Question

A first-order reaction has a half-life of 14.5 hrs. What percentage of the reactant will remain after 24 hrs?

Answer 1

Hint: Remember one thing regarding first-order reactions - the half-life of these reactions is independent of the initial concentration of their reactants. In other words, it does not depend upon the amount of reactant you are having. The half-life of the reaction is actually the time required for half of the initial concentration of the reactant to be utilized or consumed.

 Complete step by step answer:
A first-order reaction is a reaction that takes place at a rate that depends linearly only on one reactant concentration.
Mathematically,
$\left[ A \right]{\text{t = }}\left[ A \right]0{{\text{e}}^{ - kt}}$
Where, $\left[ A \right]{\text{t}}$ is the final concentration of reactant after a time $\left( t \right)$, $\left[ A \right]0$ is the initial concentration of reactant and $k$represents the rate constant.
Now, taking the natural log on both sides:
\[\ln \left( {\left[ A \right]t} \right) = \ln \left( {\left[ A \right]0{e^{{\text{ - kt}}}}} \right)\]$\dfrac{{\left[ A \right]t}}{{\left[ A \right]0}} = 31.8$
You will get:
$\ln \left( {\left[ A \right]} \right) - \ln \left( {\left[ A \right]0} \right) = - kt.\ln \left( e \right)$
$\ln \left( {\left[ A \right]0} \right) - \ln \left( {\left[ A \right]} \right) = kt$
Finally, you will obtain:
$t = \dfrac{{\ln \left( {\dfrac{{\left[ A \right]0}}{{\left[ A \right]t}}} \right)}}{k}$
Now, it is known that after one half-life i.e. t1/2, the concentration of the reactant becomes half of its initial concentration value.
$\left[ A \right]t = \dfrac{{\left[ A \right]0}}{2}$ (after t1/2)
Thus:
$\ln \left( {\dfrac{{\left[ A \right]0}}{{\left[ A \right]t}}} \right) = \ln \left( {\dfrac{{\left[ A \right]0}}{{\dfrac{{\left[ A \right]0}}{2}}}} \right) = \ln \left( 2 \right)$
At last:
$t1/2 = \dfrac{{\ln (2)}}{k} = \dfrac{{0.693}}{k}$
From this equation, first we will calculate the value of k.
$14.5 = \dfrac{{0.693}}{k}$
$k = 0.048$
Now, estimate the value of$\dfrac{{\left[ A \right]t}}{{\left[ A \right]0}}$:
We already have the formula $t = \dfrac{{\ln \left( {\dfrac{{\left[ A \right]0}}{{\left[ A \right]t}}} \right)}}{k}$
$\dfrac{{\left[ A \right]t}}{{\left[ A \right]0}} = {e^{ - kt}}$, where k = 0.048, t = 24
As a result, $\dfrac{{\left[ A \right]t}}{{\left[ A \right]0}} = 31.8$
Thus, 31.8% of the reactant will remain after 24 hrs.

Note:
The length of half-life will always be constant, independent of the concentration. That means, it takes the same time for the concentration to reduce from one point to another point. The formula for half life demonstrates that for the first order reaction, it depends solely on the reaction rate constant i.e. k.