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(a)
Interpretation:
ΔH°rxn for the three reactions involved in the combustion of candle (C21H44) 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 (ΔH°rxn) is as follows:
ΔH°rxn=∑mΔH°f (products)−∑mΔH°f (reactants)
Here, m and n are the
(a)
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Answer to Problem 6.91P
ΔH°rxn for the three reactions are –13108 kJ, –7165 kJ and –4844 kJ respectively.
Explanation of Solution
The balanced chemical equations for the combustion of paraffin (C21H44) are as follows:
C21H44(s)+32O2(g)→21CO2(g)+22H2O(g)C21H44(s)+432O2(g)→21CO(g)+22H2O(g)C21H44(s)+11O2(g)→21C(s)+22H2O(g)
The standard state of oxygen is O2(g) so the standard enthalpy of formation for O2 is zero.
The formula to calculate the standard enthalpy for first reaction (ΔH°rxn) is as follows:
ΔH°rxn=[{21ΔH°f[CO2(g)]+22ΔH°f[H2O(g)]}−{1ΔH°f[C21H44(s)]+32ΔH°f[O2(g)]}] (1)
Substitute −393.5 kJ/mol for ΔH°f[CO2(g)], −241.826 kJ/mol for ΔH°f[H2O(g)], −476 kJ/mol for ΔH°f[C21H44(s)] and 0 for ΔH°f[O2(g)] in the equation (1).
ΔH°rxn=[{(21 mol)(−393.5 kJ/mol)+(22 mol)(−241.826 kJ/mol)}−{(1 mol)(−476 kJ/mol)+(22 mol)(0)}]=–13107.672 kJ≈–13108 kJ
The formula to calculate the standard enthalpy for second reaction (ΔH°rxn) is as follows:
ΔH°rxn=[{21ΔH°f[CO(g)]+22ΔH°f[H2O(g)]}−{1ΔH°f[C21H44(s)]+432ΔH°f[O2(g)]}] (2)
Substitute −110.5 kJ/mol for ΔH°f[CO(g)], −241.826 kJ/mol for ΔH°f[H2O(g)], −476 kJ/mol for ΔH°f[C21H44(s)] and 0 for ΔH°f[O2(g)] in the equation (2).
ΔH°rxn=[{(21 mol)(−110.5 kJ/mol)+(22 mol)(−241.826 kJ/mol)}−{(1 mol)(−476 kJ/mol)+(432 mol)(0)}]=–7164.672 kJ≈–7165 kJ
The formula to calculate the standard enthalpy for third reaction (ΔH°rxn) is as follows:
ΔH°rxn=[{21ΔH°f[C(s)]+22ΔH°f[H2O(g)]}−{1ΔH°f[C21H44(s)]+11ΔH°f[O2(g)]}] (3)
Substitute 0 for ΔH°f[C(s)], −241.826 kJ/mol for ΔH°f[H2O(g)], −476 kJ/mol for ΔH°f[C21H44(s)] and 0 for ΔH°f[O2(g)] in the equation (3).
ΔH°rxn=[{(21 mol)(0)+(22 mol)(−241.826 kJ/mol)}−{(1 mol)(−476 kJ/mol)+(11 mol)(0)}]=–4844.172 kJ≈–4844 kJ
ΔH°rxn for the three reactions are –13108 kJ, –7165 kJ and –4844 kJ respectively.
(b)
Interpretation:
The heat released when a 254 g candle burns 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 heat absorbed by the surrounding.
(b)
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Answer to Problem 6.91P
The heat released when a 254 g candle burns is −1.12×104 kJ.
Explanation of Solution
The formula to calculate heat released when 254 g candle burns is as follows:
Heat released=(mass of C21H44molar mass of C21H44)(heat released per mole) (4)
Substitute 254 g for mass of C21H44, 296.56 g/mol for molar mass of C21H44 and –13107.672 kJ/mol for heat released per mole in the equation (4).
Heat released=(254 g296.56 g/mol)(–13107.672 kJ/mol)=−1.12266×104 kJ≈−1.12×104 kJ
The heat released when a 254 g candle burns is −1.12×104 kJ.
(c)
Interpretation:
The heat released when 8.00 % by mass of the candle burns incompletely and 5.00 % by mass of it undergoes soot formation 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 heat absorbed by the surrounding.
(c)
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Answer to Problem 6.91P
The heat released is −1.05×104 kJ.
Explanation of Solution
Consider 87.00 % by mass of the candle undergoes complete combustion.
The formula to calculate moles of candle (C21H44) is as follows:
Moles of C21H44=(mass of C21H44molar mass of C21H44) (5)
Substitute 254 g for mass of C21H44 and 296.56 g/mol for molar mass of C21H44 in the equation (5).
Moles of C21H44=(254 g296.56 g/mol)=0.856488 mol
The formula to calculate heat released when candle burns is as follows:
Heat released=[(mass % of complete combustion100 %)(moles of C21H44)(ΔH°rxnof complete combustion)(mass % of incomplete combustion100 %)(moles of C21H44)(ΔH°rxnof incomplete combustion)(mass % of soot formation100 %)(moles of C21H44)(ΔH°rxnof soot formation)] (6)
Substitute 87.00 % for mass % of complete combustion, 0.856488 mol for moles of C21H44, –13107.672 kJ/mol for ΔH°rxn of complete combustion, 8.00 % for mass % of incomplete combustion, –7164.672 kJ/mol for ΔH°rxn of incomplete combustion, 5.00 % for mass % of soot formation and –4844.172 kJ/mol for ΔH°rxn of soot formation in the equation (6).
Heat released=[(87.00 %100 %)(0.856488 mol)(–13107.672 kJ/mol)(8.00 %100 %)(0.856488 mol)(–7164.672 kJ/moln)(5.00 %100 %)(0.856488 mol)(–4844.172 kJ/mol)]=[(0.87)(0.856488 mol)(–13107.672 kJ/mol)(0.080)(0.856488 mol)(–7164.672 kJ/moln)(0.050)(0.856488 mol)(–4844.172 kJ/mol)]=−1.04655×104 kJ≈−1.05×104 kJ
The heat released is −1.05×104 kJ.
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