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.

Question

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Answer 1

Question

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.

(a)

Expert Solution

Answer to Problem 20.107P

Balanced equation is,

C6H12O6(s) + 6O2(g) → 6CO2(g) + 6H2O(l)C6H12O6(s) → 2C2H5OH(l) + 2CO2(g)C2H5OH(l) + 3O2(g) → 2CO2(g) + 3H2O(l)

Explanation of Solution

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

C6H12O6(s) + O2(g) → CO2(g) + H2O(l)C6H12O6(s) → C2H5OH(l) + CO2(g)C2H5OH(l) + O2(g) → CO2(g) + H2O(l)

Balancing the equation:

C6H12O6(s) + 6O2(g) → 6CO2(g) + 6H2O(l)C6H12O6(s) → 2C2H5OH(l) + 2CO2(g)C2H5OH(l) + 3O2(g) → 2CO2(g) + 3H2O(l)

The given process of respiration, glucose is oxidized completely to give CO2 and H2O.

In fermentation process, oxygen O2 is absent and glucose is broken down to ethanol and CO2.

In third reaction, ethanol is oxidized to CO2 and H2O.

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 1.00 g of glucose, the standard free energy ΔGrxno 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 G. All spontaneous process is associated with the decrease of free energy in the system. The standard free energy change (ΔG°rxn) is the difference in free energy of the reactants and products in their standard state.

ΔG°rxn=∑nΔGf°(Products)-∑nΔGf°(Reactants)

(b)

Expert Solution

Answer to Problem 20.107P

The ΔGrxno for respiration 1.00 g of glucose is -16.0 kJ/g_.

Explanation of Solution

The resiparation of glucose reaction is,

C6H12O6(s) + 6O2(g) → 6CO2(g) + 6H2O(l)

Calculate the Free enrgy change ΔGrxno

Standared Free energy change equation is,

ΔG°rxn=∑nΔGf°(Products)-∑nΔGf°(Reactants)

Calculate the free energy change for this reaction as follows,

ΔG°rxn = [(1 mol CO2) ( ΔGfo of CO2) + (6 mol H2O) ( ΔGfo of H2O)] −[(1 mol C6H12O6)(ΔG°f of C6H12O6) + (6 mol O2)(ΔG°f of O2)] ΔG°rxn = [(1 mol CO2)(−394.4 kJ/mol ) + (6 mol H2O)(−237.192 kJ/mol )]− [(1 mol C6H12O6)(−910.56 kJ/mol ) + (6 mol O2)(0 kJ/mol )] ΔG°rxn = -2878.992 kJ.

Hence, the standard free energy is ΔG°rxn =-2878.992 kJ

Next solving for resipration 1.00 g of glucose,

ΔG°rxn/ g of glucose = (1.00 g glucose) (1 mol of glucose180. 16 g glucose) (−2878.992 kJ1 mol of glucose)ΔG°rxn/ g of glucose = −15.980195ΔG°rxn/ g of glucose =-16.0 kJ/g.

Therefore, the ΔGrxno for respiration 1.00 g of glucose is -16.0 kJ/g_.

(c)

Interpretation Introduction

Interpretation:

For fermentation reaction of 1.00 g glucose, the standard free energy ΔGrxno 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 G. All spontaneous process is associated with the decrease of free energy in the system. The standard free energy change (ΔG°rxn) is the difference in free energy of the reactants and products in their standard state.

ΔG°rxn=∑nΔGf°(Products)-∑nΔGf°(Reactants)

(c)

Expert Solution

Answer to Problem 20.107P

The ΔGrxno for fermentation 1.00 g of glucose is -1.26 kJ/g_.

Explanation of Solution

The resiparation of glucose reaction is,

C6H12O6(s) → 2C2H5OH(l) + 2CO2(g)

Calculate the Free enrgy change ΔGrxno

Standared Free energy change equation is,

ΔG°rxn=∑nΔGf°(Products)-∑nΔGf°(Reactants)

Calculate the free energy change for this reaction as follows,

ΔG°rxn = [(2 mol CO2) ( ΔGfo of CO2) + (2 mol C2H5OH) ( ΔGfo of C2H5OH)] −[(1 mol C6H12O6)(ΔG°f of C6H12O6)] ΔG°rxn = [(2 mol CO2)(−394.4 kJ/mol ) + (2 mol C2H5OH)(−174.8 kJ/mol )]− [(1 mol C6H12O6)(−910.56 kJ/mol )] ΔG°rxn = -227.84 kJ.

The fermentation reaction standard free energy is ΔG°rxn =-227.84 kJ.

Next solving for fermentation 1.00 g of glucose,

ΔG°rxn/ g of glucose = (1.00 g glucose) (1 mol of glucose180. 16 g glucose) (−227.84 kJ1 mol of glucose)ΔG°rxn/ g of glucose = −1.264653641ΔG°rxn/ g of glucose =-1.26 kJ/g.

Therefore, the ΔGrxno for fermentation 1.00 g of glucose is -1.26 kJ/g_.

(d)

Interpretation Introduction

Interpretation:

For oxidation reaction of 1.00 g glucose, the standard free energy ΔGrxno 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 G. All spontaneous process is associated with the decrease of free energy in the system. The standard free energy change (ΔG°rxn) is the difference in free energy of the reactants and products in their standard state.

ΔG°rxn=∑nΔGf°(Products)-∑nΔGf°(Reactants)

(d)

Expert Solution

Answer to Problem 20.107P

The 1.00 g of ethanol oxidation process standard free energy is -14.7 kJ/g_.

Explanation of Solution

The resiparation of glucose reaction is,

C2H5OH(l) + 3O2(g) → 2CO2(g) + 3H2O(l)

Calculate the Free enrgy change ΔGrxno

Standared Free energy change equation is,

ΔG°rxn=∑nΔGf°(Products)-∑nΔGf°(Reactants)

Calculate the free energy change for this reaction as follows,

ΔG°rxn = [(2 mol CO2) ( ΔGfo of CO2) + (3 mol H2O) ( ΔGfo of H2O)] −[(1 mol C2H5OH)(ΔG°f of C2H5OH) + (3 mol O2) ( ΔGfo of O2) ] ΔG°rxn = [(2 mol CO2)(−394.4 kJ/mol ) + (3 mol H2O)(−237.192 kJ/mol )]− [(1 mol C2H5OH)(−174.8 kJ/mol ) + (1 mol O2)(0 kJ/mol ) ] ΔG°rxn = -1325.576 kJ.

Ethanol oxidation reaction standard free energy is ΔG°rxn =-1325.576 kJ.

Next solving for fermentation 1.00 g of glucose,

ΔG°rxn = (1.00 g glucose) (1 mol of glucose180. 16 g glucose)(2 mol of C2H5OH1 mol of glucose) (−1325.576 kJ1 mol of C2H5OH)ΔG°rxn= −14.71554ΔG°rxn =-14.7 kJ/g.

Therefore, the 1.00 g of ethanol oxidation process standard free energy is -14.7 kJ/g_.

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Answer 2

Question

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.

(a)

Expert Solution

Answer to Problem 20.107P

Balanced equation is,

C6H12O6(s) + 6O2(g) → 6CO2(g) + 6H2O(l)C6H12O6(s) → 2C2H5OH(l) + 2CO2(g)C2H5OH(l) + 3O2(g) → 2CO2(g) + 3H2O(l)

Explanation of Solution

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

C6H12O6(s) + O2(g) → CO2(g) + H2O(l)C6H12O6(s) → C2H5OH(l) + CO2(g)C2H5OH(l) + O2(g) → CO2(g) + H2O(l)

Balancing the equation:

C6H12O6(s) + 6O2(g) → 6CO2(g) + 6H2O(l)C6H12O6(s) → 2C2H5OH(l) + 2CO2(g)C2H5OH(l) + 3O2(g) → 2CO2(g) + 3H2O(l)

The given process of respiration, glucose is oxidized completely to give CO2 and H2O.

In fermentation process, oxygen O2 is absent and glucose is broken down to ethanol and CO2.

In third reaction, ethanol is oxidized to CO2 and H2O.

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 1.00 g of glucose, the standard free energy ΔGrxno 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 G. All spontaneous process is associated with the decrease of free energy in the system. The standard free energy change (ΔG°rxn) is the difference in free energy of the reactants and products in their standard state.

ΔG°rxn=∑nΔGf°(Products)-∑nΔGf°(Reactants)

(b)

Expert Solution

Answer to Problem 20.107P

The ΔGrxno for respiration 1.00 g of glucose is -16.0 kJ/g_.

Explanation of Solution

The resiparation of glucose reaction is,

C6H12O6(s) + 6O2(g) → 6CO2(g) + 6H2O(l)

Calculate the Free enrgy change ΔGrxno

Standared Free energy change equation is,

ΔG°rxn=∑nΔGf°(Products)-∑nΔGf°(Reactants)

Calculate the free energy change for this reaction as follows,

ΔG°rxn = [(1 mol CO2) ( ΔGfo of CO2) + (6 mol H2O) ( ΔGfo of H2O)] −[(1 mol C6H12O6)(ΔG°f of C6H12O6) + (6 mol O2)(ΔG°f of O2)] ΔG°rxn = [(1 mol CO2)(−394.4 kJ/mol ) + (6 mol H2O)(−237.192 kJ/mol )]− [(1 mol C6H12O6)(−910.56 kJ/mol ) + (6 mol O2)(0 kJ/mol )] ΔG°rxn = -2878.992 kJ.

Hence, the standard free energy is ΔG°rxn =-2878.992 kJ

Next solving for resipration 1.00 g of glucose,

ΔG°rxn/ g of glucose = (1.00 g glucose) (1 mol of glucose180. 16 g glucose) (−2878.992 kJ1 mol of glucose)ΔG°rxn/ g of glucose = −15.980195ΔG°rxn/ g of glucose =-16.0 kJ/g.

Therefore, the ΔGrxno for respiration 1.00 g of glucose is -16.0 kJ/g_.

(c)

Interpretation Introduction

Interpretation:

For fermentation reaction of 1.00 g glucose, the standard free energy ΔGrxno 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 G. All spontaneous process is associated with the decrease of free energy in the system. The standard free energy change (ΔG°rxn) is the difference in free energy of the reactants and products in their standard state.

ΔG°rxn=∑nΔGf°(Products)-∑nΔGf°(Reactants)

(c)

Expert Solution

Answer to Problem 20.107P

The ΔGrxno for fermentation 1.00 g of glucose is -1.26 kJ/g_.

Explanation of Solution

The resiparation of glucose reaction is,

C6H12O6(s) → 2C2H5OH(l) + 2CO2(g)

Calculate the Free enrgy change ΔGrxno

Standared Free energy change equation is,

ΔG°rxn=∑nΔGf°(Products)-∑nΔGf°(Reactants)

Calculate the free energy change for this reaction as follows,

ΔG°rxn = [(2 mol CO2) ( ΔGfo of CO2) + (2 mol C2H5OH) ( ΔGfo of C2H5OH)] −[(1 mol C6H12O6)(ΔG°f of C6H12O6)] ΔG°rxn = [(2 mol CO2)(−394.4 kJ/mol ) + (2 mol C2H5OH)(−174.8 kJ/mol )]− [(1 mol C6H12O6)(−910.56 kJ/mol )] ΔG°rxn = -227.84 kJ.

The fermentation reaction standard free energy is ΔG°rxn =-227.84 kJ.

Next solving for fermentation 1.00 g of glucose,

ΔG°rxn/ g of glucose = (1.00 g glucose) (1 mol of glucose180. 16 g glucose) (−227.84 kJ1 mol of glucose)ΔG°rxn/ g of glucose = −1.264653641ΔG°rxn/ g of glucose =-1.26 kJ/g.

Therefore, the ΔGrxno for fermentation 1.00 g of glucose is -1.26 kJ/g_.

(d)

Interpretation Introduction

Interpretation:

For oxidation reaction of 1.00 g glucose, the standard free energy ΔGrxno 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 G. All spontaneous process is associated with the decrease of free energy in the system. The standard free energy change (ΔG°rxn) is the difference in free energy of the reactants and products in their standard state.

ΔG°rxn=∑nΔGf°(Products)-∑nΔGf°(Reactants)

(d)

Expert Solution

Answer to Problem 20.107P

The 1.00 g of ethanol oxidation process standard free energy is -14.7 kJ/g_.

Explanation of Solution

The resiparation of glucose reaction is,

C2H5OH(l) + 3O2(g) → 2CO2(g) + 3H2O(l)

Calculate the Free enrgy change ΔGrxno

Standared Free energy change equation is,

ΔG°rxn=∑nΔGf°(Products)-∑nΔGf°(Reactants)

Calculate the free energy change for this reaction as follows,

ΔG°rxn = [(2 mol CO2) ( ΔGfo of CO2) + (3 mol H2O) ( ΔGfo of H2O)] −[(1 mol C2H5OH)(ΔG°f of C2H5OH) + (3 mol O2) ( ΔGfo of O2) ] ΔG°rxn = [(2 mol CO2)(−394.4 kJ/mol ) + (3 mol H2O)(−237.192 kJ/mol )]− [(1 mol C2H5OH)(−174.8 kJ/mol ) + (1 mol O2)(0 kJ/mol ) ] ΔG°rxn = -1325.576 kJ.

Ethanol oxidation reaction standard free energy is ΔG°rxn =-1325.576 kJ.

Next solving for fermentation 1.00 g of glucose,

ΔG°rxn = (1.00 g glucose) (1 mol of glucose180. 16 g glucose)(2 mol of C2H5OH1 mol of glucose) (−1325.576 kJ1 mol of C2H5OH)ΔG°rxn= −14.71554ΔG°rxn =-14.7 kJ/g.

Therefore, the 1.00 g of ethanol oxidation process standard free energy is -14.7 kJ/g_.

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Answer 3

Question

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.

(a)

Expert Solution

Answer to Problem 20.107P

Balanced equation is,

C6H12O6(s) + 6O2(g) → 6CO2(g) + 6H2O(l)C6H12O6(s) → 2C2H5OH(l) + 2CO2(g)C2H5OH(l) + 3O2(g) → 2CO2(g) + 3H2O(l)

Explanation of Solution

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

C6H12O6(s) + O2(g) → CO2(g) + H2O(l)C6H12O6(s) → C2H5OH(l) + CO2(g)C2H5OH(l) + O2(g) → CO2(g) + H2O(l)

Balancing the equation:

C6H12O6(s) + 6O2(g) → 6CO2(g) + 6H2O(l)C6H12O6(s) → 2C2H5OH(l) + 2CO2(g)C2H5OH(l) + 3O2(g) → 2CO2(g) + 3H2O(l)

The given process of respiration, glucose is oxidized completely to give CO2 and H2O.

In fermentation process, oxygen O2 is absent and glucose is broken down to ethanol and CO2.

In third reaction, ethanol is oxidized to CO2 and H2O.

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 1.00 g of glucose, the standard free energy ΔGrxno 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 G. All spontaneous process is associated with the decrease of free energy in the system. The standard free energy change (ΔG°rxn) is the difference in free energy of the reactants and products in their standard state.

ΔG°rxn=∑nΔGf°(Products)-∑nΔGf°(Reactants)

(b)

Expert Solution

Answer to Problem 20.107P

The ΔGrxno for respiration 1.00 g of glucose is -16.0 kJ/g_.

Explanation of Solution

The resiparation of glucose reaction is,

C6H12O6(s) + 6O2(g) → 6CO2(g) + 6H2O(l)

Calculate the Free enrgy change ΔGrxno

Standared Free energy change equation is,

ΔG°rxn=∑nΔGf°(Products)-∑nΔGf°(Reactants)

Calculate the free energy change for this reaction as follows,

ΔG°rxn = [(1 mol CO2) ( ΔGfo of CO2) + (6 mol H2O) ( ΔGfo of H2O)] −[(1 mol C6H12O6)(ΔG°f of C6H12O6) + (6 mol O2)(ΔG°f of O2)] ΔG°rxn = [(1 mol CO2)(−394.4 kJ/mol ) + (6 mol H2O)(−237.192 kJ/mol )]− [(1 mol C6H12O6)(−910.56 kJ/mol ) + (6 mol O2)(0 kJ/mol )] ΔG°rxn = -2878.992 kJ.

Hence, the standard free energy is ΔG°rxn =-2878.992 kJ

Next solving for resipration 1.00 g of glucose,

ΔG°rxn/ g of glucose = (1.00 g glucose) (1 mol of glucose180. 16 g glucose) (−2878.992 kJ1 mol of glucose)ΔG°rxn/ g of glucose = −15.980195ΔG°rxn/ g of glucose =-16.0 kJ/g.

Therefore, the ΔGrxno for respiration 1.00 g of glucose is -16.0 kJ/g_.

(c)

Interpretation Introduction

Interpretation:

For fermentation reaction of 1.00 g glucose, the standard free energy ΔGrxno 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 G. All spontaneous process is associated with the decrease of free energy in the system. The standard free energy change (ΔG°rxn) is the difference in free energy of the reactants and products in their standard state.

ΔG°rxn=∑nΔGf°(Products)-∑nΔGf°(Reactants)

(c)

Expert Solution

Answer to Problem 20.107P

The ΔGrxno for fermentation 1.00 g of glucose is -1.26 kJ/g_.

Explanation of Solution

The resiparation of glucose reaction is,

C6H12O6(s) → 2C2H5OH(l) + 2CO2(g)

Calculate the Free enrgy change ΔGrxno

Standared Free energy change equation is,

ΔG°rxn=∑nΔGf°(Products)-∑nΔGf°(Reactants)

Calculate the free energy change for this reaction as follows,

ΔG°rxn = [(2 mol CO2) ( ΔGfo of CO2) + (2 mol C2H5OH) ( ΔGfo of C2H5OH)] −[(1 mol C6H12O6)(ΔG°f of C6H12O6)] ΔG°rxn = [(2 mol CO2)(−394.4 kJ/mol ) + (2 mol C2H5OH)(−174.8 kJ/mol )]− [(1 mol C6H12O6)(−910.56 kJ/mol )] ΔG°rxn = -227.84 kJ.

The fermentation reaction standard free energy is ΔG°rxn =-227.84 kJ.

Next solving for fermentation 1.00 g of glucose,

ΔG°rxn/ g of glucose = (1.00 g glucose) (1 mol of glucose180. 16 g glucose) (−227.84 kJ1 mol of glucose)ΔG°rxn/ g of glucose = −1.264653641ΔG°rxn/ g of glucose =-1.26 kJ/g.

Therefore, the ΔGrxno for fermentation 1.00 g of glucose is -1.26 kJ/g_.

(d)

Interpretation Introduction

Interpretation:

For oxidation reaction of 1.00 g glucose, the standard free energy ΔGrxno 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 G. All spontaneous process is associated with the decrease of free energy in the system. The standard free energy change (ΔG°rxn) is the difference in free energy of the reactants and products in their standard state.

ΔG°rxn=∑nΔGf°(Products)-∑nΔGf°(Reactants)

(d)

Expert Solution

Answer to Problem 20.107P

The 1.00 g of ethanol oxidation process standard free energy is -14.7 kJ/g_.

Explanation of Solution

The resiparation of glucose reaction is,

C2H5OH(l) + 3O2(g) → 2CO2(g) + 3H2O(l)

Calculate the Free enrgy change ΔGrxno

Standared Free energy change equation is,

ΔG°rxn=∑nΔGf°(Products)-∑nΔGf°(Reactants)

Calculate the free energy change for this reaction as follows,

ΔG°rxn = [(2 mol CO2) ( ΔGfo of CO2) + (3 mol H2O) ( ΔGfo of H2O)] −[(1 mol C2H5OH)(ΔG°f of C2H5OH) + (3 mol O2) ( ΔGfo of O2) ] ΔG°rxn = [(2 mol CO2)(−394.4 kJ/mol ) + (3 mol H2O)(−237.192 kJ/mol )]− [(1 mol C2H5OH)(−174.8 kJ/mol ) + (1 mol O2)(0 kJ/mol ) ] ΔG°rxn = -1325.576 kJ.

Ethanol oxidation reaction standard free energy is ΔG°rxn =-1325.576 kJ.

Next solving for fermentation 1.00 g of glucose,

ΔG°rxn = (1.00 g glucose) (1 mol of glucose180. 16 g glucose)(2 mol of C2H5OH1 mol of glucose) (−1325.576 kJ1 mol of C2H5OH)ΔG°rxn= −14.71554ΔG°rxn =-14.7 kJ/g.

Therefore, the 1.00 g of ethanol oxidation process standard free energy is -14.7 kJ/g_.

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Answer 4

Question

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.

(a)

Expert Solution

Answer to Problem 20.107P

Balanced equation is,

C6H12O6(s) + 6O2(g) → 6CO2(g) + 6H2O(l)C6H12O6(s) → 2C2H5OH(l) + 2CO2(g)C2H5OH(l) + 3O2(g) → 2CO2(g) + 3H2O(l)

Explanation of Solution

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

C6H12O6(s) + O2(g) → CO2(g) + H2O(l)C6H12O6(s) → C2H5OH(l) + CO2(g)C2H5OH(l) + O2(g) → CO2(g) + H2O(l)

Balancing the equation:

C6H12O6(s) + 6O2(g) → 6CO2(g) + 6H2O(l)C6H12O6(s) → 2C2H5OH(l) + 2CO2(g)C2H5OH(l) + 3O2(g) → 2CO2(g) + 3H2O(l)

The given process of respiration, glucose is oxidized completely to give CO2 and H2O.

In fermentation process, oxygen O2 is absent and glucose is broken down to ethanol and CO2.

In third reaction, ethanol is oxidized to CO2 and H2O.

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 1.00 g of glucose, the standard free energy ΔGrxno 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 G. All spontaneous process is associated with the decrease of free energy in the system. The standard free energy change (ΔG°rxn) is the difference in free energy of the reactants and products in their standard state.

ΔG°rxn=∑nΔGf°(Products)-∑nΔGf°(Reactants)

(b)

Expert Solution

Answer to Problem 20.107P

The ΔGrxno for respiration 1.00 g of glucose is -16.0 kJ/g_.

Explanation of Solution

The resiparation of glucose reaction is,

C6H12O6(s) + 6O2(g) → 6CO2(g) + 6H2O(l)

Calculate the Free enrgy change ΔGrxno

Standared Free energy change equation is,

ΔG°rxn=∑nΔGf°(Products)-∑nΔGf°(Reactants)

Calculate the free energy change for this reaction as follows,

ΔG°rxn = [(1 mol CO2) ( ΔGfo of CO2) + (6 mol H2O) ( ΔGfo of H2O)] −[(1 mol C6H12O6)(ΔG°f of C6H12O6) + (6 mol O2)(ΔG°f of O2)] ΔG°rxn = [(1 mol CO2)(−394.4 kJ/mol ) + (6 mol H2O)(−237.192 kJ/mol )]− [(1 mol C6H12O6)(−910.56 kJ/mol ) + (6 mol O2)(0 kJ/mol )] ΔG°rxn = -2878.992 kJ.

Hence, the standard free energy is ΔG°rxn =-2878.992 kJ

Next solving for resipration 1.00 g of glucose,

ΔG°rxn/ g of glucose = (1.00 g glucose) (1 mol of glucose180. 16 g glucose) (−2878.992 kJ1 mol of glucose)ΔG°rxn/ g of glucose = −15.980195ΔG°rxn/ g of glucose =-16.0 kJ/g.

Therefore, the ΔGrxno for respiration 1.00 g of glucose is -16.0 kJ/g_.

(c)

Interpretation Introduction

Interpretation:

For fermentation reaction of 1.00 g glucose, the standard free energy ΔGrxno 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 G. All spontaneous process is associated with the decrease of free energy in the system. The standard free energy change (ΔG°rxn) is the difference in free energy of the reactants and products in their standard state.

ΔG°rxn=∑nΔGf°(Products)-∑nΔGf°(Reactants)

(c)

Expert Solution

Answer to Problem 20.107P

The ΔGrxno for fermentation 1.00 g of glucose is -1.26 kJ/g_.

Explanation of Solution

The resiparation of glucose reaction is,

C6H12O6(s) → 2C2H5OH(l) + 2CO2(g)

Calculate the Free enrgy change ΔGrxno

Standared Free energy change equation is,

ΔG°rxn=∑nΔGf°(Products)-∑nΔGf°(Reactants)

Calculate the free energy change for this reaction as follows,

ΔG°rxn = [(2 mol CO2) ( ΔGfo of CO2) + (2 mol C2H5OH) ( ΔGfo of C2H5OH)] −[(1 mol C6H12O6)(ΔG°f of C6H12O6)] ΔG°rxn = [(2 mol CO2)(−394.4 kJ/mol ) + (2 mol C2H5OH)(−174.8 kJ/mol )]− [(1 mol C6H12O6)(−910.56 kJ/mol )] ΔG°rxn = -227.84 kJ.

The fermentation reaction standard free energy is ΔG°rxn =-227.84 kJ.

Next solving for fermentation 1.00 g of glucose,

ΔG°rxn/ g of glucose = (1.00 g glucose) (1 mol of glucose180. 16 g glucose) (−227.84 kJ1 mol of glucose)ΔG°rxn/ g of glucose = −1.264653641ΔG°rxn/ g of glucose =-1.26 kJ/g.

Therefore, the ΔGrxno for fermentation 1.00 g of glucose is -1.26 kJ/g_.

(d)

Interpretation Introduction

Interpretation:

For oxidation reaction of 1.00 g glucose, the standard free energy ΔGrxno 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 G. All spontaneous process is associated with the decrease of free energy in the system. The standard free energy change (ΔG°rxn) is the difference in free energy of the reactants and products in their standard state.

ΔG°rxn=∑nΔGf°(Products)-∑nΔGf°(Reactants)

(d)

Expert Solution

Answer to Problem 20.107P

The 1.00 g of ethanol oxidation process standard free energy is -14.7 kJ/g_.

Explanation of Solution

The resiparation of glucose reaction is,

C2H5OH(l) + 3O2(g) → 2CO2(g) + 3H2O(l)

Calculate the Free enrgy change ΔGrxno

Standared Free energy change equation is,

ΔG°rxn=∑nΔGf°(Products)-∑nΔGf°(Reactants)

Calculate the free energy change for this reaction as follows,

ΔG°rxn = [(2 mol CO2) ( ΔGfo of CO2) + (3 mol H2O) ( ΔGfo of H2O)] −[(1 mol C2H5OH)(ΔG°f of C2H5OH) + (3 mol O2) ( ΔGfo of O2) ] ΔG°rxn = [(2 mol CO2)(−394.4 kJ/mol ) + (3 mol H2O)(−237.192 kJ/mol )]− [(1 mol C2H5OH)(−174.8 kJ/mol ) + (1 mol O2)(0 kJ/mol ) ] ΔG°rxn = -1325.576 kJ.

Ethanol oxidation reaction standard free energy is ΔG°rxn =-1325.576 kJ.

Next solving for fermentation 1.00 g of glucose,

ΔG°rxn = (1.00 g glucose) (1 mol of glucose180. 16 g glucose)(2 mol of C2H5OH1 mol of glucose) (−1325.576 kJ1 mol of C2H5OH)ΔG°rxn= −14.71554ΔG°rxn =-14.7 kJ/g.

Therefore, the 1.00 g of ethanol oxidation process standard free energy is -14.7 kJ/g_.

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Answer 5

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.

(a)

Expert Solution

Answer to Problem 20.107P

Balanced equation is,

C6H12O6(s) + 6O2(g) → 6CO2(g) + 6H2O(l)C6H12O6(s) → 2C2H5OH(l) + 2CO2(g)C2H5OH(l) + 3O2(g) → 2CO2(g) + 3H2O(l)

Explanation of Solution

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

C6H12O6(s) + O2(g) → CO2(g) + H2O(l)C6H12O6(s) → C2H5OH(l) + CO2(g)C2H5OH(l) + O2(g) → CO2(g) + H2O(l)

Balancing the equation:

C6H12O6(s) + 6O2(g) → 6CO2(g) + 6H2O(l)C6H12O6(s) → 2C2H5OH(l) + 2CO2(g)C2H5OH(l) + 3O2(g) → 2CO2(g) + 3H2O(l)

The given process of respiration, glucose is oxidized completely to give CO2 and H2O.

In fermentation process, oxygen O2 is absent and glucose is broken down to ethanol and CO2.

In third reaction, ethanol is oxidized to CO2 and H2O.

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 1.00 g of glucose, the standard free energy ΔGrxno 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 G. All spontaneous process is associated with the decrease of free energy in the system. The standard free energy change (ΔG°rxn) is the difference in free energy of the reactants and products in their standard state.

ΔG°rxn=∑nΔGf°(Products)-∑nΔGf°(Reactants)

(b)

Expert Solution

Answer to Problem 20.107P

The ΔGrxno for respiration 1.00 g of glucose is -16.0 kJ/g_.

Explanation of Solution

The resiparation of glucose reaction is,

C6H12O6(s) + 6O2(g) → 6CO2(g) + 6H2O(l)

Calculate the Free enrgy change ΔGrxno

Standared Free energy change equation is,

ΔG°rxn=∑nΔGf°(Products)-∑nΔGf°(Reactants)

Calculate the free energy change for this reaction as follows,

ΔG°rxn = [(1 mol CO2) ( ΔGfo of CO2) + (6 mol H2O) ( ΔGfo of H2O)] −[(1 mol C6H12O6)(ΔG°f of C6H12O6) + (6 mol O2)(ΔG°f of O2)] ΔG°rxn = [(1 mol CO2)(−394.4 kJ/mol ) + (6 mol H2O)(−237.192 kJ/mol )]− [(1 mol C6H12O6)(−910.56 kJ/mol ) + (6 mol O2)(0 kJ/mol )] ΔG°rxn = -2878.992 kJ.

Hence, the standard free energy is ΔG°rxn =-2878.992 kJ

Next solving for resipration 1.00 g of glucose,

ΔG°rxn/ g of glucose = (1.00 g glucose) (1 mol of glucose180. 16 g glucose) (−2878.992 kJ1 mol of glucose)ΔG°rxn/ g of glucose = −15.980195ΔG°rxn/ g of glucose =-16.0 kJ/g.

Therefore, the ΔGrxno for respiration 1.00 g of glucose is -16.0 kJ/g_.

(c)

Interpretation Introduction

Interpretation:

For fermentation reaction of 1.00 g glucose, the standard free energy ΔGrxno 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 G. All spontaneous process is associated with the decrease of free energy in the system. The standard free energy change (ΔG°rxn) is the difference in free energy of the reactants and products in their standard state.

ΔG°rxn=∑nΔGf°(Products)-∑nΔGf°(Reactants)

(c)

Expert Solution

Answer to Problem 20.107P

The ΔGrxno for fermentation 1.00 g of glucose is -1.26 kJ/g_.

Explanation of Solution

The resiparation of glucose reaction is,

C6H12O6(s) → 2C2H5OH(l) + 2CO2(g)

Calculate the Free enrgy change ΔGrxno

Standared Free energy change equation is,

ΔG°rxn=∑nΔGf°(Products)-∑nΔGf°(Reactants)

Calculate the free energy change for this reaction as follows,

ΔG°rxn = [(2 mol CO2) ( ΔGfo of CO2) + (2 mol C2H5OH) ( ΔGfo of C2H5OH)] −[(1 mol C6H12O6)(ΔG°f of C6H12O6)] ΔG°rxn = [(2 mol CO2)(−394.4 kJ/mol ) + (2 mol C2H5OH)(−174.8 kJ/mol )]− [(1 mol C6H12O6)(−910.56 kJ/mol )] ΔG°rxn = -227.84 kJ.

The fermentation reaction standard free energy is ΔG°rxn =-227.84 kJ.

Next solving for fermentation 1.00 g of glucose,

ΔG°rxn/ g of glucose = (1.00 g glucose) (1 mol of glucose180. 16 g glucose) (−227.84 kJ1 mol of glucose)ΔG°rxn/ g of glucose = −1.264653641ΔG°rxn/ g of glucose =-1.26 kJ/g.

Therefore, the ΔGrxno for fermentation 1.00 g of glucose is -1.26 kJ/g_.

(d)

Interpretation Introduction

Interpretation:

For oxidation reaction of 1.00 g glucose, the standard free energy ΔGrxno 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 G. All spontaneous process is associated with the decrease of free energy in the system. The standard free energy change (ΔG°rxn) is the difference in free energy of the reactants and products in their standard state.

ΔG°rxn=∑nΔGf°(Products)-∑nΔGf°(Reactants)

(d)

Expert Solution

Answer to Problem 20.107P

The 1.00 g of ethanol oxidation process standard free energy is -14.7 kJ/g_.

Explanation of Solution

The resiparation of glucose reaction is,

C2H5OH(l) + 3O2(g) → 2CO2(g) + 3H2O(l)

Calculate the Free enrgy change ΔGrxno

Standared Free energy change equation is,

ΔG°rxn=∑nΔGf°(Products)-∑nΔGf°(Reactants)

Calculate the free energy change for this reaction as follows,

ΔG°rxn = [(2 mol CO2) ( ΔGfo of CO2) + (3 mol H2O) ( ΔGfo of H2O)] −[(1 mol C2H5OH)(ΔG°f of C2H5OH) + (3 mol O2) ( ΔGfo of O2) ] ΔG°rxn = [(2 mol CO2)(−394.4 kJ/mol ) + (3 mol H2O)(−237.192 kJ/mol )]− [(1 mol C2H5OH)(−174.8 kJ/mol ) + (1 mol O2)(0 kJ/mol ) ] ΔG°rxn = -1325.576 kJ.

Ethanol oxidation reaction standard free energy is ΔG°rxn =-1325.576 kJ.

Next solving for fermentation 1.00 g of glucose,

ΔG°rxn = (1.00 g glucose) (1 mol of glucose180. 16 g glucose)(2 mol of C2H5OH1 mol of glucose) (−1325.576 kJ1 mol of C2H5OH)ΔG°rxn= −14.71554ΔG°rxn =-14.7 kJ/g.

Therefore, the 1.00 g of ethanol oxidation process standard free energy is -14.7 kJ/g_.

Answer 6

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.

(a)

Expert Solution

Answer to Problem 20.107P

Balanced equation is,

C6H12O6(s) + 6O2(g) → 6CO2(g) + 6H2O(l)C6H12O6(s) → 2C2H5OH(l) + 2CO2(g)C2H5OH(l) + 3O2(g) → 2CO2(g) + 3H2O(l)

Explanation of Solution

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

C6H12O6(s) + O2(g) → CO2(g) + H2O(l)C6H12O6(s) → C2H5OH(l) + CO2(g)C2H5OH(l) + O2(g) → CO2(g) + H2O(l)

Balancing the equation:

C6H12O6(s) + 6O2(g) → 6CO2(g) + 6H2O(l)C6H12O6(s) → 2C2H5OH(l) + 2CO2(g)C2H5OH(l) + 3O2(g) → 2CO2(g) + 3H2O(l)

The given process of respiration, glucose is oxidized completely to give CO2 and H2O.

In fermentation process, oxygen O2 is absent and glucose is broken down to ethanol and CO2.

In third reaction, ethanol is oxidized to CO2 and H2O.

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

Answer 7

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 8

(a)

Expert Solution

Answer to Problem 20.107P

Balanced equation is,

C6H12O6(s) + 6O2(g) → 6CO2(g) + 6H2O(l)C6H12O6(s) → 2C2H5OH(l) + 2CO2(g)C2H5OH(l) + 3O2(g) → 2CO2(g) + 3H2O(l)

Answer 9

(a)

Expert Solution

Answer 10

(a)

Expert Solution

Answer 11

Expert Solution

Answer 12

Explanation of Solution

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

C6H12O6(s) + O2(g) → CO2(g) + H2O(l)C6H12O6(s) → C2H5OH(l) + CO2(g)C2H5OH(l) + O2(g) → CO2(g) + H2O(l)

Balancing the equation:

C6H12O6(s) + 6O2(g) → 6CO2(g) + 6H2O(l)C6H12O6(s) → 2C2H5OH(l) + 2CO2(g)C2H5OH(l) + 3O2(g) → 2CO2(g) + 3H2O(l)

The given process of respiration, glucose is oxidized completely to give CO2 and H2O.

In fermentation process, oxygen O2 is absent and glucose is broken down to ethanol and CO2.

In third reaction, ethanol is oxidized to CO2 and H2O.

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

Answer 13

(b)

Interpretation Introduction

Interpretation:

For respiration reaction 1.00 g of glucose, the standard free energy ΔGrxno 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 G. All spontaneous process is associated with the decrease of free energy in the system. The standard free energy change (ΔG°rxn) is the difference in free energy of the reactants and products in their standard state.

ΔG°rxn=∑nΔGf°(Products)-∑nΔGf°(Reactants)

(b)

Expert Solution

Answer to Problem 20.107P

The ΔGrxno for respiration 1.00 g of glucose is -16.0 kJ/g_.

Explanation of Solution

The resiparation of glucose reaction is,

C6H12O6(s) + 6O2(g) → 6CO2(g) + 6H2O(l)

Calculate the Free enrgy change ΔGrxno

Standared Free energy change equation is,

ΔG°rxn=∑nΔGf°(Products)-∑nΔGf°(Reactants)

Calculate the free energy change for this reaction as follows,

ΔG°rxn = [(1 mol CO2) ( ΔGfo of CO2) + (6 mol H2O) ( ΔGfo of H2O)] −[(1 mol C6H12O6)(ΔG°f of C6H12O6) + (6 mol O2)(ΔG°f of O2)] ΔG°rxn = [(1 mol CO2)(−394.4 kJ/mol ) + (6 mol H2O)(−237.192 kJ/mol )]− [(1 mol C6H12O6)(−910.56 kJ/mol ) + (6 mol O2)(0 kJ/mol )] ΔG°rxn = -2878.992 kJ.

Hence, the standard free energy is ΔG°rxn =-2878.992 kJ

Next solving for resipration 1.00 g of glucose,

ΔG°rxn/ g of glucose = (1.00 g glucose) (1 mol of glucose180. 16 g glucose) (−2878.992 kJ1 mol of glucose)ΔG°rxn/ g of glucose = −15.980195ΔG°rxn/ g of glucose =-16.0 kJ/g.

Therefore, the ΔGrxno for respiration 1.00 g of glucose is -16.0 kJ/g_.

Answer 14

(b)

Interpretation Introduction

Interpretation:

For respiration reaction 1.00 g of glucose, the standard free energy ΔGrxno 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 G. All spontaneous process is associated with the decrease of free energy in the system. The standard free energy change (ΔG°rxn) is the difference in free energy of the reactants and products in their standard state.

ΔG°rxn=∑nΔGf°(Products)-∑nΔGf°(Reactants)

Answer 15

(b)

Expert Solution

Answer to Problem 20.107P

The ΔGrxno for respiration 1.00 g of glucose is -16.0 kJ/g_.

Answer 16

(b)

Expert Solution

Answer 17

(b)

Expert Solution

Answer 18

Expert Solution

Answer 19

Explanation of Solution

The resiparation of glucose reaction is,

C6H12O6(s) + 6O2(g) → 6CO2(g) + 6H2O(l)

Calculate the Free enrgy change ΔGrxno

Standared Free energy change equation is,

ΔG°rxn=∑nΔGf°(Products)-∑nΔGf°(Reactants)

Calculate the free energy change for this reaction as follows,

ΔG°rxn = [(1 mol CO2) ( ΔGfo of CO2) + (6 mol H2O) ( ΔGfo of H2O)] −[(1 mol C6H12O6)(ΔG°f of C6H12O6) + (6 mol O2)(ΔG°f of O2)] ΔG°rxn = [(1 mol CO2)(−394.4 kJ/mol ) + (6 mol H2O)(−237.192 kJ/mol )]− [(1 mol C6H12O6)(−910.56 kJ/mol ) + (6 mol O2)(0 kJ/mol )] ΔG°rxn = -2878.992 kJ.

Hence, the standard free energy is ΔG°rxn =-2878.992 kJ

Next solving for resipration 1.00 g of glucose,

ΔG°rxn/ g of glucose = (1.00 g glucose) (1 mol of glucose180. 16 g glucose) (−2878.992 kJ1 mol of glucose)ΔG°rxn/ g of glucose = −15.980195ΔG°rxn/ g of glucose =-16.0 kJ/g.

Therefore, the ΔGrxno for respiration 1.00 g of glucose is -16.0 kJ/g_.

Answer 20

(c)

Interpretation Introduction

Interpretation:

For fermentation reaction of 1.00 g glucose, the standard free energy ΔGrxno 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 G. All spontaneous process is associated with the decrease of free energy in the system. The standard free energy change (ΔG°rxn) is the difference in free energy of the reactants and products in their standard state.

ΔG°rxn=∑nΔGf°(Products)-∑nΔGf°(Reactants)

(c)

Expert Solution

Answer to Problem 20.107P

The ΔGrxno for fermentation 1.00 g of glucose is -1.26 kJ/g_.

Explanation of Solution

The resiparation of glucose reaction is,

C6H12O6(s) → 2C2H5OH(l) + 2CO2(g)

Calculate the Free enrgy change ΔGrxno

Standared Free energy change equation is,

ΔG°rxn=∑nΔGf°(Products)-∑nΔGf°(Reactants)

Calculate the free energy change for this reaction as follows,

ΔG°rxn = [(2 mol CO2) ( ΔGfo of CO2) + (2 mol C2H5OH) ( ΔGfo of C2H5OH)] −[(1 mol C6H12O6)(ΔG°f of C6H12O6)] ΔG°rxn = [(2 mol CO2)(−394.4 kJ/mol ) + (2 mol C2H5OH)(−174.8 kJ/mol )]− [(1 mol C6H12O6)(−910.56 kJ/mol )] ΔG°rxn = -227.84 kJ.

The fermentation reaction standard free energy is ΔG°rxn =-227.84 kJ.

Next solving for fermentation 1.00 g of glucose,

ΔG°rxn/ g of glucose = (1.00 g glucose) (1 mol of glucose180. 16 g glucose) (−227.84 kJ1 mol of glucose)ΔG°rxn/ g of glucose = −1.264653641ΔG°rxn/ g of glucose =-1.26 kJ/g.

Therefore, the ΔGrxno for fermentation 1.00 g of glucose is -1.26 kJ/g_.

Answer 21

(c)

Interpretation Introduction

Interpretation:

For fermentation reaction of 1.00 g glucose, the standard free energy ΔGrxno 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 G. All spontaneous process is associated with the decrease of free energy in the system. The standard free energy change (ΔG°rxn) is the difference in free energy of the reactants and products in their standard state.

ΔG°rxn=∑nΔGf°(Products)-∑nΔGf°(Reactants)

Answer 22

(c)

Expert Solution

Answer to Problem 20.107P

The ΔGrxno for fermentation 1.00 g of glucose is -1.26 kJ/g_.

Answer 23

(c)

Expert Solution

Answer 24

(c)

Expert Solution

Answer 25

Expert Solution

Answer 26

Explanation of Solution

The resiparation of glucose reaction is,

C6H12O6(s) → 2C2H5OH(l) + 2CO2(g)

Calculate the Free enrgy change ΔGrxno

Standared Free energy change equation is,

ΔG°rxn=∑nΔGf°(Products)-∑nΔGf°(Reactants)

Calculate the free energy change for this reaction as follows,

ΔG°rxn = [(2 mol CO2) ( ΔGfo of CO2) + (2 mol C2H5OH) ( ΔGfo of C2H5OH)] −[(1 mol C6H12O6)(ΔG°f of C6H12O6)] ΔG°rxn = [(2 mol CO2)(−394.4 kJ/mol ) + (2 mol C2H5OH)(−174.8 kJ/mol )]− [(1 mol C6H12O6)(−910.56 kJ/mol )] ΔG°rxn = -227.84 kJ.

The fermentation reaction standard free energy is ΔG°rxn =-227.84 kJ.

Next solving for fermentation 1.00 g of glucose,

ΔG°rxn/ g of glucose = (1.00 g glucose) (1 mol of glucose180. 16 g glucose) (−227.84 kJ1 mol of glucose)ΔG°rxn/ g of glucose = −1.264653641ΔG°rxn/ g of glucose =-1.26 kJ/g.

Therefore, the ΔGrxno for fermentation 1.00 g of glucose is -1.26 kJ/g_.

Answer 27

(d)

Interpretation Introduction

Interpretation:

For oxidation reaction of 1.00 g glucose, the standard free energy ΔGrxno 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 G. All spontaneous process is associated with the decrease of free energy in the system. The standard free energy change (ΔG°rxn) is the difference in free energy of the reactants and products in their standard state.

ΔG°rxn=∑nΔGf°(Products)-∑nΔGf°(Reactants)

(d)

Expert Solution

Answer to Problem 20.107P

The 1.00 g of ethanol oxidation process standard free energy is -14.7 kJ/g_.

Explanation of Solution

The resiparation of glucose reaction is,

C2H5OH(l) + 3O2(g) → 2CO2(g) + 3H2O(l)

Calculate the Free enrgy change ΔGrxno

Standared Free energy change equation is,

ΔG°rxn=∑nΔGf°(Products)-∑nΔGf°(Reactants)

Calculate the free energy change for this reaction as follows,

ΔG°rxn = [(2 mol CO2) ( ΔGfo of CO2) + (3 mol H2O) ( ΔGfo of H2O)] −[(1 mol C2H5OH)(ΔG°f of C2H5OH) + (3 mol O2) ( ΔGfo of O2) ] ΔG°rxn = [(2 mol CO2)(−394.4 kJ/mol ) + (3 mol H2O)(−237.192 kJ/mol )]− [(1 mol C2H5OH)(−174.8 kJ/mol ) + (1 mol O2)(0 kJ/mol ) ] ΔG°rxn = -1325.576 kJ.

Ethanol oxidation reaction standard free energy is ΔG°rxn =-1325.576 kJ.

Next solving for fermentation 1.00 g of glucose,

ΔG°rxn = (1.00 g glucose) (1 mol of glucose180. 16 g glucose)(2 mol of C2H5OH1 mol of glucose) (−1325.576 kJ1 mol of C2H5OH)ΔG°rxn= −14.71554ΔG°rxn =-14.7 kJ/g.

Therefore, the 1.00 g of ethanol oxidation process standard free energy is -14.7 kJ/g_.

Answer 28

(d)

Interpretation Introduction

Interpretation:

For oxidation reaction of 1.00 g glucose, the standard free energy ΔGrxno 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 G. All spontaneous process is associated with the decrease of free energy in the system. The standard free energy change (ΔG°rxn) is the difference in free energy of the reactants and products in their standard state.

ΔG°rxn=∑nΔGf°(Products)-∑nΔGf°(Reactants)

Answer 29

(d)

Expert Solution

Answer to Problem 20.107P

The 1.00 g of ethanol oxidation process standard free energy is -14.7 kJ/g_.

Answer 30

(d)

Expert Solution

Answer 31

(d)

Expert Solution

Answer 32

Expert Solution

Answer 33

Explanation of Solution

The resiparation of glucose reaction is,

C2H5OH(l) + 3O2(g) → 2CO2(g) + 3H2O(l)

Calculate the Free enrgy change ΔGrxno

Standared Free energy change equation is,

ΔG°rxn=∑nΔGf°(Products)-∑nΔGf°(Reactants)

Calculate the free energy change for this reaction as follows,

ΔG°rxn = [(2 mol CO2) ( ΔGfo of CO2) + (3 mol H2O) ( ΔGfo of H2O)] −[(1 mol C2H5OH)(ΔG°f of C2H5OH) + (3 mol O2) ( ΔGfo of O2) ] ΔG°rxn = [(2 mol CO2)(−394.4 kJ/mol ) + (3 mol H2O)(−237.192 kJ/mol )]− [(1 mol C2H5OH)(−174.8 kJ/mol ) + (1 mol O2)(0 kJ/mol ) ] ΔG°rxn = -1325.576 kJ.

Ethanol oxidation reaction standard free energy is ΔG°rxn =-1325.576 kJ.

Next solving for fermentation 1.00 g of glucose,

ΔG°rxn = (1.00 g glucose) (1 mol of glucose180. 16 g glucose)(2 mol of C2H5OH1 mol of glucose) (−1325.576 kJ1 mol of C2H5OH)ΔG°rxn= −14.71554ΔG°rxn =-14.7 kJ/g.