The fuel values of formaldehyde and formic acid is to be calculated and the compound having higher fuel value is to be predicted. Concept Introduction: The calorie content of substance is calculated by determining the energy released in the combustion of the one gram mass of substance. To determine: The fuel values of formaldehyde and formic acid and the compound having higher fuel value out of it.

Question

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Answer 1

Question

Chapter 13, Problem 13.109QP

Interpretation Introduction

Interpretation: The fuel values of formaldehyde and formic acid is to be calculated and the compound having higher fuel value is to be predicted.

Concept Introduction: The calorie content of substance is calculated by determining the energy released in the combustion of the one gram mass of substance.

To determine: The fuel values of formaldehyde and formic acid and the compound having higher fuel value out of it.

Expert Solution & Answer

Answer to Problem 13.109QP

The fuel value of formaldehyde is $−19.00 kJ/g$ and the fuel value of formic acid is $−13.064 kJ/g$ . The fuel value of formaldehyde is more than formic acid.

Explanation of Solution

Given

The enthalpy of formation of formaldehyde is $−108.6 kJ/mol$ .

The enthalpy of formation of formic acid is $−378.7 kJ/mol$ .

The balanced chemical equation for the combustion of formaldehyde is,

$CH2O+O2(g)→CO2(g)+H2O(l)$

The enthalpy change of the combustion reaction of formaldehyde is calculated by the formula,

$ΔH∘=∑nΔH∘(products)−∑mΔH∘(reactants)=(ΔHCO2(g)∘+ΔHH2O(l)∘)−(ΔHCH2O∘+ΔHO2(g)∘)$

Substitute the values of $ΔHCO2(g)∘$ , $ΔHH2O(l)∘$ , $ΔHCH2O∘$ and $ΔHO2(g)∘$ from appendix C in equation.

$ΔH∘=(1×−393.5 kJ/mol+1×−285.83 kJ/mol)−(1×−108.6 kJ/mol+1×0.0 kJ/mol)=−570.73 kJ/mol$

The chemical equation for the combustion of formic acid is,

$2CH2O2+O2(g)→2CO2(g)+2H2O(l)$

The enthalpy change of the combustion reaction of palmitic acid is calculated by the formula,

$ΔH∘=∑nΔH∘(products)−∑mΔH∘(reactants)=(2ΔHCO2(g)∘+2ΔHH2O(l)∘)−(2ΔHCH2O2∘+ΔHO2(g)∘)$

Substitute the values of $ΔHCO2(g)∘$ , $ΔHH2O(l)∘$ , $ΔHCH2O2∘$ and $ΔHO2(g)∘$ from appendix C in equation.

$ΔH∘=(2×−393.5 kJ/mol+2×−285.83 kJ/mol)−(2×−378.7 kJ/mol+1×0.0 kJ/mol)=−601.26 kJ/mol$

The heat produced per gram in the combustion of formaldehyde is calculated by dividing its enthalpy change of combustion by the molar mass.

$Fuel value=−570.73 kJ/mol30.031 g/mol=−19.00 kJ/g$

The heat produced per gram in the combustion of formic acid is calculated by dividing its enthalpy change of combustion by the molar mass.

$Fuel value=−601.26 kJ/mol46.025 g/mol=−13.064 kJ/g$

Conclusion

The fuel value of formaldehyde is $−19.00 kJ/g$ and the fuel value of formic acid is $−13.064 kJ/g$ . The fuel value of formaldehyde is more than formic acid.

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Answer 2

Question

Chapter 13, Problem 13.109QP

Interpretation Introduction

Interpretation: The fuel values of formaldehyde and formic acid is to be calculated and the compound having higher fuel value is to be predicted.

Concept Introduction: The calorie content of substance is calculated by determining the energy released in the combustion of the one gram mass of substance.

To determine: The fuel values of formaldehyde and formic acid and the compound having higher fuel value out of it.

Expert Solution & Answer

Answer to Problem 13.109QP

The fuel value of formaldehyde is $−19.00 kJ/g$ and the fuel value of formic acid is $−13.064 kJ/g$ . The fuel value of formaldehyde is more than formic acid.

Explanation of Solution

Given

The enthalpy of formation of formaldehyde is $−108.6 kJ/mol$ .

The enthalpy of formation of formic acid is $−378.7 kJ/mol$ .

The balanced chemical equation for the combustion of formaldehyde is,

$CH2O+O2(g)→CO2(g)+H2O(l)$

The enthalpy change of the combustion reaction of formaldehyde is calculated by the formula,

$ΔH∘=∑nΔH∘(products)−∑mΔH∘(reactants)=(ΔHCO2(g)∘+ΔHH2O(l)∘)−(ΔHCH2O∘+ΔHO2(g)∘)$

Substitute the values of $ΔHCO2(g)∘$ , $ΔHH2O(l)∘$ , $ΔHCH2O∘$ and $ΔHO2(g)∘$ from appendix C in equation.

$ΔH∘=(1×−393.5 kJ/mol+1×−285.83 kJ/mol)−(1×−108.6 kJ/mol+1×0.0 kJ/mol)=−570.73 kJ/mol$

The chemical equation for the combustion of formic acid is,

$2CH2O2+O2(g)→2CO2(g)+2H2O(l)$

The enthalpy change of the combustion reaction of palmitic acid is calculated by the formula,

$ΔH∘=∑nΔH∘(products)−∑mΔH∘(reactants)=(2ΔHCO2(g)∘+2ΔHH2O(l)∘)−(2ΔHCH2O2∘+ΔHO2(g)∘)$

Substitute the values of $ΔHCO2(g)∘$ , $ΔHH2O(l)∘$ , $ΔHCH2O2∘$ and $ΔHO2(g)∘$ from appendix C in equation.

$ΔH∘=(2×−393.5 kJ/mol+2×−285.83 kJ/mol)−(2×−378.7 kJ/mol+1×0.0 kJ/mol)=−601.26 kJ/mol$

The heat produced per gram in the combustion of formaldehyde is calculated by dividing its enthalpy change of combustion by the molar mass.

$Fuel value=−570.73 kJ/mol30.031 g/mol=−19.00 kJ/g$

The heat produced per gram in the combustion of formic acid is calculated by dividing its enthalpy change of combustion by the molar mass.

$Fuel value=−601.26 kJ/mol46.025 g/mol=−13.064 kJ/g$

Conclusion

The fuel value of formaldehyde is $−19.00 kJ/g$ and the fuel value of formic acid is $−13.064 kJ/g$ . The fuel value of formaldehyde is more than formic acid.

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Answer 3

Question

Chapter 13, Problem 13.109QP

Interpretation Introduction

Interpretation: The fuel values of formaldehyde and formic acid is to be calculated and the compound having higher fuel value is to be predicted.

Concept Introduction: The calorie content of substance is calculated by determining the energy released in the combustion of the one gram mass of substance.

To determine: The fuel values of formaldehyde and formic acid and the compound having higher fuel value out of it.

Expert Solution & Answer

Answer to Problem 13.109QP

The fuel value of formaldehyde is $−19.00 kJ/g$ and the fuel value of formic acid is $−13.064 kJ/g$ . The fuel value of formaldehyde is more than formic acid.

Explanation of Solution

Given

The enthalpy of formation of formaldehyde is $−108.6 kJ/mol$ .

The enthalpy of formation of formic acid is $−378.7 kJ/mol$ .

The balanced chemical equation for the combustion of formaldehyde is,

$CH2O+O2(g)→CO2(g)+H2O(l)$

The enthalpy change of the combustion reaction of formaldehyde is calculated by the formula,

$ΔH∘=∑nΔH∘(products)−∑mΔH∘(reactants)=(ΔHCO2(g)∘+ΔHH2O(l)∘)−(ΔHCH2O∘+ΔHO2(g)∘)$

Substitute the values of $ΔHCO2(g)∘$ , $ΔHH2O(l)∘$ , $ΔHCH2O∘$ and $ΔHO2(g)∘$ from appendix C in equation.

$ΔH∘=(1×−393.5 kJ/mol+1×−285.83 kJ/mol)−(1×−108.6 kJ/mol+1×0.0 kJ/mol)=−570.73 kJ/mol$

The chemical equation for the combustion of formic acid is,

$2CH2O2+O2(g)→2CO2(g)+2H2O(l)$

The enthalpy change of the combustion reaction of palmitic acid is calculated by the formula,

$ΔH∘=∑nΔH∘(products)−∑mΔH∘(reactants)=(2ΔHCO2(g)∘+2ΔHH2O(l)∘)−(2ΔHCH2O2∘+ΔHO2(g)∘)$

Substitute the values of $ΔHCO2(g)∘$ , $ΔHH2O(l)∘$ , $ΔHCH2O2∘$ and $ΔHO2(g)∘$ from appendix C in equation.

$ΔH∘=(2×−393.5 kJ/mol+2×−285.83 kJ/mol)−(2×−378.7 kJ/mol+1×0.0 kJ/mol)=−601.26 kJ/mol$

The heat produced per gram in the combustion of formaldehyde is calculated by dividing its enthalpy change of combustion by the molar mass.

$Fuel value=−570.73 kJ/mol30.031 g/mol=−19.00 kJ/g$

The heat produced per gram in the combustion of formic acid is calculated by dividing its enthalpy change of combustion by the molar mass.

$Fuel value=−601.26 kJ/mol46.025 g/mol=−13.064 kJ/g$

Conclusion

The fuel value of formaldehyde is $−19.00 kJ/g$ and the fuel value of formic acid is $−13.064 kJ/g$ . The fuel value of formaldehyde is more than formic acid.

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Answer 4

Question

Chapter 13, Problem 13.109QP

Interpretation Introduction

Interpretation: The fuel values of formaldehyde and formic acid is to be calculated and the compound having higher fuel value is to be predicted.

Concept Introduction: The calorie content of substance is calculated by determining the energy released in the combustion of the one gram mass of substance.

To determine: The fuel values of formaldehyde and formic acid and the compound having higher fuel value out of it.

Expert Solution & Answer

Answer to Problem 13.109QP

The fuel value of formaldehyde is $−19.00 kJ/g$ and the fuel value of formic acid is $−13.064 kJ/g$ . The fuel value of formaldehyde is more than formic acid.

Explanation of Solution

Given

The enthalpy of formation of formaldehyde is $−108.6 kJ/mol$ .

The enthalpy of formation of formic acid is $−378.7 kJ/mol$ .

The balanced chemical equation for the combustion of formaldehyde is,

$CH2O+O2(g)→CO2(g)+H2O(l)$

The enthalpy change of the combustion reaction of formaldehyde is calculated by the formula,

$ΔH∘=∑nΔH∘(products)−∑mΔH∘(reactants)=(ΔHCO2(g)∘+ΔHH2O(l)∘)−(ΔHCH2O∘+ΔHO2(g)∘)$

Substitute the values of $ΔHCO2(g)∘$ , $ΔHH2O(l)∘$ , $ΔHCH2O∘$ and $ΔHO2(g)∘$ from appendix C in equation.

$ΔH∘=(1×−393.5 kJ/mol+1×−285.83 kJ/mol)−(1×−108.6 kJ/mol+1×0.0 kJ/mol)=−570.73 kJ/mol$

The chemical equation for the combustion of formic acid is,

$2CH2O2+O2(g)→2CO2(g)+2H2O(l)$

The enthalpy change of the combustion reaction of palmitic acid is calculated by the formula,

$ΔH∘=∑nΔH∘(products)−∑mΔH∘(reactants)=(2ΔHCO2(g)∘+2ΔHH2O(l)∘)−(2ΔHCH2O2∘+ΔHO2(g)∘)$

Substitute the values of $ΔHCO2(g)∘$ , $ΔHH2O(l)∘$ , $ΔHCH2O2∘$ and $ΔHO2(g)∘$ from appendix C in equation.

$ΔH∘=(2×−393.5 kJ/mol+2×−285.83 kJ/mol)−(2×−378.7 kJ/mol+1×0.0 kJ/mol)=−601.26 kJ/mol$

The heat produced per gram in the combustion of formaldehyde is calculated by dividing its enthalpy change of combustion by the molar mass.

$Fuel value=−570.73 kJ/mol30.031 g/mol=−19.00 kJ/g$

The heat produced per gram in the combustion of formic acid is calculated by dividing its enthalpy change of combustion by the molar mass.

$Fuel value=−601.26 kJ/mol46.025 g/mol=−13.064 kJ/g$

Conclusion

The fuel value of formaldehyde is $−19.00 kJ/g$ and the fuel value of formic acid is $−13.064 kJ/g$ . The fuel value of formaldehyde is more than formic acid.

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Answer 5

Chapter 13, Problem 13.109QP

Interpretation Introduction

Interpretation: The fuel values of formaldehyde and formic acid is to be calculated and the compound having higher fuel value is to be predicted.

Concept Introduction: The calorie content of substance is calculated by determining the energy released in the combustion of the one gram mass of substance.

To determine: The fuel values of formaldehyde and formic acid and the compound having higher fuel value out of it.

Expert Solution & Answer

Answer to Problem 13.109QP

The fuel value of formaldehyde is $−19.00 kJ/g$ and the fuel value of formic acid is $−13.064 kJ/g$ . The fuel value of formaldehyde is more than formic acid.

Explanation of Solution

Given

The enthalpy of formation of formaldehyde is $−108.6 kJ/mol$ .

The enthalpy of formation of formic acid is $−378.7 kJ/mol$ .

The balanced chemical equation for the combustion of formaldehyde is,

$CH2O+O2(g)→CO2(g)+H2O(l)$

The enthalpy change of the combustion reaction of formaldehyde is calculated by the formula,

$ΔH∘=∑nΔH∘(products)−∑mΔH∘(reactants)=(ΔHCO2(g)∘+ΔHH2O(l)∘)−(ΔHCH2O∘+ΔHO2(g)∘)$

Substitute the values of $ΔHCO2(g)∘$ , $ΔHH2O(l)∘$ , $ΔHCH2O∘$ and $ΔHO2(g)∘$ from appendix C in equation.

$ΔH∘=(1×−393.5 kJ/mol+1×−285.83 kJ/mol)−(1×−108.6 kJ/mol+1×0.0 kJ/mol)=−570.73 kJ/mol$

The chemical equation for the combustion of formic acid is,

$2CH2O2+O2(g)→2CO2(g)+2H2O(l)$

The enthalpy change of the combustion reaction of palmitic acid is calculated by the formula,

$ΔH∘=∑nΔH∘(products)−∑mΔH∘(reactants)=(2ΔHCO2(g)∘+2ΔHH2O(l)∘)−(2ΔHCH2O2∘+ΔHO2(g)∘)$

Substitute the values of $ΔHCO2(g)∘$ , $ΔHH2O(l)∘$ , $ΔHCH2O2∘$ and $ΔHO2(g)∘$ from appendix C in equation.

$ΔH∘=(2×−393.5 kJ/mol+2×−285.83 kJ/mol)−(2×−378.7 kJ/mol+1×0.0 kJ/mol)=−601.26 kJ/mol$

The heat produced per gram in the combustion of formaldehyde is calculated by dividing its enthalpy change of combustion by the molar mass.

$Fuel value=−570.73 kJ/mol30.031 g/mol=−19.00 kJ/g$

The heat produced per gram in the combustion of formic acid is calculated by dividing its enthalpy change of combustion by the molar mass.

$Fuel value=−601.26 kJ/mol46.025 g/mol=−13.064 kJ/g$

Conclusion

The fuel value of formaldehyde is $−19.00 kJ/g$ and the fuel value of formic acid is $−13.064 kJ/g$ . The fuel value of formaldehyde is more than formic acid.

Answer 6

Chapter 13, Problem 13.109QP

Interpretation Introduction

Interpretation: The fuel values of formaldehyde and formic acid is to be calculated and the compound having higher fuel value is to be predicted.

Concept Introduction: The calorie content of substance is calculated by determining the energy released in the combustion of the one gram mass of substance.

To determine: The fuel values of formaldehyde and formic acid and the compound having higher fuel value out of it.

Expert Solution & Answer

Answer to Problem 13.109QP

The fuel value of formaldehyde is $−19.00 kJ/g$ and the fuel value of formic acid is $−13.064 kJ/g$ . The fuel value of formaldehyde is more than formic acid.

Explanation of Solution

Given

The enthalpy of formation of formaldehyde is $−108.6 kJ/mol$ .

The enthalpy of formation of formic acid is $−378.7 kJ/mol$ .

The balanced chemical equation for the combustion of formaldehyde is,

$CH2O+O2(g)→CO2(g)+H2O(l)$

The enthalpy change of the combustion reaction of formaldehyde is calculated by the formula,

$ΔH∘=∑nΔH∘(products)−∑mΔH∘(reactants)=(ΔHCO2(g)∘+ΔHH2O(l)∘)−(ΔHCH2O∘+ΔHO2(g)∘)$

Substitute the values of $ΔHCO2(g)∘$ , $ΔHH2O(l)∘$ , $ΔHCH2O∘$ and $ΔHO2(g)∘$ from appendix C in equation.

$ΔH∘=(1×−393.5 kJ/mol+1×−285.83 kJ/mol)−(1×−108.6 kJ/mol+1×0.0 kJ/mol)=−570.73 kJ/mol$

The chemical equation for the combustion of formic acid is,

$2CH2O2+O2(g)→2CO2(g)+2H2O(l)$

The enthalpy change of the combustion reaction of palmitic acid is calculated by the formula,

$ΔH∘=∑nΔH∘(products)−∑mΔH∘(reactants)=(2ΔHCO2(g)∘+2ΔHH2O(l)∘)−(2ΔHCH2O2∘+ΔHO2(g)∘)$

Substitute the values of $ΔHCO2(g)∘$ , $ΔHH2O(l)∘$ , $ΔHCH2O2∘$ and $ΔHO2(g)∘$ from appendix C in equation.

$ΔH∘=(2×−393.5 kJ/mol+2×−285.83 kJ/mol)−(2×−378.7 kJ/mol+1×0.0 kJ/mol)=−601.26 kJ/mol$

The heat produced per gram in the combustion of formaldehyde is calculated by dividing its enthalpy change of combustion by the molar mass.

$Fuel value=−570.73 kJ/mol30.031 g/mol=−19.00 kJ/g$

The heat produced per gram in the combustion of formic acid is calculated by dividing its enthalpy change of combustion by the molar mass.

$Fuel value=−601.26 kJ/mol46.025 g/mol=−13.064 kJ/g$

Conclusion

The fuel value of formaldehyde is $−19.00 kJ/g$ and the fuel value of formic acid is $−13.064 kJ/g$ . The fuel value of formaldehyde is more than formic acid.

Answer 7

Chapter 13, Problem 13.109QP

Interpretation Introduction

Interpretation: The fuel values of formaldehyde and formic acid is to be calculated and the compound having higher fuel value is to be predicted.

Concept Introduction: The calorie content of substance is calculated by determining the energy released in the combustion of the one gram mass of substance.

To determine: The fuel values of formaldehyde and formic acid and the compound having higher fuel value out of it.

Answer 8

Expert Solution & Answer

Answer to Problem 13.109QP

The fuel value of formaldehyde is $−19.00 kJ/g$ and the fuel value of formic acid is $−13.064 kJ/g$ . The fuel value of formaldehyde is more than formic acid.

Answer 9

Expert Solution & Answer

Answer 10

Expert Solution & Answer

Answer 11

Explanation of Solution

Given

The enthalpy of formation of formaldehyde is $−108.6 kJ/mol$ .

The enthalpy of formation of formic acid is $−378.7 kJ/mol$ .

The balanced chemical equation for the combustion of formaldehyde is,

$CH2O+O2(g)→CO2(g)+H2O(l)$

The enthalpy change of the combustion reaction of formaldehyde is calculated by the formula,

$ΔH∘=∑nΔH∘(products)−∑mΔH∘(reactants)=(ΔHCO2(g)∘+ΔHH2O(l)∘)−(ΔHCH2O∘+ΔHO2(g)∘)$

Substitute the values of $ΔHCO2(g)∘$ , $ΔHH2O(l)∘$ , $ΔHCH2O∘$ and $ΔHO2(g)∘$ from appendix C in equation.

$ΔH∘=(1×−393.5 kJ/mol+1×−285.83 kJ/mol)−(1×−108.6 kJ/mol+1×0.0 kJ/mol)=−570.73 kJ/mol$

The chemical equation for the combustion of formic acid is,

$2CH2O2+O2(g)→2CO2(g)+2H2O(l)$

The enthalpy change of the combustion reaction of palmitic acid is calculated by the formula,

$ΔH∘=∑nΔH∘(products)−∑mΔH∘(reactants)=(2ΔHCO2(g)∘+2ΔHH2O(l)∘)−(2ΔHCH2O2∘+ΔHO2(g)∘)$

Substitute the values of $ΔHCO2(g)∘$ , $ΔHH2O(l)∘$ , $ΔHCH2O2∘$ and $ΔHO2(g)∘$ from appendix C in equation.

$ΔH∘=(2×−393.5 kJ/mol+2×−285.83 kJ/mol)−(2×−378.7 kJ/mol+1×0.0 kJ/mol)=−601.26 kJ/mol$

The heat produced per gram in the combustion of formaldehyde is calculated by dividing its enthalpy change of combustion by the molar mass.

$Fuel value=−570.73 kJ/mol30.031 g/mol=−19.00 kJ/g$

The heat produced per gram in the combustion of formic acid is calculated by dividing its enthalpy change of combustion by the molar mass.

$Fuel value=−601.26 kJ/mol46.025 g/mol=−13.064 kJ/g$

Answer 12

Conclusion

The fuel value of formaldehyde is $−19.00 kJ/g$ and the fuel value of formic acid is $−13.064 kJ/g$ . The fuel value of formaldehyde is more than formic acid.

Answer 13

Ch. 13.2 - Prob. 1PE Ch. 13.2 - Prob. 2PE Ch. 13.2 - Prob. 3PE Ch. 13.2 - Prob. 4PE Ch. 13.3 - Prob. 5PE Ch. 13.3 - Prob. 6PE Ch. 13.6 - Prob. 7PE Ch. 13.6 - Prob. 8PE Ch. 13.7 - Prob. 9PE Ch. 13.7 - Prob. 10PE

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