posted by Anonymous .
The heat of combustion of propane is -2,220.1 kJ/mol
calculate the heat of formation, delta Hf, of propane given that delta Hf of H20= -285.3 kJ/mol and delta Hf of CO2 = -393.5 kJ/mol
C3H8 + 5O2 ==> 3CO2 + 4H2O
-2220.1 kJ/mol = (n*sum delta Hfproducts) - (n*sum delta Hfreactants).
The only unknown is delta Hf C3H8. You have the values for H2O and CO2. O2, of course, is zero.
so we multiply the sum by the total number of moles, or by moles of reactant and product separately?
The products are CO2 and H2O.
So it's [(3*deltaH CO2) + (4*deltaH H2O)] - (1*deltaH C3H8) = -2220.1 kJ and solve for deltaH propane.