Question

A first order reaction is $75\% $ complete in $72{\text{ min}}$. How long time will it take for
i. $50\% $ completion
ii. $87.5\% $ completion

Answer 1

Hint:To solve this question we must know the equation for the rate constant of first order reaction. A reaction in which the rate of the reaction is directly proportional to the concentration of one of the reactant species is known as the first order reaction. Using the equation calculate the rate constant at $50\% $ completion and $87.5\% $ completion.

Formulae Used: $k = \dfrac{{2.303}}{t}\log \dfrac{{{{\left[ a \right]}_0}}}{{\left[ a \right]}}$

Complete answer:
We know the equation for the rate constant of a first order reaction is,
$k = \dfrac{{2.303}}{t}\log \dfrac{{{{\left[ a \right]}_0}}}{{\left[ a \right]}}$
Where $k$ is the rate constant of a first order reaction,
$t$ is time,
${\left[ a \right]_0}$ is the initial concentration of the reactant,
$\left[ a \right]$ is the final concentration of the reactant.

First we will calculate the rate constant for the first order reaction as follows:
We are given that a first order reaction is $75\% $ complete. Thus, the initial concentration is ${\text{100}}$ and the final concentration is $100 - 75 = 25$. The time required for a reaction to be $75\% $ completed is $72{\text{ min}}$. Thus,
$\Rightarrow k = \dfrac{{2.303}}{{72{\text{ min}}}}\log \dfrac{{100}}{{25}}$
$\Rightarrow k = 0.0307 \times 0.6020$
$\Rightarrow k = 0.01848{\text{ mi}}{{\text{n}}^{ - 1}}$
Thus, the rate constant of a first order reaction is $0.01848{\text{ mi}}{{\text{n}}^{ - 1}}$.
We are given that a first order reaction is $50\% $ complete. Thus, the initial concentration is ${\text{100}}$ and the final concentration is $100 - 50 = 50$.
i. Calculate the time required for a first order reaction to be $50\% $ complete. Thus,
$0.01848{\text{ mi}}{{\text{n}}^{ - 1}} = \dfrac{{2.303}}{t}\log \dfrac{{100}}{{50}}$
$\Rightarrow t = \dfrac{{2.303}}{{0.01848{\text{ mi}}{{\text{n}}^{ - 1}}}}\log \dfrac{{100}}{{50}}$
$\Rightarrow t = 124.62 \times 0.3010$
$\Rightarrow t= 37.51{\text{ min}}$
Thus, the time required for a first order reaction for $50\% $ completion is $37.51{\text{ min}}$.
We are given that a first order reaction is $87.5\% $ complete. Thus, the initial concentration is ${\text{100}}$ and the final concentration is $100 - 87.5 = 12.5$.
ii. Calculate the time required for a first order reaction to be $87.5\% $ complete. Thus,
$0.01848{\text{ mi}}{{\text{n}}^{ - 1}} = \dfrac{{2.303}}{t}\log \dfrac{{100}}{{12.5}}$
$\Rightarrow t = \dfrac{{2.303}}{{0.01848{\text{ mi}}{{\text{n}}^{ - 1}}}}\log \dfrac{{100}}{{12.5}}$
$\Rightarrow t = 124.62 \times 0.9030$
$\Rightarrow t = 112.53{\text{ min}}$

Thus, the time required for a first order reaction for $87.5\% $ completion is $112.53{\text{ min}}$.

Note:
The time taken for the reactant species to reduce to half of its initial concentration is known as the half-life of the reaction. At the half-life, $50\% $ of the reaction is completed.