1. THERMODYNAMICS #1
Example of using standard heats of formation to get ∆H of reaction
Calculate the heat of reaction for the following combustion.
4 NH3 (g)+ 7 O2 (g) 4 NO2 (g)+ 6 H2O(l)
GIVEN ( from a table or appendix of a book)
∆Hf H2O(l)= -286 kJ/mol),
∆Hf O2 (g) = 0,
∆Hf NH3 = -46 kJ/mol,
∆Hf NO2= 34kJ/mol

2. ∆H= ∑ [∆Hf products] - ∑ [∆Hf reactants]
4 NH3 (g)+ 7 O2 (g) 4 NO2 (g)+ 6 H2O(l)
∆Hf H2O(l) = -286 kJ/mol),
∆Hf O2(g) = 0,
∆Hf NH3(g) = -46 kJ/mol,
∆Hf NO2(g) = 34kJ/mol
∆H= ∑ [∆Hf products] - ∑ [∆Hf reactants]
∆H =∑[ 4 ∆Hf NO2(g) + 6 ∆Hf H2O(l)] - ∑[4 ∆Hf NH3(g)]
∆H= [4(34 kJ/mol) + 6( -286 kJ/mol)] - [4(- 46kJ/mol)]
∆H= kJ

3. For each mole of NO2(g) that is produced, -1396 kJ/4 is released.
For each mole of NH3(g) that reacts,-1396 kJ/4 is released.
For each mole of H2O(l) that is produced, kJ/6 is released.