Question

For the second order reaction, if the concentration of reactant changes from $0.08{\text{M}}$ to $0.04{\text{M}}$ in 10 minutes. Calculate the time at which concentration of reactant becomes $0.01{\text{M}}$ .
A.$20\min$
B.$30\min$
C.$50\min$
D.$70\min$

Answer 1

A reaction is said to be of second order if the rate of the reaction depends upon two concentration terms.
The relationship between the rate constant ‘k’ of a second order reaction with time’t’ and concentration is given by the following equation:
{\text{k = }}\dfrac{{\text{1}}}{{\text{t}}}\left\\{ {\dfrac{{\text{1}}}{{{{\left[ {\text{A}} \right]}_{\text{t}}}}}{\text{ - }}\dfrac{{\text{1}}}{{{{\left[ {\text{A}} \right]}_{\text{0}}}}}} \right\\}
Here, ${\left[ {\text{A}} \right]_{\text{0}}}$ denotes the initial concentration of the reactant and ${\left[ {\text{A}} \right]_{\text{t}}}$ denotes the concentration of the reactant after time ‘t’.

Complete step by step answer:
Given that for a second order reaction, the concentration of the reactant changes from $0.08{\text{M}}$ to $0.04{\text{M}}$ in time equal to 10 minutes. We need to find out the value of the time at which the concentration of the reactant will be reduced to $0.01{\text{M}}$ .
Therefore, in the second order equation in the first case, after time ‘t’ equal to 10 minutes, the concentration of the reactant ${\left[ {\text{A}} \right]_{\text{t}}}$ is equal to $0.04{\text{M}}$ . The initial concentration of the reactant ${\left[ {\text{A}} \right]_{\text{0}}}$ is given to be equal to $0.08{\text{M}}$ . Substitute all these values of ‘t’, ${\left[ {\text{A}} \right]_{\text{t}}}$ and ${\left[ {\text{A}} \right]_{\text{0}}}$ in the equation for finding out the rate of the second order reaction. Hence, we have
{\text{k = }}\dfrac{{\text{1}}}{{10}}\left\\{ {\dfrac{{\text{1}}}{{0.04}}{\text{ - }}\dfrac{{\text{1}}}{{0.08}}} \right\\} \\\ \Rightarrow {\text{k = }}\dfrac{{10}}{4} - \dfrac{{10}}{8} \\\ \Rightarrow {\text{k = }}\dfrac{5}{4} \\\ \Rightarrow {\text{k = }}1.25{{\text{M}}^{{\text{ - 1}}}}{\text{mi}}{{\text{n}}^{{\text{ - 1}}}} \\\
Now, by the use of this value of the rate constant ‘k’, we can find out the time taken in the second case. Let, the time taken this time be ${{\text{t}}_{\text{1}}}$ . In the second case, the concentration of the reactant ${\left[ {\text{A}} \right]_{\text{t}}}$ is equal to $0.01{\text{M}}$ . Substitute all these values of k, ${\left[ {\text{A}} \right]_{\text{t}}}$ and ${\left[ {\text{A}} \right]_{\text{0}}}$ in the equation for calculating the time ${{\text{t}}_{\text{1}}}$ . Hence, we have
{\text{1}}{\text{.25 = }}\dfrac{{\text{1}}}{{\text{t}}}\left\\{ {\dfrac{{\text{1}}}{{0.01}}{\text{ - }}\dfrac{{\text{1}}}{{0.08}}} \right\\} \\\ \Rightarrow {\text{t}} = \dfrac{1}{{1.25}}\left\\{ {\dfrac{{\text{1}}}{{0.01}}{\text{ - }}\dfrac{{\text{1}}}{{0.08}}} \right\\} \\\ \Rightarrow {\text{t}} = \dfrac{{100}}{{125}}\left\\{ {100 \times \dfrac{7}{8}} \right\\} \\\ \Rightarrow {\text{t}} = 70\min \\\

So, the correct answer is D.

Note:
The half-life period of a chemical reaction is the time taken for half of the reaction to complete, i.e., it is the time in which the concentration of a reactant is decreased to half of its initial value. For a second order reaction, the concentration of the reactant after completion of half of the reaction is
${\left[ {\text{A}} \right]_{\text{t}}} = \dfrac{{{{\left[ {\text{A}} \right]}_0}}}{2}$
Let the half-life period for second order reaction be ${{\text{t}}_{\dfrac{{\text{1}}}{{\text{2}}}}}$ .
So, from the equation for the rate of the second order reaction, we have
{\text{k = }}\dfrac{{\text{1}}}{{{{\text{t}}_{\dfrac{1}{2}}}}}\left\\{ {\dfrac{{\text{1}}}{{\dfrac{{{{\left[ {\text{A}} \right]}_{\text{0}}}}}{2}}}{\text{ - }}\dfrac{{\text{1}}}{{{{\left[ {\text{A}} \right]}_{\text{0}}}}}} \right\\} \\\ \Rightarrow {\text{k}}{{\text{t}}_{\dfrac{1}{2}}}{\text{ = }}\dfrac{{\text{1}}}{{{{\left[ {\text{A}} \right]}_{\text{0}}}}} \\\ \Rightarrow {{\text{t}}_{\dfrac{1}{2}}} = \dfrac{{\text{1}}}{{{{\left[ {\text{A}} \right]}_{\text{0}}}{\text{k}}}} \\\