Question

Calculate the standard enthalpy of formation of propane (C3H8) given the following data:- Standard enthalpy of formation of CO2(g) = -393.5 kJ/mol- Standard enthalpy of formation of H2O(l) = -285.8 kJ/mol- Standard enthalpy of combustion of propane at constant pressure = -2220.1 kJ/mol

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 The standard enthalpy of combustion of propane is given as -2220.1 kJ/mol. We can use Hess's Law to relate the enthalpy of formation of propane to the enthalpy of formation of CO2 and H2O.H_combustion = [3H_f CO2  + 4H_f H2O ] - H_f C3H8 We are given the standard enthalpy of formation of CO2 and H2O:H_f CO2  = -393.5 kJ/molH_f H2O  = -285.8 kJ/molNow we can plug these values into the equation:-2220.1 kJ/mol = [3 -393.5 kJ/mol  + 4 -285.8 kJ/mol ] - H_f C3H8 -2220.1 kJ/mol = [-1180.5 kJ/mol - 1143.2 kJ/mol] - H_f C3H8 -2220.1 kJ/mol = -2323.7 kJ/mol + H_f C3H8 Now, we can solve for the standard enthalpy of formation of propane:H_f C3H8  = -2323.7 kJ/mol + 2220.1 kJ/molH_f C3H8  = 103.6 kJ/molTherefore, the standard enthalpy of formation of propane  C3H8  is 103.6 kJ/mol.