Question

A first order reaction is 40 % complete in 50 min. How long (minutes) will it take to 80 % completion?

Answer 1

Hint: A reaction is said to be a first order reaction if the rate of the reaction depends upon one concentration term only.
The relationship between the rate constant ‘k’ of a first order reaction with time and concentration is given by the following equation:
${\text{k}} = \dfrac{{2.303}}{{\text{t}}}\log \dfrac{{\text{a}}}{{{\text{a - x}}}}$
Here, ‘x’ represents the number of moles of the reactant decomposed, ‘a’ represents the initial concentration of the reactant and $\left( {{\text{a - x}}} \right)$ represents the concentration of the reactant after time ‘t’.

Complete step by step answer:
Given that the reaction is of first order and it takes 50 minutes for 40 percent completion.
We need to find out the time taken by the first order reaction to complete 80 percent.
Therefore, in the first order equation in the first case, at ‘t’ equal to 50 min, there is 40 percent completion which means the number of moles of the reactant decomposes is equal to 40 percent of the initial moles of the reactant ‘a’ . So:
$
  {\text{x = }}\dfrac{{40}}{{100}}{\text{a}} \\
   \Rightarrow {\text{x = 0}}{\text{.40a}} \\
 $
Thus, the concentration of the reactant after 50 min is given by:
$
  {\text{a - x}} = {\text{a - 0}}{\text{.40a}} \\
   \Rightarrow {\text{a - x}} = 0.60{\text{a}} \\
 $
By substituting all these values of time ‘t’, ‘a’ and$\left( {{\text{a - x}}} \right)$ in the first order equation, we will have the rate constant ’k’ for the given reaction as:
$
  {\text{k}} = \dfrac{{2.303}}{{50\min }}\log \dfrac{{\text{a}}}{{{\text{0}}{\text{.60a}}}} \\
   \Rightarrow {\text{k}} = \dfrac{{2.303}}{{50\min }}\log 1.67 \\
   \Rightarrow {\text{k}} = \dfrac{{2.303}}{{50\min }} \times 0.223 \\
   \Rightarrow {\text{k}} = 0.01028{\min ^{ - 1}} \\
 $
Now, using this value of k, we can find out the time taken in the second case. In the second case, there is 80 percent completion which means the number of moles of the reactant decomposes is equal to 80 percent of the initial moles of the reactant ‘a’. So:
$
  {\text{x = }}\dfrac{{80}}{{100}}{\text{a}} \\
   \Rightarrow {\text{x = 0}}{\text{.80a}} \\
 $
Let the time taken be ${\text{t'}}$ . Thus, the concentration of the reactant after ${\text{t'}}$ min is given by:
$
  {\text{a - x}} = {\text{a - 0}}{\text{.80a}} \\
   \Rightarrow {\text{a - x}} = 0.20{\text{a}} \\
 $
By substituting all these values of the rate constant k, ‘a’ and$\left( {{\text{a - x}}} \right)$ in the first order equation, we will have the time for 80 percent completion of the given reaction as:
$
  0.01028{\min ^{ - 1}} = \dfrac{{2.303}}{{{\text{t'}}}}\log \dfrac{{\text{a}}}{{{\text{0}}{\text{.20a}}}} \\
   \Rightarrow {\text{t'}} = \dfrac{{2.303}}{{0.01028{{\min }^{ - 1}}}}\log 5 \\
   \Rightarrow {\text{t'}} = \dfrac{{2.303}}{{0.01028{{\min }^{ - 1}}}} \times 0.6989 \\
   \Rightarrow {\text{t'}} = 156.52\min \\
 $

Note:
The expression for the rate constant for first order reactions from concentrations of the reactant at two different times is given by:
${\text{k}} = \dfrac{1}{{\left( {{{\text{t}}_{\text{2}}}{\text{ - }}{{\text{t}}_{\text{1}}}} \right)}}\ln \dfrac{{{{\left[ {\text{A}} \right]}_1}}}{{{{\left[ {\text{A}} \right]}_2}}}$
The half-life ${{\text{t}}_{\dfrac{1}{2}}}$ of a first order reaction is given by the expression:
${{\text{t}}_{\dfrac{1}{2}}} = \dfrac{{0.693}}{{\text{k}}}$