Chapter 14, Problem 14.88QP

(a)

Interpretation Introduction

Interpretation: The activation energy and the rate constant for the given reaction at $\text{300}\text{K}$ are to be calculated.

Concept introduction: Arrhenius equation describes the relationship between rate constant of a reaction, activation energy, absolute temperature and the frequency factor.

To determine: The activation energy for the given reaction.

Answer to Problem 14.88QP

Solution

The value of activation energy is $\underset{\_}{2.038\times 10^{4}J/mol}$.

Explanation of Solution

Explanation

The given reaction is,

${\text{NO}}_2\left(g\right)+{\text{O}}_{\text{3}}\left(g\right)\to {\text{NO}}_3\left(g\right)+{\text{O}}_{\text{2}}\left(g\right)$

The rate constant for the given reaction was determined over temperature range of $\text{40}\text{K}$ with the following results:

T ( $\text{K}$)	$k\left[{\text{M}}^{-1}{\text{s}}^{-1}\right]$
$203$	$4.14\times {10}^5$
$213$	$7.30\times {10}^5$
$223$	$1.22\times {10}^6$
$233$	$1.96\times {10}^6$
$243$	$3.02\times {10}^6$

The equation obtained after rearrangement of Arrhenius equation is,

$\ln \frac{k_1}{k_2}=\frac{E_{\text{a}}}{R}\left(\frac{1}{T_2}-\frac{1}{T_1}\right)$

Where,

• $k_1$ and $k_2$ are rate constants at temperature $T_1$ and $T_2$.
• $E_{\text{a}}$ is activation energy.
• $R$ is gas constant ( $8.314\text{J/mol}\cdot \text{K}$).

From the given data, at $T_1=203\text{K}$, the value of rate constant $k_1=4.14\times {10}^5$.

At $T_2=213\text{K}$, the value of rate constant $k_2=7.30\times {10}^5$.

Substitute the value of $k_1$, $k_2$, $T_1$, $T_2$ and $R$ in the above equation.

$\begin{array}{c}\ln \frac{4.14\times {10}^5}{7.30\times {10}^5}=\frac{E_{\text{a}}}{8.314}\left(\frac{1}{213\text{K}}-\frac{1}{203\text{K}}\right)\\ -0.567=\frac{E_{\text{a}}}{8.314}\left(-2.312\times {10}^{-4}\right)\\ E_{\text{a}}=\frac{0.567\times 8.314}{2.312\times {10}^{-4}}\\ =\underset{\_}{2.038\times 10^{4}J/mol}\end{array}$

The value of activation energy is $\underset{\_}{2.038\times 10^{4}J/mol}$.

(b)

Interpretation Introduction

To determine: The rate constant for the given reaction at $\text{300}\text{K}$.

Answer to Problem 14.88QP

Solution

The value of rate constant at $\text{300}\text{K}$ is $\underset{\_}{2.04\times 10^7{M}^{-1}{s}^{-1}}$.

Explanation of Solution

Explanation

The frequency factor $\left(A\right)$ is calculated by the formula,

$k_1=A{e}^{-E_{\text{a}}/R{T}_1}$

Substitute the value of $k_1$, $T_1$, $E_{\text{a}}$ and $R$ in the above equation.

$\begin{array}{c}4.14\times {10}^5=A{\text{e}}^{-2.038{\text{×10}}^{\text{4}}\text{J/mol}/8.314\text{J/mol}\cdot \text{K}\times 203\text{K}}\\ 4.14\times {10}^5=A\times 5.72\times {10}^{-6}\\ A=7.23\times {10}^{10}{\text{M}}^{-1}{\text{s}}^{-1}\end{array}$

Therefore, the value of frequency factor is $7.23\times {10}^{10}{\text{M}}^{-1}{\text{s}}^{-1}$.

The rate constant for the given reaction at $\text{300}\text{K}$ is calculated by the formula,

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

Where,

• $k$ is rate constant.
• $T$ is the temperature.

The value of $T$ is $\text{300}\text{K}$.

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

$\begin{array}{c}k=7.23\times {10}^{10}\times e^{-\text{2}{\text{.038×10}}^{\text{4}}\text{J/mol}/8.314\times 300\text{K}}\\ =\underset{\_}{2.04\times 10^7{M}^{-1}{s}^{-1}}\end{array}$

Therefore, the value of rate constant at $\text{300}\text{K}$ is $\underset{\_}{2.04\times 10^7{M}^{-1}{s}^{-1}}$.

Conclusion

The value of rate constant at $\text{300}\text{K}$ is $\underset{\_}{2.04\times 10^7{M}^{-1}{s}^{-1}}$.