Question

$40\% $ of a first order reaction is completed in $50{\text{ minutes}}$. How much time will it take for the completion of $80\% $ of this reaction?

Answer 1

Hint: The reaction in which the rate of the reaction depends on the concentration of one reactant only is known as a first order reaction.
The constant that relates the rate of the chemical reaction to the concentration of the reactant or the product at given temperature is known as rate constant.
The rate constant of a first order reaction is calculated using the equation,
$k = \dfrac{{2 \cdot 303}}{t}\log \dfrac{a}{{a - x}}$
Where, $k$ is the rate constant of first order reaction energy,
$t$ is the time,
$a$ is the initial concentration of the reactants,
$x$ is the concentration of the reactant used.

Complete step by step answer:
Calculate the rate constant when $40\% $ of the reaction is completed in $50{\text{ minutes}}$ using the equation as follows:
$k = \dfrac{{2 \cdot 303}}{t}\log \dfrac{a}{{a - x}}$
The reaction is $40\% $ complete. Thus, the used concentration is $40\% $ of the initial concentration.
Thus,
$x = a \times \dfrac{{40}}{{100}}$
Where, $a$ is the initial concentration of the reactants,
 $x$ is the concentration of the reactant used.
Substitute $50{\text{ minutes}}$ for the time, $a \times \dfrac{{40}}{{100}}$ for the concentration of reactant used. Thus,$k = \dfrac{{2 \cdot 303}}{{50{\text{ minutes}}}}\log \dfrac{a}{{a - a \times \dfrac{{40}}{{100}}}}$
$k = \dfrac{{2 \cdot 303}}{{50{\text{ minutes}}}}\log \dfrac{{100a}}{{100a - 40a}}$
$k = \dfrac{{2 \cdot 303}}{{50{\text{ minutes}}}}\log \dfrac{{100\not{a}}}{{60\not{a}}}$
$k = \dfrac{{2 \cdot 303}}{{50{\text{ minutes}}}}\log \dfrac{{10}}{6}$ …… (1)
Step 2:
Calculate the time required for the reaction to complete $80\% $ using the equation as follows:
$k = \dfrac{{2 \cdot 303}}{t}\log \dfrac{a}{{a - x}}$
Rearrange the equation for time. Thus,
$t = \dfrac{{2 \cdot 303}}{k}\log \dfrac{a}{{a - x}}$
The reaction is $80\% $ complete. Thus, the used concentration is $80\% $ of the initial concentration.
Thus,
$x = a \times \dfrac{{80}}{{100}}$
Where, $a$ is the initial concentration of the reactants,
$x$ is the concentration of the reactant used.
Thus,
$t = \dfrac{{2 \cdot 303}}{k}\log \dfrac{a}{{a - a \times \dfrac{{80}}{{100}}}}$
$t = \dfrac{{2 \cdot 303}}{k}\log \dfrac{{100a}}{{100a - 80a}}$
$t = \dfrac{{2 \cdot 303}}{k}\log \dfrac{{100\not{a}}}{{20\not{a}}}$
$t = \dfrac{{2 \cdot 303}}{k}\log \dfrac{{10}}{2}$
Substitute equation (1). Thus,
$t = \dfrac{{\not{{2 \cdot 303}}}}{{\dfrac{{\not{{2 \cdot 303}}}}{{50{\text{ minutes}}}}\log \dfrac{{10}}{6}}} \times \log \dfrac{{10}}{2}$
$t = \dfrac{{50{\text{ minutes}}}}{{0 \cdot 2218}} \times 0 \cdot 6989$
$t = 157 \cdot 55{\text{ minutes}}$
Thus, the time required for the reaction to complete $80\% $ is $157 \cdot 55{\text{ minutes}}$.

Note:
Calculate the rate of the reaction when the reaction is $40\% $ complete in $50{\text{ minutes}}$. Then using the rate constant value calculate the time required for $80\% $ of the reaction to complete.