Chapter 18, Problem 18.65QP

(a)

Interpretation Introduction

Interpretation:

Given the heterogeneous equilibrium reaction free energy $({\text{ΔG}}^{\text{ο}})$ value and equilibrium pressure ($Kp$) values should be calculated at ${\text{25}}^{°}\text{C}$.

Concept Introduction:

Free energy (Gibbs free energy) is the term that is used to explain the total energy content in a thermodynamic system that can be converted into work.  The free energy is represented by the letter $\text{G}$.  All spontaneous process is associated with the decrease of free energy in the system.  The standard free energy change ${\text{(ΔG}}^{\text{°}}{}_{\text{rxn}})$ is the difference in free energy of the reactants and products in their standard state.

${\text{ΔG}}^{\text{°}}{}_{\text{rxn}}\text{=}\sum {\text{nΔG}}_{\text{f}}{}^{\text{°}}\text{(Products)-}\sum {\text{nΔG}}_{\text{f}}{}^{\text{°}}\text{(Reactants)}$

Where, $\text{"n"}$ is the number of moles

    $\begin{array}{l}{\text{ΔG}}^{\text{0}}\text{=-RTln}\text{K}\\ \text{ΔG}=Freeenergy\\ {\text{ΔG}}^{\text{0}}=S\tan dard-statefreeenergy\\ \text{R}\text{=}\text{Gas}\text{Constant}(\text{0}\text{.0826}\text{l}\text{.atm/K}\text{.atm})\\ T=\text{Temprature}\text{273}\text{K}\\ \text{K=}\text{Equlibrium}\text{Constant}{\text{(K}}_{\text{P}}\text{and}{\text{K}}_{\text{C}}\text{)}\end{array}$

Explanation of Solution

Calculate the Gibbs free $({\text{ΔG}}^0)$ and partial pressure ($Kp$) values for given reaction (a).

$\begin{array}{l}(a).H_2(g)+B{r}_2(l)\rightleftharpoons 2HBr(g)\\ {\text{ΔG}}_{\text{rxn}}^{°}={\displaystyle \sum {\text{nΔG}}_{\text{f}}^{\text{°}}\text{(Products)}}\text{-}{\displaystyle \sum {\text{mΔG}}_{\text{f}}^{\text{°}}\text{(Reactants})}\\ {\text{ΔG}}^{\text{0}}{}_{\text{rxn}}=2{\text{ΔG}}_{\text{f}}^0\text{(HBr)}-\text{Δ}{\text{G}}_{\text{f}}^0{\text{(H}}_{\text{2}})-\text{Δ}{\text{G}}_{\text{f}}^0{\text{(Br}}_{\text{2}})\\ {\text{ΔG}}^{\text{0}}{}_{\text{rxn}}=2(-53.2\text{kJ/mol})-(1)(0)-(1)(0)\\ (\because \text{The respactive}{\text{ΔG}}_{\text{f}}^0\text{kJ/mol}valuestakingfrom\text{Appendix-2})\\ {\text{ΔG}}^{\text{0}}{}_{\text{rxn}}=-106.4\text{kJ/mol}\end{array}$

Substitute $\Delta G°,R,andT$ value in the fallowing equation to solve for $Kp$

$\begin{array}{l}{\text{ΔG}}^{\text{0}}\text{=-RTln}\text{K---------[1]}\\ \text{Let}\text{us consider following equation (1)}\\ {\text{ΔG}}^{°}=-106.4\text{KJ/mol}\\ \Delta {\text{G}}^{°}\text{values substituted equation (1) and rewrite the above equation}\\ \text{InK}\text{p=}\frac{{\text{ΔG}}^{°}}{\text{-RT}}=\frac{\text{-(106}{\text{.4×10}}^3\text{J/mol)}}{\text{(8}\text{.314}\text{J/K}\text{mol)(298}\text{K)}}\\ \text{InK}{}_{\text{P}}\text{=}\frac{{\text{ΔG}}^{\text{0}}}{\text{-RT}}=\frac{-106.4{\text{×10}}^3\text{J/mol}}{(2477.572)}=42.9(\because Here\text{R=}\text{8}\text{.314,}{\text{T=273+25}}^{\text{0}}\text{C=298K)}\\ \text{Kp=}{e}^{42.9}\\ \text{Kp=}4\times {10}^{18}\end{array}$

(b)

Interpretation Introduction

Interpretation:

Given the heterogeneous equilibrium reaction free energy $({\text{ΔG}}^{\text{ο}})$ value and equilibrium pressure ($Kp$) values should be calculated at ${\text{25}}^{°}\text{C}$.

Concept Introduction:

Free energy (Gibbs free energy) is the term that is used to explain the total energy content in a thermodynamic system that can be converted into work.  The free energy is represented by the letter $\text{G}$.  All spontaneous process is associated with the decrease of free energy in the system.  The standard free energy change ${\text{(ΔG}}^{\text{°}}{}_{\text{rxn}})$ is the difference in free energy of the reactants and products in their standard state.

${\text{ΔG}}^{\text{°}}{}_{\text{rxn}}\text{=}\sum {\text{nΔG}}_{\text{f}}{}^{\text{°}}\text{(Products)-}\sum {\text{nΔG}}_{\text{f}}{}^{\text{°}}\text{(Reactants)}$

Where, $\text{"n"}$ is the number of moles

    $\begin{array}{l}{\text{ΔG}}^{\text{0}}\text{=-RTln}\text{K}\\ \text{ΔG}=Freeenergy\\ {\text{ΔG}}^{\text{0}}=S\tan dard-statefreeenergy\\ \text{R}\text{=}\text{Gas}\text{Constant}(\text{0}\text{.0826}\text{l}\text{.atm/K}\text{.atm})\\ T=\text{Temprature}\text{273}\text{K}\\ \text{K=}\text{Equlibrium}\text{Constant}{\text{(K}}_{\text{P}}\text{and}{\text{K}}_{\text{C}}\text{)}\end{array}$

Explanation of Solution

Calculate the standard entropy $({\text{ΔG}}^0)$ and partial pressure ($Kp$) values for given reaction

$\begin{array}{l}\frac{\text{1}}{\text{2}}{\text{H}}_{\text{2}}(\text{g})\text{+}\frac{\text{1}}{\text{2}}{\text{Br}}_{\text{2}}\text{(l)}\rightleftharpoons \text{HBr}\text{(g)}\\ {\text{ΔG}}_{\text{rxn}}^{°}={\displaystyle \sum {\text{nΔG}}_{\text{f}}^{\text{°}}\text{(Products)}}\text{-}{\displaystyle \sum {\text{mΔG}}_{\text{f}}^{\text{°}}(Reac\tan ts)}\\ {\text{ΔG}}_{\text{rxn}}^{°}={\text{ΔG}}_{\text{f}}^{\text{°}}\text{(HBr)}-\left[\frac{1}{2}{\text{ΔG}}_{\text{f}}^{\text{°}}{\text{(H}}_{\text{2}})+\frac{1}{2}\text{Δ}{\text{G}}_{\text{f}}^0{\text{(Br}}_{\text{2}})\right]\\ {\text{ΔG}}_{\text{rxn}}^{°}=1(-53.2KJ/mol)-\left(\frac{1}{2}\right)(0)-\left(\frac{1}{2}\right)(0)\\ (\because \text{The respactive}{\text{ΔG}}_{\text{f}}^0,KJ/molvaluestakingfromequilibrium\text{Appendix-2})\\ {\text{ΔG}}_{}^{°}=-53.2\text{KJ/mol}\end{array}$

Next we calculate the partial pressure values ($Kp$)

$\begin{array}{l}{\text{ΔG}}^{°}\text{=-RTln}\text{K---------[1]}\\ \text{Let}\text{us consider following equation (1)}\\ {\text{ΔG}}^{°}=-53.2\text{KJ/mol}\\ \Delta {\text{G}}^{\text{0}}\text{values substituted equation (1) and rewrite the above equation}\\ \text{InK}\text{p=}\frac{{\text{ΔG}}^{°}}{\text{-RT}}=\frac{\text{-(-53}{\text{.2×10}}^3\text{J/mol)}}{\text{(8}\text{.314}\text{J/k}\cdot \text{mol)(298}\text{K)}}\\ \text{InK}{}_{\text{P}}\text{=}\frac{{\text{ΔG}}^{\text{0}}}{\text{-RT}}=\frac{53.2{\text{×10}}^3\text{J/mol}}{(2477.572)}=21.5\\ \text{Kp=}{e}^{21.5}\\ \text{Kp=}2\times {10}^9\end{array}$

The free energy value is ${\text{ΔG}}_{}^{°}=-53.2\text{KJ/mol}$ and equilibrium pressure value is $\text{Kp=}2\times {10}^9$