equation can be used to calculate the rate constant and verify that its units are consistent with the proposed rate law.