Question

A first order reaction takes $20$ minutes for \[25\% \] decomposition. Calculate the time when \[75\% \] of the reaction will be completed?
Given:
1. Order of the reaction is one
2. \[\log 2 = 0.3010\]
3. \[\log 3 = 0.4771\]
4. \[\log 4 = 0.6021\]

Answer 1

Hint: The integrated rate law for first order reaction relates the ratio of reactant concentrations with rate constant and time can be used. The formula and unit of k depends on the order of reaction.

Complete step by step solution:
Let’s start by writing the integrated rate law for a first order reaction with A as the reactant as follows:
$\log \left( {\dfrac{{{{\left[ {\rm{A}} \right]}_0}}}{{{{\left[ {\rm{A}} \right]}_t}}}} \right) = \dfrac{{kt}}{{2.303}}$
Here, $k$ is the rate constant, $t$ is the time, ${\left[ {\rm{A}} \right]_0}$ is the initial concentration of reactant and ${\left[ {\rm{A}} \right]_t}$ is the concentration of reactant at time $t$.
Now, as we are given it takes $20$ minutes for \[25\% \] decomposition So, we can express the initial concentration and the remaining concentration as follows:
${\left[ {\rm{A}} \right]_0} - \dfrac{{25}}{{100}}{\left[ {\rm{A}} \right]_0} = {\left[ {\rm{A}} \right]_t}$
We can further simplify this to calculate the ratio of the reactant concentrations as follows:
$
\Rightarrow {\left[ {\rm{A}} \right]_0} - \dfrac{{{{\left[ {\rm{A}} \right]}_0}}}{4} = {\left[ {\rm{A}} \right]_t}\\
\Rightarrow \dfrac{{3{{\left[ {\rm{A}} \right]}_0}}}{4} = {\left[ {\rm{A}} \right]_t}\\
\Rightarrow \dfrac{{{{\left[ {\rm{A}} \right]}_0}}}{{{{\left[ {\rm{A}} \right]}_t}}} = \dfrac{4}{3}
$
It means that one fourth of the reactant has been decomposed leaving behind three fourth of it. We can substitute this information in the above expression as follows:
$\Rightarrow \log \left( {\dfrac{4}{3}} \right) = \dfrac{{k\left( {20{\rm{ min}}} \right)}}{{2.303}}$
Let’s rearrange this expression for rate constant as follows:
$\Rightarrow k = \dfrac{{2.303}}{{20}}\log \left( {\dfrac{4}{3}} \right){\rm{ mi}}{{\rm{n}}^{ - 1}}$
We can now use the given values in the above expression to calculate the value of rate constant as follows:
$
\Rightarrow k = \dfrac{{2.303}}{{20}}\left( {\log 4 - \log 3} \right){\rm{ mi}}{{\rm{n}}^{ - 1}}\\
\Rightarrow k = \dfrac{{2.303}}{{20}}\left( {0.6021 - 0.4771} \right){\rm{ mi}}{{\rm{n}}^{ - 1}}\\
\Rightarrow k = {\rm{0}}{\rm{.01439 mi}}{{\rm{n}}^{ - 1}}
$
Now as per question, we need \[75\% \] decomposition that means three fourths of the reactant have to be decomposed leaving behind one fourth of it. We can substitute this information in the integrated rate law expression as follows:
$\Rightarrow \log \left( {\dfrac{{{{\left[ {\rm{A}} \right]}_0}}}{{\left( {1{{\left[ {\rm{A}} \right]}_0}/4} \right)}}} \right) = \dfrac{{kt}}{{2.303}}$
Finally, we can substitute the given values and the calculated value of rate constant in the above expression to calculate the required time as follows:
$
\Rightarrow \log \left( {\dfrac{{{{\left[ {\rm{A}} \right]}_0}}}{{\left( {1{{\left[ {\rm{A}} \right]}_0}/4} \right)}}} \right) = \dfrac{{kt}}{{2.303}}\\
\Rightarrow t = \dfrac{{2.303}}{{{\rm{0}}{\rm{.01439}}}}\log \left( {\dfrac{4}{1}} \right){\rm{ min}}\\
\Rightarrow t = \dfrac{{2.303}}{{{\rm{0}}{\rm{.01439}}}}\left( {\log 4 - \log 1} \right){\rm{ min}}\\
\Rightarrow t = \dfrac{{2.303}}{{{\rm{0}}{\rm{.01439}}}}\left( {0.6021 - 0} \right){\rm{ min}}\\
\Rightarrow t = {\rm{96}}{\rm{.36 \,min}}
$

Hence, the required time for three-fourth decomposition is ${\rm{96}}{\rm{.36 min}}$.

Note:
We can also calculate the ratio of concentrations by assuming the initial value to be $100$ as it becomes easier to use percent with this.