Chapter 13, Problem 13.126AP

(a)

Interpretation Introduction

Interpretation: The reaction involved in the formation of photochemical smog is given. Various questions based on this reaction are to be answered.

Concept introduction: According to the rate law the rate of reaction depends on the concentration of reactants that are involved in the reaction. Therefore, for a reaction like $\text{A}\to \text{B}$, the rate law equation is given as,

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

Where,

• $k$ is the rate constant.
• $m$ is the order of the reaction.

To determine: The possibility of the consistency of the observed order of the reaction of $\text{NO}$ and ${\text{O}}_3$ with the one that is obtained after assuming the given reaction to be a single step reaction.

Answer to Problem 13.126AP

Solution

The rate law would be consistent with the observed order of the reaction of $\text{NO}$ and ${\text{O}}_3$.

Explanation of Solution

Explanation

The given reaction is,

$\text{NO}\left(g\right)+{\text{O}}_3\left(g\right)\to {\text{NO}}_2\left(g\right)\to {\text{N}}_2\left(g\right)+{\text{O}}_{\text{2}}\left(g\right)$

If the reaction occurs in the single step, then the rate equation is

$\text{Rate}=k{\left[\text{NO}\right]}^1{\left[{\text{O}}_3\right]}^1$

Therefore, the rate law would be consistent with the observed order of the reaction of $\text{NO}$ and ${\text{O}}_3$.

(b)

Interpretation Introduction

To determine: The activation energy of the reaction.

Answer to Problem 13.126AP

Solution

The activation energy of the reaction is $\underset{\_}{6.1\times 10^{4}J/mol}$.

Explanation of Solution

Explanation

Given

The rate constants at $25°\text{C}$ is $80{\text{M}}^{-1}{\text{s}}^{-1}$.

The rate constants at $75°\text{C}$ is  $3000{\text{M}}^{-1}{\text{s}}^{-1}$.

The conversion of $°\text{C}$ into Kelvin is done as,

$°\text{C}=\left(T°\text{C}+273\right)\text{K}$

Where,

• $T°\text{C}$ is the temperature in degree Celsius.

Hence, the conversion of $25°\text{C}$ into Kelvin is,

$\begin{array}{c}25°\text{C}=\left(25+273\right)\text{K}\\ =298\text{K}\end{array}$

Similarly, the conversion of $75°\text{C}$ into Kelvin is,

$\begin{array}{c}75°\text{C}=\left(75+273\right)\text{K}\\ =348\text{K}\end{array}$

The formula for activation energy is given as,

$k=A{e}^{\frac{-E_{\text{a}}}{RT}}$

Where,

• $k$ is the rate constant of the reaction.
• $A$ is the frequency factor.
• $E_{\text{a}}$ is the activation energy.
• $R$ is the universal gas constant $\left(8.314\text{J}/\text{K}\text{mol}\right)$.
• $T$ is he temperature.

Take natural log on both the sides of the equation,

$\ln k=\ln A-\frac{E_{\text{a}}}{RT}$

The table for the plot of $\ln k$ and $1/T$ is,

$k\left({\text{M}}^{-1}{\text{s}}^{-1}\right)$	$\ln k$	$T\left(\text{K}\right)$	$\frac{1}{T\left(\text{K}\right)}$
$80$	$4.38$	$298$	$3.36\times {10}^{-3}$
$3000$	$8.00$	$348$	$2.87\times {10}^{-3}$

If a graph is plotted between $\ln k$ and $1/T$, then it comes out to be a straight line with a negative slope. Compare this equation with $y=mx+c$. The value of $y$ is $\ln k$ and the value of $m$ is $-\frac{E_{\text{a}}}{R}$ and the intercept is $\ln A$.

The slope is given as,

$\frac{y_2-y_1}{x_2-x_1}=\frac{-E_{\text{a}}}{\text{R}}$

Substitute the value of $y_1$, $y_2$, $x_1$ $x_2$ and $R$ in the above equation as,

$\begin{array}{c}\frac{y_2-y_1}{x_2-x_1}=\frac{-E_{\text{a}}}{\text{R}}\\ \frac{8.00-4.38}{2.87\times {10}^{-3}-3.36\times {10}^{-3}}=\frac{-E_{\text{a}}}{8.314}\\ E_{\text{a}}=\underset{\_}{6.1\times 10^{4}J/mol}\end{array}$

(c)

Interpretation Introduction

To determine: The rate constant of the reaction.

Answer to Problem 13.126AP

Solution

The rate constant of the reaction is $\underset{\_}{1.0\times 10^{-12}M/s}$.

Explanation of Solution

Explanation

Given

The concentration of $\text{NO}$ is $3\times {10}^{-6}\text{M}$.

The concentration of ${\text{O}}_{\text{3}}$ is $5\times {10}^{-9}\text{M}$.

The rate law expression for the given reaction is,

$\text{Rate}=k{\left[\text{NO}\right]}^1{\left[{\text{O}}_3\right]}^1$

Substitute the value of concentration of each species and rate constant in the above equation,

$\begin{array}{c}\text{Rate}=k{\left[\text{NO}\right]}^1{\left[{\text{O}}_3\right]}^1\\ =\left(80{\text{M}}^{-1}{\text{s}}^{-1}\right)\left(3\times {10}^{-6}\text{M}\right)\left(5\times {10}^{-9}\text{M}\right)\\ =\underset{\_}{1.0\times 10^{-12}M/s}\end{array}$

(d)

Interpretation Introduction

To determine: The values of rate constant at $10°\text{C}$ and $35°\text{C}$.

Answer to Problem 13.126AP

Solution

The values of rate constant are $\underset{\_}{22.0M/s}$ and $\underset{\_}{180.5M/s}$ at $10°\text{C}$ and $35°\text{C}$ respectively.

Explanation of Solution

Explanation

Given

The values of temperature are $10°\text{C}$ and $35°\text{C}$.

The formula for activation energy is given as,

$k=A{e}^{\frac{-E_{\text{a}}}{RT}}$ (1)

Where,

• $k$ is the rate constant of the reaction.
• $A$ is the frequency factor.
• $E_{\text{a}}$ is the activation energy.
• $R$ is the universal gas constant $\left(8.314\text{J}/\text{K}\text{mol}\right)$.
• $T$ is he temperature.

Take natural log on both the sides of the equation,

$\ln k=\ln A-\frac{E_{\text{a}}}{RT}$

Substitute the value of $k$, $E_{\text{a}}$ and $T$ t find out the value of $A$ in the above equation,

$\begin{array}{c}\ln k=\ln A-\frac{E_{\text{a}}}{RT}\\ \ln 80=\ln A-\frac{6.1\times {10}^4\text{J}/\text{mol}}{\left(8.314\text{J}/\text{K}\text{mol}\right)\left(298\text{K}\right)}\\ 4.38=\ln A-24.62\\ A=4.0\times {10}^{12}\end{array}$

The conversion of $°\text{C}$ into Kelvin is done as,

$°\text{C}=\left(T°\text{C}+273\right)\text{K}$

Where,

• $T°\text{C}$ is the temperature in degree Celsius.

Hence, the conversion of $10°\text{C}$ into Kelvin is,

$\begin{array}{c}10°\text{C}=\left(10+273\right)\text{K}\\ =283\text{K}\end{array}$

Similarly, the conversion of $35°\text{C}$ into Kelvin is,

$\begin{array}{c}35°\text{C}=\left(35+273\right)\text{K}\\ =308\text{K}\end{array}$

Substitute the value of $A$, $E_{\text{a}}$, $R$ and $T$ in the equation (1).

$\begin{array}{c}{k}_{10°\text{C}}=A{e}^{\frac{-E_{\text{a}}}{RT}}\\ =4.0\times {10}^{12}\times e^{\frac{-6.1\times {10}^4\text{J}/\text{mol}}{\left(8.314\text{J}/\text{K}\text{mol}\right)\left(283\text{K}\right)}}\\ =\underset{\_}{22.0M/s}\end{array}$

Similarly, Substitute the value of $A$, $E_{\text{a}}$, $R$ and $T$ in the equation (1).

$\begin{array}{c}{k}_{35°\text{C}}=A{e}^{\frac{-E_{\text{a}}}{RT}}\\ =4.0\times {10}^{12}\times e^{\frac{-6.1\times {10}^4\text{J}/\text{mol}}{\left(8.314\text{J}/\text{K}\text{mol}\right)\left(308\text{K}\right)}}\\ =\underset{\_}{180.5M/s}\end{array}$

Conclusion

1. a. The rate law would be consistent with the observed order of the reaction of $\text{NO}$ and ${\text{O}}_3$.
2. b. The activation energy of the reaction is $\underset{\_}{6.1\times 10^{4}J/mol}$.
3. c. The rate constant of the reaction is $\underset{\_}{1.0\times 10^{-12}M/s}$.
4. d. The values of rate constant are $\underset{\_}{22.0M/s}$ and $\underset{\_}{180.5M/s}$ at $10°\text{C}$ and $35°\text{C}$ respectively.