Chapter 25, Problem 25.39QP

(a)

Interpretation Introduction

Interpretation:

For the formation of $\text{NO}$, the values of ${\text{ΔG}}^{\text{o}}$, ${\text{K}}_{\text{p}}$ and ${\text{K}}_{\text{c}}$ have to be calculated.

Concept introduction:

Gibb’s free energy: The relationship between temperature and thermodynamic properties like enthalpy and entropy is given by Gibb’s free energy.  Based on the value of Gibb’s free energy the feasibility of reactions can be explained.

$\begin{array}{l}\text{G}\text{=}\text{H}\text{-}\text{TS}\\ \text{G-}\text{Gibb's}\text{free}\text{energy}\\ \text{H-}\text{enthalpy}\\ \text{T-}\text{temperature}\\ \text{S-}\text{entropy}\end{array}$

To calculate: Gibb’s free energy change for the formation of $\text{NO}$

Answer to Problem 25.39QP

Gibb’s free energy change for the given reaction

${\text{ΔG}}_{\text{rxn}}{}^{\text{o}}\left(\text{NO}\right)\text{=}\text{86}\text{.7}\raisebox{1ex}{$\text{kJ}$}\!\left/ \!\raisebox{-1ex}{$\text{mol}$}\right.$

Explanation of Solution

Formation reaction of $\text{NO}$

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

Given that, Gibb’s free energy change of the reaction at 298 K

${\text{ΔG}}^{\text{o}}\text{=}\text{173}\text{.4}\raisebox{1ex}{$\text{kJ}$}\!\left/ \!\raisebox{-1ex}{$\text{mol}$}\right.$

General equation for calculation of Gibb’s free energy change of the given reaction

$\begin{array}{l}{\text{ΔG}}_{\text{rxn}}{}^{\text{o}}\text{=}{\displaystyle \sum {\text{ΔG}}_{\text{f}}{}^{\text{o}}}\left(\text{products}\right)\text{-}{\displaystyle \sum {\text{ΔG}}_{\text{f}}{}^{\text{o}}}\left(\text{reactants}\right)\\ {\text{ΔG}}_{\text{rxn}}{}^{\text{o}}\text{-}\text{free}\text{energy}\text{change}\text{of}\text{the}\text{reaction}\\ {\text{ΔG}}_{\text{f}}{}^{\text{o}}\text{-}\text{free}\text{energy}\text{change}\text{of}\text{formation}\end{array}$

Apply the formula for the given reaction

Substances in elemental form (here oxygen and nitrogen molecule) have Gibb’s free energy of formation value is zero.

$\begin{array}{l}{\text{ΔG}}_{\text{rxn}}{}^{\text{o}}\text{=}{\text{2ΔG}}_{\text{f}}{}^{\text{o}}\left(\text{NO}\right)\text{-}\left[\left(\text{0}\text{+}\text{0}\right)\right]\\ \text{173}\text{.4}\raisebox{1ex}{$\text{kJ}$}\!\left/ \!\raisebox{-1ex}{$\text{mol}$}\right.\text{=}{\text{2ΔG}}_{\text{f}}{}^{\text{o}}\left(\text{NO}\right)\\ {\text{ΔG}}_{\text{f}}{}^{\text{o}}\left(\text{NO}\right)\text{=}\left(\frac{\text{173}\text{.4}}{\text{2}}\right)\raisebox{1ex}{$\text{kJ}$}\!\left/ \!\raisebox{-1ex}{$\text{mol}$}\right.\\ \text{=}\text{86}\text{.7}\raisebox{1ex}{$\text{kJ}$}\!\left/ \!\raisebox{-1ex}{$\text{mol}$}\right.\end{array}$

Gibb’s free energy change for the formation of $\text{NO}$ is calculated as $\text{86}\text{.7}\raisebox{1ex}{$\text{kJ}$}\!\left/ \!\raisebox{-1ex}{$\text{mol}$}\right.$

Gibb’s free energy change for the formation of $\text{NO}$ molecule is calculated by the application of general equation of free energy change.

(b)

Interpretation Introduction

Interpretation:

For the formation of $\text{NO}$, the values of ${\text{ΔG}}^{\text{o}}$, ${\text{K}}_{\text{p}}$ and ${\text{K}}_{\text{c}}$ have to be calculated.

Concept introduction:

Gibb’s free energy: The relationship between temperature and thermodynamic properties like enthalpy and entropy is given by Gibb’s free energy.  Based on the value of Gibb’s free energy the feasibility of reactions can be explained.

$\begin{array}{l}\text{G}\text{=}\text{H}\text{-}\text{TS}\\ \text{G-}\text{Gibb's}\text{free}\text{energy}\\ \text{H-}\text{enthalpy}\\ \text{T-}\text{temperature}\\ \text{S-}\text{entropy}\end{array}$

To calculate: The ${\text{K}}_{\text{p}}$ value for the given reaction

Answer to Problem 25.39QP

For the given reaction, value of ${\text{K}}_{\text{p}}$

${\text{K}}_{\text{p}}\text{=}\text{4}\text{×}{\text{10}}^{\text{-31}}$

Explanation of Solution

The relationship between ${\text{ΔG}}^{\text{o}}$ and ${\text{K}}_{\text{p}}$ is given by

$\begin{array}{l}{\text{ΔG}}^{\text{o}}\text{=}{\text{-RTlnK}}_{\text{p}}\\ \text{R}\text{-}{\text{gas constant; T - temperature; K}}_{\text{p}}\text{ - equilibrium constant at partial pressure}\end{array}$

Substitute the values of ${\text{ΔG}}^{\text{o}}$, $\text{T}$ and $\text{R}$

$\begin{array}{l}\text{173}\text{.4}\times {10}^3\raisebox{1ex}{$\text{J}$}\!\left/ \!\raisebox{-1ex}{$\text{mol}$}\right.\text{=}\text{-}\left(\text{8}\text{.314}\raisebox{1ex}{$\text{J}$}\!\left/ \!\raisebox{-1ex}{$\text{K×}\text{mol}$}\right.\right)\left(\text{298}\text{K}\right){\text{lnK}}_{\text{p}}\\ \\ {\text{K}}_{\text{p}}\text{=}\text{4}{\text{×10}}^{\text{-31}}\end{array}$

By the use of relationship between ${\text{ΔG}}^{\text{o}}$ and ${\text{K}}_{\text{p}}$.  The value of ${\text{K}}_{\text{p}}$ is calculated as $\text{4}\text{×}{\text{10}}^{\text{-31}}$

(c)

Interpretation Introduction

Interpretation:

For the formation of $\text{NO}$, the values of ${\text{ΔG}}^{\text{o}}$, ${\text{K}}_{\text{p}}$ and ${\text{K}}_{\text{c}}$ have to be calculated.

Concept introduction:

Gibb’s free energy: The relationship between temperature and thermodynamic properties like enthalpy and entropy is given by Gibb’s free energy.  Based on the value of Gibb’s free energy the feasibility of reactions can be explained.

$\begin{array}{l}\text{G}\text{=}\text{H}\text{-}\text{TS}\\ \text{G-}\text{Gibb's}\text{free}\text{energy}\\ \text{H-}\text{enthalpy}\\ \text{T-}\text{temperature}\\ \text{S-}\text{entropy}\end{array}$

To calculate: The ${\text{K}}_{\text{c}}$ value for the given reaction

Answer to Problem 25.39QP

For the given reaction value of ${\text{K}}_{\text{c}}$

${\text{K}}_{\text{c}}\text{=}\text{4}\text{×}{\text{10}}^{\text{-31}}$

Explanation of Solution

The relationship between ${\text{K}}_{\text{c}}$ and ${\text{K}}_{\text{p}}$ is given by

$\begin{array}{l}{\text{K}}_{\text{p}}\text{=}{\text{K}}_{\text{c}}{\left(\text{0}\text{.08206T}\right)}^{\text{Δn}}\\ \text{Δn}\text{-}\text{change}\text{in}\text{number}\text{of}\text{moles}\end{array}$

Here no change in number of moles, therefore, $\text{Δn}\text{=}\text{0}$

${\text{K}}_{\text{p}}\text{=}{\text{K}}_{\text{c}}\text{=}\text{4}\text{×}{\text{10}}^{\text{-31}}$

By the use of relationship between ${\text{K}}_{\text{c}}$ and ${\text{K}}_{\text{p}}$.  The value of ${\text{K}}_{\text{c}}$ is calculated as $\text{4}\text{×}{\text{10}}^{\text{-31}}$