Chapter 13, Problem 13.41QE

Interpretation Introduction

Interpretation:

Rate law and the rate constant for the gas-phase reaction of nitrogen dioxide with ozone has to be given.

Answer to Problem 13.41QE

Rate law for the given reaction is $\text{Rate}=k[\text{N}\text{}{\text{O}}_{\text{2}}][{\text{O}}_{\text{3}}]$ and rate constant is $5.25\times {10}^4\text{L/mol}\cdot \text{s}$.

Explanation of Solution

The reaction between nitrogen dioxide and ozone to produce nitrogen trioxide and oxygen is given be the equation shown below;

    ${\text{NO}}_{\text{2}}(g)+{\text{O}}_{\text{3}}(g)\to {\text{NO}}_{\text{3}}(g)+{\text{O}}_{\text{2}}(g)$

Relative concentration of the reactant is determined by dividing the concentration of each reactant by the smallest concentration 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 of ${\text{NO}}_{\text{2}}$, ${\text{O}}_{\text{3}}$ and also the relative rates of the reaction is given as follows;

Expt	Initial ${\text{[NO}}_{\text{3}}\text{]}(M)$	Initial ${\text{[O}}_{\text{3}}](M)$	Initial rate of the reaction	Relative ${\text{[NO}}_{\text{3}}\text{]}$	Relative ${\text{[O}}_{\text{3}}]$	Relative rates of the reaction
$1$	$2.0\times {10}^{-6}$	$2.0\times {10}^{-6}$	$2.1\times {10}^{-7}$	$1.0$	$1.0$	$1.0$
$2$	$3.0\times {10}^{-6}$	$2.0\times {10}^{-6}$	$3.1\times {10}^{-7}$	$1.5$	$1.0$	$1.5$
$3$	$4.0\times {10}^{-6}$	$3.0\times {10}^{-6}$	$6.2\times {10}^{-7}$	$2.0$	$1.5$	$3.0$
$4$	$4.0\times {10}^{-6}$	$4.0\times {10}^{-6}$	$8.3\times {10}^{-7}$	$2.0$	$2.0$	$4.0$

From the above table, it is found that in experiments 1, and 2, the concentration of ozone remains constant while the relative rate of the reaction increases in relative manner as the concentration of ${\text{NO}}_{\text{2}}$ increases from $1.0$ to $1.5$.  Therefore, the reaction is first order with respect to ${\text{NO}}_{\text{2}}$.

From the above table, it is found that in experiments 3, and 4, the concentration of nitrogen dioxide remains constant while the relative rate of the reaction doubles as the concentration of ${\text{O}}_{\text{3}}$ increases from $1.5$ to $2.0$.  Therefore, the reaction is first order with respect to ${\text{O}}_{\text{3}}$.

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;

    $\begin{array}{l}\text{Rate}\propto [{\text{NO}}_{\text{2}}][{\text{O}}_{\text{3}}]\\ \text{Rate}=k[{\text{NO}}_{\text{2}}][{\text{O}}_{\text{3}}]\end{array}$

Where,

    $k$ is the rate constant.

    $[{\text{NO}}_{\text{2}}]$ is the concentration of ${\text{NO}}_{\text{2}}$.

    $[{\text{O}}_{\text{3}}]$ is the concentration of ${\text{O}}_{\text{3}}$.

Rate constant:

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

    $\text{Rate}=k[{\text{NO}}_{\text{2}}][{\text{O}}_{\text{3}}]$        (1)

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

    $k=\frac{\text{Rate}}{[{\text{NO}}_{\text{2}}][{\text{O}}_{\text{3}}]}$

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

    $\begin{array}{c}k=\frac{\text{Rate}}{[{\text{NO}}_{\text{2}}][{\text{O}}_{\text{3}}]}\\ =\frac{2.1\times {10}^{-7}M\text{/s}}{[\text{2}\text{.0}\times {\text{10}}^{\text{-6}}M][\text{2}\text{.0}\times {\text{10}}^{\text{-6}}M]}\\ =\frac{2.1\times {10}^{-7}M\text{/s}}{[\text{4}\text{.0}\times {\text{10}}^{\text{-12}}{M}^2]}\\ =5.25\times {10}^4{M}^{-1}\text{/s}\\ =5.25\times {10}^4\text{L/mol}\cdot \text{s}\end{array}$

Therefore, the rate constant for the reaction is $5.25\times {10}^4\text{L/mol}\cdot \text{s}$.