Calculate the enthalpy for the combustion of 1 mole of heptane?
The combustion of heptane (C7H16) occurs via the following reaction:
C7H16(g) + 11O2(g) → 7CO2(g) + 8H2O(g)
The heat of formation values are provided in the table below:
Substance ΔH∘f (kJ/mol)
C7H16 (g) -187.9
CO2(g) -393.5
H2O(g) -241.8
Please calculate the enthalpy change for the combustion of 1 mole of heptane. Express your answer to four significant figures and include the appropriate units.
2 Answers
products − reactants:
All units in kJ/mol:
(7 x −393.5) + (8 x −241.8) − (−187.9) − (11 x 0) = −4501 kJ/mol C7H16
Almost, remember that you multiplied the standard heat values (kJ/mole) of the substances by the number of moles of the respective substance. In the end, the moles cancel.
(7 moles x −393.5 kJ/mole) + (8 moles x −241.8 kJ/mole) − (1 mole x −187.9 kJ/mole) − (11 moles x 0 kJ/mole) = −4501 kJ C7H16
So, the resulting energy should be in kJ, not kJ/mole.
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