Chapter 20, Problem 20.93P

(a)

Interpretation Introduction

Interpretation:

The given reaction has to be shown whether it is thermodynamically feasible or not.

Concept introduction:

Free energy (or) entropy change 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.

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

Where,

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

  $\Delta {{\rm H}}^{\text{o}}$ is the standard change in enthalpy of the system

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

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

Answer to Problem 20.93P

Methanol formation standard free energy values is ${\text{ΔG}}_{\text{rxn}}^{\text{o}}=\underset{\_}{-29.2kJ}$.

Explanation of Solution

The reaction for formation of methanol is,

  $CO\left(g\right)+2H_2\left(g\right)\stackrel{}{\to }C{H}_{3}OH\left(l\right)$

Standard enthalpy change is,

The enthalpy change for the reaction is calculated as follows,

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

  $\begin{array}{l}{\text{ΔH}}^{\text{°}}{}_{\text{rxn}}\text{ =}\left[\left(1\text{mol}C{H}_{3}OH\right)\left({\text{ΔH}}^{\text{°}}{}_{f}ofC{H}_{3}OH\right)\right]\\ -\left[\left(1molCO\right)\left({\text{ΔH}}^{\text{°}}{}_{f}ofCO\right)+\left(2mol{H}_2\right)\left({\text{ΔH}}^{\text{°}}{}_{f}of{H}_2\right)\right]\\ {\text{ΔH}}^{\text{°}}{}_{\text{rxn}}\text{ =}\left[\left(1\text{mol}C{H}_{3}OH\right)\left(-238.6kJ/mol\right)\right]-\\ \left[\left(1molCO\right)\left(-110.5kJ/mol\right)+\left(2mol{H}_2\right)\left(0kJ/mol\right)\right]\\ {\text{ΔH}}^{\text{°}}{}_{\text{rxn}}\text{ =}-128.1kJ\end{array}$

Hence, the enthalpy changes ${\text{(ΔH}}^{\text{°}}{}_{\text{rxn}}\text{)}$ is $\underset{\_}{-128.1kJ}$ 

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

Calculate the change in entropy for this reaction as follows,

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

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

  $\begin{array}{l}{\text{ΔS}}^{\text{°}}{}_{\text{rxn}}\text{ =}\left[\left(1\text{mol}C{H}_{3}OH\right)\left(S^{o}ofC{H}_{3}OH\right)\right]\\ -\left[\left(1molCO\right)\left({\text{ΔS}}^{\text{°}}{}_{\text{rxn}}ofCO\right)+\left(2mol{H}_2\right)\left({\text{ΔS}}^{\text{°}}{}_{\text{rxn}}of{H}_2\right)\right]\\ {\text{ΔH}}^{\text{°}}{}_{\text{rxn}}\text{ =}\left[\left(1\text{mol}C{H}_{3}OH\right)\left(127J/K\cdot mol\right)\right]-\\ \left[\left(1molCO\right)\left(197.5J/K\cdot mol\right)+\left(2mol{H}_2\right)\left(130.6127J/K\cdot mol\right)\right]\\ {\text{ΔH}}^{\text{°}}{}_{\text{rxn}}\text{ =}-331.7J/K\end{array}$

The ${\text{(ΔS}}^{\text{°}}{}_{\text{rxn}}\text{)}$ of the reaction is $\underset{\_}{-331.7J/K}$ 

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

Standared Free energy change equation iss,

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

Calcualted enthalpy and entropy  values are

  $\begin{array}{l}\Delta {{\rm H}}_{\text{rxn}}^{\text{o}}=-128.1kJ\\ \Delta {\text{S}}_{\text{rxn}}^{\text{o}}=-331.7J/K\\ {\text{T}}_{\text{1}}=273\text{+}25\text{=}298K\end{array}$

These values are plugging above standard free energy equation,

         $\begin{array}{c}{\text{ΔG}}_{\text{rxn}}^{\text{o}}=-128.1\text{kJ}\text{-}\left[\left(298\text{K}\right)\left(-331.7\text{J/K}\right)\left(1kJ/{10}^3\right)\right]\\ =-29.2534kJ\\ {\text{ΔG}}_{\text{rxn}}^{\text{o}}=-29.2kJ\end{array}$

Methanol formation standard free energy values is ${\text{ΔG}}_{\text{rxn}}^{\text{o}}=\underset{\_}{-29.2kJ}$, the negative value of $\Delta G^o$ indicates that the reaction is spontaneous and it is feasible. 

(b)

Interpretation Introduction

Interpretation:

The given methanol formation reaction has to be shown that whether it is favored at low or at high temperatures.

Concept introduction:

Any natural process or a chemical reaction taking place in a laboratory can be classified into two categories, spontaneous or nonspontaneous. Spontaneous process occurs by itself, without the influence of external energy. In spontaneous process the free energy of the system decreases and entropy of the system increases. Nonspontaneous process requires an external influence for initiation. In nonspontaneous process the free energy of the system increases but entropy in the system decreases.

In thermodynamics a process is spontaneous if it is taking place by itself without the help of external energy.  All spontaneous process will have highly energetic initial state than the final state. This indicates that while the process occurs, there is a decrease in free energy of the system. The increase in randomness also favors the spontaneity of a process.

Explanation of Solution

The reaction for formation of methanol is,

  $CO\left(g\right)+2H_2\left(g\right)\stackrel{}{\to }C{H}_{3}OH\left(l\right)$

Calcualted enthalpy and entropy  values are

  $\begin{array}{l}\Delta {{\rm H}}_{\text{rxn}}^{\text{o}}=-128.1kJ\\ \Delta {\text{S}}_{\text{rxn}}^{\text{o}}=-331.7J/K\\ {\text{T}}_{\text{1}}=273\text{+}25\text{=}298K\end{array}$

At temprature become above $298K$, this reaction is non-spontaneous. Because both enthalpy $\Delta {{\rm H}}_{\text{rxn}}^{\text{o}}$ and entropy $\Delta S_{\text{rxn}}^{\text{o}}$ values are negative, that means the reaction is favored at low temprature.

(c)

Interpretation Introduction

Interpretation:

For the given methanol oxidation reaction the ${\text{ΔG}}^{\text{o}}$ value has to calculate at ${100}^{o}C$.

Concept introduction:

Free energy (or) entropy change 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.

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

Where,

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

  $\Delta {{\rm H}}^{\text{o}}$ is the standard change in enthalpy of the system

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

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

Answer to Problem 20.93P

Methanol oxidation reaction standard free energy values is ${\text{ΔG}}_{\text{rxn}}^{\text{o}}=\underset{\_}{-182kJ}$

Explanation of Solution

The oxidation reaction of methanol is,

  $C{H}_{3}OH\left(g\right)+1/2O_2\left(g\right)\stackrel{}{\to }C{H}_{2}O\left(g\right)+H_{2}O\left(g\right)$

Standard enthalpy change is,

The enthalpy change for the reaction is calculated as follows,

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

  $\begin{array}{l}{\text{ΔH}}^{\text{°}}{}_{\text{rxn}}\text{ =}\left[\left(1\text{mol}C{H}_{2}O\right)\left({\text{ΔH}}^{\text{°}}{}_{f}ofC{H}_{2}O\right)+\left(1\text{mol}{H}_{2}O\right)\left({\text{ΔH}}^{\text{°}}{}_{f}of{H}_{2}O\right)\right]\\ -\left[\left(1molC{H}_{3}OH\right)\left({\text{ΔH}}^{\text{°}}{}_{f}ofC{H}_{3}OH\right)+\left(1/2mol{O}_2\right)\left({\text{ΔH}}^{\text{°}}{}_{f}of{O}_2\right)\right]\\ {\text{ΔH}}^{\text{°}}{}_{\text{rxn}}\text{ =}\left[\left(1\text{mol}C{H}_{2}O\right)\left(-116kJ/mol\right)+\left(1\text{mol}{H}_{2}O\right)\left(-241.826kJ/mol\right)\right]-\\ \left[\left(1molC{H}_{3}OH\right)\left(-201.2kJ/mol\right)+\left(1/2mol{O}_2\right)\left(0kJ/mol\right)\right]\\ {\text{ΔH}}^{\text{°}}{}_{\text{rxn}}\text{ =}-156.626kJ\end{array}$

Hence, the enthalpy changes ${\text{(ΔH}}^{\text{°}}{}_{\text{rxn}}\text{)}$ is $\underset{\_}{-156.626kJ}$ 

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

Calculate the change in entropy for this reaction as follows,

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

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

  $\begin{array}{l}{\text{ΔS}}^{\text{°}}{}_{\text{rxn}}\text{ =}\left[\left(1\text{mol}C{H}_{2}O\right)\left(S^{o}ofC{H}_{2}O\right)+\left(1\text{mol}{H}_{2}O\right)\left(S^{o}of{H}_{2}O\right)\right]\\ -\left[\left(1molC{H}_{3}OH\right)\left({\text{ΔS}}^{\text{°}}{}_{\text{rxn}}ofC{H}_{3}OH\right)+\left(1/2mol{O}_2\right)\left({\text{ΔS}}^{\text{°}}{}_{\text{rxn}}of{O}_2\right)\right]\\ {\text{ΔH}}^{\text{°}}{}_{\text{rxn}}\text{ =}\left[\left(1\text{mol}C{H}_{2}O\right)\left(219J/K\cdot mol\right)+\left(1\text{mol}{H}_{2}O\right)\left(188.72J/K\cdot mol\right)\right]-\\ \left[\left(1molC{H}_{3}OH\right)\left(238J/K\cdot mol\right)+\left(1/2mol{O}_2\right)\left(205.0J/K\cdot mol\right)\right]\\ {\text{ΔH}}^{\text{°}}{}_{\text{rxn}}\text{ =}-67.22J/K\end{array}$

The ${\text{(ΔS}}^{\text{°}}{}_{\text{rxn}}\text{)}$ of the reaction is $\underset{\_}{-67.22J/K}$

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

Standared Free energy change equation is,

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

Calculated enthalpy and entropy  values are

  $\begin{array}{l}\Delta {{\rm H}}_{\text{rxn}}^{\text{o}}=-156.626kJ\\ \Delta {\text{S}}_{\text{rxn}}^{\text{o}}=-67.22J/K\\ {\text{T}}_{\text{1}}=273\text{+}100\text{=}373K\end{array}$

These values are plugging above standard free energy equation,

  $\begin{array}{c}{\text{ΔG}}_{\text{rxn}}^{\text{o}}=-156.626\text{kJ}\text{-}\left[\left(373\text{K}\right)\left(67.22\text{J/K}\right)\left(1kJ/{10}^3\right)\right]\\ =-181.699kJ\\ {\text{ΔG}}_{\text{rxn}}^{\text{o}}=-182kJ\end{array}$

Therefore, the given oxidation reaction standard free energy values is ${\text{ΔG}}_{\text{rxn}}^{\text{o}}=\underset{\_}{-182kJ}$