1. 1 Hess suggested that the sum of the enthalpies (ΔH) of the steps of a reaction will equal the enthalpy of the overall reaction. Hess’s Law

2. 2 Why Does It Work? If you turn an equation around, you change the sign: If H 2 (g) + 1/2 O 2 (g)  H 2 O(g)  H=-285.5 kJ then, H 2 O(g)  H 2 (g) + 1/2 O 2 (g)  H =+285.5 kJ also, If you multiply the equation by a number, you multiply the heat by that number: 2 H 2 O(g)  H 2 (g) + O 2 (g)  H =+571.0 kJ

3. 3 How do you get good at this?

4. Find the heat of formation (∆H) of: 2 Al (s) + 3 CuO (g) → 3 Cu (g) + Al 2 O 3 (s) Using two equations from the sheet: 2Al (s) + 3/2 O 2 (g) → Al 2 O 3 (s) ∆H=-1676kJ Cu(s) + 1/2 O 2 (g) → CuO (g) ∆H = -155kJ

5. Need to flip the second equation and change the sign on ∆H and multiply it by 3 2Al (s) + 3/2 O 2 (g) → Al 2 O 3 (s) ∆H=-1676kJ 3CuO (g)→3Cu(s) + 3/2 O 2 (g) ∆H=(3)155kJ ___________________________________ 2 Al (s) + 3 CuO (g) → 3 Cu (s) + Al 2 O 3 (s) ∆H = - 1211 kJ ( oxygen can be cancelled because it exists on both sides of the reactions)

6. Calculating Heats of Reaction using Hess's Law 1) Write the overall equation for the reaction if not given. 2) Manipulate the given equations for the steps of the reaction so they add up to the overall equation. 3) Add up the equations canceling common substances in reactant and product. 4) Add up the heats of the steps = heat of overall reaction.

7. H 2 O(g) + C(s) → CO(g) +H 2 (g) ∆H=? These are the equations chosen from the sheet 1) H 2 (g) + 1/2O 2 (g) →H 2 O(g) ∆H = -242.kJ 2) C(s) +1/2 O 2 (g) → CO(g) ∆H = -110. kJ Note that the H 2 O (g) is a reactant and in step #1 it is a product. Thus step one needs to be reversed and the sign of the heat.

8. H 2 O(g) → H 2 (g) + ½ O 2 (g) ∆H = 242.kJ C(s) + ½ O 2 (g) → CO(g) ∆H = -110. kJ _________________________________ H 2 O(g) + C(s) → CO(g) +H 2 (g) ∆H=132

9. Calculate the heat of reaction for the following equation C 3 H 8 (g) + 5 O 2 (g) ---> 3 CO 2 (g)+4H 2 O (g) given the following steps in the reaction mechanism. 1)3C (s)+ 4 H 2 (g) -------> C 3 H 8 (g) 2) H 2 (g) + ½ O 2 (g) -------> H 2 O (g) 3) C (s) + O 2 (g) --------> CO 2 (g)

10. We can manipulate the equations by: a) Reversing equation #1 b) Multiplying equation #2 by 4 c) Multiplying equation #3 by 3

11. ∆H C 3 H 8 (g) -------> 3C (s)+ 4 H 2 (g) 103.8 4 H 2 (g) + 2O 2 (g) -----> 4H 2 O (g) -967.2 3C (s) + 3O 2 (g) -----> 3CO 2 (g) -1180.5 C 3 H 8 (g) + 5 O 2 (g) ---> 3 CO 2 (g)+4H 2 O (g) ∆H = - 2043.9 kJ

12. 12 Standard Heats of Formation The  H for a reaction that produces 1 mol of a compound from its elements at standard conditions Standard conditions: 25°C and 1 atm. Symbol is u The standard heat of formation of an element = 0 u This includes the diatomics

13. 13 What good are they? The heat of a reaction can be calculated by: –subtracting the heats of formation of the reactants from the products HoHo = (Products) -(Reactants)

14. 14 Examples CH 4 (g) + 2 O 2 (g)  CO 2 (g) + 2 H 2 O(g) CH 4 (g) = - 74.86 kJ/molO 2 (g) = 0 kJ/molCO 2 (g) = - 393.5 kJ/molH 2 O(g) = - 241.8 kJ/mol u  H= [-393.5 + 2(-241.8)] - [-74.68 +2 (0)]  H= - 802.4 kJ