Loose Leaf for Chemistry: The Molecular Nature of Matter and Change
Loose Leaf for Chemistry: The Molecular Nature of Matter and Change
8th Edition
ISBN: 9781260151749
Author: Silberberg Dr., Martin; Amateis Professor, Patricia
Publisher: McGraw-Hill Education
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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
Check Mark

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 CO2andH2O.

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

In third reaction, ethanol is oxidized to CO2andH2O.

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 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
Check Mark

Answer to Problem 20.107P

The ΔGrxno for respiration 1.00g of glucose is -16.0kJ/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 =[(1molCO2)(ΔGfoofCO2)+(6molH2O)(ΔGfoofH2O)][(1molC6H12O6)(ΔG°fofC6H12O6)+(6molO2)(ΔG°fofO2)]ΔG°rxn =[(1molCO2)(394.4kJ/mol)+(6molH2O)(237.192kJ/mol)][(1molC6H12O6)(910.56kJ/mol)+(6molO2)(0kJ/mol)]ΔG°rxn =-2878.992kJ.

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

Next solving for resipration 1.00g of glucose,

  ΔG°rxn/gofglucose=(1.00gglucose)(1molofglucose180.16gglucose)(2878.992kJ1molofglucose)ΔG°rxn/gofglucose=15.980195ΔG°rxn/gofglucose=-16.0kJ/g.

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

(c)

Interpretation Introduction

Interpretation:

For fermentation reaction of 1.00g 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
Check Mark

Answer to Problem 20.107P

The ΔGrxno for fermentation 1.00g of glucose is -1.26kJ/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 =[(2molCO2)(ΔGfoofCO2)+(2molC2H5OH)(ΔGfoofC2H5OH)][(1molC6H12O6)(ΔG°fofC6H12O6)]ΔG°rxn =[(2molCO2)(394.4kJ/mol)+(2molC2H5OH)(174.8kJ/mol)][(1molC6H12O6)(910.56kJ/mol)]ΔG°rxn =-227.84kJ.

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

Next solving for fermentation 1.00g of glucose,

  ΔG°rxn/gofglucose=(1.00gglucose)(1molofglucose180.16gglucose)(227.84kJ1molofglucose)ΔG°rxn/gofglucose=1.264653641ΔG°rxn/gofglucose=-1.26kJ/g.

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

(d)

Interpretation Introduction

Interpretation:

For oxidation reaction of 1.00g 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
Check Mark

Answer to Problem 20.107P

The 1.00 g of ethanol oxidation process standard free energy is -14.7kJ/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 =[(2molCO2)(ΔGfoofCO2)+(3molH2O)(ΔGfoofH2O)][(1molC2H5OH)(ΔG°fofC2H5OH)+(3molO2)(ΔGfoofO2)]ΔG°rxn =[(2molCO2)(394.4kJ/mol)+(3molH2O)(237.192kJ/mol)][(1molC2H5OH)(174.8kJ/mol)+(1molO2)(0kJ/mol)]ΔG°rxn =-1325.576kJ.

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

Next solving for fermentation 1.00g of glucose,

  ΔG°rxn=(1.00gglucose)(1molofglucose180.16gglucose)(2molofC2H5OH1molofglucose)(1325.576kJ1molofC2H5OH)ΔG°rxn=14.71554ΔG°rxn=-14.7kJ/g.

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

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Chapter 20 Solutions

Loose Leaf for Chemistry: The Molecular Nature of Matter and Change

Ch. 20.3 - Prob. 20.6AFPCh. 20.3 - Prob. 20.6BFPCh. 20.3 - Prob. 20.7AFPCh. 20.3 - Prob. 20.7BFPCh. 20.3 - Prob. 20.8AFPCh. 20.3 - Prob. 20.8BFPCh. 20.3 - Prob. B20.1PCh. 20.3 - Nonspontaneous processes like muscle contraction,...Ch. 20.4 - Use Appendix B to find K at 298 K for the...Ch. 20.4 - Use the given value of K to calculate ΔG° at 298 K...Ch. 20.4 - Prob. 20.10AFPCh. 20.4 - Prob. 20.10BFPCh. 20.4 - At 298 K, ΔG° = −33.5 kJ/mol for the formation of...Ch. 20.4 - Prob. 20.11BFPCh. 20 - Prob. 20.1PCh. 20 - Distinguish between the terms spontaneous and...Ch. 20 - State the first law of thermodynamics in terms of...Ch. 20 - State qualitatively the relationship between...Ch. 20 - Why is ΔSvap of a substance always larger than...Ch. 20 - Prob. 20.6PCh. 20 - Prob. 20.7PCh. 20 - Which of these processes are spontaneous? (a)...Ch. 20 - Prob. 20.9PCh. 20 - Which of these processes are spontaneous? (a)...Ch. 20 - Prob. 20.11PCh. 20 - Prob. 20.12PCh. 20 - Prob. 20.13PCh. 20 - Prob. 20.14PCh. 20 - Prob. 20.15PCh. 20 - Prob. 20.16PCh. 20 - Prob. 20.17PCh. 20 - Prob. 20.18PCh. 20 - Prob. 20.19PCh. 20 - Prob. 20.20PCh. 20 - Prob. 20.21PCh. 20 - Prob. 20.22PCh. 20 - Prob. 20.23PCh. 20 - Prob. 20.24PCh. 20 - Predict which substance has greater molar entropy....Ch. 20 - Without consulting Appendix B, arrange each group...Ch. 20 - Without consulting Appendix B, arrange each group...Ch. 20 - Without consulting Appendix B, arrange each group...Ch. 20 - Without consulting Appendix B, arrange each group...Ch. 20 - In the reaction depicted in the molecular scenes,...Ch. 20 - Describe the equilibrium condition in terms of the...Ch. 20 - Prob. 20.32PCh. 20 - For each reaction, predict the sign and find the...Ch. 20 - For each reaction, predict the sign and find the...Ch. 20 - Find for the combustion of ethane (C2H6) to...Ch. 20 - Find for the combustion of methane to carbon...Ch. 20 - Find for the reaction of nitrogen monoxide with...Ch. 20 - Find for the combustion of ammonia to nitrogen...Ch. 20 - Find for the formation of Cu2O(s) from its...Ch. 20 - Find for the formation of HI(g) from its...Ch. 20 - Find for the formation of CH3OH(l) from its...Ch. 20 - Find for the formation of N2O(g) from its...Ch. 20 - Sulfur dioxide is released in the combustion of...Ch. 20 - Oxyacetylene welding is used to repair metal...Ch. 20 - What is the advantage of calculating free energy...Ch. 20 - Given that ΔGsys = −TΔSuniv, explain how the sign...Ch. 20 - Is an endothermic reaction more likely to be...Ch. 20 - Explain your answers to each of the following for...Ch. 20 - With its components in their standard states, a...Ch. 20 - How can ΔS° for a reaction be relatively...Ch. 20 - Calculate ΔG° for each reaction using ...Ch. 20 - Calculate ΔG° for each reaction using ...Ch. 20 - Prob. 20.53PCh. 20 - Prob. 20.54PCh. 20 - Consider the oxidation of carbon...Ch. 20 - Consider the combustion of butane gas: Predict...Ch. 20 - For the gaseous reaction of xenon and fluorine to...Ch. 20 - For the gaseous reaction of carbon monoxide and...Ch. 20 - One reaction used to produce small quantities of...Ch. 20 - A reaction that occurs in the internal combustion...Ch. 20 - As a fuel, H2(g) produces only nonpolluting H2O(g)...Ch. 20 - The U.S. government requires automobile fuels to...Ch. 20 - If K << 1 for a reaction, what do you know about...Ch. 20 - How is the free energy change of a process related...Ch. 20 - The scenes and the graph relate to the reaction of...Ch. 20 - What is the difference between ΔG° and ΔG? Under...Ch. 20 - Calculate K at 298 K for each reaction: MgCO3(s) ⇌...Ch. 20 - Calculate ΔG° at 298 K for each reaction: 2H2S(g)...Ch. 20 - Calculate K at 298 K for each reaction: HCN(aq) +...Ch. 20 - Calculate ΔG° at 298 K for each reaction: 2NO(g) +...Ch. 20 - Use ΔH° and ΔS° values for the following process...Ch. 20 - Use ΔH° and ΔS° values to find the temperature at...Ch. 20 - Prob. 20.73PCh. 20 - Use Appendix B to determine the Ksp of CaF2. Ch. 20 - For the reaction I2(g) + Cl2(g) ⇌ 2ICl(g),...Ch. 20 - For the reaction CaCO3(s) ⇌ CaO(s) + CO2(g),...Ch. 20 - The Ksp of PbCl2 is 1.7×10−5 at 25°C. What is ΔG°?...Ch. 20 - Prob. 20.78PCh. 20 - The equilibrium constant for the...Ch. 20 - The formation constant for the reaction Ni2+(aq) +...Ch. 20 - Prob. 20.81PCh. 20 - Prob. 20.82PCh. 20 - High levels of ozone (O3) cause rubber to...Ch. 20 - A BaSO4 slurry is ingested before the...Ch. 20 - According to advertisements, “a diamond is...Ch. 20 - Prob. 20.86PCh. 20 - Prob. 20.87PCh. 20 - Prob. 20.88PCh. 20 - Is each statement true or false? If false, correct...Ch. 20 - Prob. 20.90PCh. 20 - Prob. 20.91PCh. 20 - Prob. 20.92PCh. 20 - Prob. 20.93PCh. 20 - Write a balanced equation for the gaseous...Ch. 20 - Prob. 20.95PCh. 20 - Hydrogenation is the addition of H2 to double (or...Ch. 20 - Prob. 20.97PCh. 20 - Prob. 20.98PCh. 20 - Prob. 20.99PCh. 20 - Prob. 20.100PCh. 20 - From the following reaction and data, find (a) S°...Ch. 20 - Prob. 20.102PCh. 20 - Prob. 20.103PCh. 20 - Prob. 20.104PCh. 20 - Prob. 20.105PCh. 20 - Prob. 20.106PCh. 20 - Prob. 20.107PCh. 20 - Consider the formation of ammonia: N2(g) + 3H2(g)...Ch. 20 - Kyanite, sillimanite, and andalusite all have the...Ch. 20 - Prob. 20.110PCh. 20 - Prob. 20.111P
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