Question

The rate of a first order reaction is $0.04{\text{ }}\dfrac{{{\text{mol}}}}{{{\text{L}}{\text{.s}}}}$ at 10 seconds and $0.03{\text{ }}\dfrac{{{\text{mol}}}}{{{\text{L}}{\text{.s}}}}$ at 20 seconds after initiation of the reaction. The half life period of the reaction is :
A ) 24.1 s
B ) 34.1 s
C ) 44.1 s
D ) 54.1 s

Answer 1

The half life period is the time required to reduce the concentration to one half the original value. The relationship between the half life period and the rate constant for the first order reaction is as shown below:
${t_{1/2}} = \dfrac{{0.693}}{k}$
The expression for the rate constant k is as shown below:
$k = \dfrac{{2.303}}{t}{\log _{10}}\dfrac{{{{\left[ {\text{A}} \right]}_1}}}{{{{\left[ {\text{A}} \right]}_2}}}$

Complete step by step answer:
For the first order reaction, the relationship between the rate of the reaction and the initial concentration is as given below:

${\text{Rate = k}}\left[ {\text{A}} \right]{\text{ }}$

The initial conditions include the rate of $0.04{\text{ }}\dfrac{{{\text{mol}}}}{{{\text{L}}{\text{.s}}}}$ at 10 seconds.
The final conditions include the rate of $0.03{\text{ }}\dfrac{{{\text{mol}}}}{{{\text{L}}{\text{.s}}}}$ at 20 seconds.
For initial conditions

$${{\text{R}}_1}{\text{ = k}}{\left[ {\text{A}} \right]_1} \\\ 0.04{\text{ }}\dfrac{{{\text{mol}}}}{{{\text{L}}{\text{.s}}}}{\text{ = k}}{\left[ {\text{A}} \right]_1}{\text{ }}...{\text{ }}...{\text{ }}\left( 1 \right) \\\$$

For final conditions

$${{\text{R}}_2}{\text{ = k}}{\left[ {\text{A}} \right]_2} \\\ 0.03{\text{ }}\dfrac{{{\text{mol}}}}{{{\text{L}}{\text{.s}}}}{\text{ = k}}{\left[ {\text{A}} \right]_2}{\text{ }}...{\text{ }}...{\text{ }}\left( 2 \right) \\\$$

Divide equation (1) with equation (2)

$$\dfrac{{0.04{\text{ }}\dfrac{{{\text{mol}}}}{{{\text{L}}{\text{.s}}}}}}{{0.03{\text{ }}\dfrac{{{\text{mol}}}}{{{\text{L}}{\text{.s}}}}}} = \dfrac{{{\text{k}}{{\left[ {\text{A}} \right]}_1}}}{{{\text{k}}{{\left[ {\text{A}} \right]}_2}}} \\\ \dfrac{4}{3} = \dfrac{{{{\left[ {\text{A}} \right]}_1}}}{{{{\left[ {\text{A}} \right]}_2}}}{\text{ }}...{\text{ }}...\left( 3 \right) \\\$$

The initial concentration be ${\left[ {\text{A}} \right]_1}$ at 10 seconds.
The final concentration be ${\left[ {\text{A}} \right]_2}$ at 20 seconds.
The difference between two time intervals 10 s, and 20 s is the time period t.

$$t = 20{\text{ s }} - {\text{ 10 s}} \\\ = 10{\text{ s}} \\\$$

Write the expression for the rate constant k:
$k = \dfrac{{2.303}}{t}{\log _{10}}\dfrac{{{{\left[ {\text{A}} \right]}_1}}}{{{{\left[ {\text{A}} \right]}_2}}}$
Substitute 10 s for t and the ratio of concentrations determined in the equation (3) in the above equation

$$k = \dfrac{{2.303}}{t}{\log _{10}}\dfrac{{{{\left[ {\text{A}} \right]}_1}}}{{{{\left[ {\text{A}} \right]}_2}}} \\\ = \dfrac{{2.303}}{{10{\text{ s}}}}{\log _{10}}\dfrac{4}{3} \\\ = 2.855 \times {10^{ - 2}}{\text{ }}{{\text{s}}^{ - 1}} \\\$$

Hence, the rate constant k is $2.855 \times {10^{ - 2}}{\text{ }}{{\text{s}}^{ - 1}}$.

The half life period is the time required to reduce the concentration to one half the original value.
Write the relationship between the half life period and the rate constant for the first order reaction, substitute the value of the rate constant and calculate the half life period:

$${t_{1/2}} = \dfrac{{0.693}}{k} \\\ = \dfrac{{0.693}}{{2.855 \times {{10}^{ - 2}}{\text{ }}{{\text{s}}^{ - 1}}}} \\\ = 24.1{\text{ s}} \\\$$

Hence, the option A ) 24.1 s is the correct answer.

Note:
For the first order reaction, the rate of the reaction is directly proportional to the reactant concentration. With decrease in the reactant concentration, over a period of time, the rate of the reaction (and the rate constant) decreases. The unit of rate constant is the reciprocal of the unit of time.