#### First-Order Reactions

We can derive an equation for determining the half-life of a first-order reaction from the alternate form of the integrated rate law as follows:

$$ln \left( \frac{[A]_0}{[A]_t} \right)=kt \\

t=ln \left( \frac{[A]_0}{[A]_t} \right)\times \frac{1}{k}$$

If we set the time *t* equal to the half-life, $t_{\frac{1}{2}}$, the corresponding concentration of *A* at this time is equal to one-half of its initial concentration (i.e. $[A]_{t_{1/2}} = \frac{1}{2} [A]_0$:

Substituting these terms into the rearranged integrated rate law and simplifying yields the equation for half-life:

$$t_{\frac{1}{2}} =ln\frac{[A]_0}{\frac{1}{2}[A]_0}\times \frac{1}{k}\\

t_{\frac{1}{2}} =ln{2}\times \frac{1}{k}$$

$$ t_{\frac{1}{2}} = \frac{ln{2}}{k} \approx \frac{0.693}{k} \label{eq1}\tag{1}$$

We can see that the half-life of a first-order reaction is inversely proportional to the rate constant *k*. A fast reaction (shorter half-life) will have a larger *k*; a slow reaction (longer half-life) will have a smaller *k*.

### Calculation of a First-order Rate Constant using Half-Life

Calculate the rate constant for the first-order decomposition of hydrogen peroxide in water at 40 °C, using the data given in the figure below.

**Solution**:

The half-life for the decomposition of H_{2}O_{2} is 2.16 × 10^{4} s:

$$ t_{\frac{1}{2}} \approx \frac{0.693}{k} \\

k=\frac{0.693}{ t_{\frac{1}{2}} }=\frac{0.693}{2.16\times 10^4 \;s}\\

=3.21\times 10^{-5}\;s^{-1}$$

### Check Your Learning

The first-order radioactive decay of iodine-131 exhibits a rate constant of 0.138 d^{−1}. What is the half-life for this decay?

## Answer:

Using Equation \ref{eq1}: $$ t_{\frac{1}{2}} = \frac{ln{2}}{k} \\ t_{\frac{1}{2}} = \frac{ln{2}}{0.138\: \text{d}^{-1}} \\ t_{\frac{1}{2}} = 5.02_{2806}\: \text{d} \\ t_{\frac{1}{2}} = 5.02\: \text{days} $$*Note on significant figures: the “2” in “ln 2” in this formula is another perfect number: we*

**define**the half life as the point where the ratio of concentrations is exactly $\frac{1}{2}$. Therefore this “2” does not affect the number of significant figures. If you used the approximate version of the formula, the value of ln(2) is abbreviated to 0.693: which is**not**an exact number anymore and would affect the number of significant figures in your final answer.