Chapter 18, Problem 18.27QP

(a)

Interpretation Introduction

Interpretation: From the given data of ${\text{ΔG}}_{\text{f}}^{\text{o}}$, ${\text{ΔG}}^{\text{o}}\text{and}{\text{K}}_{\text{p}}$ have to be calculated.

Concept Introduction:

Gibbs free energy: The amount of energy that is required to perform the maximum amount of reversible work by a system at constant temperature and pressure.

$\text{Symbol = G, Unit = Joule }\left(\text{J}\right)\text{/mol}\text{.}$

${\text{ΔG}}^{\text{o}}$-Standard Gibbs free energy change.

Gibbs free energy change of formation is the change of Gibbs free energy that is needed for the formation of 1 mole of the substance in its standard state from its constituent elements.

${\text{ΔG}}_{\text{f}}^{\text{o}}$-Standard Gibbs free energy change of formation.

Relation between $\Delta G^o$and$\Delta G_f^o$:

${\text{ΔG}}_{}^{\text{o}}\text{=}{\displaystyle \sum \text{mΔ}{\text{G}}_{\text{f}}^{\text{o}}\left(\text{products}\right)}\text{-}{\displaystyle \sum \text{nΔ}{\text{G}}_{\text{f}}^{\text{o}}\left(\text{reactants}\right)}$

$\begin{array}{l}\text{where,}\\ {\displaystyle \sum \text{mΔ}{\text{G}}_{\text{f}}^{\text{o}}\left(\text{products}\right)}\text{is}\text{the}\text{total}\text{sum}\text{of}\text{Gibbs}\text{free}\text{energy}\text{change}\text{of}\text{products}\text{.}\\ {\displaystyle \sum \text{nΔ}{\text{G}}_{\text{f}}^{\text{o}}\left(\text{reactants}\right)}\text{is}\text{the}\text{total}\text{sum}\text{of}\text{Gibbs}\text{free}\text{energy}\text{change}\text{of}\text{reactants}\text{. }\\ {\text{ΔG}}_{\text{f}}^{\text{o}}\text{is}\text{the}\text{standard}\text{change}\text{in}\text{the}\text{gibbs}\text{free}\text{energy}\text{of}\text{formation}\text{.}\end{array}$

$\begin{array}{l}\text{m}\text{is the number of moles of the products}\text{.}\\ \text{n}\text{is the number of moles of the reactants}\text{.}\end{array}$

Relation between $\Delta G^o$and$K_p$:

${\text{ΔG}}_{}^{\text{o}}\text{=}{\text{-RTlnK}}_{\text{p}}$

Where,

$\begin{array}{c}\text{R=}\text{Universal}\text{gas}\text{constant}\text{.}\\ \text{=}\text{8}\text{.314}{\text{JK}}^{-1}{\text{mol}}^{\text{-1}}\\ \text{T}\text{=}\text{Temperature}\text{.}\\ {\text{K}}_{\text{p}}\text{=}\text{Equilibrium}\text{constant}\text{.}\end{array}$

Answer to Problem 18.27QP

${\text{ΔG}}_{}^{\text{o}}\text{=}\text{39}000\text{J/mol}$

${\text{K}}_{\text{p}}=1.4578\times {10}^{-07}$

Explanation of Solution

The given reaction is:

${\text{PCl}}_{\text{5}}\left(\text{g}\right)\stackrel{}{\rightleftharpoons }{\text{PCl}}_{\text{3}}\left(\text{g}\right)\text{+}{\text{Cl}}_{\text{2}}\left(\text{g}\right)$

Given:

${\text{ΔG}}_{\text{f}}^{\text{o}}$ for ${\text{Cl}}_{\text{2}}\left(\text{g}\right)$ is 0.

${\text{ΔG}}_{\text{f}}^{\text{o}}$ for ${\text{PCl}}_{\text{3}}\left(\text{g}\right)$ is $-286\text{kJ/mol}$

${\text{ΔG}}_{\text{f}}^{\text{o}}$ for ${\text{PCl}}_5\left(\text{g}\right)$ is $-325\text{kJ/mol}$

Substituting these values in the formula shown below:

$\begin{array}{c}{\text{ΔG}}_{}^{\text{o}}\text{=}{\displaystyle \sum \text{mΔ}{\text{G}}_{\text{f}}^{\text{o}}\left(\text{products}\right)}\text{-}{\displaystyle \sum \text{nΔ}{\text{G}}_{\text{f}}^{\text{o}}\left(\text{reactants}\right)}\\ =\left[\left(\text{1}\times \text{-286}\text{kJ/mol}\right)+\left(\text{1}\times \text{0}\text{kJ/mol}\right)\right]-\left(\text{1}\times \text{-325}\text{kJ/mol}\right)\\ =\text{-286}\text{kJ/mol}+\text{325}\text{kJ/mol}\\ \text{=}\text{39}\text{kJ/mol}\end{array}$

Therefore, ${\text{ΔG}}_{}^{\text{o}}\text{=}\text{39}\text{kJ/mol}$

Converting ${\text{ΔG}}_{}^{\text{o}}$ from $\text{kJ/mol}$ to $\text{J/mol}$:

$\begin{array}{c}\text{1}\text{kJ}=\text{1000}\text{J}\\ \text{39}\text{kJ/mol}=\frac{\text{1000}\text{J}}{\text{1}\text{kJ}}\times \text{39}\text{kJ/mol}\\ =\text{39}000\text{J/mol}\end{array}$

Therefore, ${\text{ΔG}}_{}^{\text{o}}\text{=}\text{39}000\text{J/mol}$

Given: $\text{T}\text{=}\text{Temperature=}{\text{25}}^{\text{o}}\text{C}$

Converting the temperature from ${}^{\text{o}}\text{C}$ into Kelvin:

$\begin{array}{c}\text{T}={\text{25}}^{\text{o}}\text{C}+273\\ =\text{298}\text{K}\end{array}$

It is known that ${\text{ΔG}}_{}^{\text{o}}\text{=}{\text{-RTlnK}}_{\text{p}}$

This expression is rearranged as follows to find ${\text{K}}_{\text{p}}$:

$\begin{array}{c}{\text{K}}_{\text{p}}=\exp \left(-\frac{{\text{ΔG}}_{}^{\text{o}}}{\text{RT}}\right)\\ =\exp \left(-\frac{\text{39}000\text{J/mol}}{\text{8}\text{.314}{\text{JK}}^{-1}{\text{mol}}^{\text{-1}}\times \text{298}\text{K}}\right)\\ =1.4578\times {10}^{-07}\end{array}$

Therefore, ${\text{K}}_{\text{p}}=1.4578\times {10}^{-07}$

Conclusion

From the given data of ${\text{ΔG}}_{\text{f}}^{\text{o}}$, ${\text{ΔG}}^{\text{o}}\text{and}{\text{K}}_{\text{p}}$ have been calculated.

(b)

Interpretation Introduction

Interpretation: From the given data of partial pressures, $\text{ΔG}$ has to be calculated.

Concept Introduction:

Partial pressure: The hypothetical pressure occupied by a gas as alone from the entire volume of the original mixture of gases at the same temperature is said to be the partial pressure of that gas.

Gibbs free energy: The amount of energy that is required to perform the maximum amount of reversible work by a system at constant temperature and pressure.

$\text{Symbol = G, Unit = Joule }\left(\text{J}\right)\text{/mol}\text{.}$

$\text{ΔG}$- Gibbs free energy change.

Relation between $\Delta G^o$and$K_p$:

$\text{ΔG}\text{=}\text{-RTlnK}$

Where,

$\begin{array}{c}\text{R=}\text{Universal}\text{gas}\text{constant}\text{.}\\ \text{=}\text{8}\text{.314}{\text{JK}}^{-1}{\text{mol}}^{\text{-1}}\\ \text{T}\text{=}\text{Temperature}\text{.}\\ \text{K=}\text{Equilibrium}\text{constant}\text{.}\end{array}$

The equilibrium constant for gaseous reactions can be either as ${\text{K}}_{\text{p}}$ or as ${\text{K}}_{\text{c}}$. ${\text{K}}_{\text{p}}$ is defined as equilibrium constant of a reaction in terms of partial pressures whereas ${\text{K}}_{\text{c}}$ is defined as equilibrium constant of a reaction in terms of concentration.

Answer to Problem 18.27QP

$\text{ΔG}\text{=}-92268.25{\text{Jmol}}^{\text{-1}}$

Explanation of Solution

The given reaction is:

${\text{PCl}}_{\text{5}}\left(\text{g}\right)\stackrel{}{\rightleftharpoons }{\text{PCl}}_{\text{3}}\left(\text{g}\right)\text{+}{\text{Cl}}_{\text{2}}\left(\text{g}\right)$

Given data:

$\begin{array}{c}{\text{P}}_{{\text{PCl}}_{\text{5}}}\text{=}\text{0}\text{.0029}\text{atm}\\ {\text{P}}_{{\text{PCl}}_{\text{3}}}\text{=}\text{0}\text{.27}\text{atm}\\ {\text{P}}_{{\text{Cl}}_{\text{2}}}\text{=}\text{0}\text{.40}\text{atm}\end{array}$

Finding ${\text{K}}_{\text{p}}$:

$\begin{array}{c}{\text{K}}_{\text{p}}=\frac{{\text{P}}_{{\text{PCl}}_{\text{3}}}\times {\text{P}}_{{\text{Cl}}_{\text{2}}}}{{\text{P}}_{{\text{PCl}}_{\text{5}}}}\\ =\frac{\text{0}\text{.27}\text{atm}\times \text{0}\text{.40}\text{atm}}{\text{0}\text{.0029}\text{atm}}\\ =37.2414\end{array}$

Therefore, ${\text{K}}_{\text{p}}=37.2414$

Given: $\text{T}\text{=}\text{Temperature=}{\text{25}}^{\text{o}}\text{C}$

Converting the temperature from ${}^{\text{o}}\text{C}$ into Kelvin:

$\begin{array}{c}\text{T}={\text{25}}^{\text{o}}\text{C}+273\\ =\text{298}\text{K}\end{array}$

It is known that $\text{ΔG}\text{=}\text{-RTlnK}$

The equilibrium constant in the expression of $\text{ΔG}$ is considered to be the equilibrium constant in terms of partial pressure which means $\text{K}$ is considered as ${\text{K}}_{\text{p}}$.

Finding $\text{ΔG}$ by substituting ${\text{K}}_{\text{p}}$ in the expression of $\text{ΔG}$:

$\begin{array}{c}\text{ΔG}\text{=}\text{-RTlnK}\\ =-\text{8}\text{.314}{\text{JK}}^{-1}{\text{mol}}^{\text{-1}}\times \text{298}\text{K}\times 37.2414\\ =-92268.25{\text{Jmol}}^{\text{-1}}\end{array}$

Therefore, $\text{ΔG}\text{=}-92268.25{\text{Jmol}}^{\text{-1}}$