Question

A first- order is $20\%$ complete in 10 minutes. Calculate the rate constant of the reaction
(II) Time taken for the reaction to go to $75\%$ completion.

Answer 1

To answer this question, you must recall the rate law equation for a first order reaction. Rate constant of a first order reaction does not depend on the concentration of the reactant. We shall substitute the values in the given equation.
Formula used: $kt = \ln \left( {\dfrac{{{a_i}}}{{{a_t}}}} \right)$
Where, $k$ is the rate constant of the first order reaction under consideration
${a_i}$ is the initial concentration of the reactants in the reaction mixture
And ${a_t}$ is the concentration of reactants in the reaction mixture at time $t$

Complete step by step answer:
It is given in the question that the time taken for completion of $20\%$of the reaction $= 20{\text{ minutes}}$
Let $x$ be the number of moles of reactant used up in the reaction. Assuming the initial concentration of the reactants in the reaction mixture as $100$, we get
$x = 20$
Using the rate law equation, $kt = \ln \left( {\dfrac{{{a_i}}}{{{a_t}}}} \right) = \ln \left( {\dfrac{{{a_i}}}{{{a_i} - x}}} \right)$
Substituting the values, we get,
$k = \dfrac{{2.303}}{{10}}\log \left( {\dfrac{{100}}{{100 - 20}}} \right)$
$\Rightarrow k = 2.2 \times {10^{ - 2}}{\text{ mi}}{{\text{n}}^{ - 1}}$
(II) Now, let the time taken for $75\%$completion of the reaction be $T$.
Here, $75\%$ of reactant is consumed in the reaction, so $x = 75$
Substituting the values of $x$ and rate constant into the rate law equation, we get,
$T = \dfrac{{2.303}}{{2.2 \times {{10}^{ - 2}}{\text{ }}}}\log \left( {\dfrac{{100}}{{100 - 75}}} \right)$
$\therefore T = 63{\text{ min}}$

Hence, the time taken for $75\%$completion of the reaction is $63{\text{ minutes}}$.
Note:
For a first order reaction, half-life is independent of the initial concentration of the reactant.
The concentration of reactant decreases exponentially with time in a first order equation.
If we are given the value of ${C_0}$ and ${C_t}$ at different time instants, the value of $k$ can be calculated for different time instants by using the first order law. If the reaction for which the data is given is a first order reaction, then all values of $k$ will be approximately equal to each other.