Chapter 25, Problem 25.63QP

Interpretation Introduction

Interpretation:

For the formation of ${\text{NO}}_{\text{2}}$, the values of ${\text{ΔG}}^{\text{o}}$, ${\text{K}}_{\text{p}}$ and ${\text{K}}_{\text{c}}$ have to be calculated.

Answer to Problem 25.63QP

• Gibb’s free energy change for the given reaction

${\text{ΔG}}^{\text{o}}\text{=}\text{-198}\text{.3}\raisebox{1ex}{$\text{kJ}$}\!\left/ \!\raisebox{-1ex}{$\text{mol}$}\right.$

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

${\text{K}}_{\text{p}}\text{=}\text{6}\text{×}{\text{10}}^{\text{34}}$

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

${\text{K}}_{\text{C}}\text{=}\text{6}\text{×}{\text{10}}^{\text{34}}$

Explanation of Solution

To calculate: Gibb’s free energy change for the formation of ${\text{NO}}_{\text{2}}$

Formation reaction of ${\text{NO}}_{\text{2}}$

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

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

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

Apply the formula for the given reaction

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

$\begin{array}{l}{\text{ΔG}}^{\text{o}}\text{=}{\text{ΔG}}_{\text{f}}{}^{\text{o}}\left({\text{NO}}_{\text{2}}\right)\text{+}{\text{ΔG}}_{\text{f}}{}^{\text{o}}\left({\text{O}}_{\text{2}}\right)\text{-}\left[{\text{ΔG}}_{\text{f}}{}^{\text{o}}\left(\text{NO}\right)\text{+}{\text{ΔG}}_{\text{f}}{}^{\text{o}}\left({\text{O}}_{\text{3}}\right)\right]\\ {\text{ΔG}}^{\text{o}}\text{=}\left(\left(\text{1}\right)\left(\text{51}\text{.8}\right)\text{+}\left(\text{0}\right)\text{-}\left[\left(\text{1}\right)\left(\text{86}\text{.7}\right)\text{+}\left(\text{1}\right)\left(\text{163}\text{.4}\right)\right]\right)\raisebox{1ex}{$\text{kJ}$}\!\left/ \!\raisebox{-1ex}{$\text{mol}$}\right.\\ \text{=}\text{-}\text{198}\text{.3}\raisebox{1ex}{$\text{kJ}$}\!\left/ \!\raisebox{-1ex}{$\text{mol}$}\right.\end{array}$

Gibb’s free energy change for the formation of ${\text{NO}}_{\text{2}}$ is calculated as $\text{-198}\text{.3}\raisebox{1ex}{$\text{kJ}$}\!\left/ \!\raisebox{-1ex}{$\text{mol}$}\right.$

Gibb’s free energy change for the formation of ${\text{NO}}_{\text{2}}$ molecule is calculated by the application of general equation of free energy change.

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

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

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

Modify the above equation

${\text{lnK}}_{\text{p}}\text{=}\frac{{\text{-ΔG}}^{\text{o}}}{\text{RT}}$

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

$\begin{array}{l}{\text{lnK}}_{\text{p}}\text{=}\text{-}\frac{\text{198}\text{.3}\text{×}{\text{10}}^{\text{3}}\raisebox{1ex}{$\text{J}$}\!\left/ \!\raisebox{-1ex}{$\text{mol}$}\right.}{\left(\text{8}\text{.314}\raisebox{1ex}{$\text{J}$}\!\left/ \!\raisebox{-1ex}{$\text{K×}\text{mol}$}\right.\right)\left(\text{298}\text{K}\right)}\\ {\text{K}}_{\text{p}}\text{=}\text{6}{\text{×10}}^{\text{34}}\end{array}$

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 $\text{6}\text{×}{\text{10}}^{\text{34}}$

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

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, $\text{Δn}\text{=}\text{0}$

${\text{K}}_{\text{p}}\text{=}{\text{K}}_{\text{c}}\text{=}\text{6}\text{×}{\text{10}}^{\text{34}}$

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 $\text{6}\text{×}{\text{10}}^{\text{34}}$

Conclusion

For the formation of ${\text{NO}}_{\text{2}}$, the values of ${\text{ΔG}}^{\text{o}}$, ${\text{K}}_{\text{p}}$ and ${\text{K}}_{\text{c}}$ were calculated.