1. Standard Enthalpy Changes =  H o P = 1 bar (0.997 atm) T = 298K, unless otherwise specified n = 1 mole for key compound

2. Std. Enthalpy

3. Hess’s Law

4. Hess’s Law with Non- Standard Temps

5. Enthalpy of Reaction Enthalpy is an extensive property It depends upon how much mass you have. The enthalpy change for a reaction is equal in magnitude, but opposite in sign, to H for the reverse reaction. Enthalpy change for a reaction depends on the state of the reactants and products. If a reactant is liquid and the product gas, there is less heat available to transfer to the surroundings Enthalpy of H 2 O (g) is greater than enthalpy of H 2 O (l)

6. Enthalpy of Reaction 2H 2 O (l)  2H 2 O (g) ΔH = +88kJ CH 4 (g) + 2O 2 (g)  CO 2 (g) + 2H 2 O (l) ΔH = - 890kJ What if H2O was gas instead of liquid? Break up the equation into 2 Heat equations: 1. CH 4 (g) + 2O 2 (g)  CO 2 (g) + 2H 2 O (l) ΔH = -890kJ 2. 2H 2 O (l)  2H 2 O (g) ΔH = +88kJ 3. Add the 2 equations together keeping track of states: 4. CH 4 (g) + 2O 2 (g)  CO 2 (g) + 2H 2 O (g) ΔH rxn = -802kJ

7. Enthalpy of Reactions Example: H 2 (g) + 1/2 O 2 (g)  H 2 O(l)  H = - 285 kJ/mol This process can be divided into smaller steps such as 1. H 2 (g) + 1/2 O 2 (g)  H 2 O(g)  H = -241 kJ/mol 2. H 2 O(g)  H 2 O(l)  H = -44 kJ/mol 3. Add rxns and enthalpys: 4. H 2 (g) + 1/2 O 2 (g)  H 2 O(l)  H = -285 kJ/mol

8. Adjusting for Temperatures: If you have 2 objects at different temperatures, you have to use specific heat to calculate heat gain or lost: q = C s x m x ΔT q = heat C s = specific heat m = mass ΔT = temperature change. Rearrange the equation to solve for temperature change. Heat lost has to equal heat gained somewhere.

9. Heat of Reaction and Hess’s Law Calculate ΔH for the reaction: 2C (s) + H 2 (g)  C 2 H 2 (g) Givens: C 2 H 2 (g) + 5/2 O 2 (g)  2CO 2 (g) + H 2 O (l) ΔH = -1299.6kJ C (s) + O 2 (g)  CO 2 (g) ΔH = -393.5 kJ H 2 (g) + ½ O 2 (g)  H 2 O (l) ΔH = -285.8 kJ First reaction has to be reversed!

10. Enthalpy of Reaction and Hess’s Law 2CO 2 (g) + H 2 O (l)  C 2 H 2 (g) + 5/2 O 2 (g) ΔH = +1299.6kJ 2(C (s) + O 2 (g)  CO 2 (g) ) ΔH = -393.5 kJ H 2 (g) + ½ O 2 (g)  H 2 O (l) ΔH = -285.8 kJ Net Reaction: 2C (s) + H 2 (g)  C 2 H 2 (g)