1. EXAMPLE: How much heat is required to heat 10.0 g of ice at -15.0 o C to steam at 127.0 o C? q overall = q ice + q fusion + q water + q boil + q steam q= (10.0g  2.09J/g o C  15.0 o C) + (10.0g  333J/g) + (10.0g  4.18J/g o C  100.0 o C) + (10.0g  2260J/g) + (10.0g  2.03J/g o C  27.0 o C) q = (314 + 3.33×10 3 + 4.18×10 3 + 2.26×10 4 + 548)J = 30.9 kJ

2. Heat Flow in Reactions exothermic –reaction that gives off energy q < 0 isolated system  E=0 heat released by reaction raises the temperature of the solvent constant T, heat is released to the surroundings endothermic – reaction that absorbs energy q > 0

3. Expansion Type Work w = -P  V system does work P P V initial VV  V = V final - V initial q p = +2kJ

4. Do 250 J of work to compress a gas, 180 J of heat are released by the gas What is  E for the gas? 1.430 J 2.70 J 3.-70 J 4.-180 J 5.-250 J 0 0 130 10

5. Enthalpy H  E = q + w at constant V, w expansion = 0  E = q v at constant P, w expansion = -P  V  E = q p - P  V Define  PV) =  P  V at constant P Hence  H = q p

6. Enthalpy Enthalpy  heat at constant pressure or the heat of reaction q p =  H = H products - H reactants Exothermic Reaction  H = (H products - H reactants ) < 0 2 H 2(g) + O 2(g)  2 H 2 O (l)  H < 0 Endothermic Reaction  H = (H products - H reactants ) > 0 2 H 2 O (l)  2 H 2(g) + O 2(g)  H > 0

7. State Functions  H and  E along with  P,  T,  V (or P, T, V) and many others are state functions. They are the same no matter what path we take for the change. q and w are not state functions, they depend on which path we take between two points. initial final EE q w q w  E=E final -E initial q and w can be anything

8. Path Independent Energy Changes

9. Which day would you like OWL quizzes due (4 AM) 1234567891011121314151617181920 2122232425262728293031323334353637383940 4142434445464748495051525354555657585960 6162636465666768697071727374757677787980 81828384858687888990919293949596979899100 101102103104105106107108109110111112113114115116117118119120 121122123124125126127128129130 1.Monday 2.Tuesday 3.Wednesday 4.Thursday 5.Friday 10 0 0 130

10. Stepwise Energy Changes in Reactions

11. Laws of Thermochemistry 1. The magnitude of  is directly proportional to the amount of reaction.  H is for 1 mole of reaction as written 2 H 2(g) + O 2(g)  2 H 2 O (l)  H = -571.6 kJ H 2(g) + ½ O 2(g)  H 2 O (l)  H = -285.8 kJ Can have ½ mole O 2 just not ½ molecule

12. Laws of Thermochemistry 2.  H for a reaction is equal in magnitude but opposite in sign to  H for the reverse reaction. H 2(g) + ½ O 2(g)  H 2 O (l)  H = -285.8 kJ H 2 O (l)  H 2(g) + ½ O 2(g)  H = +285.8 kJ

13. Laws of Thermochemistry 3. The value of  H for the reaction is the same whether it occurs directly or in a series of steps.  H overall =  H 1 +  H 2 +  H 3 + · · · also called Hess’ Law

14. Enthalpy Diagram H 2 (g) + ½ O 2 (g)  H 2 O(l)  H = -285.8 kJ H 2 O(l)  H 2 O(g)  H = +44.0 kJ H 2 (g) + ½ O 2 (g)  H 2 O(g)  H = -241.8 kJ

15. Given 3 CO + 3/2 O 2  3 CO 2  H = -849 kJ What is  H for CO 2  CO + ½ O 2 ? 10 0 0 130 1.-283 kJ 2.+283 kJ 3.+849 kJ 4.-2547 kJ 5.+2547 kJ

16. Energy and Stoichiometry Since  H is per mole of reaction we can relate heat to amount of reaction Given C 2 H 6 + 7/2 O 2  2 CO 2 + 3 H 2 O  H=-1559.7 kJ If 632.5 kJ are released to surroundings what mass of H 2 O is formed? 632.5 kJ released means  H = -632.5 kJ for this much H 2 O

17. Bomb Calorimeter measure q v q rxn + q cal = 0 q rxn = -q cal q rxn = - c cal  T  E rxn = q rxn /moles rxn  E rxn ≈  H rxn  H =  E +  (PV)  H =  E + RT  n gas @298K RT = 2.5 kJ/mol

18. “Coffee Cup” Calorimeter q p Photo by George Lisensky

19. Measuring  H When 25.0 mL of 1.0 M H 2 SO 4 are added to 50.0 mL of 1.0 M KOH, both initially at 24.6  C the temperature rises to 33.9  C. What is  H for H 2 SO 4 + 2 KOH  K 2 SO 4 + 2 H 2 O ? (Assume d = 1.00 g/mL, c = 4.18 J/g.  C) q soln = mc  T m = (25.0 + 50.0)mL×1.00g/mL = 75.0 g

20. Measuring  H cont. q=mc  T q soln = 75.0 g × 4.18 J/g.  C × (33.9-24.6)  C q soln = 2916 J q rxn + q soln = 0 q rxn = -2916 J  H rxn = q rxn /moles rxn

21. Measuring  H cont How many moles rxn? 1 mol rxn / 1 mol H 2 SO 4 1 mol rxn / 2 mol KOH Stoichiometric mixture so 0.025 mol rxn

22. Measuring  H cont  H rxn = q rxn /moles rxn  H rxn = -2916 J / 0.025 mol rxn  H rxn = -116622 J / mol rxn  H rxn = -117 kJ  H is per mole of reaction as written

23. If excess Al is added to 50 mL of 0.250 M H 2 SO 4 how many moles of the following reaction occur? 2 Al + 3 H 2 SO 4  Al 2 (SO 4 ) 3 + 3 H 2 10 0 0 130 1.0.0125 mol 2.0.0375 mol 3.0.025 mol 4.0.00625 mol 5.0.00417 mol

24. Hess’s Law Can find  H for an unknown, or hard to measure, reaction by summing measured  H values of known reactions.

25. EXAMPLE  H for formation of CO cannot readily be measured since a mixture of CO and CO 2 is always formed. C (s) + ½ O 2 (g)  CO (g)  H = ? C (s) + O 2 (g)  CO 2 (g)  H 1 = -393.5 kJ CO (g) + ½ O 2 (g)  CO 2 (g)  H 2 = -283.0 kJ C (s) + ½ O 2 (g)  CO (g)  H =  H 1 -  H 2  H =  H 1 -  H 2 = -393.5 – (-283.0) = -110.5 kJ

26. Standard Enthalpy of Formation the enthalpy associated with the formation of 1 mol of a substance from its constituent elements under standard state conditions at the specified temperature For an element this is a null reaction O 2 (g)  O 2 (g)  H = 0  H  f = 0 for all elements in their standard states

27. For which one of these reactions is ΔH º rxn = ΔH º f ? 1.N 2 (g) + 3 H 2 (g)  2 NH 3 (g) 2.C(graphite) + 2 H 2 (g)  CH 4 (g) 3.C(diamond) + O 2 (g)  CO 2 (g) 4.CO(g) + ½ O 2 (g)  CO 2 (g) 5.H 2 (g) + Cl 2 (g)  2 HCl(g) 10 0 0 130

28. Calculation of  H o  H o =  mols   H f o products –  mols   H f o reactants We can always convert products and reactants to the elements. Hess’s law says  H is the same whether we go directly from reactants to products or go via elements

29. Example What is the value of  H rxn for the reaction: 2 C 6 H 6(l) + 15 O 2(g)  12 CO 2(g) + 6 H 2 O (g) from Appendix J Text C 6 H 6(l)  H f o = + 49.0 kJ/mol O 2(g)  H f o = 0 CO 2(g)  H f o = - 393.5 H 2 O (g)  H f o = - 241.8  H rxn  mols   H f o  product –  mols  H f o  reactants

30. Example What is the value of  H rx for the reaction: 2 C 6 H 6(l) + 15 O 2(g)  12 CO 2(g) + 6 H 2 O (g) from Appendix J Text C 6 H 6(l)  H f o = + 49.0 kJ/mol; O 2(g)  H f o = 0 CO 2(g)  H f o = - 393.5; H 2 O (g)  H f o = - 241.8  H rxn  mols  H f o  product -  mols  H f o  reactants  H rxn  - 393.5) + 6(- 241.8)  product -  2(+ 49.0 ) + 15(0)  reactants kJ/mol = - 6.2708  10 3 kJ

31. Fossil Fuels coal petroleum natural gas

32. Energy Resources in the U.S.

33. Caloric Value of Some Foods