1. Chapter 12 12-1: Chemical Reactions That Involve Heat Suggested Reading: Pages 381 - 387

2. Chemical Reactions Involve ENERGY Changes in ENERGY result from bonds being broken and new bonds being formed.

3. Requires ENERGY Breaking Bonds

4. Releases ENERGY Bond Formation

5. Both absorption and release of energy occurs. In a chemical reaction We detect the net result. Measure the temperature of the surroundings.

6. System: Reactants and Products System & Surroundings Surroundings: Solvent, container, atmosphere above the reaction, etc.

7. The study of the changes in heat in chemical reactions. Thermochemistry

8. Types of Reactions

9. RELEASE HEAT! Exothermic Reactions C 3 H 8 (g) + 5O 2 (g) 3CO 2 (g) + 4H 2 O (g) + 2043 kJ Combustion reactions are exothermic! HEAT is listed as a product in the reaction! HEAT ENERGY

10. Exothermic Reactions Energy needed to break bonds must be LESS THAN the energy released when new bonds are formed. Surroundings will have a higher temperature after the reaction!

11. NaOH (s) Na + (aq) + OH - (aq) Beginning Temp of Surroundings: 25.4 ° C Ending Temp of Surroundings: 29.5 ° C Change in Temp of Surroundings: +4.1 ° C +4.1 ° C Means heat was GIVEN OFF EXOTHERMIC

12. NaOH (s) Na + (aq) + OH - (aq) +4.1 ° C Calculate HEAT per MOLE if you have 5.0 grams of NaOH to start. 5.0 g NaOH 40 g NaOH 1 mol NaOH 0.125 mol NaOH = +4.1 ° C 0.125 mol NaOH = + 33 ° C per mole NaOH Means HEAT is given off.

13. ABSORB HEAT! Endothermic Reactions C (s) + H 2 O (g) + 113 kJ CO (g) + H 2 (g) HEAT is listed as a reactant in the reaction! HEAT ENERGY

14. Endothermic Reactions The energy needed to break bonds is GREATER THAN the energy released when new bonds are formed. Surroundings will have a lower temperature after the reaction!

15. Exo & Endo Demos

16. Fireworks

17. More about fireworks

18. Section 2 Heat & Enthalpy Changes

19. ENTHALPYENTHALPY b Heat content of a system at constant pressure. b Heat absorbed or released in a reaction depends on the difference in enthalpy.

20. ENTHALPYENTHALPY b Represented by capital H. b  (delta) means a change or difference. b  H = change in enthalpy.

21. ENTHALPY CHANGE  H  H RXN = H PRODUCTS - H REACTANTS Heat of Reaction ΔH

22. Exothermic Reactions ENTHALPY CHANGE IS NEGATIVE Heat content of products Is LOWER! Heat content of reactants Is HIGHER HEAT IS RELEASED!

23. Endothermic Reactions ENTHALPY CHANGE IS POSITIVE Heat content of reactants Is LOWER! Heat content of products Is HIGHER HEAT IS ABSORBED!

24. ENTHALPY PROBLEMS b Similar to “Stoich” problems. b The change in energy depends on the number of moles of the reactants.

25. Calculate the amount of heat released when 5.50 g of methane reacts with excess oxygen. CH 4 (g) + 2O 2 (g)  CO 2 (g) + 2H 2 O(g) ΔH°= -890kJ 5.50 g CH 4 1 mol CH 4 16 g CH 4 890 kJ 1 mol CH 4 = 306 kJ

26. How much heat will be released when 6.44 g of sulfur reacts with excess oxygen? 2S + 3O 2  2SO 3 ΔH°= -791.4 kJ 6.44 g S 1 mol S 32 g S 791.4 kJ 2 mol S = 79.6 kJ

27. Suggested Reading: Online Pages 539-542 Hess’s Law

28. HESS’S LAW b If a series of reactions are added together, the enthalpy change for the net reaction will be the sum of the enthalpy changes for the individual steps.

29. Hess’s Law Start Finish Enthalpy is Path independent. Both lines accomplished the same result, they went from start to finish. Net result = same.

30. Hess’s law can be used to determine the enthalpy change for a reaction that cannot be measured directly!

31.  H NET =  H 1 +  H 2 N 2 (g) + O 2 (g)  2NO(g) ΔH 1 = +181kJ 2NO (g) + O 2 (g)  2NO 2 (g) ΔH 2 = - 113kJ ADD THEM UP ALEGBRAICALLY

32. N 2 (g) + O 2 (g)  2NO(g) ΔH 1 = +181kJ 2NO (g) + O 2 (g)  2NO 2 (g) ΔH 2 = - 113kJ N 2 (g) + 2O 2 (g) First, add up the chemical equations. + 2NO(g)  2NO(g) + 2NO 2 (g)

33. Notice that 2NO(g) is on both the reactants and products side and can be cancelled out. N 2 (g) + O 2 (g)  2NO(g) ΔH 1 = +181kJ 2NO (g) + O 2 (g)  2NO 2 (g) ΔH 2 = - 113kJ N 2 (g) + 2O 2 (g) + 2NO(g)  2NO(g) + 2NO 2 (g)

34. Write the net equation: N 2 (g) + 2O 2 (g) + 2NO(g)  2NO(g) + 2NO 2 (g) N 2 (g) + 2O 2 (g)  2NO 2 (g)

35.  H NET =  H 1 +  H 2 ΔH 1 = +181kJ ΔH 2 = -113kJ Apply Hess’s Law to calculate the enthalpy for the reaction.

36.  H NET =  H 1 +  H 2 ΔH NET = (+181kJ) + (- 113kJ) ΔH NET = +68kJ Overall, the formation of NO 2 from N 2 and O 2 is an endothermic process, although one of the steps is exothermic.

37. ΔH Reaction Progress N 2 (g) + 2O 2 (g) 2NO(g) + O 2 (g) 2NO 2 (g) ΔH NET = +68kJ ΔH 1 = +181kJ ΔH 2 = -113kJ

38. RULES for Hess’s Law Problems 1.If the coefficients are multiplied by a factor, then the enthalpy value MUST also be multiplied by the same factor. 2.If an equation is reversed, the sign of ΔH MUST also be reversed.

39. C(s) + ½O 2 (g)  CO(g) ΔH 1 = -110.5kJ CO(g) + ½O 2 (g)  CO 2 (g) ΔH 2 = -283.0kJ C(s) + O 2 (g) + CO(g)  CO(g) + CO 2 (g) Practice Problem: #1 C(s) + O 2 (g)  CO 2 (g)  H NET =  H 1 +  H 2  H NET = (-110.5kJ) + (-283.0kJ)  H NET = -393.5kJ Net Equation

40. C 2 H 5 OH(l) + 3O 2 (g)  2CO 2 (g) + 3H 2 O(g) Practice Problem: #3 CH 3 OCH 3 (l) + 3O 2 (g)  2CO 2 (g) + 3H 2 O(g) ΔH 1 = -1234.7kJ ΔH 2 = -1328.3kJ You have to REVERSE equation 2 to get the NET equation. DON’T forget to change the sign Of ΔH 2

41. C 2 H 5 OH(l) + 3O 2 (g)  2CO 2 (g) + 3H 2 O(g) Practice Problem: #3 2CO 2 (g) + 3H 2 O(g)  CH 3 OCH 3 (l) + 3O 2 (g) ΔH 1 = -1234.7kJ ΔH 2 = +1328.3kJ C 2 H 5 OH(l) + 3O 2 (g) + 2CO 2 (g) + 3H 2 O(g)  2CO 2 (g) + 3H 2 O(g) + CH 3 OCH 3 (l) + 3O 2 (g) Net Equation C 2 H 5 OH(l)  CH 3 OCH 3 (l)

42. Net Equation C 2 H 5 OH(l)  CH 3 OCH 3 (l)  H NET =  H 1 +  H 2  H NET = (-1234.7kJ) + (+1328.3kJ)  H NET = +93.6kJ

43. H 2 (g) + F 2 (g)  2HF(g) ΔH 1 = -542.2kJ 2H 2 (g) + O 2 (g)  2H 2 O(g) ΔH 2 = -571.6kJ Practice Problem: #5 You have to REVERSE equation 2 to get the NET equation. DON’T forget to change the sign Of ΔH 2

44. H 2 (g) + F 2 (g)  2HF(g) ΔH 1 = -542.2kJ 2H 2 O(g)  2H 2 (g) + O 2 (g) ΔH 2 = +571.6kJ Practice Problem: #5 You will need to multiply the first equation by 2. DON’T forget to multiply the ΔH by 2 also.

45. 2H 2 (g) + 2F 2 (g)  4HF(g) ΔH 1 = - 1084.4kJ 2H 2 O(g)  2H 2 (g) + O 2 (g) ΔH 2 = +571.6kJ Practice Problem: #5 Net Equation 2H 2 (g) + 2F 2 (g) + 2H 2 O(g)  4HF(g) + 2H 2 (g) + O 2 (g) 2F 2 (g) + 2H 2 O(g)  4HF(g) + O 2 (g)  H NET =  H 1 +  H 2  H NET = (-1084.4kJ) + (+571.6kJ)  H NET = -512.8kJ

46. Suggested Online Review Pages 530- 545 Section 4 Calorimetry

47. CalorimetryCalorimetry b The study of heat flow and heat measurement.

48. Heat Capacity b The amount of heat needed to raise the temperature of an object by 1 Celsius degree.

49. Specific Heat b The heat capacity of 1 gram of a substance b Specific Heat of liquid water is 4.184 J/gºC b 4.184 J = 1 Calorie

50. b 1 Calorie = 1000 calories = 1 kilocalorie b 1 Food Calorie = 1000 calories

51. Calorimetry Experiments b Determine the heats of reaction (ENTHALPY CHANGES) by making accurate measurements of temperature changes using a calorimeter.

52. b Scientists use q to denote measurements made in a calorimeter. b Heat transferred in a reaction is EQUAL, but OPPOSITE in sign to heat absorbed by the surroundings. b q rxn = - q sur

53. q sur = m x C p x (T f –T i ) Mass of Water Specific heat of Water Temperature change Heat Gain/Lost

54. Practice Problem: #1 When a 12.8g sample of KCl dissolves in 75.0g of water in a calorimeter, the temperature drops from 31.0ºC to 21.6ºC. Calculate  H for the process. KCl(s)  K + (aq) + Cl - (aq) First, calculate q sur and then calculate  H.

55. Practice Problem: #1 When a 12.8g sample of KCl dissolves in 75.0g of water in a calorimeter, the temperature drops from 31.0ºC to 21.6ºC. Calculate  H for the process. KCl(s)  K + (aq) + Cl - (aq) q sur = m x C p x (T f –T i ) q sur = 75.0g x 4.184 J/gºC x (21.6 ºC –31.0 ºC ) q sur = -2949.72 J

56. q sur is negative (as expected) based on the temperature drop of the water. KCl(s)  K + (aq) + Cl - (aq) q rxn = - q sur = +2949.72 J q rxn represents the heat absorbed due to the reaction of 12.8g KCl. Now you must convert the KCl to moles.

57. Convert grams KCl to moles. 12.8 g KCl 74 g KCl 1 mol KCl = 0.173 mol KCl

58. Calculate  H for the reaction = 0.173 mol KCl KCl(s)  K + (aq) + Cl - (aq) HH +2949.72 J x 1 mol KCl Coefficient from balanced equation HH = +17050 J HH = +17.1 kJ  H Must be positive because it was an endothermic reaction!

59. Practice Problem: #2 What is the specific heat of aluminum if the temperature of a 28.4 g sample of aluminum is increased by 8.1ºC when 207 J of heat is added? q sur = m x C p x (T f –T i ) 207J = 28.4g x C p x 8.1 º C Cp =Cp = 28.4 g x 8.1 º C 207J = 0.90 J/g º C