# The half life of a first order reaction is one hour. During what time interval will the concentration be reduced to 7/8 of its initial value?

Jan 17, 2018

8 min

#### Explanation:

For a 1st order reaction:

$\textsf{A \rightarrow \text{products}}$

$\textsf{{\left[A\right]}_{t} = {\left[A\right]}_{0} {e}^{- k t}}$

$\textsf{k}$ is the rate constant which is related to the half - life:

$\textsf{k = \frac{0.693}{t} _ \left(\frac{1}{2}\right) = \frac{0.693}{1} = 0.693 \textcolor{w h i t e}{x} {\text{hr}}^{-} 1}$

$\therefore$$\textsf{\frac{7}{8} \cancel{{\left[A\right]}_{0}} = \cancel{{\left[A\right]}_{0}} {e}^{- k t}}$

Taking natural logs of both sides$\Rightarrow$

$\textsf{\ln \left(\frac{7}{8}\right) = - k t}$

$\textsf{t = - \ln \left(\frac{7}{8}\right) \times \frac{1}{k}}$

$\textsf{t = 0.133 \textcolor{w h i t e}{x} \text{hr}}$

$\textsf{t = 0.133 \times 60 = 8 \textcolor{w h i t e}{x} \text{min}}$