Chapter 13, Problem 13.90QE

(a)

Interpretation Introduction

Interpretation:

The order of the decomposition of nitrogen dioxide has to be determined.

Explanation of Solution

The time and the pressure of the reactant is given as shown below;

Time ($\text{s}$)	Pressure of ${\text{NO}}_{\text{2}}$ at $\text{310}\text{K}$	Pressure of ${\text{NO}}_{\text{2}}$ at $\text{315}\text{K}$
$0$	$24.0$	$24.0$
$1$	$18.1$	$15.2$
$2$	$13.7$	$9.7$
$3$	$10.3$	$6.1$
$4$	$7.8$	$3.9$
$5$	$5.9$	$2.5$
$6$	$4.5$	$1.6$
$7$	$3.4$	$1.0$
$8$	$2.6$	$0.6$
$9$	$1.9$	$0.4$
$10$	$1.5$	$0.3$

Graph of concentration and time is plotted considering the time in x-axis and pressure of the reactant at $\text{310}\text{K}$ in y-axis.  If the graph obtained is a straight line then it means the reaction is of zero-order.

As the obtained graph is not a straight line, the reaction is not a zero order.

If the plot of graph with time and $\ln [concentration]$ is a straight line, then the reaction is of first-order.

Time ($\text{s}$)	Pressure of ${\text{NO}}_{\text{2}}$ at $\text{310}\text{K}$	${\text{ln[P}}_{{\text{NO}}_{\text{2}}}]$ at $\text{310}\text{K}$
$0$	$24.0$	$3.178$
$1$	$18.1$	$2.895$
$2$	$13.7$	$2.617$
$3$	$10.3$	$2.332$
$4$	$7.8$	$2.054$
$5$	$5.9$	$1.774$
$6$	$4.5$	$1.504$
$7$	$3.4$	$1.223$
$8$	$2.6$	$0.955$
$9$	$1.9$	$0.641$
$10$	$1.5$	$0.405$

As the obtained graph is a straight line, the reaction is a first order.

(b)

Interpretation Introduction

Interpretation:

Rate constant for the decomposition of nitrogen dioxide has to be determined.

Explanation of Solution

The time and the pressure of the reactant is given as shown below;

Time ($\text{s}$)	Pressure of ${\text{NO}}_{\text{2}}$ at $\text{310}\text{K}$	Pressure of ${\text{NO}}_{\text{2}}$ at $\text{315}\text{K}$
$0$	$24.0$	$24.0$
$1$	$18.1$	$15.2$
$2$	$13.7$	$9.7$
$3$	$10.3$	$6.1$
$4$	$7.8$	$3.9$
$5$	$5.9$	$2.5$
$6$	$4.5$	$1.6$
$7$	$3.4$	$1.0$
$8$	$2.6$	$0.6$
$9$	$1.9$	$0.4$
$10$	$1.5$	$0.3$

Graph of concentration and time is plotted considering the time in x-axis and pressure of the reactant at $\text{310}\text{K}$ in y-axis.  If the graph obtained is a straight line then it means the reaction is of zero-order.

As the obtained graph is not a straight line, the reaction is not a zero order.

If the plot of graph with time and $\ln [concentration]$ is a straight line, then the reaction is of first-order.

Time ($\text{s}$)	Pressure of ${\text{NO}}_{\text{2}}$ at $\text{310}\text{K}$	${\text{ln[P}}_{{\text{NO}}_{\text{2}}}]$ at $\text{310}\text{K}$
$0$	$24.0$	$3.178$
$1$	$18.1$	$2.895$
$2$	$13.7$	$2.617$
$3$	$10.3$	$2.332$
$4$	$7.8$	$2.054$
$5$	$5.9$	$1.774$
$6$	$4.5$	$1.504$
$7$	$3.4$	$1.223$
$8$	$2.6$	$0.955$
$9$	$1.9$	$0.641$
$10$	$1.5$	$0.405$

As the obtained graph is a straight line, the reaction is a first order.  Therefore, the rate law can be given as shown below;

    $\text{Rate}=k[{\text{P}}_{{\text{NO}}_{\text{2}}}]$

Rate constant for the reaction at $\text{310}\text{K}$:

The rate constant of the first order reaction can be calculated as shown below;

    $\begin{array}{c}{k}_{310\text{K}}=-\text{Slope}\\ =-\frac{(1.5-24.0)}{(10-0)\text{s}}\\ =\frac{22.5}{10}{\text{s}}^{-1}\\ =2.25{\text{s}}^{-1}\end{array}$

The rate constant of the reaction at $\text{310}\text{K}$ is $2.25{\text{s}}^{-1}$.

Rate constant for the reaction at $\text{315}\text{K}$:

The rate constant of the first order reaction can be calculated as shown below;

    $\begin{array}{c}{k}_{315\text{K}}=-\text{Slope}\\ =-\frac{(0.3-24.0)}{(10-0)\text{s}}\\ =\frac{23.7}{10}{\text{s}}^{-1}\\ =2.37{\text{s}}^{-1}\end{array}$

The rate constant of the reaction at $\text{310}\text{K}$ is $2.37{\text{s}}^{-1}$.

(c)

Interpretation Introduction

Interpretation:

Activation energy has to be calculated for the decomposition of nitrogen dioxide.

Concept Introduction:

Activation energy is the minimum amount of energy that has to be possessed by the reactant species in order to produce products.  Activation energy is represented as $E_a$.

Answer to Problem 13.90QE

Activation energy of the reaction is $\text{8}\text{.427}\text{kJ/mol}$.

Explanation of Solution

The rate constants and temperature are as follows;

Rate constant ${\text{s}}^{-1}$	Temperature $\text{(K)}$
$2.25$	$310$
$2.37$	$315$

Activation energy and the rate constants for a reaction at two different temperatures is related by the equation as follows;

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

Where,

    $k_1$ is the rate constant of the reaction at temperature $T_1$.

    $k_2$ is the rate constant of the reaction at temperature $T_2$.

    $R$ is the gas constant.

    $E_a$ is the activation energy.

Rearranging equation (1) in order to obtain activation energy, the equation is given as shown below;

    $E_a=-\frac{R{T}_1{T}_2}{(T_2-T_1)}\ln \left(\frac{k_1}{k_2}\right)$        (2)

Substituting the first and third entry from the table given above in equation (2), the activation energy can be calculated as follows;

    $\begin{array}{c}{E}_a=-\frac{8.314\text{J/mol}\cdot \text{K}\times \text{(310K})(\text{315}\text{K})}{315\text{K}-310\text{K}}\ln \left(\frac{2.25}{2.37}\right)\\ =-\frac{8.314\text{J/mol}\times \text{97650}\text{K}}{\text{5}\text{K}}\ln (0.9493)\\ =-162372\text{J/mol}\times \text{(-0}\text{.0519)}\\ =8427.12\text{J/mol}\\ \text{=8}\text{.427}\text{kJ/mol}\end{array}$

Therefore, the activation energy of the reaction is $\text{8}\text{.427}\text{kJ/mol}$.