Chemistry: The Molecular Nature of Matter and Change
Chemistry: The Molecular Nature of Matter and Change
8th Edition
ISBN: 9781259631757
Author: Martin Silberberg Dr., Patricia Amateis Professor
Publisher: McGraw-Hill Education
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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
Check Mark

Answer to Problem 6.103P

The standard enthalpies of reaction (1) and (2) are 657.0kJ and 32.9kJ 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 910.9kJ/mol for ΔHf°[SiO2(s)], 92.31kJ/mol for ΔHf°[HCl(g)] and 139.5kJ for ΔHrxn° and 241.826kJ/mol for ΔHf°[H2O(g)] in the equation (2).

(1mol)ΔHf°[SiCl4(g)]=[(1 mol)(910.9kJ/mol)+(4 mol)(92.31kJ/mol)((139.5kJ)+(2 mol)(241.826kJ/mol))]ΔHf°[SiCl4(g)]=656.988kJ1mol=656.988kJ/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 656.988kJ/mol for ΔHf°[SiCl4(g)], 0 for ΔHf°[Si(s)] and 0 for ΔHf°[Cl2(g)] in the equation (3).

ΔHrxn°=[{(1mol)(656.988kJ/mol)}{(1mol)(0)+(2mol)(0)}]=656.988kJ657.0kJ

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

ΔHrxn°=[{(1mol)(656.988kJ/mol)+(2mol)(110.5kJ/mol)}{(1mol)(910.9kJ/mol)+(2mol)(0)+(2mol)(0)}]=32.912kJ32.9kJ

Conclusion

The standard enthalpies of reaction (1) and (2) are 657.0kJ and 32.9kJ 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
Check Mark

Answer to Problem 6.103P

ΔHrxn° for the fourth reaction is 106.6kJ.

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

ΔH4°=(32.912kJ)+(139.5kJ)=106.588kJ106.6kJ

Conclusion

ΔHrxn° for the fourth reaction is 106.6kJ.

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Chapter 6 Solutions

Chemistry: The Molecular Nature of Matter and Change

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