Chapter 22, Problem 22.69P

(a)

Interpretation Introduction

Interpretation:

The chemical equation of nitric oxide with ozone and its reverse reaction has to be written.

Explanation of Solution

In the stratosphere, nitric oxide destroys ozone by forming nitrogen dioxide and oxygen.  The chemical equation is given by

    $\begin{array}{l}{\text{NO}}_{\text{(g)}}{\text{ + O}}_{\text{3(g)}}\underset{}{\overset{}{\rightleftarrows }}{\text{ NO}}_{\text{2(g) }}{\text{+ O}}_{\text{2(g)}}\\ \\ {\text{NO}}_{\text{2(g)}}{\text{ + O}}_{\text{2(g)}}\underset{}{\overset{}{\rightleftarrows }}{\text{ NO}}_{\text{(g)}}{\text{ + O}}_{\text{3(g)}}\text{ (reverse)}\end{array}$

(b)

Interpretation Introduction

Interpretation:

The general rate law for given reaction has to be written.

Explanation of Solution

The rate of the reaction depends on concentration of the reactants and the order of the reaction translates to the exponent in the rate law.

    $\begin{array}{l}{\text{NO}}_{\text{(g)}}{\text{ + O}}_{\text{3(g)}}\underset{}{\overset{}{\rightleftarrows }}{\text{ NO}}_{\text{2(g) }}{\text{+ O}}_{\text{2(g)}}{\text{Rate}}_{\text{forward}}{\text{ = k}}_{\text{f}}{\text{[NO][O}}_{\text{3}}\text{]}\\ \\ {\text{NO}}_{\text{2(g)}}{\text{ + O}}_{\text{2(g)}}\underset{}{\overset{}{\rightleftarrows }}{\text{ NO}}_{\text{(g)}}{\text{ + O}}_{\text{3(g)}}\text{ (reverse)}{\text{Rate}}_{\text{reverse}}{\text{ = k}}_{\text{r}}{\text{[NO}}_{\text{2}}{\text{][O}}_{\text{2}}\text{]}\end{array}$

(c)

Interpretation Introduction

Interpretation:

The standard free energy of given reaction has to be written.

Concept Introduction: 

Free energy, ${\text{ΔG}}^{\text{ο}}$, is the measure of the ability of a reaction to proceed spontaneously. ${\text{ΔG}}^{\text{ο}}$ can be calculated with the relationship.

    ${\text{ΔG}}^{\text{ο}}{}_{\text{rxn}}\text{=}{\displaystyle \sum _{}^{}{\text{mΔG}}_{{}^{\text{f}}\text{(products)}}^{\text{o}}\text{-}}{\displaystyle \sum _{}^{}{\text{mΔG}}_{{}^{\text{f}}\text{(reactants)}}^{\text{o}}}$

    ${\text{ΔG}}^{\text{ο}}{}_{\text{rxn}}\text{=}{\text{ΔH}}_{\text{rxn}}^{\text{o}}\text{-}{\text{TΔS}}_{\text{rxn}}^{\text{o}}$

Explanation of Solution

The standard Gibb’s free energy is calculated by substituting change in enthalpy and change in entropy of the reaction.

    $\begin{array}{l}{\text{ΔH}}^{\text{ο}}{}_{\text{rxn}}\text{=}{\displaystyle \sum _{}^{}{\text{mΔH}}_{{}^{\text{f}}\text{(products)}}^{\text{o}}\text{-}}{\displaystyle \sum _{}^{}{\text{mΔH}}_{{}^{\text{f}}\text{(reactants)}}^{\text{o}}}\\ \\ {\text{ΔH}}^{\text{ο}}{}_{\text{rxn}}\text{=}{\text{[(1 mol NO}}_{\text{2}}\text{)(33}{\text{.2 kJ/mol) + (1 mol O}}_{\text{2}}\text{)(0 kJ/mol)]}\\ \\ \text{-}\text{[(1 mol NO)(90}{\text{.29 kJ/mol) + (1 mol O}}_{\text{3}}\text{)(143 kJ/mol)]}\\ \\ {\text{ΔH}}^{\text{ο}}{}_{\text{rxn}}\text{=}\text{-200}\text{.09 kJ}\\ \\ {\text{ΔS}}^{\text{ο}}{}_{\text{rxn}}\text{=}{\text{[(1 mol NO}}_{\text{2}}\text{)(239}\text{.9 J/mol}\text{.K)+}{\text{(1 mol O}}_{\text{2}}\text{)(205}\text{.0 J/mol}\text{.K)]-}\\ \\ \text{[(1 mol NO)(210}\text{.65 J/mol}\text{.K)}\text{+}{\text{(1 mol O}}_{\text{3}}\text{)(238}\text{.82 J/mol}\text{.K)]}\\ \\ {\text{ΔS}}^{\text{ο}}{}_{\text{rxn}}\text{=}\text{-4}\text{.57 J/K}\\ \\ {\text{ΔG}}^{\text{ο}}{}_{\text{rxn}}\text{=}{\text{ΔH}}_{\text{rxn}}^{\text{o}}\text{-}{\text{TΔS}}_{\text{rxn}}^{\text{o}}\text{=}\text{(280}\text{. K)(-4}{\text{.57J/K)(1kJ/10}}^{\text{3}}\text{J)}\text{=}\text{-199kJ}\end{array}$

(d)

Interpretation Introduction

Interpretation:

The ratio of rate constants is consistent with equilibrium constant at given temperature has to be calculated.

Concept Introduction: 

Free energy, ${\text{ΔG}}^{\text{ο}}$, is the measure of the ability of a reaction to proceed spontaneously. ${\text{ΔG}}^{\text{ο}}$ can be calculated with the relationship.

    ${\text{ΔG}}^{\text{ο}}{}_{\text{rxn}}\text{=}{\displaystyle \sum _{}^{}{\text{mΔG}}_{{}^{\text{f}}\text{(products)}}^{\text{o}}\text{-}}{\displaystyle \sum _{}^{}{\text{mΔG}}_{{}^{\text{f}}\text{(reactants)}}^{\text{o}}}$

    ${\text{ΔG}}^{\text{ο}}{}_{\text{rxn}}\text{=}{\text{ΔH}}_{\text{rxn}}^{\text{o}}\text{-}{\text{TΔS}}_{\text{rxn}}^{\text{o}}$

Explanation of Solution

At equilibrium, the forward rate equals the reverse rate and their ratio equals the equilibrium constant: ${\text{k}}_{\text{f}}{\text{/k}}_{\text{r}}{\text{ =K}}_{\text{eq}}$.  Both $\text{ΔG}\text{and}\text{Keq}$ express the extent to which a reaction proceeds and are related by the equation $\text{ΔG=}{\text{–RTlnK}}_{\text{eq}}$

    $\begin{array}{l}\text{ln K =}\frac{{\text{ΔG}}^{\text{o}}}{\text{-RT}}\text{=}\left(\frac{\text{-198}\text{.810 kJ/mol}}{\text{-(8}\text{.314J/mol}\text{.K)(280K)}}\right)\left(\frac{\text{1000J}}{\text{1kJ}}\right)\text{=}\text{85}\text{.402591}\\ \\ \text{K}\text{=}\text{1}{\text{.2x10}}^{\text{37}}\end{array}$

Therefore, the ratio of rate constants is $\text{1}{\text{.2x10}}^{\text{37}}$ and the forward rate is much faster than the reverse rate.