# What is first order half life derivation?

Jul 19, 2018

Well, let us begin from the rate law... Is the FIRST-order half-life dependent on concentration?

The first-order rate law for the reaction

$A \to B$

is

$r \left(t\right) = k \left[A\right] = - \frac{d \left[A\right]}{\mathrm{dt}}$

where:

• $r \left(t\right)$ is the rate as a function of time, which we take to be the initial rate in $\text{M/s}$
• $k$ is the rate constant. What are the units? You should know this.
• $\left[A\right]$ is the concentration of a reactant $A$ in $\text{M}$.
• $\frac{d \left[A\right]}{\mathrm{dt}}$ is the rate of disappearance of reactant $A$, a NEGATIVE value. But the negative sign forces the rate to be positive, fitting for a FORWARD reaction.

By separation of variables:

$- k \mathrm{dt} = \frac{1}{\left[A\right]} d \left[A\right]$

Integrate on the left from time zero to time $t$, and the right from initial concentration ${\left[A\right]}_{0}$ to current concentration $\left[A\right]$. We obtain:

$- k {\int}_{0}^{t} \mathrm{dt} = {\int}_{{\left[A\right]}_{0}}^{\left[A\right]} \frac{1}{\left[A\right]} d \left[A\right]$

$- k t = \ln \left[A\right] - \ln {\left[A\right]}_{0}$

Therefore, the first-order integrated rate law is:

$\textcolor{g r e e n}{\ln \left[A\right] = - k t + \ln {\left[A\right]}_{0}}$

As we should recognize, the half-life is when the concentration drops by half. Hence, we set ${\left[A\right]}_{1 / 2} \equiv 0.5 {\left[A\right]}_{0}$ to get:

$\ln 0.5 {\left[A\right]}_{0} = - k {t}_{1 / 2} + \ln {\left[A\right]}_{0}$

$\ln 0.5 {\left[A\right]}_{0} - \ln {\left[A\right]}_{0} = - k {t}_{1 / 2}$

$\ln \setminus \frac{0.5 \cancel{{\left[A\right]}_{0}}}{\cancel{{\left[A\right]}_{0}}} = - k {t}_{1 / 2}$

$- \ln 0.5 = \ln 2 = k {t}_{1 / 2}$

As a result, the half-life is given by...

$\textcolor{b l u e}{\overline{\underline{|}} \stackrel{\text{ ")(" "t_(1//2) = (ln2)/k" }}{|}}$