Question

A first order reaction is $50\%$ complete in $25 minutes$ . Calculate the time for $80\%$ complete of the reaction.

Answer 1

In order to this question, to calculate the required time as per the question, we will first apply the first order reaction formula to find the rate constant of the reaction, and then we will apply the first order equation to find the time for the completion of $80\%$ of the reaction.

Complete answer: As we know, the first-order reaction is a chemical reaction in which the rate varies according to changes in only one of the reactants' concentration.
For a first order reaction,
$\because {t_{\dfrac{1}{2}}} = \dfrac{{0.693}}{k}$ (formula of half-life reaction)
where, ${t_{\dfrac{1}{2}}}$ is the time for half-life reaction, (given- $25 minutes$ )
and, $k$ is the rate constant for the reaction.
$\Rightarrow 25 = \dfrac{{0.693}}{k} \\\ \Rightarrow k = \dfrac{{0.693}}{{25}} = 0.028{\min ^{ - 1}} \\\$
Now, we will find the time for $80\%$ complete of the reaction by using the first order equation-
$K = \dfrac{{2.303}}{t}\log \dfrac{{[{R_ \circ }]}}{R}$
From the definition of first order reaction, ${R_ \circ }$ and $R$ is the rate that varies during first order reaction.
Change in rate is $80\%$ .
$\Rightarrow t = \dfrac{{2.303}}{{0.028}}\log \dfrac{{[{R_ \circ }]}}{{0.2[{R_ \circ }]}}\,\,\,\,\,(\because [R] = \dfrac{{100 - 80}}{{100}}[{R_ \circ }]) \\\ \Rightarrow t = 0.0645 \times \log 5 \\\ \therefore t = 57.49\min \\\$
Hence, the required time for $80\%$ complete of the reaction is $57.49 minutes$ .

Note:
Integrated Rate Law for a First-Order Reaction Integrated rate expressions can be used to derive the rate constant of a reaction in an experimental setting. The differential rate rule for a first-order reaction must be rearranged as follows in order to obtain the integral form of the rate expression for the first-order reaction.