1. The problem statement, all variables and given/known data 4 C3H5N3O9 -> 12 CO2+10 H2O+6 N2 -- H rxn = -5678kJ Standard Heat of Formation of CO2=-394KJ Standard Heat of Formation of H2O=-242KJ Calculate the standard enthalpy of formation for nitroglycerin. 2. Relevant equations Product minus reactant (I think) 3. The attempt at a solution I thought you would swap this equation which gives 5678Kj. Then you have 12 moles of carbon dioxide and 10 moles of water (nitrogen is zero). Thus: 5678- (-4728 - 2420) I think I'm way off though.