Chapter 20, Problem 20.107P

(a)

Interpretation Introduction

Interpretation:

The given reaction equation has to be balanced.

Concept introduction:

Balanced equation: A balanced chemical equation is an equation which contains same elements in same number on both the sides (reactant and product side) of the chemical equation thereby obeying the law of conservation of mass.

The equation for a reaction, which has same number of atoms and charge of the ions in both reactants and product sides, is known as balanced equation.

Answer to Problem 20.107P

Balanced equation is,

  $\begin{array}{c}{C}_6{H}_{12}{O}_6\left(s\right)+6O_2\left(g\right)\underset{}{\to }6C{O}_2\left(g\right)+6H_{2}O\left(l\right)\\ \\ C_6{H}_{12}{O}_6\left(s\right)\underset{}{\to }2C_2{H}_{5}OH\left(l\right)+2C{O}_2\left(g\right)\\ \\ C_2{H}_{5}OH\left(l\right)+3O_2\left(g\right)\stackrel{}{\to }2C{O}_2\left(g\right)+3H_{2}O\left(l\right)\end{array}$

Explanation of Solution

The given statement of equation can be completed by writing as follows,

  $\begin{array}{c}{C}_6{H}_{12}{O}_6\left(s\right)+O_2\left(g\right)\underset{}{\to }C{O}_2\left(g\right)+H_{2}O\left(l\right)\\ \\ C_6{H}_{12}{O}_6\left(s\right)\underset{}{\to }{C}_2{H}_{5}OH\left(l\right)+C{O}_2\left(g\right)\\ \\ C_2{H}_{5}OH\left(l\right)+O_2\left(g\right)\stackrel{}{\to }C{O}_2\left(g\right)+H_{2}O\left(l\right)\end{array}$

Balancing the equation:

  $\begin{array}{c}{C}_6{H}_{12}{O}_6\left(s\right)+6O_2\left(g\right)\underset{}{\to }6C{O}_2\left(g\right)+6H_{2}O\left(l\right)\\ \\ C_6{H}_{12}{O}_6\left(s\right)\underset{}{\to }2C_2{H}_{5}OH\left(l\right)+2C{O}_2\left(g\right)\\ \\ C_2{H}_{5}OH\left(l\right)+3O_2\left(g\right)\stackrel{}{\to }2C{O}_2\left(g\right)+3H_{2}O\left(l\right)\end{array}$

The given process of respiration, glucose is oxidized completely to give $C{O}_{2}and{H}_{2}O$.

In fermentation process, oxygen $O_2$ is absent and glucose is broken down to ethanol and $C{O}_2$.

In third reaction, ethanol is oxidized to $C{O}_{2}and{H}_{2}O$.

A balanced equation will have same elements, and same number of atoms of each side of the reaction.

(b)

Interpretation Introduction

Interpretation:

For respiration reaction $1.00g$ of glucose, the standard free energy ${\text{ΔG}}_{\text{rxn}}^{\text{o}}$ value should be calculated.

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.

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

Answer to Problem 20.107P

The ${\text{ΔG}}_{\text{rxn}}^{\text{o}}$ for respiration $1.00g$ of glucose is $\underset{\_}{-16.0kJ/g}.$

Explanation of Solution

The resiparation of glucose reaction is,

  $C_6{H}_{12}{O}_6\left(s\right)+6O_2\left(g\right)\underset{}{\to }6C{O}_2\left(g\right)+6H_{2}O\left(l\right)$

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

Standared Free energy change equation is,

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

Calculate the free energy change for this reaction as follows,

  $\begin{array}{l}{\text{ΔG}}^{\text{°}}{}_{\text{rxn}}\text{ =}\left[\left(\text{1}\text{mol}C{O}_2\right)\left({\text{ΔG}}_f^{o}ofC{O}_2\right)+\left(6\text{mol}{H}_{2}O\right)\left({\text{ΔG}}_f^{o}of{H}_{2}O\right)\right]\\ \\ -\left[\left(1mol{C}_6{H}_{12}{O}_6\right)\left({\text{ΔG}}^{\text{°}}{}_{f}of{C}_6{H}_{12}{O}_6\right)+\left(6mol{O}_2\right)\left({\text{ΔG}}^{\text{°}}{}_{f}of{O}_2\right)\right]\\ \\ {\text{ΔG}}^{\text{°}}{}_{\text{rxn}}\text{ =}\left[\left(\text{1}\text{mol}C{O}_2\right)\left(-394.4kJ/mol\right)+\left(6\text{mol}{H}_{2}O\right)\left(-237.192kJ/mol\right)\right]-\\ \\ \left[\left(1mol{C}_6{H}_{12}{O}_6\right)\left(-910.56kJ/mol\right)+\left(6mol{O}_2\right)\left(0kJ/mol\right)\right]\\ \\ {\text{ΔG}}^{\text{°}}{}_{\text{rxn}}\text{ =}-2878.992kJ.\end{array}$

Hence, the standard free energy is ${\text{ΔG}}^{\text{°}}{}_{rxn}\text{ =}-2878.992kJ$

Next solving for resipration $1.00g$ of glucose,

  $\begin{array}{l}{\text{ΔG}}^{\text{°}}{}_{rxn}/gof\text{glucose}\text{=}\left(1.00g\text{glucose}\right)\left(\frac{1molof\text{glucose}}{180.16g\text{glucose}}\right)\left(\frac{-2878.992kJ}{1molof\text{glucose}}\right)\\ \\ {\text{ΔG}}^{\text{°}}{}_{rxn}/gof\text{glucose}\text{=}-15.980195\\ \\ {\text{ΔG}}^{\text{°}}{}_{rxn}/gof\text{glucose}\text{=}-16.0kJ/g.\end{array}$

Therefore, the ${\text{ΔG}}_{\text{rxn}}^{\text{o}}$ for respiration $1.00g$ of glucose is $\underset{\_}{-16.0kJ/g}.$

(c)

Interpretation Introduction

Interpretation:

For fermentation reaction of $1.00g$ glucose, the standard free energy ${\text{ΔG}}_{\text{rxn}}^{\text{o}}$ value should be calculated.

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.

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

Answer to Problem 20.107P

The ${\text{ΔG}}_{\text{rxn}}^{\text{o}}$ for fermentation $1.00g$ of glucose is $\underset{\_}{-1.26kJ/g}.$

Explanation of Solution

The resiparation of glucose reaction is,

  $C_6{H}_{12}{O}_6\left(s\right)\underset{}{\to }2C_2{H}_{5}OH\left(l\right)+2C{O}_2\left(g\right)$

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

Standared Free energy change equation is,

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

Calculate the free energy change for this reaction as follows,

  $\begin{array}{l}{\text{ΔG}}^{\text{°}}{}_{\text{rxn}}\text{ =}\left[\left(2\text{mol}C{O}_2\right)\left({\text{ΔG}}_f^{o}ofC{O}_2\right)+\left(2\text{mol}{C}_2{H}_{5}OH\right)\left({\text{ΔG}}_f^{o}of{C}_2{H}_{5}OH\right)\right]\\ \\ -\left[\left(1mol{C}_6{H}_{12}{O}_6\right)\left({\text{ΔG}}^{\text{°}}{}_{f}of{C}_6{H}_{12}{O}_6\right)\right]\\ \\ {\text{ΔG}}^{\text{°}}{}_{\text{rxn}}\text{ =}\left[\left(2\text{mol}C{O}_2\right)\left(-394.4kJ/mol\right)+\left(2\text{mol}{C}_2{H}_{5}OH\right)\left(-174.8kJ/mol\right)\right]-\\ \\ \left[\left(1mol{C}_6{H}_{12}{O}_6\right)\left(-910.56kJ/mol\right)\right]\\ \\ {\text{ΔG}}^{\text{°}}{}_{\text{rxn}}\text{ =}-227.84kJ.\end{array}$

The fermentation reaction standard free energy is ${\text{ΔG}}^{\text{°}}{}_{rxn}\text{ =}-227.84kJ.$

Next solving for fermentation $1.00g$ of glucose,

  $\begin{array}{l}{\text{ΔG}}^{\text{°}}{}_{rxn}/gof\text{glucose}\text{=}\left(1.00g\text{glucose}\right)\left(\frac{1molof\text{glucose}}{180.16g\text{glucose}}\right)\left(\frac{-227.84kJ}{1molof\text{glucose}}\right)\\ \\ {\text{ΔG}}^{\text{°}}{}_{rxn}/gof\text{glucose}\text{=}-1.264653641\\ \\ {\text{ΔG}}^{\text{°}}{}_{rxn}/gof\text{glucose}\text{=}-1.26kJ/g.\end{array}$

Therefore, the ${\text{ΔG}}_{\text{rxn}}^{\text{o}}$ for fermentation $1.00g$ of glucose is $\underset{\_}{-1.26kJ/g}.$

(d)

Interpretation Introduction

Interpretation:

For oxidation reaction of $1.00g$ glucose, the standard free energy ${\text{ΔG}}_{\text{rxn}}^{\text{o}}$ value should be calculated.

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.

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

Answer to Problem 20.107P

The 1.00 g of ethanol oxidation process standard free energy is $\underset{\_}{-14.7kJ/g}.$

Explanation of Solution

The resiparation of glucose reaction is,

  $C_2{H}_{5}OH\left(l\right)+3O_2\left(g\right)\stackrel{}{\to }2C{O}_2\left(g\right)+3H_{2}O\left(l\right)$

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

Standared Free energy change equation is,

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

Calculate the free energy change for this reaction as follows,

  $\begin{array}{l}{\text{ΔG}}^{\text{°}}{}_{\text{rxn}}\text{ =}\left[\left(2\text{mol}C{O}_2\right)\left({\text{ΔG}}_f^{o}ofC{O}_2\right)+\left(3\text{mol}{H}_{2}O\right)\left({\text{ΔG}}_f^{o}of{H}_{2}O\right)\right]\\ \\ -\left[\left(1mol{C}_2{H}_{5}OH\right)\left({\text{ΔG}}^{\text{°}}{}_{f}of{C}_2{H}_{5}OH\right)+\left(3\text{mol}{O}_2\right)\left({\text{ΔG}}_f^{o}of{O}_2\right)\right]\\ \\ {\text{ΔG}}^{\text{°}}{}_{\text{rxn}}\text{ =}\left[\left(2\text{mol}C{O}_2\right)\left(-394.4kJ/mol\right)+\left(3\text{mol}{H}_{2}O\right)\left(-237.192kJ/mol\right)\right]-\\ \\ \left[\left(1mol{C}_2{H}_{5}OH\right)\left(-174.8kJ/mol\right)+\left(1mol{O}_2\right)\left(0kJ/mol\right)\right]\\ \\ {\text{ΔG}}^{\text{°}}{}_{\text{rxn}}\text{ =}-1325.576kJ.\end{array}$

Ethanol oxidation reaction standard free energy is ${\text{ΔG}}^{\text{°}}{}_{rxn}\text{ =}-1325.576kJ.$

Next solving for fermentation $1.00g$ of glucose,

  $\begin{array}{l}{\text{ΔG}}^{\text{°}}{}_{rxn}\text{=}\left(1.00g\text{glucose}\right)\left(\frac{1molof\text{glucose}}{180.16g\text{glucose}}\right)\left(\frac{2molof{C}_2{H}_{5}OH}{1molof\text{glucose}}\right)\left(\frac{-1325.576kJ}{1molof{C}_2{H}_{5}OH}\right)\\ \\ {\text{ΔG}}^{\text{°}}{}_{rxn}\text{=}-14.71554\\ \\ {\text{ΔG}}^{\text{°}}{}_{rxn}\text{=}-14.7kJ/g.\end{array}$

Therefore, the 1.00 g of ethanol oxidation process standard free energy is $\underset{\_}{-14.7kJ/g}.$