Chapter 20, Problem 20.54P

Interpretation Introduction

Interpretation:

For the given set of reactions in problem 20.52 the standard free energy $\Delta G^o$ value have to be calculated using the $\Delta H_f^{o}and{S}^o$ values.

Concept introduction:

Enthalpy is the amount energy absorbed or released in a process.

The enthalpy change in a system $\text{(Δ}{{\rm H}}_{\text{sys}}\text{)}$ can be calculated by the following equation.

  ${\text{ΔH}}_{\text{rxn}}\text{ = }{\displaystyle \sum {\text{ΔH}}^{\text{°}}{}_{\text{produdcts}}}\text{-}{\displaystyle \sum {\text{ΔH}}^{\text{°}}{}_{\text{reactants}}}$

Where,

  ${\text{ΔH}}_{\text{r}}^{\text{o}}{}_{\left(\text{reactants}\right)}$ is the standard enthalpy of the reactants

  ${\text{ΔH}}_{\text{r}}^{\text{o}}{}_{\left(\text{produdcts}\right)}$ is the standard enthalpy of the products

Entropy is the measure of randomness in the system.  Standard entropy change in a reaction is the difference in entropy of the products and reactants.  ${\text{(ΔS}}^{\text{°}}{}_{\text{rxn}}\text{)}$ can be calculated by the following equation.

  ${\text{ΔS}}^{\text{°}}{}_{\text{rxn}}\text{ = }{\displaystyle \sum \text{m }}{\text{S}}^{\text{°}}{}_{\text{Products}}-{\displaystyle \sum \text{n}}{\text{S}}^{\text{°}}{}_{\text{reactants}}$

Where,

  ${\text{S}}^{\text{°}}{}_{\text{reactants}}$ is the standard entropy of the reactants

  ${\text{S}}^{\text{°}}{}_{\text{Products}}$ is the standard entropy of the products

Standard entropy change in a reaction and entropy change in the system are same.

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 G.  All spontaneous process is associated with the decrease of free energy in the system.  The equation given below helps us to calculate the change in free energy in a system.

  $\text{ΔG = }\Delta {\rm H}\text{- T}\Delta \text{S}$

Where,

  $\text{ΔG }$ is the change in free energy of the system

  $\Delta {\rm H}$ is the change in enthalpy of the system

  $\text{T}$ is the absolute value of the temperature

  $\Delta \text{S}$ is the change in entropy in the system

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.

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

Where,

  ${\text{nΔG}}_{\text{f}}{{}^{\text{°}}}_{\left(\text{Reactants}\right)}$ is the standard entropy of the reactants

  ${\text{mΔG}}_{\text{f}}{{}^{\text{°}}}_{\left(\text{products}\right)}$ is the standard free energy of the products

Answer to Problem 20.54P

Reaction-A the standard free energy value is $\Delta G_{rxn}^o=\underset{\_}{2.4kJ}$

Reaction-B the standard free energy value is $\Delta G_{rxn}^o=\underset{\_}{-48.4kJ}$

Reaction-C the standard free energy value is $\Delta G_{rxn}^o=\underset{\_}{91.2kJ}$

Explanation of Solution

Reaction-A

Given chemical reaction is,

         $H_2\left(g\right)+I_2\left(g\right)\stackrel{}{\to }2HI\left(g\right)$

The number of particle also decreases, indicating a decrease in entropy.

So, given reaction $\Delta G_f^o$ value is zero.

Standard enthalpy change is,

The enthalpy change for the reaction is calculated as follows,

  ${\text{ΔH}}^{\text{°}}{}_{\text{rxn}}\text{ = }{\displaystyle \sum \text{m }}\Delta {\text{H}}^{\text{°}}{}_{\text{f(Products)}}-{\displaystyle \sum \text{n}}\Delta {\text{H}}^{\text{°}}{}_{\text{f(reactants)}}$

  $\begin{array}{l}{\text{ΔH}}^{\text{°}}{}_{\text{rxn}}\text{ =}\left[\left(2\text{mol}HI\right)\left({\text{ΔH}}^{\text{°}}{}_{f}ofHI\right)\right]\\ -\left[\left(1mol{H}_2\right)\left({\text{ΔH}}^{\text{°}}{}_{f}of{H}_2\right)+\left(\text{1}mol{I}_2\right)\left({\text{ΔH}}^{\text{°}}{}_{f}of{I}_2\right)\right]\\ {\text{ΔH}}^{\text{°}}{}_{\text{rxn}}\text{ =}\left[\left(2\text{mol}HI\right)\left(25.9kJ/mol\right)\right]-\\ \left[\left(1mol{H}_2\right)\left(0kJ/mol\right)+\left(1mol{I}_2\right)\left(0kJ/mol\right)\right]\\ {\text{ΔH}}^{\text{°}}{}_{\text{rxn}}\text{ =}-51.8kJ\end{array}$

The enthalpy change is negative.

Hence, the enthalpy ${\text{(ΔH}}^{\text{°}}{}_{\text{rxn}}\text{)}$ value is $\underset{\_}{-51.8kJ}$

Entropy change  ${\text{ΔS}}^{\text{°}}{}_{system}$

Standard entropy change equation is,

  ${\text{ΔS}}^{\text{°}}{}_{\text{rxn}}\text{ = }{\displaystyle \sum \text{m }}{\text{S}}^{\text{°}}{}_{\text{Products}}-{\displaystyle \sum \text{n}}{\text{S}}^{\text{°}}{}_{\text{reactants}}$

Where, $\left(m\right)and\left(n\right)$ are the stoichiometric co-efficient.

  $\begin{array}{l}{\text{ΔS}}^{\text{°}}{}_{\text{rxn}}\text{ =}\left[\left(2\text{mol}HI\right)\left(206.33J/mol\times K\right)\right]-\\ \left[\left(2mol{H}_2\right)\left(130.6J/mol\times K\right)+\left(1mol{I}_2\right)\left(116.14J/mol\times K\right)\right]\\ {\text{ΔS}}^{\text{°}}{}_{\text{rxn}}\text{ =}165.92J/K\end{array}$

Therefore, the ${\text{(ΔS}}^{\text{°}}{}_{\text{rxn}}\text{)}$ of the reaction is $\underset{\_}{165.92J/K}$ 

Next calculate the Free enrgy change  $\Delta G_{rxn}^o$

Standared Free energy change equation iss,

  ${\text{ΔG}}_{\text{rxn}}^{\text{o}}\text{ = }\Delta {{\rm H}}_{\text{rxn}}^{\text{o}}\text{- T}\Delta {\text{S}}_{\text{rxn}}^{\text{o}}$

Free energy change  $\Delta G_f^o$

Calcualted enthalpy and entropy  values are

  $\Delta {{\rm H}}_{\text{rxn}}^{\text{o}}=-51.8kJ$

  $\Delta {\text{S}}_{\text{rxn}}^{\text{o}}=165.92J/K$

These values are plugging above standard free energy equation,

   $\begin{array}{l}{\text{ΔG}}_{\text{rxn}}^{\text{o}}=51.8\text{kJ}\text{-}\left[\left(298\text{K}\right)\left(165.92\text{J/K}\right)\left(1kJ/{10}^{3}J\right)\right]\\ {\text{ΔG}}_{\text{rxn}}^{\text{o}}=2.3558kJ\end{array}$

Therefore, the standard free energy value is $\Delta G_{rxn}^o=\underset{\_}{2.4kJ}$

Reaction-B

Given reaction is,

         $Mn{O}_2\left(s\right)+2CO\left(g\right)\stackrel{}{\to }Mn\left(s\right)+4C{O}_2\left(g\right)$

The number of particle also increases, indicating a increase in entropy.

Standard enthalpy change is,

The enthalpy change for the reaction is calculated as follows,

  ${\text{ΔH}}^{\text{°}}{}_{\text{rxn}}\text{ = }{\displaystyle \sum \text{m }}\Delta {\text{H}}^{\text{°}}{}_{\text{f(Products)}}-{\displaystyle \sum \text{n}}\Delta {\text{H}}^{\text{°}}{}_{\text{f(reactants)}}$

  $\begin{array}{l}\Delta {{\rm H}}_{\text{rxn}}^{\text{o}}\text{ =}\left[\left(1\text{mol}Mn\right)\left(0kJ/mol\right)+\left(2\text{mol}C{O}_2\right)\left(-393.5kJ/mol\right)\right]-\\ \left[\left(1molMn{O}_2\right)\left(-520.9kJ/mol\right)+\left(2molCO\right)\left(-110.5kJ/mol\right)\right]\\ \Delta {{\rm H}}_{\text{rxn}}^{\text{o}}\text{ =}-45.1kJ\end{array}$

The enthalpy change is negative.

Hence, the enthalpy ${\text{(ΔH}}^{\text{°}}{}_{\text{rxn}}\text{)}$ value is $\underset{\_}{-45.1kJ}$

Entropy change  ${\text{ΔS}}^{\text{°}}{}_{system}$

Standard entropy change equation is,

  ${\text{ΔS}}^{\text{°}}{}_{\text{rxn}}\text{ = }{\displaystyle \sum \text{m }}{\text{S}}^{\text{°}}{}_{\text{Products}}-{\displaystyle \sum \text{n}}{\text{S}}^{\text{°}}{}_{\text{reactants}}$

Where, $\left(m\right)and\left(n\right)$ are the stoichiometric co-efficient.

$\begin{array}{l}{\text{ΔS}}_{\text{rxn}}^{\text{o}}\text{ =}\left[\left(1\text{mol}Mn\right)\left(31.8J/mol\cdot K\right)+\left(2\text{mol}C{O}_2\right)\left(213.7J/mol\cdot K\right)\right]-\\ \left[\left(1molMn{O}_2\right)\left(53.1J/mol\cdot K\right)+\left(2molCO\right)\left(197.5J/mol\cdot K\right)\right]\\ {\text{ΔS}}_{\text{rxn}}^{\text{o}}\text{ =}11.1J/K\end{array}$

Therefore, the ${\text{(ΔS}}^{\text{°}}{}_{\text{rxn}}\text{)}$ of the reaction is $\underset{\_}{-91.28J/K}$ 

Next calculate the Free energy change  $\Delta G_{rxn}^o$

Standared Free energy change equation iss,

  ${\text{ΔG}}_{\text{rxn}}^{\text{o}}\text{ = }\Delta {{\rm H}}_{\text{rxn}}^{\text{o}}\text{- T}\Delta {\text{S}}_{\text{rxn}}^{\text{o}}$

Free enrgy change  $\Delta G_f^o$

Calcualted enthalpy and entropy  values are

  $\Delta {{\rm H}}_{\text{rxn}}^{\text{o}}=-45.1kJ$

  $\Delta {\text{S}}_{\text{rxn}}^{\text{o}}=-91.28J/K$

These values are plugging above standard free energy equation,

   $\begin{array}{l}{\text{ΔG}}_{\text{rxn}}^{\text{o}}=-45.1\text{kJ}\text{-}\left[\left(298\text{K}\right)\left(-11.1\text{J/K}\right)\left(1kJ/{10}^{3}J\right)\right]\\ {\text{ΔG}}_{\text{rxn}}^{\text{o}}=-48.4078kJ\end{array}$

Therefore, the standard free energy value is $\Delta G_{rxn}^o=\underset{\_}{-48.4kJ}$

Reaction-C

Given reaction is,

         $N{H}_{4}Cl\left(s\right)\stackrel{}{\to }N{H}_3\left(g\right)+HCl\left(g\right)$

Standard enthalpy change is,

The enthalpy change for the reaction is calculated as follows,

  ${\text{ΔH}}^{\text{°}}{}_{\text{rxn}}\text{ = }{\displaystyle \sum \text{m }}\Delta {\text{H}}^{\text{°}}{}_{\text{f(Products)}}-{\displaystyle \sum \text{n}}\Delta {\text{H}}^{\text{°}}{}_{\text{f(reactants)}}$

  $\begin{array}{l}\Delta {{\rm H}}_{\text{rxn}}^{\text{o}}\text{ =}\left[\left(1\text{mol}N{H}_3\right)\left(-45.9kJ/mol\right)+\left(1\text{mol}HCl\right)\left(-92.3kJ/mol\right)\right]-\\ \left[\left(1molN{H}_{4}Cl\right)\left(-314.4kJ/mol\right)\right]\\ \Delta {{\rm H}}_{\text{rxn}}^{\text{o}}\text{ =}176.2kJ\end{array}$

The enthalpy change is positive.

Hence, the enthalpy ${\text{(ΔH}}^{\text{°}}{}_{\text{rxn}}\text{)}$ value is $\underset{\_}{176.2kJ}$

Entropy change ${\text{ΔS}}^{\text{°}}{}_{system}$

Standard entropy change equation is,

  ${\text{ΔS}}^{\text{°}}{}_{\text{rxn}}\text{ = }{\displaystyle \sum \text{m }}{\text{S}}^{\text{°}}{}_{\text{Products}}-{\displaystyle \sum \text{n}}{\text{S}}^{\text{°}}{}_{\text{reactants}}$

Where, $\left(m\right)and\left(n\right)$ are the stoichiometric co-efficient.

$\begin{array}{l}{\text{ΔS}}_{\text{rxn}}^{\text{o}}\text{ =}\left[\left(1\text{mol}N{H}_3\right)\left(193J/mol\cdot K\right)+\left(1\text{mol}HCl\right)\left(186.79J/mol\cdot K\right)\right]-\\ \left[\left(1molN{H}_{4}Cl\right)\left(94.6J/mol\cdot K\right)\right]\\ {\text{ΔS}}_{\text{rxn}}^{\text{o}}\text{ =}285.19J/K\end{array}$

Therefore, the ${\text{(ΔS}}^{\text{°}}{}_{\text{rxn}}\text{)}$ of the reaction is $\underset{\_}{285.19J/K}$ 

Finally calculate the Free enrgy change  $\Delta G_{rxn}^o$

Standared Free energy change equation iss,

  ${\text{ΔG}}_{\text{rxn}}^{\text{o}}\text{ = }\Delta {{\rm H}}_{\text{rxn}}^{\text{o}}\text{- T}\Delta {\text{S}}_{\text{rxn}}^{\text{o}}$

Free energy change  $\Delta G_f^o$

Calcualted enthalpy and entropy  values are

  $\Delta {{\rm H}}_{\text{rxn}}^{\text{o}}=176.2kJ$

  $\Delta {\text{S}}_{\text{rxn}}^{\text{o}}=285.19J/K$

These values are plugging above standard free energy equation,

       $\begin{array}{l}{\text{ΔG}}_{\text{rxn}}^{\text{o}}=176.2\text{kJ}\text{-}\left[\left(298\text{K}\right)\left(285.19\text{J/K}\right)\left(1kJ/{10}^{3}J\right)\right]\\ {\text{ΔG}}_{\text{rxn}}^{\text{o}}=91.213kJ\end{array}$

Therefore, the standard free energy value is $\Delta G_{rxn}^o=\underset{\_}{91.2kJ}$