Chapter 20, Problem 20.57QP

Assume that the formation of nitrogen dioxide:

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

is an elementary reaction. (a) Write the rate law for this reaction. (b) A sample of air at a certain temperature is contaminated with 2.0 ppm of NO by volume. Under these conditions, can the rate law be simplified? If so, write the simplified rate law. (c) Under the conditions described in (b), the half-life of the reaction has been estimated to be 6.4 × 10³ min. What would the half-life be if the initial concentration of NO were 10 ppm?

Interpretation Introduction

Interpretation:

The rate law for given reaction has to be written.

Concept Introduction:

Rate law:

The rate of a chemical reaction is directly proportional to the initial concentration of reactants.  The rate law for a general reaction is given as

$\begin{array}{l}\text{A}\text{+}\text{B}\underset{}{\overset{}{\to }}\text{product}\\ \\ \text{Rate}\text{=}\text{k}\text{[A][B]}\\ \\ \text{k}\text{is}\text{the}\text{rate}\text{consant}\end{array}$

Explanation of Solution

Given,

$2N{O}_{(g)}+O_2{}_{(g)}\underset{}{\overset{}{\to }}2N{O}_2{}_{(g)}$

The rate law is written as rate constant multiplied by product of concentration of nitrogen oxide and oxygen with coefficients raised to the power.  The rate law is given as

$\mathrm{rate}=k{\left[NO\right]}^2\left[O_2\right]$

Interpretation Introduction

Interpretation:

The simplified rate law for given reaction when air is contaminated with $\text{2}\text{.0ppm}$ of nitrogen oxide has to be written.

Concept Introduction:

Rate law:

The rate of a chemical reaction is directly proportional to the initial concentration of reactants.  The rate law for a general reaction is given as

$\begin{array}{l}\text{A}\text{+}\text{B}\underset{}{\overset{}{\to }}\text{product}\\ \\ \text{Rate}\text{=}\text{k}\text{[A][B]}\\ \\ \text{k}\text{is}\text{the}\text{rate}\text{consant}\end{array}$

Explanation of Solution

Given,

$2N{O}_{(g)}+O_2{}_{(g)}\underset{}{\overset{}{\to }}2N{O}_2{}_{(g)}$

The rate law is written as rate constant multiplied by product of concentration of nitrogen oxide and oxygen with coefficients raised to the power.  The rate law is given as

$\mathrm{rate}=k{\left[NO\right]}^2\left[O_2\right]$

This reaction is appeared as a higher order reaction but actually it follows the second order kinetics, this type of reaction is known as pseudo second order reaction.  Because the concentration of oxygen is larger than nitrogen oxide the reaction becomes pseudo order reaction.  The simplified rate law is given as

$\begin{array}{l}rate=k^{\text{'}}{\left[NO\right]}^2\\ \\ k^{\text{'}}=k[O_2]\end{array}$

Interpretation Introduction

Interpretation:

The half-life of nitrogen oxide has to be calculated if the initial concentration of $\text{NO}$ is $\text{10ppm}$.

Concept Introduction:

Half-life:

The time required for half of a reactant to be reduced in a reaction is said to be half-life.

• Half-life of a reaction is represented by the symbol as ${\text{t}}_{\text{1/2}}$.
• Half-life is discovered by Ernst Rutherford's in 1907 from the original term half-life period.
• For the second order reaction ${\text{t}}_{\text{1/2}}\text{=}\frac{\text{1}}{{\text{k[A}}_{\text{o}}\text{]}}$

Rate law:

The rate of a chemical reaction is directly proportional to the initial concentration of reactants.  The rate law for a general reaction is given as

$\begin{array}{l}\text{A}\text{+}\text{B}\underset{}{\overset{}{\to }}\text{product}\\ \\ \text{Rate}\text{=}\text{k}\text{[A][B]}\\ \\ \text{k}\text{is}\text{the}\text{rate}\text{consant}\end{array}$

Answer to Problem 20.57QP

The half-life of nitrogen oxide from the given initial concentration of nitrogen oxide is $1.3\times 10^{3}min$

Explanation of Solution

Given,

$\begin{array}{l}2N{O}_{(g)}+O_2{}_{(g)}\underset{}{\overset{}{\to }}2N{O}_2{}_{(g)}{t}_{1/2}=6.4\times 10^{3}min\\ \\ [NO]=10ppm\end{array}$

The rate law is written as rate constant multiplied by product of concentration of nitrogen oxide and oxygen with coefficients raised to the power.  The rate law is given as

$\mathrm{rate}=k{\left[NO\right]}^2\left[O_2\right]$

The rate law of a pseudo order reaction is given as

$\begin{array}{l}rate=k^{\text{'}}{\left[NO\right]}^2\\ \\ k^{\text{'}}=k[O_2]\end{array}$

Half-life for a second order reaction is given as,

$\begin{array}{l}{t}_{1/2}=\frac{1}{k{\left[A\right]}_0}\\ \\ {\left[A\right]}_{0}isinitialconcentrationofthereac\tan t.\end{array}$

The half-life of nitrogen oxide is determined as,

$\begin{array}{l}\frac{(t_{1/2})}{{(t_{1/2})}_2}=\frac{\left[{(A_0)}_2\right]}{\left[{(A_0)}_1\right]}\\ \\ \frac{6.4\times 10^{3}min}{{(t_{1/2})}_2}=\frac{10ppm}{2ppm}\\ \\ {(}_{t_{1/2}}=1.3\times 10^{3}min\end{array}$