Chapter 13, Problem 13.107QP

Interpretation Introduction

Interpretation:

The concentration of ${\text{NO}}_{\text{2}}$ after $\text{2}{\text{.5×10}}^{\text{2}}\text{sec}$ and half-life period has to be calculated.

Concept introduction:

Integrated rate law for second order reactions:

Taking in the example of following reaction,

$\text{aA}\to \text{products}$

And the reaction follows second order rate law,

Then the relationship between the concentration of $\text{A}$ and time can be mathematically expressed as,

$\frac{\text{1}}{{\left[\text{A}\right]}_{\text{t}}}\text{=}\text{kt+}\frac{\text{1}}{{\left[\text{A}\right]}_{\text{0}}}$

The above expression is called as integrated rate for second order reactions.

Half life for second order reactions:

In second order reaction, the half-life is inversely proportional to the initial concentration of the reactant (A).

The half-life of second order reaction can be calculated using the equation,

${\text{t}}_{\text{1/2}}\text{=}\frac{\text{1}}{\text{(k}{\left[\text{A}\right]}_{\text{0}}\text{)}}$

Since the reactant will be consumed in lesser amount of time, these reactions will have shorter half-life.

Answer to Problem 13.107QP

The concentration of ${\text{NO}}_{\text{2}}$ after $\text{2}{\text{.5×10}}^{\text{2}}\text{sec}$ is $\text{4}{\text{.7×10}}^{\text{-3}}\text{M}$ .

The half life period of the reaction is $\text{26}\text{s}$.

Explanation of Solution

To calculate the concentration of ${\text{NO}}_{\text{2}}$ after $\text{2}{\text{.5×10}}^{\text{2}}\text{sec}$

The integrated rate law for second order reaction is,

$\frac{\text{1}}{{\left[\text{A}\right]}_{\text{t}}}\text{=}\text{kt+}\frac{\text{1}}{{\left[\text{A}\right]}_{\text{0}}}$

Given,

Concentration of ${\text{NO}}_{\text{2}}\text{(A)=0}\text{.050}\text{M}$

Rate constant = $\text{0}\text{.775}\text{L/}\left(\text{mol}\text{.s}\right)$

After $\text{2}{\text{.5×10}}^{\text{2}}\text{sec}$ concentration of ${\text{NO}}_{\text{2}}$ becomes,

$\begin{array}{l}\frac{\text{1}}{{\left[\text{A}\right]}_{\text{t}}}\text{=0}\text{.755}\text{L/}\left(\text{mol}\text{.s}\right)\left(\text{2}{\text{.5×10}}^{\text{2}}\text{s}\right)\text{+}\frac{\text{1}}{\text{0}\text{.050mol/L}}\\ \\ \frac{\text{1}}{{\left[\text{A}\right]}_{\text{t}}}\text{=193}\text{L/mol+20}\text{.0L/mol=213}\text{.}\text{L/mol}\\ \\ \frac{\text{1}}{{\left[\text{A}\right]}_{\text{t}}}\text{=}\frac{\text{1}}{\text{213}\text{L/mol}}\text{=4}{\text{.67×10}}^{\text{-3}}\text{=4}{\text{.7×10}}^{\text{-3}}\text{M}\end{array}$

The concentration of ${\text{NO}}_{\text{2}}$ after $\text{2}{\text{.5×10}}^{\text{2}}\text{sec}$ = $\text{4}{\text{.7×10}}^{\text{-3}}\text{M}$.

To calculate the half life of the reaction

The half-life of second order reaction can be calculated using the equation,

${\text{t}}_{\text{1/2}}\text{=}\frac{\text{1}}{\text{(k}{\left[\text{A}\right]}_{\text{0}}\text{)}}$

Given,

Concentration of ${\text{NO}}_{\text{2}}\text{(A)=0}\text{.050}\text{M}$

Rate constant = $\text{0}\text{.775}\text{L/}\left(\text{mol}\text{.s}\right)$

Then, the half life period is calculated as,

$\begin{array}{l}{\text{t}}_{\text{1/2}}\text{=}\frac{\text{1}}{\left(\text{0}\text{.755}\text{L}\left(\text{mol}\text{.s}\right)\right)\left(\text{0}\text{.050}\text{mol/L}\right)}\\ \\ {\text{t}}_{\text{1/2}}\text{=25}\text{.80=26}\text{s}\end{array}$

The half-life period of the reaction = $\text{26}\text{s}$.

Conclusion

The concentration of ${\text{NO}}_{\text{2}}$ after $\text{2}{\text{.5×10}}^{\text{2}}\text{sec}$ and half-life period was calculated using the integrated law and half-life period for second order reactions and were found to be $\text{4}{\text{.7×10}}^{\text{-3}}\text{M}$ and $\text{26}\text{s}$.