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What is the enthalpy change for the combustion of 1 mole of propane (C3H8) given the enthalpies of formation of CO2, H2O, and C3H8? C3H8(g) + 5O2(g) → 3CO2(g) + 4H2O(g)    ΔH° = ?

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To calculate the enthalpy change  H  for the combustion of 1 mole of propane, we need the enthalpies of formation  Hf  for all the reactants and products involved in the reaction. The enthalpy change for the reaction can be calculated using Hess's Law:H =  Hf products  -  Hf reactants The standard enthalpies of formation for CO2, H2O, and C3H8 are as follows:Hf CO2  = -393.5 kJ/molHf H2O  = -241.8 kJ/molHf C3H8  = -103.8 kJ/molHf O2  = 0 kJ/mol  since O2 is in its elemental form Now, we can plug these values into Hess's Law equation:H = [3  Hf CO2  + 4  Hf H2O ] - [Hf C3H8  + 5  Hf O2 ]H = [3   -393.5 kJ/mol  + 4   -241.8 kJ/mol ] - [ -103.8 kJ/mol  + 5   0 kJ/mol ]H =  -1180.5 kJ + -967.2 kJ  -  -103.8 kJ H = -2147.7 kJ + 103.8 kJH = -2043.9 kJ/molSo, the enthalpy change for the combustion of 1 mole of propane  C3H8  is -2043.9 kJ/mol.

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