The standard enthalpy of formation of benzene and acetylene, and hence, the enthalpy change for the formation of benzene from acetylene are to be calculated. Concept introduction: The standard enthalpy for a reaction is the amount of enthalpy that occursunderstandard conditions The standard enthalpy of a reaction is determined by using the equation given below: Δ H ° rxn = ∑ n Δ H f ° ( products ) − ∑ m Δ H f ° ( reactants ) Here, the stoichiometric coefficients are represented by m for the reactants and n for the productsand the enthalpy of formation understandard conditions isrepresented by Δ H f ° . The value of the enthalpy of formation of an element is zero at its most stable state.

Question

Want to see more full solutions like this?

Answer 1

Question

Definition Definition Number that is expressed before molecules, ions, and atoms such that it balances out the number of components present on either section of the equation in a chemical reaction. Stoichiometric coefficients can be a fraction or a whole number and are useful in determining the mole ratio among the reactants and products. In any equalized chemical equation, the number of components on either side of the equation will be the same.

Chapter 5, Problem 102AP

Interpretation Introduction

Interpretation:

The standard enthalpy of formation of benzene and acetylene, and hence, the enthalpy change for the formation of benzene from acetylene are to be calculated.

Concept introduction:

The standard enthalpy for a reaction is the amount of enthalpy that occursunderstandard conditions

The standard enthalpy of a reaction is determined by using the equation given below:

ΔH°rxn=∑nΔHf°(products)−∑mΔHf°(reactants)

Here, the stoichiometric coefficients are represented by m for the reactants and n for the productsand the enthalpy of formation understandard conditions isrepresented by ΔHf°.

The value of the enthalpy of formation of an element is zero at its most stable state.

Expert Solution & Answer

Answer to Problem 102AP

Solution: ΔH°f[C6H6(l)]=49.0 kJ/mol, ΔH°f[C2H2(g)]=226.6 kJ/mol, and ΔH°rxn=−630.8 kJ/mol.

Explanation of Solution

Given information:

ΔH°[C2H2(s)]=−1299.4 kJ/molΔH°[C6H6(s)]=−3267.4 kJ/mol

The given reaction is as follows:

3C2H2(g)→C6H6(l)

The reaction for the combustion of acetylene isas follows:

C2H2(g)+52O2(g) →2CO2(g)+H2O(l) ΔH° = −1299.4 kJ/mol

The reaction for the combustion of benzene isas follows:

C6H6(l)+152O2(g) →6CO2(g)+3H2O(l) ΔH° = −3267.4 kJ/mol

From appendix 2,

ΔH°f[CO2(g)]=−393.5 kJ/molΔH°f[H2O(l)]=−285.8 kJ/molΔH°f[O2(g)]=0 kJ/mol

The oxygen gas is in its most stable form, so its enthalpy of formation is zero.

Calculate the standard enthalpy for the combustion of acetylene as follows:

ΔH°rxn={2ΔH°f[CO2(g)]+ΔH°f[H2O(l)]}−{ΔH°f[C2H2(g)]+52ΔH°f[O2(g)]}

Substitute −393.5 kJ/mol for ΔH°f[CO2(g)], −285.8 kJ/mol for ΔH°f[H2O(l)], −1299.4 kJ/mol for ΔH°rxn, and 0 kJ/mol for ΔH°f[O2(g)] in the above equation

−1299.4 kJ/mol={2(−393.5 kJ/mol)+(−285.8 kJ/mol)}−{ΔH°f[C2H2(g)]+52(0)}=[−787 kJ/mol−285.8 kJ/mol]−ΔH°f[C2H2(g)]=(−1072.8 kJ/mol)−ΔH°f[C2H2(g)]

Rearrange the above equation for ΔH°f[C2H2(g)] as follows:

ΔH°f[C2H2(g)]=−1072.8 kJ/mol+1299.4 kJ/mol=226.6 kJ/mol

Calculate the standard enthalpy for thecombustion of benzene as follows:

ΔH°rxn={6ΔH°f[CO2(g)]+3ΔH°f[H2O(l)]}−{ΔH°f[C6H6(l)]+152ΔH°f[O2(g)]}

Substitute −393.5 kJ/mol for ΔH°f[CO2(g)], −285.8 kJ/mol for ΔH°f[H2O(l)], −3267.4 kJ/mol for ΔH°rxn, and 0 kJ/mol for ΔH°f[O2(g)] in the above equation

−3267.4 kJ/mol={6(−393.5 kJ/mol)+3(−285.8 kJ/mol)}−{ΔH°f[C6H6(l)]+152(0)}=[−2361 kJ/mol−857.4 kJ/mol]−ΔH°f[C6H6(l)]=(−3218.4 kJ/mol)−ΔH°f[C6H6(l)]

Rearrange the above equation for ΔH°f[C6H6(l)] as follows:

ΔH°f[C6H6(l)]=−3218.4 kJ/mol+3267.4 kJ/mol=49 kJ/mol

Calculate the enthalpy of the reaction for the formation of benzene from acetylene as follows:

ΔH°rxn={ΔH°f[C6H6(l)]}−{3ΔH°f[C2H2(g)]}

Substitute 49 kJ/mol for ΔH°f[C6H6(l)] and 226.6 kJ/mol for ΔH°f[C2H2(g)] in the above equation

ΔH°rxn={(49 kJ/mol)}−{3(226.6 kJ/mol)}=49 kJ/mol−679.8 kJ/mol=−630.8 kJ/mol

Conclusion

Theenthalpy change for theformation of benzene from acetylene is −630.8 kJ/mol.

Want to see more full solutions like this?

Subscribe now to access step-by-step solutions to millions of textbook problems written by subject matter experts!

Students have asked these similar questions

BF3, Boron Trifluoride, known to contain three covalent boron-fluorine bonds. suggest and illustrate all of the processes as well as their energetical consequences for the formation of BF3 from its elements.

Draw the mechanism of the reaction.

9. Draw all of the possible Monochlorination Products that would Result From the Free Radical Chlormation OF 23,4-TRIMethyl Pentane b. Calculate the To Yield For the major • Product given the Following Relative Restritus For 1° 2° and 30 Hydrogens toward Free Radical Chloration 5.0: 38 : 1 30 2° 1° C. what would be the major product in the Free Radical brominator Of the Same Molecule. Explain your Reasoning.

Answer 2

Question

Definition Definition Number that is expressed before molecules, ions, and atoms such that it balances out the number of components present on either section of the equation in a chemical reaction. Stoichiometric coefficients can be a fraction or a whole number and are useful in determining the mole ratio among the reactants and products. In any equalized chemical equation, the number of components on either side of the equation will be the same.

Chapter 5, Problem 102AP

Interpretation Introduction

Interpretation:

The standard enthalpy of formation of benzene and acetylene, and hence, the enthalpy change for the formation of benzene from acetylene are to be calculated.

Concept introduction:

The standard enthalpy for a reaction is the amount of enthalpy that occursunderstandard conditions

The standard enthalpy of a reaction is determined by using the equation given below:

ΔH°rxn=∑nΔHf°(products)−∑mΔHf°(reactants)

Here, the stoichiometric coefficients are represented by m for the reactants and n for the productsand the enthalpy of formation understandard conditions isrepresented by ΔHf°.

The value of the enthalpy of formation of an element is zero at its most stable state.

Expert Solution & Answer

Answer to Problem 102AP

Solution: ΔH°f[C6H6(l)]=49.0 kJ/mol, ΔH°f[C2H2(g)]=226.6 kJ/mol, and ΔH°rxn=−630.8 kJ/mol.

Explanation of Solution

Given information:

ΔH°[C2H2(s)]=−1299.4 kJ/molΔH°[C6H6(s)]=−3267.4 kJ/mol

The given reaction is as follows:

3C2H2(g)→C6H6(l)

The reaction for the combustion of acetylene isas follows:

C2H2(g)+52O2(g) →2CO2(g)+H2O(l) ΔH° = −1299.4 kJ/mol

The reaction for the combustion of benzene isas follows:

C6H6(l)+152O2(g) →6CO2(g)+3H2O(l) ΔH° = −3267.4 kJ/mol

From appendix 2,

ΔH°f[CO2(g)]=−393.5 kJ/molΔH°f[H2O(l)]=−285.8 kJ/molΔH°f[O2(g)]=0 kJ/mol

The oxygen gas is in its most stable form, so its enthalpy of formation is zero.

Calculate the standard enthalpy for the combustion of acetylene as follows:

ΔH°rxn={2ΔH°f[CO2(g)]+ΔH°f[H2O(l)]}−{ΔH°f[C2H2(g)]+52ΔH°f[O2(g)]}

Substitute −393.5 kJ/mol for ΔH°f[CO2(g)], −285.8 kJ/mol for ΔH°f[H2O(l)], −1299.4 kJ/mol for ΔH°rxn, and 0 kJ/mol for ΔH°f[O2(g)] in the above equation

−1299.4 kJ/mol={2(−393.5 kJ/mol)+(−285.8 kJ/mol)}−{ΔH°f[C2H2(g)]+52(0)}=[−787 kJ/mol−285.8 kJ/mol]−ΔH°f[C2H2(g)]=(−1072.8 kJ/mol)−ΔH°f[C2H2(g)]

Rearrange the above equation for ΔH°f[C2H2(g)] as follows:

ΔH°f[C2H2(g)]=−1072.8 kJ/mol+1299.4 kJ/mol=226.6 kJ/mol

Calculate the standard enthalpy for thecombustion of benzene as follows:

ΔH°rxn={6ΔH°f[CO2(g)]+3ΔH°f[H2O(l)]}−{ΔH°f[C6H6(l)]+152ΔH°f[O2(g)]}

Substitute −393.5 kJ/mol for ΔH°f[CO2(g)], −285.8 kJ/mol for ΔH°f[H2O(l)], −3267.4 kJ/mol for ΔH°rxn, and 0 kJ/mol for ΔH°f[O2(g)] in the above equation

−3267.4 kJ/mol={6(−393.5 kJ/mol)+3(−285.8 kJ/mol)}−{ΔH°f[C6H6(l)]+152(0)}=[−2361 kJ/mol−857.4 kJ/mol]−ΔH°f[C6H6(l)]=(−3218.4 kJ/mol)−ΔH°f[C6H6(l)]

Rearrange the above equation for ΔH°f[C6H6(l)] as follows:

ΔH°f[C6H6(l)]=−3218.4 kJ/mol+3267.4 kJ/mol=49 kJ/mol

Calculate the enthalpy of the reaction for the formation of benzene from acetylene as follows:

ΔH°rxn={ΔH°f[C6H6(l)]}−{3ΔH°f[C2H2(g)]}

Substitute 49 kJ/mol for ΔH°f[C6H6(l)] and 226.6 kJ/mol for ΔH°f[C2H2(g)] in the above equation

ΔH°rxn={(49 kJ/mol)}−{3(226.6 kJ/mol)}=49 kJ/mol−679.8 kJ/mol=−630.8 kJ/mol

Conclusion

Theenthalpy change for theformation of benzene from acetylene is −630.8 kJ/mol.

Want to see more full solutions like this?

Subscribe now to access step-by-step solutions to millions of textbook problems written by subject matter experts!

Answer 3

Question

Definition Definition Number that is expressed before molecules, ions, and atoms such that it balances out the number of components present on either section of the equation in a chemical reaction. Stoichiometric coefficients can be a fraction or a whole number and are useful in determining the mole ratio among the reactants and products. In any equalized chemical equation, the number of components on either side of the equation will be the same.

Chapter 5, Problem 102AP

Interpretation Introduction

Interpretation:

The standard enthalpy of formation of benzene and acetylene, and hence, the enthalpy change for the formation of benzene from acetylene are to be calculated.

Concept introduction:

The standard enthalpy for a reaction is the amount of enthalpy that occursunderstandard conditions

The standard enthalpy of a reaction is determined by using the equation given below:

ΔH°rxn=∑nΔHf°(products)−∑mΔHf°(reactants)

Here, the stoichiometric coefficients are represented by m for the reactants and n for the productsand the enthalpy of formation understandard conditions isrepresented by ΔHf°.

The value of the enthalpy of formation of an element is zero at its most stable state.

Expert Solution & Answer

Answer to Problem 102AP

Solution: ΔH°f[C6H6(l)]=49.0 kJ/mol, ΔH°f[C2H2(g)]=226.6 kJ/mol, and ΔH°rxn=−630.8 kJ/mol.

Explanation of Solution

Given information:

ΔH°[C2H2(s)]=−1299.4 kJ/molΔH°[C6H6(s)]=−3267.4 kJ/mol

The given reaction is as follows:

3C2H2(g)→C6H6(l)

The reaction for the combustion of acetylene isas follows:

C2H2(g)+52O2(g) →2CO2(g)+H2O(l) ΔH° = −1299.4 kJ/mol

The reaction for the combustion of benzene isas follows:

C6H6(l)+152O2(g) →6CO2(g)+3H2O(l) ΔH° = −3267.4 kJ/mol

From appendix 2,

ΔH°f[CO2(g)]=−393.5 kJ/molΔH°f[H2O(l)]=−285.8 kJ/molΔH°f[O2(g)]=0 kJ/mol

The oxygen gas is in its most stable form, so its enthalpy of formation is zero.

Calculate the standard enthalpy for the combustion of acetylene as follows:

ΔH°rxn={2ΔH°f[CO2(g)]+ΔH°f[H2O(l)]}−{ΔH°f[C2H2(g)]+52ΔH°f[O2(g)]}

Substitute −393.5 kJ/mol for ΔH°f[CO2(g)], −285.8 kJ/mol for ΔH°f[H2O(l)], −1299.4 kJ/mol for ΔH°rxn, and 0 kJ/mol for ΔH°f[O2(g)] in the above equation

−1299.4 kJ/mol={2(−393.5 kJ/mol)+(−285.8 kJ/mol)}−{ΔH°f[C2H2(g)]+52(0)}=[−787 kJ/mol−285.8 kJ/mol]−ΔH°f[C2H2(g)]=(−1072.8 kJ/mol)−ΔH°f[C2H2(g)]

Rearrange the above equation for ΔH°f[C2H2(g)] as follows:

ΔH°f[C2H2(g)]=−1072.8 kJ/mol+1299.4 kJ/mol=226.6 kJ/mol

Calculate the standard enthalpy for thecombustion of benzene as follows:

ΔH°rxn={6ΔH°f[CO2(g)]+3ΔH°f[H2O(l)]}−{ΔH°f[C6H6(l)]+152ΔH°f[O2(g)]}

Substitute −393.5 kJ/mol for ΔH°f[CO2(g)], −285.8 kJ/mol for ΔH°f[H2O(l)], −3267.4 kJ/mol for ΔH°rxn, and 0 kJ/mol for ΔH°f[O2(g)] in the above equation

−3267.4 kJ/mol={6(−393.5 kJ/mol)+3(−285.8 kJ/mol)}−{ΔH°f[C6H6(l)]+152(0)}=[−2361 kJ/mol−857.4 kJ/mol]−ΔH°f[C6H6(l)]=(−3218.4 kJ/mol)−ΔH°f[C6H6(l)]

Rearrange the above equation for ΔH°f[C6H6(l)] as follows:

ΔH°f[C6H6(l)]=−3218.4 kJ/mol+3267.4 kJ/mol=49 kJ/mol

Calculate the enthalpy of the reaction for the formation of benzene from acetylene as follows:

ΔH°rxn={ΔH°f[C6H6(l)]}−{3ΔH°f[C2H2(g)]}

Substitute 49 kJ/mol for ΔH°f[C6H6(l)] and 226.6 kJ/mol for ΔH°f[C2H2(g)] in the above equation

ΔH°rxn={(49 kJ/mol)}−{3(226.6 kJ/mol)}=49 kJ/mol−679.8 kJ/mol=−630.8 kJ/mol

Conclusion

Theenthalpy change for theformation of benzene from acetylene is −630.8 kJ/mol.

Want to see more full solutions like this?

Subscribe now to access step-by-step solutions to millions of textbook problems written by subject matter experts!

Answer 4

Question

Definition Definition Number that is expressed before molecules, ions, and atoms such that it balances out the number of components present on either section of the equation in a chemical reaction. Stoichiometric coefficients can be a fraction or a whole number and are useful in determining the mole ratio among the reactants and products. In any equalized chemical equation, the number of components on either side of the equation will be the same.

Chapter 5, Problem 102AP

Interpretation Introduction

Interpretation:

The standard enthalpy of formation of benzene and acetylene, and hence, the enthalpy change for the formation of benzene from acetylene are to be calculated.

Concept introduction:

The standard enthalpy for a reaction is the amount of enthalpy that occursunderstandard conditions

The standard enthalpy of a reaction is determined by using the equation given below:

ΔH°rxn=∑nΔHf°(products)−∑mΔHf°(reactants)

Here, the stoichiometric coefficients are represented by m for the reactants and n for the productsand the enthalpy of formation understandard conditions isrepresented by ΔHf°.

The value of the enthalpy of formation of an element is zero at its most stable state.

Expert Solution & Answer

Answer to Problem 102AP

Solution: ΔH°f[C6H6(l)]=49.0 kJ/mol, ΔH°f[C2H2(g)]=226.6 kJ/mol, and ΔH°rxn=−630.8 kJ/mol.

Explanation of Solution

Given information:

ΔH°[C2H2(s)]=−1299.4 kJ/molΔH°[C6H6(s)]=−3267.4 kJ/mol

The given reaction is as follows:

3C2H2(g)→C6H6(l)

The reaction for the combustion of acetylene isas follows:

C2H2(g)+52O2(g) →2CO2(g)+H2O(l) ΔH° = −1299.4 kJ/mol

The reaction for the combustion of benzene isas follows:

C6H6(l)+152O2(g) →6CO2(g)+3H2O(l) ΔH° = −3267.4 kJ/mol

From appendix 2,

ΔH°f[CO2(g)]=−393.5 kJ/molΔH°f[H2O(l)]=−285.8 kJ/molΔH°f[O2(g)]=0 kJ/mol

The oxygen gas is in its most stable form, so its enthalpy of formation is zero.

Calculate the standard enthalpy for the combustion of acetylene as follows:

ΔH°rxn={2ΔH°f[CO2(g)]+ΔH°f[H2O(l)]}−{ΔH°f[C2H2(g)]+52ΔH°f[O2(g)]}

Substitute −393.5 kJ/mol for ΔH°f[CO2(g)], −285.8 kJ/mol for ΔH°f[H2O(l)], −1299.4 kJ/mol for ΔH°rxn, and 0 kJ/mol for ΔH°f[O2(g)] in the above equation

−1299.4 kJ/mol={2(−393.5 kJ/mol)+(−285.8 kJ/mol)}−{ΔH°f[C2H2(g)]+52(0)}=[−787 kJ/mol−285.8 kJ/mol]−ΔH°f[C2H2(g)]=(−1072.8 kJ/mol)−ΔH°f[C2H2(g)]

Rearrange the above equation for ΔH°f[C2H2(g)] as follows:

ΔH°f[C2H2(g)]=−1072.8 kJ/mol+1299.4 kJ/mol=226.6 kJ/mol

Calculate the standard enthalpy for thecombustion of benzene as follows:

ΔH°rxn={6ΔH°f[CO2(g)]+3ΔH°f[H2O(l)]}−{ΔH°f[C6H6(l)]+152ΔH°f[O2(g)]}

Substitute −393.5 kJ/mol for ΔH°f[CO2(g)], −285.8 kJ/mol for ΔH°f[H2O(l)], −3267.4 kJ/mol for ΔH°rxn, and 0 kJ/mol for ΔH°f[O2(g)] in the above equation

−3267.4 kJ/mol={6(−393.5 kJ/mol)+3(−285.8 kJ/mol)}−{ΔH°f[C6H6(l)]+152(0)}=[−2361 kJ/mol−857.4 kJ/mol]−ΔH°f[C6H6(l)]=(−3218.4 kJ/mol)−ΔH°f[C6H6(l)]

Rearrange the above equation for ΔH°f[C6H6(l)] as follows:

ΔH°f[C6H6(l)]=−3218.4 kJ/mol+3267.4 kJ/mol=49 kJ/mol

Calculate the enthalpy of the reaction for the formation of benzene from acetylene as follows:

ΔH°rxn={ΔH°f[C6H6(l)]}−{3ΔH°f[C2H2(g)]}

Substitute 49 kJ/mol for ΔH°f[C6H6(l)] and 226.6 kJ/mol for ΔH°f[C2H2(g)] in the above equation

ΔH°rxn={(49 kJ/mol)}−{3(226.6 kJ/mol)}=49 kJ/mol−679.8 kJ/mol=−630.8 kJ/mol

Conclusion

Theenthalpy change for theformation of benzene from acetylene is −630.8 kJ/mol.

Want to see more full solutions like this?

Subscribe now to access step-by-step solutions to millions of textbook problems written by subject matter experts!

Answer 5

Chapter 5, Problem 102AP

Interpretation Introduction

Interpretation:

The standard enthalpy of formation of benzene and acetylene, and hence, the enthalpy change for the formation of benzene from acetylene are to be calculated.

Concept introduction:

The standard enthalpy for a reaction is the amount of enthalpy that occursunderstandard conditions

The standard enthalpy of a reaction is determined by using the equation given below:

ΔH°rxn=∑nΔHf°(products)−∑mΔHf°(reactants)

Here, the stoichiometric coefficients are represented by m for the reactants and n for the productsand the enthalpy of formation understandard conditions isrepresented by ΔHf°.

The value of the enthalpy of formation of an element is zero at its most stable state.

Expert Solution & Answer

Answer to Problem 102AP

Solution: ΔH°f[C6H6(l)]=49.0 kJ/mol, ΔH°f[C2H2(g)]=226.6 kJ/mol, and ΔH°rxn=−630.8 kJ/mol.

Explanation of Solution

Given information:

ΔH°[C2H2(s)]=−1299.4 kJ/molΔH°[C6H6(s)]=−3267.4 kJ/mol

The given reaction is as follows:

3C2H2(g)→C6H6(l)

The reaction for the combustion of acetylene isas follows:

C2H2(g)+52O2(g) →2CO2(g)+H2O(l) ΔH° = −1299.4 kJ/mol

The reaction for the combustion of benzene isas follows:

C6H6(l)+152O2(g) →6CO2(g)+3H2O(l) ΔH° = −3267.4 kJ/mol

From appendix 2,

ΔH°f[CO2(g)]=−393.5 kJ/molΔH°f[H2O(l)]=−285.8 kJ/molΔH°f[O2(g)]=0 kJ/mol

The oxygen gas is in its most stable form, so its enthalpy of formation is zero.

Calculate the standard enthalpy for the combustion of acetylene as follows:

ΔH°rxn={2ΔH°f[CO2(g)]+ΔH°f[H2O(l)]}−{ΔH°f[C2H2(g)]+52ΔH°f[O2(g)]}

Substitute −393.5 kJ/mol for ΔH°f[CO2(g)], −285.8 kJ/mol for ΔH°f[H2O(l)], −1299.4 kJ/mol for ΔH°rxn, and 0 kJ/mol for ΔH°f[O2(g)] in the above equation

−1299.4 kJ/mol={2(−393.5 kJ/mol)+(−285.8 kJ/mol)}−{ΔH°f[C2H2(g)]+52(0)}=[−787 kJ/mol−285.8 kJ/mol]−ΔH°f[C2H2(g)]=(−1072.8 kJ/mol)−ΔH°f[C2H2(g)]

Rearrange the above equation for ΔH°f[C2H2(g)] as follows:

ΔH°f[C2H2(g)]=−1072.8 kJ/mol+1299.4 kJ/mol=226.6 kJ/mol

Calculate the standard enthalpy for thecombustion of benzene as follows:

ΔH°rxn={6ΔH°f[CO2(g)]+3ΔH°f[H2O(l)]}−{ΔH°f[C6H6(l)]+152ΔH°f[O2(g)]}

Substitute −393.5 kJ/mol for ΔH°f[CO2(g)], −285.8 kJ/mol for ΔH°f[H2O(l)], −3267.4 kJ/mol for ΔH°rxn, and 0 kJ/mol for ΔH°f[O2(g)] in the above equation

−3267.4 kJ/mol={6(−393.5 kJ/mol)+3(−285.8 kJ/mol)}−{ΔH°f[C6H6(l)]+152(0)}=[−2361 kJ/mol−857.4 kJ/mol]−ΔH°f[C6H6(l)]=(−3218.4 kJ/mol)−ΔH°f[C6H6(l)]

Rearrange the above equation for ΔH°f[C6H6(l)] as follows:

ΔH°f[C6H6(l)]=−3218.4 kJ/mol+3267.4 kJ/mol=49 kJ/mol

Calculate the enthalpy of the reaction for the formation of benzene from acetylene as follows:

ΔH°rxn={ΔH°f[C6H6(l)]}−{3ΔH°f[C2H2(g)]}

Substitute 49 kJ/mol for ΔH°f[C6H6(l)] and 226.6 kJ/mol for ΔH°f[C2H2(g)] in the above equation

ΔH°rxn={(49 kJ/mol)}−{3(226.6 kJ/mol)}=49 kJ/mol−679.8 kJ/mol=−630.8 kJ/mol

Conclusion

Theenthalpy change for theformation of benzene from acetylene is −630.8 kJ/mol.

Answer 6

Chapter 5, Problem 102AP

Interpretation Introduction

Interpretation:

The standard enthalpy of formation of benzene and acetylene, and hence, the enthalpy change for the formation of benzene from acetylene are to be calculated.

Concept introduction:

The standard enthalpy for a reaction is the amount of enthalpy that occursunderstandard conditions

The standard enthalpy of a reaction is determined by using the equation given below:

ΔH°rxn=∑nΔHf°(products)−∑mΔHf°(reactants)

Here, the stoichiometric coefficients are represented by m for the reactants and n for the productsand the enthalpy of formation understandard conditions isrepresented by ΔHf°.

The value of the enthalpy of formation of an element is zero at its most stable state.

Expert Solution & Answer

Answer to Problem 102AP

Solution: ΔH°f[C6H6(l)]=49.0 kJ/mol, ΔH°f[C2H2(g)]=226.6 kJ/mol, and ΔH°rxn=−630.8 kJ/mol.

Explanation of Solution

Given information:

ΔH°[C2H2(s)]=−1299.4 kJ/molΔH°[C6H6(s)]=−3267.4 kJ/mol

The given reaction is as follows:

3C2H2(g)→C6H6(l)

The reaction for the combustion of acetylene isas follows:

C2H2(g)+52O2(g) →2CO2(g)+H2O(l) ΔH° = −1299.4 kJ/mol

The reaction for the combustion of benzene isas follows:

C6H6(l)+152O2(g) →6CO2(g)+3H2O(l) ΔH° = −3267.4 kJ/mol

From appendix 2,

ΔH°f[CO2(g)]=−393.5 kJ/molΔH°f[H2O(l)]=−285.8 kJ/molΔH°f[O2(g)]=0 kJ/mol

The oxygen gas is in its most stable form, so its enthalpy of formation is zero.

Calculate the standard enthalpy for the combustion of acetylene as follows:

ΔH°rxn={2ΔH°f[CO2(g)]+ΔH°f[H2O(l)]}−{ΔH°f[C2H2(g)]+52ΔH°f[O2(g)]}

Substitute −393.5 kJ/mol for ΔH°f[CO2(g)], −285.8 kJ/mol for ΔH°f[H2O(l)], −1299.4 kJ/mol for ΔH°rxn, and 0 kJ/mol for ΔH°f[O2(g)] in the above equation

−1299.4 kJ/mol={2(−393.5 kJ/mol)+(−285.8 kJ/mol)}−{ΔH°f[C2H2(g)]+52(0)}=[−787 kJ/mol−285.8 kJ/mol]−ΔH°f[C2H2(g)]=(−1072.8 kJ/mol)−ΔH°f[C2H2(g)]

Rearrange the above equation for ΔH°f[C2H2(g)] as follows:

ΔH°f[C2H2(g)]=−1072.8 kJ/mol+1299.4 kJ/mol=226.6 kJ/mol

Calculate the standard enthalpy for thecombustion of benzene as follows:

ΔH°rxn={6ΔH°f[CO2(g)]+3ΔH°f[H2O(l)]}−{ΔH°f[C6H6(l)]+152ΔH°f[O2(g)]}

Substitute −393.5 kJ/mol for ΔH°f[CO2(g)], −285.8 kJ/mol for ΔH°f[H2O(l)], −3267.4 kJ/mol for ΔH°rxn, and 0 kJ/mol for ΔH°f[O2(g)] in the above equation

−3267.4 kJ/mol={6(−393.5 kJ/mol)+3(−285.8 kJ/mol)}−{ΔH°f[C6H6(l)]+152(0)}=[−2361 kJ/mol−857.4 kJ/mol]−ΔH°f[C6H6(l)]=(−3218.4 kJ/mol)−ΔH°f[C6H6(l)]

Rearrange the above equation for ΔH°f[C6H6(l)] as follows:

ΔH°f[C6H6(l)]=−3218.4 kJ/mol+3267.4 kJ/mol=49 kJ/mol

Calculate the enthalpy of the reaction for the formation of benzene from acetylene as follows:

ΔH°rxn={ΔH°f[C6H6(l)]}−{3ΔH°f[C2H2(g)]}

Substitute 49 kJ/mol for ΔH°f[C6H6(l)] and 226.6 kJ/mol for ΔH°f[C2H2(g)] in the above equation

ΔH°rxn={(49 kJ/mol)}−{3(226.6 kJ/mol)}=49 kJ/mol−679.8 kJ/mol=−630.8 kJ/mol

Conclusion

Theenthalpy change for theformation of benzene from acetylene is −630.8 kJ/mol.

Answer 7

Chapter 5, Problem 102AP

Interpretation Introduction

Interpretation:

The standard enthalpy of formation of benzene and acetylene, and hence, the enthalpy change for the formation of benzene from acetylene are to be calculated.

Concept introduction:

The standard enthalpy for a reaction is the amount of enthalpy that occursunderstandard conditions

The standard enthalpy of a reaction is determined by using the equation given below:

ΔH°rxn=∑nΔHf°(products)−∑mΔHf°(reactants)

Here, the stoichiometric coefficients are represented by m for the reactants and n for the productsand the enthalpy of formation understandard conditions isrepresented by ΔHf°.

The value of the enthalpy of formation of an element is zero at its most stable state.

Answer 8

Expert Solution & Answer

Answer to Problem 102AP

Solution: ΔH°f[C6H6(l)]=49.0 kJ/mol, ΔH°f[C2H2(g)]=226.6 kJ/mol, and ΔH°rxn=−630.8 kJ/mol.

Answer 9

Expert Solution & Answer

Answer 10

Expert Solution & Answer

Answer 11

Explanation of Solution

Given information:

ΔH°[C2H2(s)]=−1299.4 kJ/molΔH°[C6H6(s)]=−3267.4 kJ/mol

The given reaction is as follows:

3C2H2(g)→C6H6(l)

The reaction for the combustion of acetylene isas follows:

C2H2(g)+52O2(g) →2CO2(g)+H2O(l) ΔH° = −1299.4 kJ/mol

The reaction for the combustion of benzene isas follows:

C6H6(l)+152O2(g) →6CO2(g)+3H2O(l) ΔH° = −3267.4 kJ/mol

From appendix 2,

ΔH°f[CO2(g)]=−393.5 kJ/molΔH°f[H2O(l)]=−285.8 kJ/molΔH°f[O2(g)]=0 kJ/mol

The oxygen gas is in its most stable form, so its enthalpy of formation is zero.

Calculate the standard enthalpy for the combustion of acetylene as follows:

ΔH°rxn={2ΔH°f[CO2(g)]+ΔH°f[H2O(l)]}−{ΔH°f[C2H2(g)]+52ΔH°f[O2(g)]}

Substitute −393.5 kJ/mol for ΔH°f[CO2(g)], −285.8 kJ/mol for ΔH°f[H2O(l)], −1299.4 kJ/mol for ΔH°rxn, and 0 kJ/mol for ΔH°f[O2(g)] in the above equation

−1299.4 kJ/mol={2(−393.5 kJ/mol)+(−285.8 kJ/mol)}−{ΔH°f[C2H2(g)]+52(0)}=[−787 kJ/mol−285.8 kJ/mol]−ΔH°f[C2H2(g)]=(−1072.8 kJ/mol)−ΔH°f[C2H2(g)]

Rearrange the above equation for ΔH°f[C2H2(g)] as follows:

ΔH°f[C2H2(g)]=−1072.8 kJ/mol+1299.4 kJ/mol=226.6 kJ/mol

Calculate the standard enthalpy for thecombustion of benzene as follows:

ΔH°rxn={6ΔH°f[CO2(g)]+3ΔH°f[H2O(l)]}−{ΔH°f[C6H6(l)]+152ΔH°f[O2(g)]}

Substitute −393.5 kJ/mol for ΔH°f[CO2(g)], −285.8 kJ/mol for ΔH°f[H2O(l)], −3267.4 kJ/mol for ΔH°rxn, and 0 kJ/mol for ΔH°f[O2(g)] in the above equation