Question

Determine the standard enthalpy of formation for propane (C3H8) using the following values: - C(graphite) + O2(g) → CO2(g) ΔH° = -393.5 kJ/mol- 2H2(g) + O2(g) → 2H2O(l) ΔH° = -572.4 kJ/mol- C3H8(g) + 5O2(g) → 3CO2(g) + 4H2O(l) ΔH° = -2220 kJ/mol

Answer 1

To determine the standard enthalpy of formation for propane  C3H8 , we need to find the enthalpy change for the formation reaction:C3H8 g  = 3C graphite  + 4H2 g We can do this by manipulating the given reactions and their enthalpies to match the formation reaction of propane. First, we need to reverse the first reaction and multiply it by 3 to get 3 moles of CO2:3CO2 g   3C graphite  + 3O2 g  H = 3 * 393.5 kJ/mol = 1180.5 kJ/molNext, we need to reverse the second reaction and multiply it by 2 to get 4 moles of H2:2H2O l   2H2 g  + O2 g  H = 2 * 572.4 kJ/mol = 1144.8 kJ/molNow, we can add these two manipulated reactions and the given propane combustion reaction:3CO2 g  + 4H2O l  H = -2220 kJ/mol3CO2 g   3C graphite  + 3O2 g  H = 1180.5 kJ/mol2H2O l   2H2 g  + O2 g  H = 1144.8 kJ/mol----------------------------------------------C3H8 g  = 3C graphite  + 4H2 g  H = -2220 + 1180.5 + 1144.8H = -2220 + 1180.5 + 1144.8 = 105.3 kJ/molTherefore, the standard enthalpy of formation for propane  C3H8  is 105.3 kJ/mol.