Answer to Question #130205 in General Chemistry for A

Question

Hi,

The combustion of heptane, C7H16, occurs via the reaction C7H16(g)+11O2(g)→7CO2(g)+8H2O(g)
with heat of formation values given by the following table:
Substance ΔH∘f(kJ/mol)
C7H16 (g)=-187.9
CO2(g)= -393.5
H2O(g)=-241.8
Please Calculate the enthalpy for the combustion of 1 mole of heptane.
Please Express your answer to four significant figures and include the appropriate units.Thank you so much

Answer 1

The enthalpy for the reaction of combustion of 1 mole of heptane is calculated by the formula:

$\Delta\Eta r=\Sigma(n\Delta\Eta)products - \Sigma(n\Delta\Eta)reactants$

$\Delta\Eta r=n(CO2)*\Delta\Eta(CO2) + n(H2O)*\Delta\Eta(H2O)-n(C7H16)*\Delta\Eta(C7H16) - n(O2)*\Delta\Eta(O2)$

$\Delta\Eta r=7*(-393.5)+8*(-241.8)-1*(-187.9)-11*0$

$\Delta\Eta r=-4501 kJ/mol$