1. Dr. S. M. Condren Chapter 6 Thermochemistry

2. Dr. S. M. Condren Thermite Reaction

3. Dr. S. M. Condren Thermite Reaction

4. Dr. S. M. Condren Terminology Energy capacity to do work Kinetic Energy energy that something has because it is moving Potential Energy energy that something has because of its position

5. Dr. S. M. Condren Kinetic Energy

6. Dr. S. M. Condren Chemical Potential Energy

7. Dr. S. M. Condren Chemical Potential Energy

8. Dr. S. M. Condren Internal Energy The sum of the individual energies of all nanoscale particles (atoms, ions, or molecules) in that sample. E = 1/2mc 2 The total internal energy of a sample of matter depends on temperature, the type of particles, and how many of them there are in the sample.

9. Dr. S. M. Condren Energy Units calorie - energy required to heat 1-g of water 1 o C Calorie - unit of food energy; 1 Cal = 1-kcal = 1000-cal Joule - 1-cal = 4.184 J = 1-kg*m 2 /sec 2

10. Dr. S. M. Condren Law of Conservation of Energy energy can neither be created nor destroyed the total amount of energy in the universe is a constant energy can be transformed from one form to another

11. Dr. S. M. Condren First Law of Thermodynamics the amount of heat transferred into a system plus the amount of work done on the system must result in a corresponding increase of internal energy in the system

12. Dr. S. M. Condren Thermochemistry Terminology system => that part of the universe under investigation surroundings => the rest of the universe universe = system + surroundings

13. Dr. S. M. Condren Thermodynamic System

14. Dr. S. M. Condren Energy Transfer Energy is always transferred from the hotter to the cooler sample Heat – the energy that flows into or out of a system because of a difference in temperature between the thermodynamic system and its surroundings

15. Dr. S. M. Condren Thermochemistry Terminology state properties => properties which depend only on the initial and final states => properties which are path independent non-state properties => properties which are path dependent state properties => E non-state properties => q & w

16. Dr. S. M. Condren Thermochemistry Terminology exothermic - reaction that gives off energy endothermic - reaction that absorbs energy chemical energy - energy associated with a chemical reaction thermochemistry - the quantitative study of the heat changes accompanying chemical reactions thermodynamics - the study of energy and its transformations

17. Dr. S. M. Condren 2 H 2(g) + O 2(g) --> 2 H 2 O (g) + heat and light Energy & Chemistry This can be set up to provide ELECTRIC ENERGY in a. in a fuel cell. Oxidation: 2 H 2 ---> 4 H+ + 4 e- 2 H 2 ---> 4 H+ + 4 e-Reduction: 4 e- + O2 + 2 H 2 O ---> 4 OH- 4 e- + O2 + 2 H 2 O ---> 4 OH-

18. Dr. S. M. Condren Energy & Chemistry

19. Dr. S. M. Condren Enthalpy heat at constant pressure q p =  H = H products - H reactants Exothermic Reaction  H = (H products - H reactants ) < 0 H 2 O (l) -----> H 2 O (s)  H < 0 Endothermic Reaction  H = (H products - H reactants ) > 0 H 2 O (l) -----> H 2 O (g)  H > 0

20. Dr. S. M. Condren Enthalpy H = E + PV  H =  E + P  V  E =  H – P  V Where text uses U for internal energy

21. Dr. S. M. Condren Pressure-Volume Work

22. Dr. S. M. Condren First Law of Thermodynamics heat => q internal energy => E internal energy change =>  E work => w  E = q - w (Engineering convention)

23. Dr. S. M. Condren Specific Heat the amount of heat necessary to raise the temperature of 1 gram of the substance 1 o C independent of mass substance dependent s.h. Specific Heat of Water = 4.184 J/g o C

24. Dr. S. M. Condren Heat q = m * s.h. *  t whereq => heat, J m => mass, g s.h. => specific heat, J/g* o C  t = change in temperature, o C

25. Dr. S. M. Condren Molar Heat Capacity the heat necessary to raise the temperature of one mole of substance by 1 o C substance dependent C

26. Dr. S. M. Condren Heat Capacity the heat necessary to raise the temperature 1 o C mass dependent substance dependent C

27. Dr. S. M. Condren Heat Capacity C = m X s.h. whereC => heat capacity, J/ o C m => mass, g s.h. => specific heat, J/g o C

28. Dr. S. M. Condren Plotted are graphs of heat absorbed versus temperature for two systems. Which system has the larger heat capacity? A, B

29. Dr. S. M. Condren Heat Transfer q lost = - q gained (m X s.h. X  t) lost = - (m X s.h. X  t) gained

30. Dr. S. M. Condren EXAMPLE If 100. g of iron at 100.0 o C is placed in 200. g of water at 20.0 o C in an insulated container, what will the temperature, o C, of the iron and water when both are at the same temperature? The specific heat of iron is 0.106 cal/g o C. (100.g*0.106cal/g o C*(T f - 100.) o C) = q lost - q gained = (200.g*1.00cal/g o C*(T f - 20.0) o C) 10.6(T f - 100. o C) = - 200.(T f - 20.0 o C) 10.6T f - 1060 o C = - 200.T f + 4000 o C (10.6 + 200.)T f = (1060 + 4000) o C T f = (5060/211.) o C = 24.0 o C

31. Dr. S. M. Condren 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 =  H ice +  H fusion +  H water +  boil. +  steam q =  H ice +  H fusion +  H water +  boil. +  steam

32. Dr. S. M. Condren Heat Transfer

33. Dr. S. M. Condren 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 =  H ice +  H fusion +  H water +  boil. +  steam q= (10.0g*2.09J/g o C*((0.0 – (-15.0)) o C)) Mass of the ice specific heat of ice Temperature change {

34. Dr. S. M. Condren 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 =  H ice +  H fusion +  H water +  boil. +  steam q= (10.0g*2.09J/g o C*15.0 o C) + (10.0g*333J/g) Mass of ice Heat of fusion Melting of ice occurs at a constant temperature

35. Dr. S. M. Condren 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 =  H ice +  H fusion +  H water +  boil. +  steam q= (10.0g*2.09J/g o C*15.0 o C) + (10.0g*333J/g) + (10.0g*4.18J/g o C*((100.0-0.00) o C)) Mass of water Specific heat of liquid water Temperature change of the liquid water

36. Dr. S. M. Condren 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 =  H ice +  H fusion +  H water +  boil. +  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) Mass of waterHeat of vaporization Boiling of water occurs at a constant temperature

37. Dr. S. M. Condren 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 =  H ice +  H fusion +  H water +  boil. +  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*((127.0-100.0) o C)) Mass of steam Specific heat of steam Temperature change for the steam

38. Dr. S. M. Condren 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 =  H ice +  H fusion +  H water +  boil. +  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 )J + 3.33X10 3 + 4.18X10 3 + 2.26X10 4 + 548 = 30.96 kJ

39. Dr. S. M. Condren Spreadsheet of Previous Problem

40. Dr. S. M. Condren Bomb Calorimeter Parr calorimeter

41. Dr. S. M. Condren EXAMPLE A 1.000g sample of a particular compound produced 11.0 kJ of heat. The temperature of the calorimeter and 3000 g of water was raised 0.629 o C. How much heat is gained by the calorimeter? heat gained = - heat lost heat calorimeter + heat water = heat reaction heat calorimeter = heat reaction - heat water

42. Dr. S. M. Condren EXAMPLE A 1.000g sample of a particular compound produced 11.0 kJ of heat. The temperature of the calorimeter and 3000 g of water was raised 0.629 o C. How much heat is gained by the calorimeter? heat calorimeter = heat reaction - heat water heat = 11.0 kJ - ((3.00kg)(0.629 o C)(4.184kJ/kg o C)) = 3.1 kJ

43. Dr. S. M. Condren Example What is the mass of water equivalent of the heat absorbed by the calorimeter? #g = (3.1 kJ/0.629 o C)(1.00kg* o C/4.184kJ) = 6.5 x 10 2 g

44. Dr. S. M. Condren Example A 1.000 g sample of ethanol was burned in the sealed bomb calorimeter described above. The temperature of the water rose from 24.284 o C to 26.225 o C. Determine the heat for the reaction. m = (3000 + "647")g H 2 O q = m X s.h. X  t = (3647g)(4.184J/g o C)(1.941 o C) = 29.61 kJ

45. Dr. S. M. Condren When graphite is burned to yield CO 2, 394 kJ of energy are released per mole of C atoms burned. When C 60 is burned to yield CO 2 approximately 435 kJ of energy is released per mole of carbon atoms burned. Would the buckyball-to-graphite conversion be exothermic or endothermic? exothermic, endothermic

46. Dr. S. M. Condren Laws of Thermochemistry 1. The magnitude of  is directly proportional to the amount of reactant or product. s --> l  H => heat of fusion l --> g  H => heat of vaporization

47. Dr. S. M. Condren Laws of Thermochemistry 2.  H for a reaction is equal in magnitude but opposite in sign to  H for the reverse reaction. H 2 O (l) -----> H 2 O (s)  H < 0 H 2 O (s) -----> H 2 O (l)  H > 0

48. Dr. S. M. Condren 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 + · · ·

49. Dr. S. M. Condren Hess' Law a relation stating that the heat flow in a reaction which is the sum of a series of reactions is equal to the sum of the heat flows in those reactions

50. Dr. S. M. Condren EXAMPLE CH 4(g) + 2 O 2(g) -----> CO 2(g) + 2 H 2 O (l) CH 4(g) -----> C (s) + 2 H 2(g)  H 1 2 O 2(g) -----> 2 O 2(g)  H 2 C (s) + O 2(g) -----> CO 2(g)  H 3 2 H 2(g) + O 2(g) -----> 2 H 2 O (l)  H 4 --------------------------------------------- CH 4(g) + 2 O 2(g) -----> CO 2(g) + 2 H 2 O (l)  H overall =  H 1 +  H 2 +  H 3 +  H 4

51. Dr. S. M. Condren Standard Enthalpy of Formation the enthalpy associated with the formation of a substance from its constituent elements under standard state conditions

52. Dr. S. M. Condren Calculation of  H o  H o =  c*  H f o products -  c*  H f o reactants

53. Dr. S. M. Condren 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 rx  c*  H f o  product -  c*  H f o  reactants

54. Dr. S. M. Condren 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 rx  c*  H f o  product -  c*  H f o  reactants  H rx  - 393.5) + 6(- 241.8)  product -  2(+ 49.0 ) + 15(0)  reactants kJ/mol = - 6.271 x 10 3 kJ

55. Dr. S. M. Condren Example What is the value of  H rx for the reaction: Fe 2 O 3(s) + 2 Al (s) --> 2 Fe (l) + Al 2 O 3(s) from Appendix J Text Fe 2 O 3(s)  H f o = -825.5 kJ/mol; Al (s)  H f o = 0 kJ/mol Al 2 O 3(s)  H f o = - 1675.7 kJ/mol; Fe (l)  H f o = +12.4 kJ/mol  H rx  c*  H f o  product -  c*  H f o  reactants  H rx  - 1675.7) + 2(- 12.4)  product -  1(-825.5 ) + 2(0)  reactants kJ/mol = - 2.4764 x 10 3 kJ

56. Dr. S. M. Condren Thermite Reaction on Saturday ~150g Fe 2 O 3 ~1 mol Fe 2 O 3 ~2.5x10 6 J ~2.5MJ

57. Dr. S. M. Condren Fossil Fuels natural gas coal petroleum

58. Dr. S. M. Condren Energy Sources

59. Dr. S. M. Condren Based on 1998 Data