1. Thermodynamics Heat, disorder, spontaneity

2. Energy The capacity to perform work –often measured as heat

3. Energy A tub is filled with water at 35°C –Dip a cup into the water and fill it. –What is the temperature of the water in the cup?

4. Energy Which amount of water, that in the tub or in the cup, can melt the greater amount of ice during the same time frame?

5. Energy Two substances may have the same temperature but different amounts of heat energy.

6. Energy Temperature is the measure of average KE of a substance

7. Energy Heat is the measure of the total energy transferred from an object with a higher temperature to an object with a lower temperature.

8. Energy Heat is measured in either Joules (J) or calories (cal) A calorie is defined as the amount of heat needed to raise 1 g of water 1°C. 1 cal = 4.18 J

9. Energy Graph the following data for two experiments on the same hand-drawn graph.

10. Time for ice to melt… 0°C00 0°C (ice disappears)25190 25°C33250 50°C41310 75°C49370 100°C (water begins to boil) 57429 100°C (water disappears) 2261701 Temperature 1 cube (s) 8 cubes (s) Time

11. Energy Temperature (°C) Time (s)

12. Specific Heat Capacity the amount of heat energy required to raise the temperature of one gram of a substance by one degree Celsius Measured in J/g°C or cal/g°C

13. Specific Heat Capacity When a substance’s SHC (or C) is greater, more heat is required to make that substance equal in temperature to a substance with a lesser SHC

14. Specific Heat Capacity Which has the greater SHC, silicone or iron?  heat = (  T)(mass)(SHC)

15. Heating Curve for H 2 O Heat (cal) Temperature (°C) G H G

16. Heating Curve for H2O BC has value of 80 cal/g –Known as the heat of fusion (s  l) or heat of solidification (l  s) DE has a value of 540 cal/g –Known as the heat of vaporization (l  g) or heat of condensation (g  l)

17. Heating Curve for H2O G has a value of 0°C –known as the melting point or the freezing point H has a value of 100°C –Known as the boiling point or the condensation point

18. Calorimetry Measurement of heat energy Two types of calorimeters –Constant pressure (coffee-cup calorimeter) –Constant volume (bomb calorimeter)

19. Biological Calorimetry Nutrients –Carbohydrates –Proteins –Lipids –Water –Vitamins –minerals

20. Biological Calorimetry Carbohydrates –4 kcal/g or 17 kJ/g Proteins –4 kcal/g or 17 kJ/g Lipids –9 kcal/g or 38 kJ/g

21. Heat of Reaction  H rxn amount of heat absorbed or released in a chemical reaction If absorbed, it is a reactant and the process is endothermic If released, it is a product and the process is exothermic

22. Heat of Reaction Deviations –  H formation is amount of heat absorbed or released during synthesis of one mole of an element or compound at 298 K and 1atm of pressure

23. Heat of Reaction Deviations –  H solution is amount of heat absorbed or released when a substance dissolves in a solvent

24. Heat of Reaction Deviations –  H combustion is amount of heat released when a substance reacts with O 2 to form CO 2 and H 2 O

25. Heat of Reaction Is part of the stoichiometry of a reaction…the heat of combustion of methane is 803 kJ CH 4 (g) + 2O 2 (g)  CO 2 (g) + 2H 2 O(g) + 803 kJ If there were 5 moles of CH 4 present, how many kJ would be produced?

26. Heat Energy Practice Problems 1.How many kJ are released by a reaction that raises the temperature of 1.00 kg of water in a coffee-cup calorimeter from 25.0°C to 27.0°C? Psst…you know the SHC of water

27. 2. A swimming pool measures 6.0 m x 12.0 m and is 3.0 m deep all around. The pool is filled with water at a temperature of 20.0°C. How many kJ must be released by the pool’s heater to raise the water temperature to 25.0°C? Psst…the density of water is 1 g/cm 3, you know the SHC of water, and 1 m = 100 cm, so 1 m 3 would equal how many cm 3 ?

28. 3. Gaseous butane, C 4 H 10, is burned in lots of lighters. Write the balanced equation for the complete combustion of butane. Butane’s heat of combustion is 2878 kJ. How many kJ of heat energy would be released by the combustion of 10.0 g of butane?

29. 4. Use the table on the next slide to calculate the number of kiloJoules provided by the fat in one serving of each of the following foods: a. french fries b. cheeseburger

30. 4. (continued) French fries (3.4 oz.) 32036.34.417.1 Cheeseburge r (4.1 oz.) 31031.215.013.8 Food (amt.) kcal carb(g) prot(g) fat(g)

31. 5. Is more energy released when 428 g of H 2 or 428 g of isooctane, C 8 H 18, react with an excess of oxygen? Psst…balance the equations. 2H 2 + O 2  2H 2 O + 484 kJ 2C 8 H 18 + 25O 2  16CO 2 + 18H 2 O + 4893 kJ

32. 1.8.36 kJ 2.4.51 x 10 6 kJ 3.248 kJ 4.a. 650 kJ b. 524 kJ 5. 428 g H 2 will release 5.13 x 10 4 kJ while 428 g of C 8 H 18 will release 9.16 x 10 3 kJ. So, the 428 g H 2 will release more energy.

33. Activation Energy Rxn progress (s) Energy (kJ) EaEa HH Reactants Products

34. Activation Energy Rxn progress (s) Energy (kJ) EaEa Uncatalyzed Catalyzed HH

35. Enthalpy Enthalpy can be equated with heat energy represented by H  H is also known as change in enthalpy

36. Hess’s Law states that in going from a particular set of reactants to a particular set of products, the change in enthalpy is the same whether the reaction takes place in one step or in a series of steps.

37. Finding  H using Hess’s Law If a reaction is reversed, the sign of  H is also reversed. If the coefficients in a balanced equation are multiplied by an integer, then the value of  H is multiplied by that same integer.

38. Hess’s Law Consider the following reaction: N 2 (g) + 2O 2 (g)  2NO 2 (g) It does not necessarily occur as we see it. It can, in fact, occur in a few additive steps, known as elementary steps.

39. Hess’s Law Plausible elementary steps: a.N 2 (g) + 2O 2 (g)  2NO(g)  H = 180 kJ b. 2NO(g) + O 2 (g)  2NO 2 (g)  H = -112 kJ

40. Hess’s Law a.N 2 (g) + O 2 (g)  2NO(g)  H = 180 kJ b. 2NO(g) + O 2 (g)  2NO 2 (g)  H = -112 kJ N 2 (g) + 2O 2 (g) + 2NO(g)  2NO(g) + 2NO 2 (g) N 2 (g) + 2O 2 (g)  + 2NO 2 (g)  H = 68 kJ So, the reaction is endothermic.

41. Hess’s Law Two forms of carbon are graphite and diamond. Using the enthalpies of combustion for graphite and diamond as your elementary steps, calculate the  H for the conversion of graphite to diamond and state whether it is an endo- or exothermic process. C (graphite) (s)  C (diamond) (s)  H = ?

42. Hess’s Law The elementary steps are: a. C (graphite) (s) + O 2 (g)  CO 2 (g)  H = -394kJ b. C (diamond) (s) + O 2 (g)  CO 2 (g)  H = -396kJ

43. Hess’s Law a. C (graphite) (s) + O 2 (g)  CO 2 (g)  H = -394kJ b. CO 2 (g)  C (diamond) (s) + O 2 (g)  H = +396kJ C (graphite) (s) + O 2 (g) + CO 2 (g)  CO 2 (g) + C (diamond) (s) + O 2 (g) C (graphite) (s)  C (diamond) (s)  H = 2kJ So, it is endothermic.

44. Finding  H using standard heats of formation  H = ∑  H f ° products − ∑  H f ° reactants Use pages Zumdahl Chemistry II textbook All elements in their natural states will have  H f ° equal to zero.

45. Finding  H using standard heats of formation Find the  H of the following reaction using  H f ° values: N 2 (g) + 2O 2 (g)  2NO 2 (g)

46. Finding  H using standard heats of formation (2mol NO 2 x 34kJ) mol (1mol N 2 x 0kJ) + (2mol O 2 x 0kJ) mol mol 68kJ; the reaction is endothermic

47. Finding  H using standard heats of formation Find the  H of the reaction which converts graphite to diamond using  H f ° values.

48. Entropy Is the measure of disorder or chaos present in a substance. Chemical reactions may result in increasing disorder or decreasing disorder. Represented by S…thus, change in entropy is  S

49. Entropy When there are more moles of products than reactants, entropy usually increases. When phase changes from more organized to less organized, entropy increases. If  S is positive, entropy increases; if negative, entropy decreases.

50. Finding  S using standard entropy values  S = ∑  S ° products − ∑  S ° reactants Use Zumdahl Chemistry II textbook

51. Finding  S using standard entropy values Find the  S of the following reaction using S ° values: N 2 (g) + 2O 2 (g)  2NO 2 (g)

52. Finding  S using standard entropy values (2mol NO2 x 240J) K·mol (1mol N2 x 192J) + (2mol O2 x 205J) K·mol K·mol -122 J/K; entropy is decreasing

53. Finding  S using standard entropy values Find the  S of the reaction which converts graphite to diamond using S ° values.

54. Spontaneity refers to whether a reaction will happen without outside intervention or not. It says nothing about how quickly the reaction will happen only that it will or will not occur.

55. Free Energy is symbolized by G and is used to determine the spontaneity of a reaction  G =  H T  S

56. Free Energy If  G is positive, it is a nonspontaneous process and is known as an endergonic reaction. If  G is negative, it is a spontaneous process and is known as an exergonic reaction.

57. Free Energy Find the  G of the following reaction at 25°C : N 2 (g) + 2O 2 (g)  2NO 2 (g)

58. Free Energy  G = 68 kJ (298 K)(-0.122 kJ) K  G = 104 kJ; the reaction is nonspontaneous or endergonic

59. Free Energy Find the  G of the reaction which converts graphite to diamond at 100°C and state whether it is spontaneous or not.

60. More Practice Problems… 6.Acetylene gas, C 2 H 2, is used in some welding applications and can be made via the following reaction: 2C(s) + H 2 (g)  C 2 H 2 (g) Determine its  H using the elementary steps on the following slide.

61. a. C 2 H 2 (g) + 2½O 2 (g)  2CO 2 (g) + H 2 O(l)  H = -1300kJ b. C(s) + O 2 (g)  CO 2 (g)  H = -394kJ c. H 2 (g) + ½O 2 (g)  H 2 O(l)  H = -286kJ

62. 7. Think about photosynthesis… recall that carbon dioxide gas reacts with water to produce solid glucose (C 6 H 12 O 6 ) and oxygen gas. Write a balanced equation, and determine the  H,  S, and  G at 25°C. State whether the reaction is endo- or exothermic, whether entropy increases or decreases, and whether it spontaneous or not.

63. 6.  H = +226 kJ 7.  H = +2802 kJ; endothermic  S = -262 J/K; entropy decreases  G = +2880 kJ; nonspontaneous