CHEMISTRY >CUSTOM<
CHEMISTRY >CUSTOM<
8th Edition
ISBN: 9781309097182
Author: SILBERBERG
Publisher: MCG/CREATE
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Chapter 6, Problem 6.87P

(a)

Interpretation Introduction

Interpretation:

A balanced equation for the combustion of stearic acid is to be written.

Concept introduction:

The standard enthalpy of formation (ΔHf°) is the enthalpy change of the formation of 1mol of compound from its elements and the reactants and products, in this case, is also present in their standard state.

Standard state includes 1atm pressure for gases, 1M concentration for substances in the aqueous phase and the most stable form of substance at 1atm pressure and 298K temperature. For example, metal like sodium exists in solid state and molecular elements like halogen exist in a gaseous state.

(a)

Expert Solution
Check Mark

Answer to Problem 6.87P

A balanced equation for the combustion of stearic acid is as follows:

C18H36O2(s)+26O2(g)18CO2(g)+18H2O(g)

Explanation of Solution

Combustion reaction involves the reaction of stearic acid (C18H36O2) with an oxygen molecule to form carbon dioxide and the water molecule

One mole of stearic acid (C18H36O2) reacts with twenty-four moles of O2 to form eighteen moles of CO2 and eighteen moles of H2O.

A balanced equation for the combustion of stearic acid is as follows:

C18H36O2(s)+26O2(g)18CO2(g)+18H2O(g)

Conclusion

A balanced equation for the combustion of stearic acid is as follows:

C18H36O2(s)+26O2(g)18CO2(g)+18H2O(g)

(b)

Interpretation Introduction

Interpretation:

ΔHrxn° for the combustion of stearic acid is 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.

(b)

Expert Solution
Check Mark

Answer to Problem 6.87P

ΔHrxn° for the combustion of stearic acid is 10488kJ.

Explanation of Solution

The given balanced chemical equation is as follows:

C18H36O2(s)+26O2(g)18CO2(g)+18H2O(g)

One mole of stearic acid (C18H36O2) reacts with twenty-four moles of O2 to form eighteen moles of CO2 and eighteen moles of H2O.

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

ΔHrxn°=[{18ΔHf°[CO2(l)]+18ΔHf°[H2O(g)]}{ΔHf°[C18H36O2]+26ΔHf°[O2(g)]}] (1)

Substitute 393.5kJ/mol for ΔHf°[CO2(l)], 241.826kJ/mol for ΔHf°[H2O(g)], 948kJ/mol for ΔHf°[C18H36O2] and 0 for ΔHf°[O2(g)] in the equation (1).

ΔHrxn°=[(18mol)(393.5kJ/mol)+(18mol)(241.826kJ/mol){(1mol)(948kJ/mol)+(26mol)(0)}]=10478.868kJ10488kJ

Conclusion

ΔHrxn° for the combustion of stearic acid is 10488kJ.

(c)

Interpretation Introduction

Interpretation:

The heat that is released in kJ and kcal when 1.00 g of stearic acid is burned completely is to be calculated.

Concept introduction:

Heat (q) is the energy transferred as a consequence of the difference of the temperature between the system and surrounding.

The heat released by the system is negative and the heat taken by the system is positive.

The amount of heat released by the system is equal to the amount of amount absorbed by the surrounding.

The expression to convert kcal into kJ is:

1kcal=4.184kJ

(c)

Expert Solution
Check Mark

Answer to Problem 6.87P

The heat released in kJ and kcal is 36.9kJ and 8.81kcal respectively.

Explanation of Solution

The formula to calculate the moles of stearic acid (C18H36O2) is as follows:

Moles of C18H36O2=( given mass of C18H36O2molar mass of C18H36O2) (2)

Substitute 1.00 g for given mass of C18H36O2 and 284.47g/mol for molar mass of C18H36O2 in the equation (2).

Moles of C18H36O2=( 1.00 g284.47g/mol)=0.00351530mol

The heat released by combustion of 1mol of C18H36O2 is 10488kJ.

Therefore, the heat released by combustion of 0.00351530mol of C18H36O2 is as follows:

Heatreleased=(0.00351530moll)(10488kJ1mol)=36.8631kJ36.9kJ

The heat released by combustion of 0.00351530mol of C18H36O2 in kcal is as follows:

Heatreleased=36.8631kJ(1kcal4.184kJ)=8.81168kcal8.81kcal

Conclusion

The heat released in kJ and kcal is 36.9kJ and 8.81kcal respectively.

(d)

Interpretation Introduction

Interpretation:

Whether the data that a candy containing 11.0 g fat and provides 100 Cal energy is consistent with the answer calculated in part (c) is to be determined.

Concept introduction:

Heat (q) is the energy transferred as a consequence of the difference of the temperature between the system and surrounding.

The heat released by the system is negative and the heat taken by the system is positive.

The amount of heat released by the system is equal to the amount of amount absorbed by the surrounding.

The conversion factor to convert nutritional calorie (Calorie) into kcal is:

1Calorie=1kcal

(d)

Expert Solution
Check Mark

Answer to Problem 6.87P

Yes, the data is consistent with the given package information.

Explanation of Solution

The heat released by 1g of fat is 8.81168kcal.

The formula to calculate heat released by 11g of fat is as follows:

Heat released=(mass of fat)(8.81168kcal1g fat) (3)

Substitute 11g for the mass of fat in the equation (3).

Heat released=(11g)(8.81168kcal1g fat)=96.9286kcal96.9kcal

The heat released in Cal is,

Heat released=96.9kcal(1Cal1kcal)=96.9Cal

The calculated value is close to 100Calorie. Therefore, the calculated value is consistent with the package information.

Conclusion

The heat released in Cal is 96.9Cal. Therefore, the calculated value is consistent with the package information.

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Chapter 6 Solutions

CHEMISTRY >CUSTOM<

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