Question

Here's a precise problem for a chemistry student:Calculate the standard enthalpy of formation of propane (C3H8) using the following information: ΔH°f CO2 = -393.5 kJ/molΔH°f H2O(l) = -285.8 kJ/molΔH°f C3H8(g) = ?

Answer 1

To calculate the standard enthalpy of formation of propane  C3H8 , we can use the following combustion reaction:C3H8 g  + 5O2 g   3CO2 g  + 4H2O l Now, we can apply Hess's Law, which states that the enthalpy change of a reaction is the same whether it occurs in one step or several steps. We can use the given standard enthalpies of formation for CO2 and H2O to find the enthalpy change for the combustion of propane.Hcombustion = [3Hf CO2  + 4Hf H2O ] - Hf C3H8 We are given the standard enthalpies of formation for CO2 and H2O:Hf CO2  = -393.5 kJ/molHf H2O  = -285.8 kJ/molNow, we can plug these values into the equation:Hcombustion = [3 -393.5 kJ/mol  + 4 -285.8 kJ/mol ] - Hf C3H8 The standard enthalpy of combustion for propane  Hcombustion  is -2219.2 kJ/mol. Now we can solve for Hf C3H8 :-2219.2 kJ/mol = [3 -393.5 kJ/mol  + 4 -285.8 kJ/mol ] - Hf C3H8 -2219.2 kJ/mol =  -1180.5 kJ/mol - 1143.2 kJ/mol  - Hf C3H8 Hf C3H8  = -1180.5 kJ/mol - 1143.2 kJ/mol + 2219.2 kJ/molHf C3H8  = -104.5 kJ/molSo, the standard enthalpy of formation of propane  C3H8  is -104.5 kJ/mol.