Chapter 13, Problem 13.99QE

(a)

Interpretation Introduction

Interpretation:

Rate law for the reaction of nitrogen monoxide with hydrogen has to be given.

Answer to Problem 13.99QE

Rate law for the given reaction is $\text{Rate}=k{[{\text{P}}_{\text{NO}}]}^2{[{\text{P}}_{{\text{H}}_{\text{2}}}]}^1$.

Explanation of Solution

The reaction between nitrogen monoxide and hydrogen to produce nitrogen and water is given be the equation shown below;

    $\text{2NO}+2{\text{H}}_{\text{2}}\to {\text{N}}_{\text{2}}(g)+{\text{2H}}_{\text{2}}\text{O}(g)$

Relative concentration of the reactant is determined by dividing the pressure of each reactant by the smallest pressure of the reactant.  Relative rate of the reaction is determined by dividing the rate of the reaction by the smallest rate that is obtained from the experimental data.

The relative concentration in terms of pressure of $\text{NO}$, ${\text{H}}_{\text{2}}$ and also the relative rates of the reaction is given as follows;

Initial ${\text{[P}}_{{\text{H}}_{\text{2}}}\text{]}(\text{torr})$	Initial ${\text{[P}}_{\text{NO}}\text{]}(\text{torr})$	Initial rate of the reaction	Relative ${\text{[P}}_{{\text{H}}_{\text{2}}}\text{]}$	Relative ${\text{[P}}_{\text{NO}}]$	Relative rates of the reaction
$289$	$400$	$0.160$	$1.97$	$2.63$	$6.4$
$205$	$400$	$0.110$	$1.39$	$2.63$	$4.4$
$147$	$400$	$0.079$	$1.00$	$2.63$	$3.16$
$400$	$359$	$0.150$	$2.72$	$2.36$	$6.0$
$400$	$300$	$0.103$	$2.72$	$1.97$	$4.12$
$400$	$152$	$0.025$	$2.72$	$1.00$	$1.00$

From the above table, it is found that in experiments 1, 2 and 3, the concentration of nitrogen monoxide remains constant while the relative rate of the reaction increases in relative manner as the concentration of ${\text{H}}_{\text{2}}$ increases from $3.16$ to $6.4$.  Therefore, the reaction is first order with respect to ${\text{H}}_{\text{2}}$.

From the above table, it is found that in experiments 4, 5 and 6, the concentration of hydrogen remains constant while the relative rate of the reaction doubles as the concentration of $\text{NO}$ increases from $1.00$ to $6.0$.  Therefore, the reaction is second order with respect to $\text{NO}$.

Rate law:

Rate law is the relationship between the concentration of the reactants and the rate of the reaction.  The rate law equation is given as the rate of the reaction that is directly proportional to the product of the reactant concentration that is raised to the power of the respective reactant coefficient.  Therefore, the rate law for the given reaction is as follows;

    $\begin{array}{l}\text{Rate}\propto {[{\text{P}}_{\text{NO}}]}^2{[{\text{H}}_{\text{2}}]}^1\\ \text{Rate}=k{[{\text{P}}_{\text{NO}}]}^2{[{\text{P}}_{{\text{H}}_{\text{2}}}]}^1\end{array}$

Where,

    $k$ is the rate constant.

    $[{\text{P}}_{\text{NO}}]$ is the pressure of $\text{NO}$.

    $[{\text{P}}_{{\text{H}}_{\text{2}}}]$ is the pressure of ${\text{H}}_{\text{2}}$.

(b)

Interpretation Introduction

Interpretation:

Rate constant for the reaction of nitrogen monoxide with hydrogen has to be given.

Answer to Problem 13.99QE

Rate constant for the given reaction is $3.46\times {10}^{-9}{\text{torr}}^{-2}\cdot {\text{s}}^{-1}$.

Explanation of Solution

The rate law for the given reaction is as follows;

    $\begin{array}{l}\text{Rate}\propto {[{\text{P}}_{\text{NO}}]}^2{[{\text{H}}_{\text{2}}]}^1\\ \text{Rate}=k{[{\text{P}}_{\text{NO}}]}^2{[{\text{P}}_{{\text{H}}_{\text{2}}}]}^1\end{array}$

Where,

    $k$ is the rate constant.

    $[{\text{P}}_{\text{NO}}]$ is the pressure of $\text{NO}$.

    $[{\text{P}}_{{\text{H}}_{\text{2}}}]$ is the pressure of ${\text{H}}_{\text{2}}$.

Rate constant:

The rate constant for the reaction can be calculated from the rate law using the initial pressure of the reactants as shown below;

    $\text{Rate}=k{[{\text{P}}_{\text{NO}}]}^2{[{\text{P}}_{{\text{H}}_{\text{2}}}]}^1$        (1)

Rearranging the above equation in order to calculate the rate constant;

    $k=\frac{\text{Rate}}{{[{\text{P}}_{\text{NO}}]}^2{[{\text{P}}_{{\text{H}}_{\text{2}}}]}^1}$

Substituting the values for rate and the pressure of the reactants in the above equation, the rate constant of the reaction is calculated as shown below;

    $\begin{array}{c}k=\frac{\text{Rate}}{{[{\text{P}}_{\text{NO}}]}^2{[{\text{P}}_{{\text{H}}_{\text{2}}}]}^1}\\ =\frac{0.160\text{torr/s}}{(289torr){(400torr)}^2}\\ =3.46\times {10}^{-9}{\text{torr}}^{-2}\cdot {\text{s}}^{-1}\end{array}$

Therefore, the rate constant for the reaction is $3.46\times {10}^{-9}{\text{torr}}^{-2}\cdot {\text{s}}^{-1}$.

(c)

Interpretation Introduction

Interpretation:

Activation energy has to be calculated for the reaction of nitrogen monoxide with hydrogen.

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 13.99QE

Activation energy of the reaction is $\text{179}\text{.2}\text{kJ/mol}$.

Explanation of Solution

The relative rate constants and temperature are as follows;

Rate constant	Temperature $\text{(K)}$
$1.00$	$956$
$18.8$	$1099$

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;

    $\begin{array}{c}{E}_a=-\frac{8.314\text{J/mol}\cdot \text{K}\times \text{(956}\text{K})(\text{1099}\text{K})}{1099\text{K}-956\text{K}}\ln \left(\frac{1.00}{18.8}\right)\\ =-\frac{8.314\text{J/mol}\times \text{1050644}\text{K}}{\text{143}\text{K}}\ln (0.05319)\\ =-162372\text{J/mol}\times \text{(-2}\text{.9338)}\\ =179209.1\text{J/mol}\\ \text{=179}\text{.2}\text{kJ/mol}\end{array}$

Therefore, the activation energy of the reaction is $\text{179}\text{.2}\text{kJ/mol}$.