Chapter 13, Problem 13.87QP

(a)

Interpretation Introduction

Interpretation: The activation energy; the frequency factor for the reaction 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 13.87QP

Solution

The value of activation energy is $\underset{\_}{3.14\times 10^{5}J/mol}$.

Explanation of Solution

Explanation

The given reaction is,

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

The given data is,

T ( $\text{K}$)	$k\left[{\text{M}}^{-1/2}{\text{s}}^{-1}\right]$
$2000$	$318$
$2100$	$782$
$2200$	$1770$
$2300$	$3733$
$2400$	$7396$

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 $T_1=2000\text{K}$, the value of rate constant $k_1=318$.

At $T_2=2100\text{K}$, the value of rate constant $k_2=782$.

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

$\begin{array}{c}\ln \frac{318}{782}=\frac{E_{\text{a}}}{8.314}\left(\frac{1}{2100\text{K}}-\frac{1}{2000\text{K}}\right)\\ -0.899=\frac{E_{\text{a}}}{8.314}\left(-2.380\times {10}^{-5}\right)\\ E_{\text{a}}=\frac{0.899\times 8.314}{2.380\times {10}^{-5}}\\ =\underset{\_}{3.14\times 10^{5}J/mol}\end{array}$

The value of activation energy is $\underset{\_}{3.14\times 10^{5}J/mol}$.

(b)

Interpretation Introduction

To determine: The frequency factor for the reaction.

Answer to Problem 13.87QP

Solution

The frequency factor for the reaction is $\underset{\_}{5.03\times 10^{10}{M}^{-1/2}{s}^{-1}}$.

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.

$\begin{array}{c}318=A{\text{e}}^{-3.14{\text{×10}}^5\text{J/mol}/8.314\text{J/mol}\cdot \text{K}\times 2000\text{K}}\\ 318=A\times 6.317\times {10}^{-9}\\ A=\underset{\_}{5.03\times 10^{10}{M}^{-1/2}{s}^{-1}}\end{array}$

Therefore, the value of frequency factor is $\underset{\_}{5.03\times 10^{10}{M}^{-1/2}{s}^{-1}}$.

(c)

Interpretation Introduction

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

Answer to Problem 13.87QP

Solution

The value of rate constant at $\text{300}\text{K}$ is

Explanation of Solution

Explanation

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.

$\begin{array}{c}k=5.03\times {10}^{10}\times e^{-3.14{\text{×10}}^5\text{J/mol}/8.314\times 300\text{K}}\\ =5.03\times 2.32\times {10}^{-55}\\ =\underset{\_}{1.166\times 10^{-56}{M}^{-1/2}{s}^{-1}}\end{array}$

Therefore, the value of rate constant at $\text{300}\text{K}$ is $\underset{\_}{1.166\times 10^{-56}{M}^{-1/2}{s}^{-1}}$.

Conclusion

The value of rate constant at $\text{300}\text{K}$ is $\underset{\_}{1.166\times 10^{-56}{M}^{-1/2}{s}^{-1}}$.