Chapter 14, Problem 14.24QP

(a)

Interpretation Introduction

Interpretation:

The time taken for the concentration of $N{O}_2$ to decrease from $0.62M$ to $0.28M$ has to be calculated.

Concept Introduction:

Intergrated rate eqaution for a second order reaction is,

$\frac{1}{\left[A\right]}=\frac{1}{{\left[A\right]}_0}+kt$

$\left[A\right]$ is the concentration of reactant A at time t ${\left[A\right]}_0$ is the  initial concentration of reactant and k is the rate constant.

Answer to Problem 14.24QP

The time taken for the concentration of $N{O}_2$ to decrease from $0.62M$ to $0.28M$ is $3.63\text{s}$

Explanation of Solution

Given data:

• The second order reaction is ${\text{2NO}}_{\text{2}}\left(\text{g}\right)\to \text{2NO}\left(\text{g}\right){\text{+O}}_{\text{2}}\left(\text{g}\right)$
• The second order rate constant is $\text{k}\text{=}\text{0}\text{.54/M}\text{.s}$
• $\begin{array}{l}\left[N{O}_2\right]=0.28M\\ {\left[N{O}_2\right]}_{\circ }=0.62M\end{array}$

Intergrated rate eqaution for a second order reaction is,

$\begin{array}{l}\frac{1}{\left[N{O}_2\right]}=\frac{1}{{\left[N{O}_2\right]}_0}+kt\\ t=\frac{\frac{1}{\left[N{O}_2\right]}-\frac{1}{{\left[N{O}_2\right]}_0}}{k}\end{array}$

Substituting all the known values in this expression:

$\begin{array}{c}t=\frac{\frac{1}{\left[N{O}_2\right]}-\frac{1}{{\left[N{O}_2\right]}_0}}{k}\\ t=\frac{\frac{1}{0.28M}-\frac{1}{0.62M}}{\text{0}\text{.54/M}\text{.s}}\\ =3.63\text{s}\end{array}$

Therefore, the time taken is $t=3.63\text{s}$.

Conclusion

The time taken for the concentration of $N{O}_2$ to decrease from $0.62M$ to $0.28M$ has been calculated.

(b)

Interpretation Introduction

Interpretation:

The half-lives of ${\text{NO}}_{\text{2}}$ at the given concentrations have to be calculated.

Concept Introduction:

Half-life for a second order reaction is

${\text{t}}_{\raisebox{1ex}{$\text{1}$}\!\left/ \!\raisebox{-1ex}{$\text{2}$}\right.}\text{=}\frac{\text{1}}{\text{k}{\left[\text{A}\right]}_{\text{0}}}$

$\begin{array}{l}\text{where,}\\ {\left[\text{A}\right]}_{\text{0}}\text{is}\text{the}\text{initial}\text{concentration}\text{of}\text{reactant}\text{A}\\ {\text{t}}_{\raisebox{1ex}{$\text{1}$}\!\left/ \!\raisebox{-1ex}{$\text{2}$}\right.}\text{is}\text{the}\text{half}\text{-life}\text{period}\text{.}\\ \text{k}\text{is}\text{the}\text{rate}\text{constant}\text{.}\end{array}$

Answer to Problem 14.24QP

The half-life period when ${\left[{\text{NO}}_{\text{2}}\right]}_{\text{0}}\text{=}\text{0}\text{.62}\text{M}$ is $\text{2}\text{.99 s}$

The half-life period when ${\left[{\text{NO}}_{\text{2}}\right]}_{\text{0}}\text{=}\text{0}\text{.28}\text{M}$ is $\text{6}\text{.61 s}$

Explanation of Solution

The half-life when period ${\left[{\text{NO}}_{\text{2}}\right]}_{\text{0}}\text{=}\text{0}\text{.62}\text{M}$ can be calculated as follows:

$\begin{array}{l}{\text{t}}_{\raisebox{1ex}{$\text{1}$}\!\left/ \!\raisebox{-1ex}{$\text{2}$}\right.}\text{=}\frac{\text{1}}{\text{k}{\left[\text{A}\right]}_{\text{0}}}\\ {\text{t}}_{\raisebox{1ex}{$\text{1}$}\!\left/ \!\raisebox{-1ex}{$\text{2}$}\right.}\text{=}\frac{\text{1}}{\left(\text{0}\text{.54/M}\text{.s}\right)\left(\text{0}\text{.62}\text{M}\right)}\because \text{Given:}\text{k}\text{=}\text{0}\text{.54/M}\text{.s}\\ {\text{t}}_{\raisebox{1ex}{$\text{1}$}\!\left/ \!\raisebox{-1ex}{$\text{2}$}\right.}\text{=2}\text{.99 s}\end{array}$

Therefore, the half-life when period ${\left[{\text{NO}}_{\text{2}}\right]}_{\text{0}}\text{=}\text{0}\text{.62}\text{M}$${\text{t}}_{\raisebox{1ex}{$\text{1}$}\!\left/ \!\raisebox{-1ex}{$\text{2}$}\right.}\text{=2}\text{.99 s}$.

The half-life when period ${\left[{\text{NO}}_{\text{2}}\right]}_{\text{0}}\text{=}\text{0}\text{.28}\text{M}$ can be calculated as follows:

$\begin{array}{l}{\text{t}}_{\raisebox{1ex}{$\text{1}$}\!\left/ \!\raisebox{-1ex}{$\text{2}$}\right.}\text{=}\frac{\text{1}}{\text{k}{\left[\text{A}\right]}_{\text{0}}}\\ {\text{t}}_{\raisebox{1ex}{$\text{1}$}\!\left/ \!\raisebox{-1ex}{$\text{2}$}\right.}\text{=}\frac{\text{1}}{\left(\text{0}\text{.54/M}\text{.s}\right)\left(\text{0}\text{.28}\text{M}\right)}\because \text{Given:}\text{k}\text{=}\text{0}\text{.54/M}\text{.s}\\ {\text{t}}_{\raisebox{1ex}{$\text{1}$}\!\left/ \!\raisebox{-1ex}{$\text{2}$}\right.}\text{=}\text{6}\text{.61 s}\end{array}$

Therefore, the half-life when period ${\left[{\text{NO}}_{\text{2}}\right]}_{\text{0}}\text{=}\text{0}\text{.28}\text{M}$${\text{t}}_{\raisebox{1ex}{$\text{1}$}\!\left/ \!\raisebox{-1ex}{$\text{2}$}\right.}\text{=6}\text{.61 s}$.

Conclusion

The half-lives of ${\text{NO}}_{\text{2}}$ at the given concentrations have been calculated.