Question

Please show all work and derivations

A "barrel of oil" in international trade has a volume of 42 US gallons, or 159 L. How much heat would be released if 159 L of pure n-octane (C8H18) was burned completely at constant pressure, and with products and reactant at 298 K. The standard enthalpy of combustion of n-octane is -5471 kJ/mol and its density is 0.703 g/cm^3.

Answer 1

Total oil = 159 L = 159000 ml = 159000 cm^3
                              = 159000 cm^3 x 0.703 g/cm^3
                              = 111777 g / 114.23 g/mol = 978.52 mol

C8H18(l) + 25/2 O2(g) ---> 8 CO2(g) + 9 H2O(l) DHo = -5471 kJ

therefore 1 mole octane give 8 mole CO2 and 9 mole H2O

      978.52 mole octane give = 8 x 978.52 mol CO2 and 9 x 978.52 mol H2O

Given : DHfo [CO2(g)] = -393.5 kJ/mol and DHfo[H2O(l)] = -285.8 kJ/mol.
Calculate the enthalpy of formation of one mole of liquid octane.

Since the heat of formation for 1 mole of octane is -5471 kJ for 1 mole.

According to the equation above, 1 mole of octane will give you 8 moles of CO2 and 9 moles of H2O.

The heat of formation for O2 is zero because it naturally exists and has a bond energy of 0.

Recall that heat of formation is = heat of formation of products - heat of formation of reactants

            = ((8 x 978.52 mol CO2 x DHfo -393.5 kJ/mol ) + ( 9 x 978.52 mol H2O x DHfo -285.8 kJ/mol))

                                                            - (978.52 mole octane x 5471 kJ for 1 mole)

            = 243847.184 kJ