Chapter 21, Problem 84AP

Interpretation Introduction

Interpretation:

The rate law for given reaction to be written and simplificationf rate law to be checked, along with the half-lifeaccording to the given conditions, is to be determined.

Concept introduction:

The study of reaction rates, rearrangement of atoms, and the effect of various variables in a chemical reaction is known as chemical kinetics

A half-life is the quantity of time it takes for half of a substance to undergo some specified process.

The half-life for the second order reaction is as follows:

$t_{\frac{1}{2}}=\frac{1}{k{\left[\text{A}\right]}_0}$

The rate law for a chemical reactionis defined as an equation that relates the reaction rate with the concentrations or partial pressures of the reactants.

The rate law is given by:

$\text{r=k}{\left[\text{A}\right]}^x{\left[\text{B}\right]}^y$

Answer to Problem 84AP

Solution:

a) $\text{Rate}=k{\left[\text{NO}\right]}^2\left[{\text{O}}_2\right]$

b) $\text{Rate}=k\text{'}{\left[\text{NO}\right]}^2$

c) ${\left(t_{\frac{1}{2}}\right)}_2=1.3\times {10}^3\text{ min}$

Explanation of Solution

a) The rate law for the reaction

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

The rate law for a given reaction is as follows:

$\text{Rate}=k{\left[\text{NO}\right]}^2\left[{\text{O}}_2\right]$

Here, $k$ is the rate constant for the reaction.

It suggests that two molecules of NO are colliding with a molecule of oxygen and forming nitrogen dioxide. This reaction has two or more elementary steps.

Explanation:

b) The rate law can be simplified is to be checked and if so simplified rate law to be written.

A sample of air at a certain temperature is contaminated with $2.0\text{ ppm}$ of $\left[\text{NO}\right]$ by volume.

In thegivenreaction, two molecules of $\left[\text{NO}\right]$ are colliding with a molecule of oxygen, but the size of $\left[{\text{O}}_2\right]$ is larger than that of $\left[\text{NO}\right]$, due to which it acts as a pseudo-second-order reaction.

The rate law is given as follows:

$\text{Rate}=k\text{'}{\left[\text{NO}\right]}^2$

Here, $k\text{'}$ is equal to $k\left[{\text{O}}_2\right]$ and $k,k\text{'}$ are the rate constantsfor the reaction.

Given information: A sample of air at a certain temperature is contaminated with $2.0\text{ ppm}$ of $\left[\text{NO}\right]$ by volume, and the half-life of the reaction has been estimated to be $6.4\times {10}^3\text{}\text{min}$. The initial concentration of $\left[\text{NO}\right]$ is $10\text{ppm}$.

Explanation:

c) The half-lifeto be estimated if the initial concentration of NO were 10 ppm

As this reaction is a second-order reaction, the half-life time for this reaction is written as follows:

$t_{\frac{1}{2}}=\frac{1}{k{\left[\text{A}\right]}_0}$

Here, $t_{\frac{1}{2}}$ is the half-life time of reaction, $k$ is rate constant, and ${\left[\text{A}\right]}_0$ is the initial concentration of the reactant.

The sample is contaminated with $2.0\text{ ppm}$ of $\left[\text{NO}\right]$ by volume $\left({\left[{\left(\text{A}\right)}_0\right]}_1\right)$ and the half-life of the reaction has been estimated to be $6.4\times {10}^3\text{}\text{min}$ ${\left(t_{\frac{1}{2}}\right)}_1$.

The expression of half-life, by substituting these values in the above equation, can be written as follows:

$\begin{array}{c}{\left(t_{\frac{1}{2}}\right)}_1=\frac{1}{k{\left[{\left(\text{A}\right)}_0\right]}_1}\\ 6.4\times {10}^3\text{}\text{min}=\frac{1}{k{\left[2.0\text{ ppm}\right]}_{}}\end{array}$

The initial concentration of $\left[\text{NO}\right]$ $\left({\left[{\left(\text{A}\right)}_0\right]}_2\right)$ is $10\text{ppm}$. The expression of half-life, by substituting these values in the above equation, can be written as follows:

$\begin{array}{c}{\left(t_{\frac{1}{2}}\right)}_2=\frac{1}{k{\left[{\left(\text{A}\right)}_0\right]}_2}\\ {\left(t_{\frac{1}{2}}\right)}_2=\frac{1}{k{\left[\text{10 ppm}\right]}_2}\end{array}$

By dividing ${\left(t_{\frac{1}{2}}\right)}_1$ and ${\left(t_{\frac{1}{2}}\right)}_2$ as follows:

$\frac{{\left(t_{\frac{1}{2}}\right)}_1}{{\left(t_{\frac{1}{2}}\right)}_2}=\frac{\frac{1}{k{\left[{\left(\text{A}\right)}_0\right]}_1}}{\frac{1}{k{\left[{\left(\text{A}\right)}_0\right]}_2}}$

$\frac{{\left(t_{\frac{1}{2}}\right)}_1}{{\left(t_{\frac{1}{2}}\right)}_2}=\frac{\cancel{k}{\left[{\left(\text{A}\right)}_0\right]}_2}{\cancel{k}{\left[{\left(\text{A}\right)}_0\right]}_1}$

Substitute the values of ${\left(t_{\frac{1}{2}}\right)}_1$, $\left({\left[{\left(\text{A}\right)}_0\right]}_1\right)$, and $\left({\left[{\left(\text{A}\right)}_0\right]}_2\right)$ in theabove expression, as follows:

$\begin{array}{c}\frac{6.4\times {10}^3\text{}\text{min}}{{\left(t_{\frac{1}{2}}\right)}_2}=\frac{10\cancel{\text{ppm}}}{2.0\cancel{\text{ppm}}}\\ {\left(t_{\frac{1}{2}}\right)}_2=1.3\times {10}^3\text{ min}\end{array}$

Therefore, the half-life for an initial concentration of $\left[\text{NO}\right]$ of $10\text{ppm}$ is $1.3\times {10}^3\text{ min}$.