Half Life


The half-life of a reaction (\(t_{1/2}\)) is the time required for an initial reactant concentration, \([A]_0\) to decrease by one-half. Therefore,

\[[\mathrm{A}]_0 = \frac{1}{2}[\mathrm{A}]_0 ~~~ \mathrm{at} ~~ t_{1/2} \]

The half-life equations for a zeroth, first, and second order reaction can be derived from the corresponding integrated rate laws using the relationship given above.


Derivations


0th-order

Beginning with the zeroth-order integrated rate law, change \(t \rightarrow t_{1/2}\) and \([\mathrm{A}_t] \rightarrow \frac{1}{2}[\mathrm{A}]_0~\) and solve for \(t_{1/2}\).

Rearrange

\[\begin{align*} [\mathrm{A}]_t &= -kt + [\mathrm{A}]_0 \\ \frac{1}{2}[\mathrm{A}]_0 &= -kt_{1/2} + [\mathrm{A}]_0 \\ kt_{1/2} &= [\mathrm{A}]_0 - \frac{1}{2}[\mathrm{A}]_0 \\ t_{1/2} &= \dfrac{[\mathrm{A}]_0}{2k} \end{align*}\]

\[ t_{1/2} = \dfrac{[\mathrm{A}]_0}{2k}\]


1st-order

Beginning with the first-order integrated rate law, change \(t \rightarrow t_{1/2}\) and \([\mathrm{A}_t] \rightarrow \frac{1}{2}[\mathrm{A}_0]~\) and solve for \(t_{1/2}\).

Rearrange

\[\begin{align*} \ln\left ( \dfrac{[\mathrm{A}]_t}{[\mathrm{A}]_0} \right ) &= -kt \\[1.5ex] \ln\left ( \dfrac{\frac{1}{2}[\mathrm{A}]_0}{[\mathrm{A}]_0} \right ) &= -kt_{1/2} \\[1.5ex] \ln\dfrac{1}{2} &= -kt_{1/2} \\[1.5ex] t_{1/2} &= -\dfrac{\ln\frac{1}{2}}{k} \\[1.5ex] &= \dfrac{\ln 2}{k}\\[1.5ex] & = \dfrac{0.693}{k} \end{align*}\]

\[t_{1/2} = \dfrac{\ln 2}{k} \approx \dfrac{0.693}{k}\]


2nd-order

Beginning with the second-order integrated rate law, change \(t \rightarrow t_{1/2}\) and \([\mathrm{A}_t] \rightarrow \frac{1}{2}[\mathrm{A}_0]~\) and solve for \(t_{1/2}\).

Rearrange

\[\begin{align*} \dfrac{1}{[\mathrm{A}]_t} &= kt + \dfrac{1}{[\mathrm{A}]_0} \\ \dfrac{1}{\frac{1}{2}[\mathrm{A}]_0} &= kt_{1/2} + \dfrac{1}{[\mathrm{A}]_0} \\ \dfrac{2}{[\mathrm{A}]_0} - \dfrac{1}{[\mathrm{A}]_0}&= kt_{1/2} \\ \dfrac{1}{[\mathrm{A}]_0}&= kt_{1/2} \\ \dfrac{1}{k[\mathrm{A}]_0}&= t_{1/2} \\ t_{1/2} &= \dfrac{1}{k[\mathrm{A}]_0} \end{align*}\]

\[t_{1/2} = \dfrac{1}{k[\mathrm{A}]_0}\]


Practice Problems


Problem 1

What is the half life (in h) one-reactant first-order reaction with a rate constant of 3.24 × 10–4 s–1 at 55.0 °C?

Solution

\[\begin{align*} t_{1/2} &= \dfrac{0.693}{k} \\ &= \dfrac{0.693}{3.24\times 10^{-4}~\mathrm{s}^{-1}}\\ &= 2.139\times 10^{3}~\mathrm{s} \left ( \dfrac{\mathrm{min}}{60~\mathrm{s}} \right ) \left ( \dfrac{\mathrm{h}}{60~\mathrm{min}} \right ) \\ &= 0.594~\mathrm{h} \end{align*}\]


Problem 2

A reaction with a rate constant of 4.24 × 10–3 M s–1 is found to be zeroth order. What is the half life of the reaction (in min) if the initial concentration of the reactant is 8.25 M?

Solution

\[\begin{align*} t_{1/2} &= \dfrac{[\mathrm{A}]_0}{2k}\\ &= \dfrac{8.25~M}{2(4.24\times 10^{-3}~M~\mathrm{s}^{-1})} \left ( \dfrac{\mathrm{min}}{60~\mathrm{s}} \right ) \\[1.5ex] &= 16.2~\mathrm{min} \end{align*}\]


Timed Assessments


Half-Life

VIDEO

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