The standard enthalpies of reaction (1) and (2) are to be determined. Concept introduction: The standard enthalpy of reaction is calculated by the summation of standard enthalpy of formation of the product minus the summation of standard enthalpy of formation of product at the standard conditions. The formula to calculate the standard enthalpy of reaction ( Δ H rxn ° ) is as follows: Δ H rxn ° = ∑ m Δ H f (products) ° − ∑ m Δ H f (reactants) ° Here, m and n are the stoichiometric coefficients of reactants and product in the balanced chemical equation.

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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 6, Problem 6.103P

(a)

Interpretation Introduction

Interpretation:

The standard enthalpies of reaction (1) and (2) are to be determined.

Concept introduction:

The standard enthalpy of reaction is calculated by the summation of standard enthalpy of formation of the product minus the summation of standard enthalpy of formation of product at the standard conditions. The formula to calculate the standard enthalpy of reaction (ΔHrxn°) is as follows:

ΔHrxn°=∑mΔHf (products)°−∑mΔHf (reactants)°

Here, m and n are the stoichiometric coefficients of reactants and product in the balanced chemical equation.

(a)

Expert Solution

Answer to Problem 6.103P

The standard enthalpies of reaction (1) and (2) are –657.0 kJ and 32.9 kJ respectively.

Explanation of Solution

The balanced chemical equation for the reaction(3) is as follows:

SiCl4(g)+2H2O(g)→SiO2(s)+4HCl(g)

The formula to calculate the standard enthalpy of reaction (3) (ΔHrxn°) is as follows:

ΔHrxn°=[{ΔHf°[SiO2(s)]+4ΔHf°[HCl(g)]}−{ΔHf°[SiCl4(g)]+2ΔHf°[H2O(g)]}] (1)

Rearrange equation (1) to calculate ΔHf° of SiCl4(g) is as follows:

ΔHf°[SiCl4(g)]=[{ΔHf°[SiO2(s)]+4ΔHf°[HCl(g)]}−(ΔHrxn°+2ΔHf°[H2O(g)])] (2)

Substitute −910.9 kJ/mol for ΔHf°[SiO2(s)], −92.31 kJ/mol for ΔHf°[HCl(g)] and −139.5 kJ for ΔHrxn° and −241.826 kJ/mol for ΔHf°[H2O(g)] in the equation (2).

(1 mol)ΔHf°[SiCl4(g)]=[(1 mol)(−910.9 kJ/mol)+(4 mol)(−92.31 kJ/mol)−((−139.5 kJ)+(2 mol)(−241.826 kJ/mol))]ΔHf°[SiCl4(g)]=−656.988 kJ1 mol=−656.988 kJ/mol

The balanced chemical equation for the reaction(1) is as follows:

Si(s)+2Cl2(g)→SiCl4(g)

The formula to calculate the standard enthalpy for first reaction (ΔHrxn°) is as follows:

ΔHrxn°=[{ΔHf°[SiCl4(g)]}−{1ΔHf°[Si(s)]+2ΔHf°[Cl2(g)]}] (3)

Substitute −656.988 kJ/mol for ΔHf°[SiCl4(g)], 0 for ΔHf°[Si(s)] and 0 for ΔHf°[Cl2(g)] in the equation (3).

ΔHrxn°=[{(1 mol)(−656.988 kJ/mol)}−{(1 mol)(0)+(2 mol)(0)}]=−656.988 kJ≈–657.0 kJ

The balanced chemical equation for the reaction(2) is as follows:

SiO2(s)+2C(graphite)+2Cl2(g)→SiCl4(g)+2CO(g)

The formula to calculate the standard enthalpy for second reaction (ΔHrxn°) is as follows:

ΔHrxn°=[{ΔHf°[SiCl4(g)]+2ΔHf°[CO(g)]}−{1ΔHf°[SiO2(s)]+2ΔHf°[C(graphite)]+2ΔHf°[Cl2(g)]}] (4)

Substitute −656.988 kJ/mol for ΔHf°[SiCl4(g)], −110.5 kJ/mol for ΔHf°[CO(g)], −910.9 kJ/mol for ΔHf°[SiO2(s)], 0 for ΔHf°[C(graphite)] and 0 for ΔHf°[Cl2(g)] in the equation (4).

ΔHrxn°=[{(1 mol)(−656.988 kJ/mol)+(2 mol)(−110.5 kJ/mol)}−{(1 mol)(−910.9 kJ/mol)+(2 mol)(0)+(2 mol)(0)}]=32.912 kJ≈32.9 kJ

Conclusion

The standard enthalpies of reaction (1) and (2) are –657.0 kJ and 32.9 kJ respectively.

(b)

Interpretation Introduction

Interpretation:

ΔHrxn° for the fourth reaction that is sum of the reaction(2) and (3) is to be determined.

Concept introduction:

Hess’s law is used to calculate the enthalpy change of an overall reaction that can be derived as a sum of two or more reaction. According to Hess’s law ΔH of an overall reaction is equal to the sum of the enthalpy change for each individual reaction. ΔHoverall rxn=ΔH1+ΔH2+.......+ΔHn

Enthalpy is a state function so the value depends upon the initial state and final state not on the path so ΔH of an overall reaction can be calculated by the addition or subtraction of the individual steps whose ΔH is known.

(b)

Expert Solution

Answer to Problem 6.103P

ΔHrxn° for the fourth reaction is −106.6 kJ.

Explanation of Solution

The enthalpy change of the following reaction is ΔH2°

SiO2(s)+2C(graphite)+2Cl2(g)→SiCl4(g)+2CO(g) (5)

The enthalpy change of the following reaction is ΔH3°.

SiCl4(g)+2H2O(g)→SiO2(s)+4HCl(g) (6)

Add equation (5) and (6).

SiO2(s)+2C(graphite)+2Cl2(g)→SiCl4(g)+2CO(g)SiCl4(g)+2H2O(g)→SiO2(s)+4HCl(g)_2C(graphite)+2Cl2(g)+2H2O(g)→2CO(g)+4HCl(g) (8)

The enthalpy change of the final reaction (8) is ΔH4°.

The expression to calculate ΔH4° is as follows:

ΔH4°=ΔH2°+ΔH3° (9)

Substitute 32.912 kJ for ΔH2° and −139.5 kJ for ΔH3° in the equation (9).

ΔH4°=(32.912 kJ)+(−139.5 kJ)=−106.588 kJ≈−106.6 kJ

Conclusion

ΔHrxn° for the fourth reaction is −106.6 kJ.

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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 6, Problem 6.103P

(a)

Interpretation Introduction

Interpretation:

The standard enthalpies of reaction (1) and (2) are to be determined.

Concept introduction:

The standard enthalpy of reaction is calculated by the summation of standard enthalpy of formation of the product minus the summation of standard enthalpy of formation of product at the standard conditions. The formula to calculate the standard enthalpy of reaction (ΔHrxn°) is as follows:

ΔHrxn°=∑mΔHf (products)°−∑mΔHf (reactants)°

Here, m and n are the stoichiometric coefficients of reactants and product in the balanced chemical equation.

(a)

Expert Solution

Answer to Problem 6.103P

The standard enthalpies of reaction (1) and (2) are –657.0 kJ and 32.9 kJ respectively.

Explanation of Solution

The balanced chemical equation for the reaction(3) is as follows:

SiCl4(g)+2H2O(g)→SiO2(s)+4HCl(g)

The formula to calculate the standard enthalpy of reaction (3) (ΔHrxn°) is as follows:

ΔHrxn°=[{ΔHf°[SiO2(s)]+4ΔHf°[HCl(g)]}−{ΔHf°[SiCl4(g)]+2ΔHf°[H2O(g)]}] (1)

Rearrange equation (1) to calculate ΔHf° of SiCl4(g) is as follows:

ΔHf°[SiCl4(g)]=[{ΔHf°[SiO2(s)]+4ΔHf°[HCl(g)]}−(ΔHrxn°+2ΔHf°[H2O(g)])] (2)

Substitute −910.9 kJ/mol for ΔHf°[SiO2(s)], −92.31 kJ/mol for ΔHf°[HCl(g)] and −139.5 kJ for ΔHrxn° and −241.826 kJ/mol for ΔHf°[H2O(g)] in the equation (2).

(1 mol)ΔHf°[SiCl4(g)]=[(1 mol)(−910.9 kJ/mol)+(4 mol)(−92.31 kJ/mol)−((−139.5 kJ)+(2 mol)(−241.826 kJ/mol))]ΔHf°[SiCl4(g)]=−656.988 kJ1 mol=−656.988 kJ/mol

The balanced chemical equation for the reaction(1) is as follows:

Si(s)+2Cl2(g)→SiCl4(g)

The formula to calculate the standard enthalpy for first reaction (ΔHrxn°) is as follows:

ΔHrxn°=[{ΔHf°[SiCl4(g)]}−{1ΔHf°[Si(s)]+2ΔHf°[Cl2(g)]}] (3)

Substitute −656.988 kJ/mol for ΔHf°[SiCl4(g)], 0 for ΔHf°[Si(s)] and 0 for ΔHf°[Cl2(g)] in the equation (3).

ΔHrxn°=[{(1 mol)(−656.988 kJ/mol)}−{(1 mol)(0)+(2 mol)(0)}]=−656.988 kJ≈–657.0 kJ

The balanced chemical equation for the reaction(2) is as follows:

SiO2(s)+2C(graphite)+2Cl2(g)→SiCl4(g)+2CO(g)

The formula to calculate the standard enthalpy for second reaction (ΔHrxn°) is as follows:

ΔHrxn°=[{ΔHf°[SiCl4(g)]+2ΔHf°[CO(g)]}−{1ΔHf°[SiO2(s)]+2ΔHf°[C(graphite)]+2ΔHf°[Cl2(g)]}] (4)

Substitute −656.988 kJ/mol for ΔHf°[SiCl4(g)], −110.5 kJ/mol for ΔHf°[CO(g)], −910.9 kJ/mol for ΔHf°[SiO2(s)], 0 for ΔHf°[C(graphite)] and 0 for ΔHf°[Cl2(g)] in the equation (4).

ΔHrxn°=[{(1 mol)(−656.988 kJ/mol)+(2 mol)(−110.5 kJ/mol)}−{(1 mol)(−910.9 kJ/mol)+(2 mol)(0)+(2 mol)(0)}]=32.912 kJ≈32.9 kJ

Conclusion

The standard enthalpies of reaction (1) and (2) are –657.0 kJ and 32.9 kJ respectively.

(b)

Interpretation Introduction

Interpretation:

ΔHrxn° for the fourth reaction that is sum of the reaction(2) and (3) is to be determined.

Concept introduction:

Hess’s law is used to calculate the enthalpy change of an overall reaction that can be derived as a sum of two or more reaction. According to Hess’s law ΔH of an overall reaction is equal to the sum of the enthalpy change for each individual reaction. ΔHoverall rxn=ΔH1+ΔH2+.......+ΔHn

Enthalpy is a state function so the value depends upon the initial state and final state not on the path so ΔH of an overall reaction can be calculated by the addition or subtraction of the individual steps whose ΔH is known.

(b)

Expert Solution

Answer to Problem 6.103P

ΔHrxn° for the fourth reaction is −106.6 kJ.

Explanation of Solution

The enthalpy change of the following reaction is ΔH2°

SiO2(s)+2C(graphite)+2Cl2(g)→SiCl4(g)+2CO(g) (5)

The enthalpy change of the following reaction is ΔH3°.

SiCl4(g)+2H2O(g)→SiO2(s)+4HCl(g) (6)

Add equation (5) and (6).

SiO2(s)+2C(graphite)+2Cl2(g)→SiCl4(g)+2CO(g)SiCl4(g)+2H2O(g)→SiO2(s)+4HCl(g)_2C(graphite)+2Cl2(g)+2H2O(g)→2CO(g)+4HCl(g) (8)

The enthalpy change of the final reaction (8) is ΔH4°.

The expression to calculate ΔH4° is as follows:

ΔH4°=ΔH2°+ΔH3° (9)

Substitute 32.912 kJ for ΔH2° and −139.5 kJ for ΔH3° in the equation (9).

ΔH4°=(32.912 kJ)+(−139.5 kJ)=−106.588 kJ≈−106.6 kJ

Conclusion

ΔHrxn° for the fourth reaction is −106.6 kJ.

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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 6, Problem 6.103P

(a)

Interpretation Introduction

Interpretation:

The standard enthalpies of reaction (1) and (2) are to be determined.

Concept introduction:

The standard enthalpy of reaction is calculated by the summation of standard enthalpy of formation of the product minus the summation of standard enthalpy of formation of product at the standard conditions. The formula to calculate the standard enthalpy of reaction (ΔHrxn°) is as follows:

ΔHrxn°=∑mΔHf (products)°−∑mΔHf (reactants)°

Here, m and n are the stoichiometric coefficients of reactants and product in the balanced chemical equation.

(a)

Expert Solution

Answer to Problem 6.103P

The standard enthalpies of reaction (1) and (2) are –657.0 kJ and 32.9 kJ respectively.

Explanation of Solution

The balanced chemical equation for the reaction(3) is as follows:

SiCl4(g)+2H2O(g)→SiO2(s)+4HCl(g)

The formula to calculate the standard enthalpy of reaction (3) (ΔHrxn°) is as follows:

ΔHrxn°=[{ΔHf°[SiO2(s)]+4ΔHf°[HCl(g)]}−{ΔHf°[SiCl4(g)]+2ΔHf°[H2O(g)]}] (1)

Rearrange equation (1) to calculate ΔHf° of SiCl4(g) is as follows:

ΔHf°[SiCl4(g)]=[{ΔHf°[SiO2(s)]+4ΔHf°[HCl(g)]}−(ΔHrxn°+2ΔHf°[H2O(g)])] (2)

Substitute −910.9 kJ/mol for ΔHf°[SiO2(s)], −92.31 kJ/mol for ΔHf°[HCl(g)] and −139.5 kJ for ΔHrxn° and −241.826 kJ/mol for ΔHf°[H2O(g)] in the equation (2).

(1 mol)ΔHf°[SiCl4(g)]=[(1 mol)(−910.9 kJ/mol)+(4 mol)(−92.31 kJ/mol)−((−139.5 kJ)+(2 mol)(−241.826 kJ/mol))]ΔHf°[SiCl4(g)]=−656.988 kJ1 mol=−656.988 kJ/mol

The balanced chemical equation for the reaction(1) is as follows:

Si(s)+2Cl2(g)→SiCl4(g)

The formula to calculate the standard enthalpy for first reaction (ΔHrxn°) is as follows:

ΔHrxn°=[{ΔHf°[SiCl4(g)]}−{1ΔHf°[Si(s)]+2ΔHf°[Cl2(g)]}] (3)

Substitute −656.988 kJ/mol for ΔHf°[SiCl4(g)], 0 for ΔHf°[Si(s)] and 0 for ΔHf°[Cl2(g)] in the equation (3).

ΔHrxn°=[{(1 mol)(−656.988 kJ/mol)}−{(1 mol)(0)+(2 mol)(0)}]=−656.988 kJ≈–657.0 kJ

The balanced chemical equation for the reaction(2) is as follows:

SiO2(s)+2C(graphite)+2Cl2(g)→SiCl4(g)+2CO(g)

The formula to calculate the standard enthalpy for second reaction (ΔHrxn°) is as follows:

ΔHrxn°=[{ΔHf°[SiCl4(g)]+2ΔHf°[CO(g)]}−{1ΔHf°[SiO2(s)]+2ΔHf°[C(graphite)]+2ΔHf°[Cl2(g)]}] (4)

Substitute −656.988 kJ/mol for ΔHf°[SiCl4(g)], −110.5 kJ/mol for ΔHf°[CO(g)], −910.9 kJ/mol for ΔHf°[SiO2(s)], 0 for ΔHf°[C(graphite)] and 0 for ΔHf°[Cl2(g)] in the equation (4).

ΔHrxn°=[{(1 mol)(−656.988 kJ/mol)+(2 mol)(−110.5 kJ/mol)}−{(1 mol)(−910.9 kJ/mol)+(2 mol)(0)+(2 mol)(0)}]=32.912 kJ≈32.9 kJ

Conclusion

The standard enthalpies of reaction (1) and (2) are –657.0 kJ and 32.9 kJ respectively.

(b)

Interpretation Introduction

Interpretation:

ΔHrxn° for the fourth reaction that is sum of the reaction(2) and (3) is to be determined.

Concept introduction:

Hess’s law is used to calculate the enthalpy change of an overall reaction that can be derived as a sum of two or more reaction. According to Hess’s law ΔH of an overall reaction is equal to the sum of the enthalpy change for each individual reaction. ΔHoverall rxn=ΔH1+ΔH2+.......+ΔHn

Enthalpy is a state function so the value depends upon the initial state and final state not on the path so ΔH of an overall reaction can be calculated by the addition or subtraction of the individual steps whose ΔH is known.

(b)

Expert Solution

Answer to Problem 6.103P

ΔHrxn° for the fourth reaction is −106.6 kJ.

Explanation of Solution

The enthalpy change of the following reaction is ΔH2°

SiO2(s)+2C(graphite)+2Cl2(g)→SiCl4(g)+2CO(g) (5)

The enthalpy change of the following reaction is ΔH3°.

SiCl4(g)+2H2O(g)→SiO2(s)+4HCl(g) (6)

Add equation (5) and (6).

SiO2(s)+2C(graphite)+2Cl2(g)→SiCl4(g)+2CO(g)SiCl4(g)+2H2O(g)→SiO2(s)+4HCl(g)_2C(graphite)+2Cl2(g)+2H2O(g)→2CO(g)+4HCl(g) (8)

The enthalpy change of the final reaction (8) is ΔH4°.

The expression to calculate ΔH4° is as follows:

ΔH4°=ΔH2°+ΔH3° (9)

Substitute 32.912 kJ for ΔH2° and −139.5 kJ for ΔH3° in the equation (9).

ΔH4°=(32.912 kJ)+(−139.5 kJ)=−106.588 kJ≈−106.6 kJ

Conclusion

ΔHrxn° for the fourth reaction is −106.6 kJ.

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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 6, Problem 6.103P

(a)

Interpretation Introduction

Interpretation:

The standard enthalpies of reaction (1) and (2) are to be determined.

Concept introduction:

The standard enthalpy of reaction is calculated by the summation of standard enthalpy of formation of the product minus the summation of standard enthalpy of formation of product at the standard conditions. The formula to calculate the standard enthalpy of reaction (ΔHrxn°) is as follows:

ΔHrxn°=∑mΔHf (products)°−∑mΔHf (reactants)°

Here, m and n are the stoichiometric coefficients of reactants and product in the balanced chemical equation.

(a)

Expert Solution

Answer to Problem 6.103P

The standard enthalpies of reaction (1) and (2) are –657.0 kJ and 32.9 kJ respectively.

Explanation of Solution

The balanced chemical equation for the reaction(3) is as follows:

SiCl4(g)+2H2O(g)→SiO2(s)+4HCl(g)

The formula to calculate the standard enthalpy of reaction (3) (ΔHrxn°) is as follows:

ΔHrxn°=[{ΔHf°[SiO2(s)]+4ΔHf°[HCl(g)]}−{ΔHf°[SiCl4(g)]+2ΔHf°[H2O(g)]}] (1)

Rearrange equation (1) to calculate ΔHf° of SiCl4(g) is as follows:

ΔHf°[SiCl4(g)]=[{ΔHf°[SiO2(s)]+4ΔHf°[HCl(g)]}−(ΔHrxn°+2ΔHf°[H2O(g)])] (2)

Substitute −910.9 kJ/mol for ΔHf°[SiO2(s)], −92.31 kJ/mol for ΔHf°[HCl(g)] and −139.5 kJ for ΔHrxn° and −241.826 kJ/mol for ΔHf°[H2O(g)] in the equation (2).

(1 mol)ΔHf°[SiCl4(g)]=[(1 mol)(−910.9 kJ/mol)+(4 mol)(−92.31 kJ/mol)−((−139.5 kJ)+(2 mol)(−241.826 kJ/mol))]ΔHf°[SiCl4(g)]=−656.988 kJ1 mol=−656.988 kJ/mol

The balanced chemical equation for the reaction(1) is as follows:

Si(s)+2Cl2(g)→SiCl4(g)

The formula to calculate the standard enthalpy for first reaction (ΔHrxn°) is as follows:

ΔHrxn°=[{ΔHf°[SiCl4(g)]}−{1ΔHf°[Si(s)]+2ΔHf°[Cl2(g)]}] (3)

Substitute −656.988 kJ/mol for ΔHf°[SiCl4(g)], 0 for ΔHf°[Si(s)] and 0 for ΔHf°[Cl2(g)] in the equation (3).

ΔHrxn°=[{(1 mol)(−656.988 kJ/mol)}−{(1 mol)(0)+(2 mol)(0)}]=−656.988 kJ≈–657.0 kJ

The balanced chemical equation for the reaction(2) is as follows:

SiO2(s)+2C(graphite)+2Cl2(g)→SiCl4(g)+2CO(g)

The formula to calculate the standard enthalpy for second reaction (ΔHrxn°) is as follows:

ΔHrxn°=[{ΔHf°[SiCl4(g)]+2ΔHf°[CO(g)]}−{1ΔHf°[SiO2(s)]+2ΔHf°[C(graphite)]+2ΔHf°[Cl2(g)]}] (4)

Substitute −656.988 kJ/mol for ΔHf°[SiCl4(g)], −110.5 kJ/mol for ΔHf°[CO(g)], −910.9 kJ/mol for ΔHf°[SiO2(s)], 0 for ΔHf°[C(graphite)] and 0 for ΔHf°[Cl2(g)] in the equation (4).

ΔHrxn°=[{(1 mol)(−656.988 kJ/mol)+(2 mol)(−110.5 kJ/mol)}−{(1 mol)(−910.9 kJ/mol)+(2 mol)(0)+(2 mol)(0)}]=32.912 kJ≈32.9 kJ

Conclusion

The standard enthalpies of reaction (1) and (2) are –657.0 kJ and 32.9 kJ respectively.

(b)

Interpretation Introduction

Interpretation:

ΔHrxn° for the fourth reaction that is sum of the reaction(2) and (3) is to be determined.

Concept introduction:

Hess’s law is used to calculate the enthalpy change of an overall reaction that can be derived as a sum of two or more reaction. According to Hess’s law ΔH of an overall reaction is equal to the sum of the enthalpy change for each individual reaction. ΔHoverall rxn=ΔH1+ΔH2+.......+ΔHn

Enthalpy is a state function so the value depends upon the initial state and final state not on the path so ΔH of an overall reaction can be calculated by the addition or subtraction of the individual steps whose ΔH is known.

(b)

Expert Solution

Answer to Problem 6.103P

ΔHrxn° for the fourth reaction is −106.6 kJ.

Explanation of Solution

The enthalpy change of the following reaction is ΔH2°

SiO2(s)+2C(graphite)+2Cl2(g)→SiCl4(g)+2CO(g) (5)

The enthalpy change of the following reaction is ΔH3°.

SiCl4(g)+2H2O(g)→SiO2(s)+4HCl(g) (6)

Add equation (5) and (6).

SiO2(s)+2C(graphite)+2Cl2(g)→SiCl4(g)+2CO(g)SiCl4(g)+2H2O(g)→SiO2(s)+4HCl(g)_2C(graphite)+2Cl2(g)+2H2O(g)→2CO(g)+4HCl(g) (8)

The enthalpy change of the final reaction (8) is ΔH4°.

The expression to calculate ΔH4° is as follows:

ΔH4°=ΔH2°+ΔH3° (9)

Substitute 32.912 kJ for ΔH2° and −139.5 kJ for ΔH3° in the equation (9).

ΔH4°=(32.912 kJ)+(−139.5 kJ)=−106.588 kJ≈−106.6 kJ

Conclusion

ΔHrxn° for the fourth reaction is −106.6 kJ.

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Answer 5

Chapter 6, Problem 6.103P

(a)

Interpretation Introduction

Interpretation:

The standard enthalpies of reaction (1) and (2) are to be determined.

Concept introduction:

The standard enthalpy of reaction is calculated by the summation of standard enthalpy of formation of the product minus the summation of standard enthalpy of formation of product at the standard conditions. The formula to calculate the standard enthalpy of reaction (ΔHrxn°) is as follows:

ΔHrxn°=∑mΔHf (products)°−∑mΔHf (reactants)°

Here, m and n are the stoichiometric coefficients of reactants and product in the balanced chemical equation.

(a)

Expert Solution

Answer to Problem 6.103P

The standard enthalpies of reaction (1) and (2) are –657.0 kJ and 32.9 kJ respectively.

Explanation of Solution

The balanced chemical equation for the reaction(3) is as follows:

SiCl4(g)+2H2O(g)→SiO2(s)+4HCl(g)

The formula to calculate the standard enthalpy of reaction (3) (ΔHrxn°) is as follows:

ΔHrxn°=[{ΔHf°[SiO2(s)]+4ΔHf°[HCl(g)]}−{ΔHf°[SiCl4(g)]+2ΔHf°[H2O(g)]}] (1)

Rearrange equation (1) to calculate ΔHf° of SiCl4(g) is as follows:

ΔHf°[SiCl4(g)]=[{ΔHf°[SiO2(s)]+4ΔHf°[HCl(g)]}−(ΔHrxn°+2ΔHf°[H2O(g)])] (2)

Substitute −910.9 kJ/mol for ΔHf°[SiO2(s)], −92.31 kJ/mol for ΔHf°[HCl(g)] and −139.5 kJ for ΔHrxn° and −241.826 kJ/mol for ΔHf°[H2O(g)] in the equation (2).

(1 mol)ΔHf°[SiCl4(g)]=[(1 mol)(−910.9 kJ/mol)+(4 mol)(−92.31 kJ/mol)−((−139.5 kJ)+(2 mol)(−241.826 kJ/mol))]ΔHf°[SiCl4(g)]=−656.988 kJ1 mol=−656.988 kJ/mol

The balanced chemical equation for the reaction(1) is as follows:

Si(s)+2Cl2(g)→SiCl4(g)

The formula to calculate the standard enthalpy for first reaction (ΔHrxn°) is as follows:

ΔHrxn°=[{ΔHf°[SiCl4(g)]}−{1ΔHf°[Si(s)]+2ΔHf°[Cl2(g)]}] (3)

Substitute −656.988 kJ/mol for ΔHf°[SiCl4(g)], 0 for ΔHf°[Si(s)] and 0 for ΔHf°[Cl2(g)] in the equation (3).

ΔHrxn°=[{(1 mol)(−656.988 kJ/mol)}−{(1 mol)(0)+(2 mol)(0)}]=−656.988 kJ≈–657.0 kJ

The balanced chemical equation for the reaction(2) is as follows:

SiO2(s)+2C(graphite)+2Cl2(g)→SiCl4(g)+2CO(g)

The formula to calculate the standard enthalpy for second reaction (ΔHrxn°) is as follows:

ΔHrxn°=[{ΔHf°[SiCl4(g)]+2ΔHf°[CO(g)]}−{1ΔHf°[SiO2(s)]+2ΔHf°[C(graphite)]+2ΔHf°[Cl2(g)]}] (4)

Substitute −656.988 kJ/mol for ΔHf°[SiCl4(g)], −110.5 kJ/mol for ΔHf°[CO(g)], −910.9 kJ/mol for ΔHf°[SiO2(s)], 0 for ΔHf°[C(graphite)] and 0 for ΔHf°[Cl2(g)] in the equation (4).

ΔHrxn°=[{(1 mol)(−656.988 kJ/mol)+(2 mol)(−110.5 kJ/mol)}−{(1 mol)(−910.9 kJ/mol)+(2 mol)(0)+(2 mol)(0)}]=32.912 kJ≈32.9 kJ

Conclusion

The standard enthalpies of reaction (1) and (2) are –657.0 kJ and 32.9 kJ respectively.

(b)

Interpretation Introduction

Interpretation:

ΔHrxn° for the fourth reaction that is sum of the reaction(2) and (3) is to be determined.

Concept introduction:

Hess’s law is used to calculate the enthalpy change of an overall reaction that can be derived as a sum of two or more reaction. According to Hess’s law ΔH of an overall reaction is equal to the sum of the enthalpy change for each individual reaction. ΔHoverall rxn=ΔH1+ΔH2+.......+ΔHn

Enthalpy is a state function so the value depends upon the initial state and final state not on the path so ΔH of an overall reaction can be calculated by the addition or subtraction of the individual steps whose ΔH is known.

(b)

Expert Solution

Answer to Problem 6.103P

ΔHrxn° for the fourth reaction is −106.6 kJ.

Explanation of Solution

The enthalpy change of the following reaction is ΔH2°

SiO2(s)+2C(graphite)+2Cl2(g)→SiCl4(g)+2CO(g) (5)

The enthalpy change of the following reaction is ΔH3°.

SiCl4(g)+2H2O(g)→SiO2(s)+4HCl(g) (6)

Add equation (5) and (6).

SiO2(s)+2C(graphite)+2Cl2(g)→SiCl4(g)+2CO(g)SiCl4(g)+2H2O(g)→SiO2(s)+4HCl(g)_2C(graphite)+2Cl2(g)+2H2O(g)→2CO(g)+4HCl(g) (8)

The enthalpy change of the final reaction (8) is ΔH4°.

The expression to calculate ΔH4° is as follows:

ΔH4°=ΔH2°+ΔH3° (9)

Substitute 32.912 kJ for ΔH2° and −139.5 kJ for ΔH3° in the equation (9).

ΔH4°=(32.912 kJ)+(−139.5 kJ)=−106.588 kJ≈−106.6 kJ

Conclusion

ΔHrxn° for the fourth reaction is −106.6 kJ.

Answer 6

Chapter 6, Problem 6.103P

(a)

Interpretation Introduction

Interpretation:

The standard enthalpies of reaction (1) and (2) are to be determined.

Concept introduction:

The standard enthalpy of reaction is calculated by the summation of standard enthalpy of formation of the product minus the summation of standard enthalpy of formation of product at the standard conditions. The formula to calculate the standard enthalpy of reaction (ΔHrxn°) is as follows:

ΔHrxn°=∑mΔHf (products)°−∑mΔHf (reactants)°

Here, m and n are the stoichiometric coefficients of reactants and product in the balanced chemical equation.

(a)

Expert Solution

Answer to Problem 6.103P

The standard enthalpies of reaction (1) and (2) are –657.0 kJ and 32.9 kJ respectively.

Explanation of Solution

The balanced chemical equation for the reaction(3) is as follows:

SiCl4(g)+2H2O(g)→SiO2(s)+4HCl(g)

The formula to calculate the standard enthalpy of reaction (3) (ΔHrxn°) is as follows:

ΔHrxn°=[{ΔHf°[SiO2(s)]+4ΔHf°[HCl(g)]}−{ΔHf°[SiCl4(g)]+2ΔHf°[H2O(g)]}] (1)

Rearrange equation (1) to calculate ΔHf° of SiCl4(g) is as follows:

ΔHf°[SiCl4(g)]=[{ΔHf°[SiO2(s)]+4ΔHf°[HCl(g)]}−(ΔHrxn°+2ΔHf°[H2O(g)])] (2)

Substitute −910.9 kJ/mol for ΔHf°[SiO2(s)], −92.31 kJ/mol for ΔHf°[HCl(g)] and −139.5 kJ for ΔHrxn° and −241.826 kJ/mol for ΔHf°[H2O(g)] in the equation (2).

(1 mol)ΔHf°[SiCl4(g)]=[(1 mol)(−910.9 kJ/mol)+(4 mol)(−92.31 kJ/mol)−((−139.5 kJ)+(2 mol)(−241.826 kJ/mol))]ΔHf°[SiCl4(g)]=−656.988 kJ1 mol=−656.988 kJ/mol

The balanced chemical equation for the reaction(1) is as follows:

Si(s)+2Cl2(g)→SiCl4(g)

The formula to calculate the standard enthalpy for first reaction (ΔHrxn°) is as follows:

ΔHrxn°=[{ΔHf°[SiCl4(g)]}−{1ΔHf°[Si(s)]+2ΔHf°[Cl2(g)]}] (3)

Substitute −656.988 kJ/mol for ΔHf°[SiCl4(g)], 0 for ΔHf°[Si(s)] and 0 for ΔHf°[Cl2(g)] in the equation (3).

ΔHrxn°=[{(1 mol)(−656.988 kJ/mol)}−{(1 mol)(0)+(2 mol)(0)}]=−656.988 kJ≈–657.0 kJ

The balanced chemical equation for the reaction(2) is as follows:

SiO2(s)+2C(graphite)+2Cl2(g)→SiCl4(g)+2CO(g)

The formula to calculate the standard enthalpy for second reaction (ΔHrxn°) is as follows:

ΔHrxn°=[{ΔHf°[SiCl4(g)]+2ΔHf°[CO(g)]}−{1ΔHf°[SiO2(s)]+2ΔHf°[C(graphite)]+2ΔHf°[Cl2(g)]}] (4)

Substitute −656.988 kJ/mol for ΔHf°[SiCl4(g)], −110.5 kJ/mol for ΔHf°[CO(g)], −910.9 kJ/mol for ΔHf°[SiO2(s)], 0 for ΔHf°[C(graphite)] and 0 for ΔHf°[Cl2(g)] in the equation (4).

ΔHrxn°=[{(1 mol)(−656.988 kJ/mol)+(2 mol)(−110.5 kJ/mol)}−{(1 mol)(−910.9 kJ/mol)+(2 mol)(0)+(2 mol)(0)}]=32.912 kJ≈32.9 kJ

Conclusion

The standard enthalpies of reaction (1) and (2) are –657.0 kJ and 32.9 kJ respectively.

Answer 7

Chapter 6, Problem 6.103P

(a)

Interpretation Introduction

Interpretation:

The standard enthalpies of reaction (1) and (2) are to be determined.

Concept introduction:

The standard enthalpy of reaction is calculated by the summation of standard enthalpy of formation of the product minus the summation of standard enthalpy of formation of product at the standard conditions. The formula to calculate the standard enthalpy of reaction (ΔHrxn°) is as follows:

ΔHrxn°=∑mΔHf (products)°−∑mΔHf (reactants)°

Here, m and n are the stoichiometric coefficients of reactants and product in the balanced chemical equation.

Answer 8

(a)

Expert Solution

Answer to Problem 6.103P

The standard enthalpies of reaction (1) and (2) are –657.0 kJ and 32.9 kJ respectively.

Answer 9

(a)

Expert Solution

Answer 10

(a)

Expert Solution

Answer 11

Expert Solution

Answer 12

Explanation of Solution

The balanced chemical equation for the reaction(3) is as follows:

SiCl4(g)+2H2O(g)→SiO2(s)+4HCl(g)

The formula to calculate the standard enthalpy of reaction (3) (ΔHrxn°) is as follows:

ΔHrxn°=[{ΔHf°[SiO2(s)]+4ΔHf°[HCl(g)]}−{ΔHf°[SiCl4(g)]+2ΔHf°[H2O(g)]}] (1)

Rearrange equation (1) to calculate ΔHf° of SiCl4(g) is as follows:

ΔHf°[SiCl4(g)]=[{ΔHf°[SiO2(s)]+4ΔHf°[HCl(g)]}−(ΔHrxn°+2ΔHf°[H2O(g)])] (2)

Substitute −910.9 kJ/mol for ΔHf°[SiO2(s)], −92.31 kJ/mol for ΔHf°[HCl(g)] and −139.5 kJ for ΔHrxn° and −241.826 kJ/mol for ΔHf°[H2O(g)] in the equation (2).

(1 mol)ΔHf°[SiCl4(g)]=[(1 mol)(−910.9 kJ/mol)+(4 mol)(−92.31 kJ/mol)−((−139.5 kJ)+(2 mol)(−241.826 kJ/mol))]ΔHf°[SiCl4(g)]=−656.988 kJ1 mol=−656.988 kJ/mol

The balanced chemical equation for the reaction(1) is as follows:

Si(s)+2Cl2(g)→SiCl4(g)

The formula to calculate the standard enthalpy for first reaction (ΔHrxn°) is as follows:

ΔHrxn°=[{ΔHf°[SiCl4(g)]}−{1ΔHf°[Si(s)]+2ΔHf°[Cl2(g)]}] (3)

Substitute −656.988 kJ/mol for ΔHf°[SiCl4(g)], 0 for ΔHf°[Si(s)] and 0 for ΔHf°[Cl2(g)] in the equation (3).

ΔHrxn°=[{(1 mol)(−656.988 kJ/mol)}−{(1 mol)(0)+(2 mol)(0)}]=−656.988 kJ≈–657.0 kJ

The balanced chemical equation for the reaction(2) is as follows:

SiO2(s)+2C(graphite)+2Cl2(g)→SiCl4(g)+2CO(g)

The formula to calculate the standard enthalpy for second reaction (ΔHrxn°) is as follows:

ΔHrxn°=[{ΔHf°[SiCl4(g)]+2ΔHf°[CO(g)]}−{1ΔHf°[SiO2(s)]+2ΔHf°[C(graphite)]+2ΔHf°[Cl2(g)]}] (4)

Substitute −656.988 kJ/mol for ΔHf°[SiCl4(g)], −110.5 kJ/mol for ΔHf°[CO(g)], −910.9 kJ/mol for ΔHf°[SiO2(s)], 0 for ΔHf°[C(graphite)] and 0 for ΔHf°[Cl2(g)] in the equation (4).

ΔHrxn°=[{(1 mol)(−656.988 kJ/mol)+(2 mol)(−110.5 kJ/mol)}−{(1 mol)(−910.9 kJ/mol)+(2 mol)(0)+(2 mol)(0)}]=32.912 kJ≈32.9 kJ

Answer 13

Conclusion

The standard enthalpies of reaction (1) and (2) are –657.0 kJ and 32.9 kJ respectively.

Answer 14

(b)

Interpretation Introduction

Interpretation:

ΔHrxn° for the fourth reaction that is sum of the reaction(2) and (3) is to be determined.

Concept introduction:

Hess’s law is used to calculate the enthalpy change of an overall reaction that can be derived as a sum of two or more reaction. According to Hess’s law ΔH of an overall reaction is equal to the sum of the enthalpy change for each individual reaction. ΔHoverall rxn=ΔH1+ΔH2+.......+ΔHn

Enthalpy is a state function so the value depends upon the initial state and final state not on the path so ΔH of an overall reaction can be calculated by the addition or subtraction of the individual steps whose ΔH is known.

(b)

Expert Solution

Answer to Problem 6.103P

ΔHrxn° for the fourth reaction is −106.6 kJ.

Explanation of Solution

The enthalpy change of the following reaction is ΔH2°

SiO2(s)+2C(graphite)+2Cl2(g)→SiCl4(g)+2CO(g) (5)

The enthalpy change of the following reaction is ΔH3°.

SiCl4(g)+2H2O(g)→SiO2(s)+4HCl(g) (6)

Add equation (5) and (6).

SiO2(s)+2C(graphite)+2Cl2(g)→SiCl4(g)+2CO(g)SiCl4(g)+2H2O(g)→SiO2(s)+4HCl(g)_2C(graphite)+2Cl2(g)+2H2O(g)→2CO(g)+4HCl(g) (8)

The enthalpy change of the final reaction (8) is ΔH4°.

The expression to calculate ΔH4° is as follows:

ΔH4°=ΔH2°+ΔH3° (9)

Substitute 32.912 kJ for ΔH2° and −139.5 kJ for ΔH3° in the equation (9).

ΔH4°=(32.912 kJ)+(−139.5 kJ)=−106.588 kJ≈−106.6 kJ

Conclusion

ΔHrxn° for the fourth reaction is −106.6 kJ.

Answer 15

(b)

Interpretation Introduction

Interpretation:

ΔHrxn° for the fourth reaction that is sum of the reaction(2) and (3) is to be determined.

Concept introduction:

Hess’s law is used to calculate the enthalpy change of an overall reaction that can be derived as a sum of two or more reaction. According to Hess’s law ΔH of an overall reaction is equal to the sum of the enthalpy change for each individual reaction. ΔHoverall rxn=ΔH1+ΔH2+.......+ΔHn

Enthalpy is a state function so the value depends upon the initial state and final state not on the path so ΔH of an overall reaction can be calculated by the addition or subtraction of the individual steps whose ΔH is known.

Answer 16

(b)

Expert Solution

Answer to Problem 6.103P

ΔHrxn° for the fourth reaction is −106.6 kJ.

Answer 17

(b)

Expert Solution

Answer 18

(b)

Expert Solution

Answer 19

Expert Solution

Answer 20

Explanation of Solution

The enthalpy change of the following reaction is ΔH2°

SiO2(s)+2C(graphite)+2Cl2(g)→SiCl4(g)+2CO(g) (5)

The enthalpy change of the following reaction is ΔH3°.

SiCl4(g)+2H2O(g)→SiO2(s)+4HCl(g) (6)

Add equation (5) and (6).

SiO2(s)+2C(graphite)+2Cl2(g)→SiCl4(g)+2CO(g)SiCl4(g)+2H2O(g)→SiO2(s)+4HCl(g)_2C(graphite)+2Cl2(g)+2H2O(g)→2CO(g)+4HCl(g) (8)

The enthalpy change of the final reaction (8) is ΔH4°.

The expression to calculate ΔH4° is as follows:

ΔH4°=ΔH2°+ΔH3° (9)

Substitute 32.912 kJ for ΔH2° and −139.5 kJ for ΔH3° in the equation (9).

ΔH4°=(32.912 kJ)+(−139.5 kJ)=−106.588 kJ≈−106.6 kJ

Answer 21

Conclusion

ΔHrxn° for the fourth reaction is −106.6 kJ.

Answer 22

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