1.7: Half-life

The half-life, $t_{\frac{1}{2}}$ , is the time it takes for half the initial reactant concentration to react.
For a first-order reaction $A\rightarrow B$ , the half-life is $t_{\frac{1}{2}}=\frac{ln2}{k}$
For a second-order reaction $2A\rightarrow B$ , the half-life is $t_{\frac{1}{2}}=\frac{1}{2k_{1}[A]_0}$ or $t_{\frac{1}{2}}=\frac{1}{k_{2}[A]_0}$ .
For a zeroth-order reaction $A\rightarrow B$ , the half-life is $t_{\frac{1}{2}}=\frac{[A]_0}{2k}$ .

The half life, $t_\frac{1}{2}$ , is the amount of time it takes for the initial concentration, $[A]_{0}$ , to decrease to half its concentration such that $[A]_{t_{\frac{1}{2}}}=\frac{1}{2}[A]_0$ .

For a first-order reaction (see section 1.4),

$ln\frac{[A]}{[A]_0}=-kt$ .

To solve for the half-life, we replace $t$ with $t_\frac{1}{2}$ and $[A]$ with $\frac{1}{2}[A]_{0}$

$ln\frac{\frac{1}{2}[A]_0}{[A]_0}=-kt_{\frac{1}{2}}$ .

Solving for the half-life gives

$t_{\frac{1}{2}}=-\frac{ln\frac{1}{2}}{k}$ .

Noting that $ln{\frac{1}{2}}=ln2^{-1}=-ln2$ , the first-order half-life formula is more commonly written as

$t_{\frac{1}{2}}=\frac{ln2}{k}$ (Equation 1).

Note that Equation 1 does not depend on $[A]_{0}$ , so the half-life is independent of the sample size.

For a second-order reaction, the half-life formula will differ slightly depending on the convention (see section 1.5), but the the calculated value will be the same. Below, the convention with $rate=\frac{d[B]}{dt}=-\frac{1}{2}\frac{d[A]}{dt}=k_{1}[A]^{2}$ is used on the left and the convention $rate=-\frac{d[A]}{dt}=2\frac{d[B]}{dt}=k_{2}[A]^{2}$ is used on the right. As above, we'll (1) start with the integrated rate law, (2) replace $t$ with $t_\frac{1}{2}$ and $[A]$ with $\frac{1}{2}[A]_{0}$ , and finally (3) solve for $t_\frac{1}{2}$ .

1. $\frac{1}{[A]}=2k_{1}t+\frac{1}{[A]_0}$ or $\frac{1}{[A]}=k_{2}t+\frac{1}{[A]_0}$

2. $\frac{1}{\frac{1}{2}[A]_0}=2k_{1}t_{\frac{1}{2}}+\frac{1}{[A]_0}$ or $\frac{1}{\frac{1}{2}[A]_0}=k_{2}t_{\frac{1}{2}}+\frac{1}{[A]_0}$

$\frac{2}{[A]_0}=2k_{1}t_{\frac{1}{2}}+\frac{1}{[A]_0}$ or $\frac{2}{[A]_0}=k_{2}t_{\frac{1}{2}}+\frac{1}{[A]_0}$

3. $t_{\frac{1}{2}}=\frac{1}{2k_{1}[A]_0}$ or $t_{\frac{1}{2}}=\frac{1}{k_{2}[A]_0}$ (Equations 2a and 2b)

You can determine which version of this equation to use based on whether the rate in a question is defined as $rate=\frac{d[B]}{dt}=-\frac{1}{2}\frac{d[A]}{dt}=k_{1}[A]^{2}$ or $rate=-\frac{d[A]}{dt}=2\frac{d[B]}{dt}=k_{2}[A]^{2}$ .

For a zeroth-order reaction (see Section 1.6), the integrated rate law is

$[A]_t-[A]_0=-kt$

Replacing $[A]_t$ with $\frac{1}{2}[A]_0$ and $t$ with $t_{\frac{1}{2}}$ gives

$\frac{1}{2}[A]_0-[A]_0=-kt_{\frac{1}{2}}$

Solving for $t_{\frac{1}{2}}$ gives the zeroth-order half-life equation

$t_{\frac{1}{2}}=\frac{[A]_0}{2k}$ (Equation 3)

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Last updated on February 6, 2019 @5:21 pm

1.7: Half-life

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