Chapter 12, Problem 87AE

For the reaction

${\text{2N}}_{\text{2}}{\text{O}}_{\text{5}}\left(g\right)\to 4{\text{NO}}_2\left(g\right)+{\text{O}}_2\left(g\right)$

the following data were collected, where

$\text{Rate}\text{= }-\frac{\Delta \left[{\text{N}}_{\text{2}}{\text{O}}_{\text{5}}\right]}{\Delta t}$

Time (s)	T = 338 K [N2O5]	T=318K [N2O5]
0	1.00 × 10−1 M	1.00 × 10−1 M
100.	6.14 × 10−2 M	9.54 × 10−2 M
300	2.33 × 10−2 M	8.63 × 10−2 M
600.	5.41 × 1O−3 M	7.43 × 10−2 M
900.	1.26 × 1O−3 M	6.39 × 10−2 M

Calculate E_a for tills reaction.

Interpretation Introduction

Interpretation: The decomposition reaction of $\left({\text{N}}_{\text{2}}{\text{O}}_{\text{5}}\right)$, its rate law and concentration of $\left({\text{N}}_{\text{2}}{\text{O}}_{\text{5}}\right)$ at different temperature and time is given. The value of $E_{\text{a}}$ is to be calculated.

Concept introduction: The change observed in the concentration of a reactant or a product per unit time is known as the rate of the particular reaction. The differential rate law provides the rate of a reaction at specific reaction concentrations.

The half-life of the first order reaction is calculated using the formula,

$t_{\frac{\text{1}}{\text{2}}}=\frac{\text{0}\text{.693}}{k}$

To determine: The value of $E_{\text{a}}$ for the given reaction.

Answer to Problem 87AE

Answer

The value of $E_{\text{a}}$ for the given reaction is $\underset{\_}{1.01\times 10^{5}J/mol}$.

Explanation of Solution

Explanation

The reaction that takes place is,

${\text{2N}}_{\text{2}}{\text{O}}_{\text{5}}\left(g\right)\stackrel{}{\to }{\text{4NO}}_{\text{2}}\left(g\right)+{\text{O}}_{\text{2}}\left(g\right)$

The value of $\text{ln}\left({\text{N}}_{\text{2}}{\text{O}}_{\text{5}}\right)$ at $\text{338}\text{K}$ is calculated using the given table. That is,

Time (s)	$\begin{array}{c}T=\text{338}\text{K}\\ \left[{\text{N}}_{\text{2}}{\text{O}}_{\text{5}}\right]\end{array}$	$\begin{array}{c}T=\text{338}\text{K}\\ \text{ln}\left[{\text{N}}_{\text{2}}{\text{O}}_{\text{5}}\right]\end{array}$
$\text{0}$	$\text{1}\text{.00}\times {\text{10}}^{-\text{1}}\text{M}$	$-\text{2}\text{.302}$
$\text{100}$	$\text{6}\text{.14}\times {\text{10}}^{-\text{2}}\text{M}$	$-\text{2}\text{.79}$
$\text{300}$	$\text{2}\text{.33}\times {\text{10}}^{-\text{2}}\text{M}$	$-\text{3}\text{.75}$
$\text{600}$	$\text{5}\text{.41}\times {\text{10}}^{-\text{3}}\text{M}$	$-\text{5}\text{.21}$
$\text{900}$	$\text{1}\text{.26}\times {\text{10}}^{-\text{3}}\text{M}$	$-\text{6}\text{.67}$

The graph is plotted for $\text{ln}\left[{\text{N}}_{\text{2}}{\text{O}}_{\text{5}}\right]$ versus time at $T=\text{338}\text{K}$ is,

Since, the obtained graph is a straight line, hence, the order of the reaction is first. The slope of graph is $-\text{0}\text{.0048}$.

The integral rate law equation of first order reaction is,

$\text{ln}\left[{\text{N}}_{\text{2}}{\text{O}}_{\text{2}}\right]=-k_{\text{1}}t+\text{ln}{\left[{\text{N}}_{\text{2}}{\text{O}}_{\text{2}}\right]}_{\text{0}}$ (1)

Where,

• $k_{\text{1}}$ is the rate constant at $\text{338}\text{K}$.
• ${\left[{\text{N}}_{\text{2}}{\text{O}}_{\text{2}}\right]}_{\text{0}}$ is the initial concentration of reactant.
• $t$ is the time.

The equation (1) is similar to the equation of a straight line, that is,

$y=mx+c$ (2)

Where,

• $y$ is the y-intercept.
• $x$ is the x-intercept.
• $m$ is the slope.
• $c$ is a constant.

Compare equation (1) and (2).

$m=-k_{\text{1}}$

Substitute the value of slope in the above equation.

$\begin{array}{c}m=-k_{\text{1}}\\ k_{\text{1}}=-\left(-\text{0}\text{.0048}{\text{s}}^{-1}\right)\\ =\underset{\_}{0.0048s^{-1}}\end{array}$

The value of rate constant at $T=\text{318}\text{K}$ is $\underset{\_}{4.977\times 10^{-4}{s}^{-1}}$.

The value of $\text{ln}\left({\text{N}}_{\text{2}}{\text{O}}_{\text{5}}\right)$ at $\text{318}\text{K}$ is calculated using the given table. That is,

Time (s)	$\begin{array}{c}T=\text{318}\text{K}\\ \left[{\text{N}}_{\text{2}}{\text{O}}_{\text{5}}\right]\end{array}$	$\begin{array}{c}T=\text{318}\text{K}\\ \ln \left[{\text{N}}_{\text{2}}{\text{O}}_{\text{5}}\right]\end{array}$
$\text{0}$	$\text{1}\text{.00}\times {\text{10}}^{-\text{1}}\text{M}$	$-\text{2}\text{.302}$
$\text{100}$	$\text{9}\text{.54}\times {\text{10}}^{-\text{2}}\text{M}$	$-\text{2}\text{.349}$
$\text{300}$	$\text{8}\text{.63}\times {\text{10}}^{-\text{2}}\text{M}$	$-\text{2}\text{.449}$
$\text{600}$	$\text{7}\text{.43}\times {\text{10}}^{-\text{2}}\text{M}$	$-\text{2}\text{.599}$
$\text{900}$	$\text{6}\text{.39}\times {\text{10}}^{-\text{2}}\text{M}$	$-\text{2}\text{.750}$

The graph is plotted for $\text{ln}\left[{\text{N}}_{\text{2}}{\text{O}}_{\text{5}}\right]$ versus time at $T=\text{318}\text{K}$ is,

Since, the obtained graph is a straight line, hence, the order of the reaction is first. The slope of graph is $-\text{0}\text{.0004977}$.

The integral rate law equation of first order reaction is,

$\text{ln}\left[{\text{N}}_{\text{2}}{\text{O}}_{\text{2}}\right]=-k_{\text{2}}t+\text{ln}{\left[{\text{N}}_{\text{2}}{\text{O}}_{\text{2}}\right]}_{\text{0}}$ (3)

Where,

• $k_{\text{2}}$ is the rate constant at $\text{318}\text{K}$.
• ${\left[{\text{N}}_{\text{2}}{\text{O}}_{\text{2}}\right]}_{\text{0}}$ is the initial concentration of reactant.
• $t$ is the time.

The equation (3) is similar to the equation of a straight line, that is,

$y=mx+c$ (4)

Where,

• $y$ is the y-intercept.
• $x$ is the x-intercept.
• $m$ is the slope.
• $c$ is a constant.

Compare equation (3) and (4).

$m=-k_{\text{2}}$

Substitute the value of slope in the above equation.

$\begin{array}{c}m=-k_{\text{2}}\\ k_{\text{2}}=-\left(-\text{0}\text{.0004977}{\text{s}}^{-1}\right)\\ =\underset{\_}{4.977\times 10^{-4}{s}^{-1}}\end{array}$

The value of $E_{\text{a}}$ for the given reaction is $\underset{\_}{1.01\times 10^{5}J/mol}$.

The value of rate constant at $T=\text{338}\text{K}$ is $\text{0}\text{.0048}{\text{s}}^{-\text{1}}$.

The value of rate constant at $T=\text{318}\text{K}$ is $\text{4}\text{.977}\times {\text{10}}^{-\text{4}}{\text{s}}^{-\text{1}}$.

Formula

The Arrhenius equation is,

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

Where,

• $E_{\text{a}}$ is the activation energy.
• $R$ is the universal gas constant $(\text{8}\text{.314}\text{J}/\text{K}\cdot \text{mol})$.
• $T_1$ is the absolute temperature $\left(\text{338}\text{K}\right)$.
• $T_2$ is the absolute temperature $\left(\text{318}\text{K}\right)$.

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

$\begin{array}{c}\ln \frac{k_2}{k_1}=\frac{E_{\text{a}}}{R}\left(\frac{1}{T_2}-\frac{1}{T_1}\right)\\ \ln \left(\frac{\text{4}\text{.977}\times {10}^{-4}{\text{s}}^{-1}}{0.048{\text{s}}^{-1}}\right)=\frac{E_{\text{a}}}{(\text{8}\text{.314}\text{J}/\text{K}\cdot \text{mol})}\left(\frac{1}{\left(\text{318}\text{K}\right)}-\frac{1}{\left(\text{338}\text{K}\right)}\right)\\ E_{\text{a}}=\underset{\_}{1.01\times 10^{5}J/mol}\end{array}$

Conclusion

Conclusion

The calculated value of $E_{\text{a}}$ for the given reaction is $\underset{\_}{1.01\times 10^{5}J/mol}$