1. Energy: Standard Enthalpy of Formation and Reaction – Direct & Hess’s Law Enthalpy Enthalpy (H): heat flow for a chemical reaction. q rxn @ constant P =  H = H products – H reactants exothermic:q <  H H products < H reactants endothermic:q >  H H products > H reactants

2. Energy: Standard Enthalpy of Formation and Reaction – Direct & Hess’s Law Standard Enthalpy of Formation and Reaction

3. Energy: Standard Enthalpy of Formation and Reaction – Direct & Hess’s Law aA + bB →  cC + dD Standard Enthalpy of Formation and Reaction standard-state conditions = 25 o C & 1 atm

4. Energy: Standard Enthalpy of Formation and Reaction – Direct & Hess’s Law Example 6.10 Standard Enthalpy of Reactions: Direct Method 2Al (s) + 1Fe 2 O 3 (s) → 1Al 2 O 3 (g) + 2Fe (l)

5. Energy: Standard Enthalpy of Formation and Reaction – Direct & Hess’s Law 2Al (s) + 1Fe 2 O 3 (s)  →  1Al 2 O 3 (g) + 2Fe (l) Al (s) 0 kJ/mol Al 2 O 3(s) –1669.8kJ/mol Fe (l) 12.40kJ/molFe 2 O 3 (s)–822.2kJ/mol Example 6.10 (con’t) H = 0 kJ/mol for elements at 25 o C (e.g., Fe(s) = 0 kJ/mol; Fe(l)  0)

6. Energy: Standard Enthalpy of Formation and Reaction – Direct & Hess’s Law Example 6.10 (con’t) 2Al (s) + 1Fe 2 O 3 (s) →  1Al 2 O 3 (g) + 2Fe (l)

7. INDIRECT METHOD (Hess’s Law) Hess’s Law is used is convenient for obtaining values of  H for reactions that are difficult to carry out in a calorimeter. E.g., C (s) + ½O 2 (g)  CO (g) Nearly impossible to carry out in a calorimeter because when C is burned, CO 2 is almost always produced.

9. Energy: Standard Enthalpy of Formation and Reaction – Direct & Hess’s Law For H2 + Cl2  2HCl  H = –185 kJ –185 kJ..............sign = exothermic coefficients represent moles (the -185 kJ refers to ONE MOLE of reaction) states of matter (phases) MUST be written standard temp = 25 o C (not STP: 0 o C)

10. Energy: Standard Enthalpy of Formation and Reaction – Direct & Hess’s Law Rules: (For H2 + Cl2  2HCl  H = –185 kJ 1.Magnitude of DH is directly proportional to the amount of reactant or product. -185 kJ1 mol H2-185 kJ 1 mol Cl2 -185 kJ2 mol HCl 2.Reverse the sign for  H for the reverse reaction.  H for a reaction is the same, whether carried out in one step or several (State Function)

11. Energy: Standard Enthalpy of Formation and Reaction – Direct & Hess’s Law Example 6.09 Standard Enthalpy of Reactions: Hess’s Law 2C (graphite) + 1H 2 (g) →  1C 2 H 2 (g) (a) C (graphite) + O 2 (g) → CO 2 (g) –393.5 (b) H 2 (g) + ½O 2 (g) → H 2 O (l) –285.8 (c) 2C 2 H 2 (g) + 5O 2 (g) → 4CO 2 (g) + 2H 2 O (l) –2598.8 The values for H are determined by experiment, so they need to be given to you; either directly or in a table. ΔH kJ/mol

12. Energy: Standard Enthalpy of Formation and Reaction – Direct & Hess’s Law Example 6.09 Standard Enthalpy of Reactions: Hess’s Law 2C (graphite) + 1H 2 (g) → 1C 2 H 2 (g) (a) C (graphite) + O 2 (g) →  CO 2 (g) –393.5 Concentrate on aligning the substances in the step reactions to be on the same sides as the original equation. Notice: There’s 2 moles of C(graphite) in the original equation but only 1 mole in (a).......................... (b) H 2 (g) + ½O 2 (g) → H 2 O (l) –285.8 (c) 2C 2 H 2 (g) + 5O 2 (g) → 4CO 2 (g) + 2H 2 O (l) –2598.8 ΔH kJ/mol

13. Energy: Standard Enthalpy of Formation and Reaction – Direct & Hess’s Law Example 6.09 Standard Enthalpy of Reactions: Hess’s Law 2C (graphite) + 1H 2 (g) →  1C 2 H 2 (g) (2a) 2C (graphite) + 2O 2 (g) →  2CO 2 (g) –787.0 Double (a) to match the original equation. And ΔH doubles, too (–393.5 * 2 = –787.0 kJ/mol) (b) H 2 (g) + ½O 2 (g) → H 2 O (l) –285.8 (c) 2C 2 H 2 (g) + 5O 2 (g) → 4CO 2 (g) + 2H 2 O (l) –2598.8 ΔH kJ/mol

14. Energy: Standard Enthalpy of Formation and Reaction – Direct & Hess’s Law Example 6.09 Standard Enthalpy of Reactions: Hess’s Law 2C (graphite) + 1H 2 (g) →  1C 2 H 2 (g) (2a) 2C (graphite) + 2O 2 (g) →  2CO 2 (g) –787.0 H2 is a reactant in both the original equation and in step (b). And, there’s 1 mole H2 in the original equation and 1 mol H2 in step (b). (b) H 2 (g) + ½O 2 (g) → H 2 O (l) –285.8 (c) 2C 2 H 2 (g) + 5O 2 (g) → 4CO 2 (g) + 2H 2 O (l) –2598.8 ΔH kJ/mol

15. Energy: Standard Enthalpy of Formation and Reaction – Direct & Hess’s Law Example 6.09 Standard Enthalpy of Reactions: Hess’s Law 2C (graphite) + 1H 2 (g) →  1C 2 H 2 (g) (b) H 2 (g) + ½O 2 (g) → H 2 O (l) –285.8 (c) 2C 2 H 2 (g) + 5O 2 (g) → 4CO 2 (g) + 2H 2 O (l) –2598.8 (2a) 2C (graphite) + 2O 2 (g) →  2CO 2 (g) –787.0 Align (c) C2H2 with original equation as a product by reversing step (c).......... ΔH kJ/mol

16. Energy: Standard Enthalpy of Formation and Reaction – Direct & Hess’s Law Example 6.09 Standard Enthalpy of Reactions: Hess’s Law 2C (graphite) + 1H 2 (g) →  1C 2 H 2 (g) (–c) 4CO 2 (g) + 2H 2 O (l) → 2C 2 H 2 (g) + 5O 2 (l) +2598.8 (b) H 2 (g) + ½O 2 (g) → H 2 O (l) –285.8 (2a) 2C (graphite) + 2O 2 (g) →  2CO 2 (g) –787.0 Reverse (c) so it on the same side as the original equation. Include reversing the sign of ΔH, too. Shown by (–c). However, there 2 moles of C2H2 in step (c) but only 1 mole in the original equation. So, we need to halve step (c).......................................................... ΔH kJ/mol

17. Energy: Standard Enthalpy of Formation and Reaction – Direct & Hess’s Law Example 6.09 Standard Enthalpy of Reactions: Hess’s Law 2C (graphite) + 1H 2 (g) →  1C 2 H 2 (g) (b) H 2 (g) + ½O 2 (g) → H 2 O (l) –285.8 Divide step (–c), including ΔH, by 2 to match the original equation’s 1 mole C2H2. (shown by (–c/2) (2a) 2C (graphite) + 2O 2 (g) →  2CO 2 (g) –787.0 ΔH kJ/mol

18. Energy: Standard Enthalpy of Formation and Reaction – Direct & Hess’s Law Example 6.09 Standard Enthalpy of Reactions: Hess’s Law 2C (graphite) + 1H 2 (g) →  1C 2 H 2 (g) (b) 1H 2 (g) + ½O 2 (g) → 1H 2 O (l) –285.8 Determine the net equation for both substances & ΔH by adding them together. (2a) 2C (graphite) + 2O 2 (g) →  2CO 2 (g) –787.0 CO 2 : 2 → 2 O 2 : 2 + ½ → 5/2 H 2 O: 1 → 1  H = 226.6 kJ/mol ΔH kJ/mol

19. Energy: Standard Enthalpy of Formation and Reaction – Direct & Hess’s Law Example 6.09 Standard Enthalpy of Reactions: Hess’s Law 2C (graphite) + 1H 2 (g) →  1C 2 H 2 (g) (-c/2) → 1C 2 H 2 (g) +1299.4 (b) + 1H 2 (g) –285.8 Net equation of steps = original equation  Δ H = +226.6 kJ/mol (2a) 2C (graphite) –787.0 ΔH kJ/mol 2C (graphite) + 1H 2 (g) →  1C 2 H 2 (g) +226.6 kJ/mol

20. Energy: Standard Enthalpy of Formation and Reaction – Direct & Hess’s Law Example 6.09 Standard Enthalpy of Reactions: Hess’s Law 2C (graphite) + 1H 2 (g) →  1C 2 H 2 (g) ΔH = +226.6 kJ/mol

21. 330_06_05 Energy: Calorimetry Fin