1. John E. McMurry Robert C. Fay Lecture Notes Alan D. Earhart Southeast Community College Lincoln, NE General Chemistry: Atoms First Chapter 8 Thermochemistry: Chemical Energy Copyright © 2010 Pearson Prentice Hall, Inc.

2. Chapter 8/2 Energy and Its Conservation Energy: The capacity to supply heat or do work. Kinetic Energy (E K ): The energy of motion. Potential Energy (E P ): Stored energy.

3. Copyright © 2010 Pearson Prentice Hall, Inc.Chapter 8/3 Energy and Its Conservation Law of Conservation of Energy: Energy cannot be created or destroyed; it can only be converted from one form to another.

4. Energy and Its Conservation Total Energy = E K + E P

5. Copyright © 2010 Pearson Prentice Hall, Inc.Chapter 8/5 Energy and Its Conservation Thermal Energy: The kinetic energy of molecular motion, measured by finding the temperature of an object. Heat: The amount of thermal energy transferred from one object to another as the result of a temperature difference between the two.

6. Copyright © 2010 Pearson Prentice Hall, Inc.Chapter 8/6 Energy and Its Conservation First Law of Thermodynamics: Energy cannot be created or destroyed; it can only be converted from one form to another.

7. Internal Energy and State Functions First Law of Thermodynamics: The total internal energy of an isolated system is constant. ∆E = E final - E initial

8. Internal Energy and State Functions First Law of Thermodynamics: The total internal energy of an isolated system is constant. ∆E = E final - E initial

9. Internal Energy and State Functions First Law of Thermodynamics: The total internal energy of an isolated system is constant. ∆E = E final - E initial

10. Copyright © 2010 Pearson Prentice Hall, Inc.Chapter 8/10 Internal Energy and State Functions CO 2 (g) + 2H 2 O(g) + 802 kJ energyCH 4 (g) + 2O 2 (g) ∆E = E final - E initial = -802 kJ 802 kJ is released when 1 mole of methane, CH 4, reacts with 2 moles of oxygen to produce 1 mole of carbon dioxide and two moles of water.

11. Internal Energy and State Functions State Function: A function or property whose value depends only on the present state, or condition, of the system, not on the path used to arrive at that state.

12. Expansion Work w = F x d Work = Force x Distance

13. Expansion Work 3CO 2 (g) + 4H 2 O(g)C 3 H 8 (g) + 5O 2 (g) 7 mol of gas6 mol of gas

14. Expansion Work Expansion Work: Work done as the result of a volume change in the system. Also called pressure-volume or PV work.

15. Copyright © 2010 Pearson Prentice Hall, Inc.Chapter 8/15 Energy and Enthalpy Constant Pressure: ∆E = q + w w = work = -P∆V q = heat transferred q = ∆E + P∆V q P = ∆E + P∆V Constant Volume (∆V = 0):q V = ∆E

17. Copyright © 2010 Pearson Prentice Hall, Inc.Chapter 8/17 Energy and Enthalpy = H products - H reactants ∆H∆H Enthalpy change or Heat of reaction (at constant pressure) q P = ∆E + P∆V = ∆H = H final - H initial Enthalpy is a state function whose value depends only on the current state of the system, not on the path taken to arrive at that state.

18. Copyright © 2010 Pearson Prentice Hall, Inc.Chapter 8/18 The Thermodynamic Standard State 3CO 2 (g) + 4H 2 O(g)C 3 H 8 (g) + 5O 2 (g) 3CO 2 (g) + 4H 2 O(l)C 3 H 8 (g) + 5O 2 (g) ∆H = -2220 kJ Thermodynamic Standard State: Most stable form of a substance at 1 atm pressure and at a specified temperature, usually 25 °C; 1 M concentration for all substances in solution. 3CO 2 (g) + 4H 2 O(g)C 3 H 8 (g) + 5O 2 (g) ∆H = -2044 kJ ∆H° = -2044 kJ

19. Copyright © 2010 Pearson Prentice Hall, Inc.Chapter 8/19 Enthalpies of Physical and Chemical Change Enthalpy of Fusion (∆H fusion ): The amount of heat necessary to melt a substance without changing its temperature. Enthalpy of Vaporization (∆H vap ): The amount of heat required to vaporize a substance without changing its temperature. Enthalpy of Sublimation (∆H subl ): The amount of heat required to convert a substance from a solid to a gas without going through a liquid phase.

20. Chapter 8/20 Enthalpies of Physical and Chemical Change

21. Copyright © 2010 Pearson Prentice Hall, Inc.Chapter 8/21 Enthalpies of Physical and Chemical Change BaCl 2 (aq) + 2NH 3 (aq) + 10H 2 O(l) Ba(OH) 2 8H 2 O(s) + 2NH 4 Cl(s) ∆H° = +80.3 kJ 2Fe(s) + Al 2 O 3 (s)2Al(s) + Fe 2 O 3 (s) ∆H° = -852 kJ Endothermic: Heat (enthalpy) flows from the surroundings to the system. Exothermic: Heat (enthalpy) flows from the system to the surroundings.

22. Copyright © 2010 Pearson Prentice Hall, Inc.Chapter 8/22 Enthalpies of Physical and Chemical Change 2Al(s) + Fe 2 O 3 (s)2Fe(s) + Al 2 O 3 (s) ∆H° = +852 kJ 2Fe(s) + Al 2 O 3 (s)2Al(s) + Fe 2 O 3 (s) ∆H° = -852 kJ Endothermic: Heat (enthalpy) flows from the surroundings to the system. Exothermic: Heat (enthalpy) flows from the system to the surroundings.

23. Calorimetry and Heat Capacity Measure the heat flow at constant pressure (∆H).

24. Calorimetry and Heat Capacity Measure the heat flow at constant volume (∆E).

25. Copyright © 2010 Pearson Prentice Hall, Inc.Chapter 8/25 Calorimetry and Heat Capacity Specific Heat: The amount of heat required to raise the temperature of 1 g of a substance by 1 °C. q = (specific heat) x (mass of substance) x ∆T Molar Heat Capacity (C m ): The amount of heat required to raise the temperature of 1 mol of a substance by 1 °C. q = (C m ) x (moles of substance) x ∆T Heat Capacity (C): The amount of heat required to raise the temperature of an object or substance a given amount. q = C x ∆T

26. Calorimetry and Heat Capacity

27. Copyright © 2010 Pearson Prentice Hall, Inc.Chapter 8/27 Calorimetry and Heat Capacity Assuming that a can of soda has the same specific heat as water, calculate the amount of heat (in kilojoules) transferred when one can (about 350 g) is cooled from 25 °C to 3 °C. Specific heat = 4.18 g °C J Temperature change = 25 °C - 3 °C = 22 °C Mass = 350 g q = (specific heat) x (mass of substance) x ∆T

28. Copyright © 2010 Pearson Prentice Hall, Inc.Chapter 8/28 Calorimetry and Heat Capacity Calculate the amount of heat transferred. = 32 000 J 22 °C 1000 J 1 kJ x 350 g x Heat evolved = = 32 kJ g °C 4.18 J x 32 000 J

29. Copyright © 2010 Pearson Prentice Hall, Inc.Chapter 8/29 Hess’s Law Haber Process: Multiple-Step Process N2H4(g)N2H4(g)2H 2 (g) + N 2 (g) Hess’s Law: The overall enthalpy change for a reaction is equal to the sum of the enthalpy changes for the individual steps in the reaction. 2NH 3 (g)3H 2 (g) + N 2 (g) ∆H° = -92.2 kJ ∆H° 1 = ? 2NH 3 (g)N 2 H 4 (g) + H 2 (g) ∆H° 1+2 = -92.2 kJ 2NH 3 (g)3H 2 (g) + N 2 (g) ∆H° 2 = -187.6 kJ

30. ∆H° 1 = ∆H° 1+2 - ∆H° 2 Hess’s Law = -92.2 kJ - (-187.6 kJ) = 95.4 kJ ∆H° 1 + ∆H° 2 = ∆H° 1+2

31. Copyright © 2010 Pearson Prentice Hall, Inc.Chapter 8/31 Standard Heats of Formation Standard Heat of Formation (∆H° f ): The enthalpy change for the formation of 1 mol of a substance in its standard state from its constituent elements in their standard states. ∆H° f = -74.8 kJ CH 4 (g)C(s) + 2H 2 (g) Standard states 1 mol of 1 substance

32. Chapter 8/32 Standard Heats of Formation

33. Copyright © 2010 Pearson Prentice Hall, Inc.Chapter 8/33 Standard Heats of Formation cC + dDaA + bB ReactantsProducts ∆H° = ∆H° f (Products) - ∆H° f (Reactants) ∆H° = [c ∆H° f (C) + d ∆H° f (D)] - [a ∆H° f (A) + b ∆H° f (B)]

34. Copyright © 2010 Pearson Prentice Hall, Inc.Chapter 8/34 Standard Heats of Formation Using standard heats of formation, calculate the standard enthalpy of reaction for the photosynthesis of glucose (C 6 H 12 O 6 ) and O 2 from CO 2 and liquid H 2 O. C 6 H 12 O 6 (s) + 6O 2 (g)6CO 2 (g) + 6H 2 O(l) ∆H° = ? H° = [∆H° f (C 6 H 12 O 6 (s))] - [6 ∆H° f (CO 2 (g)) + 6 ∆H° f (H 2 O(l))] = 2802.5 kJ [(6 mol)(-393.5 kJ/mol) + (6 mol)(-285.8 kJ/mol)] ∆H° = [(1 mol)(-1273.3 kJ/mol)] -

35. Copyright © 2010 Pearson Prentice Hall, Inc.Chapter 8/35 Standard Heats of Formation 1.Reverse the “reaction” and reverse the sign on the standard enthalpy change. C(s) + O 2 (g)CO 2 (g) Why does the calculation “work”? ∆H° = 393.5 kJ Becomes CO 2 (g)C(s) + O 2 (g) ∆H° = -393.5 kJ C 6 H 12 O 6 (s) + 6O 2 (g)6CO 2 (g) + 6H 2 O(l) ∆H° = 2802.5 kJ (3)(1)(2)

36. Copyright © 2010 Pearson Prentice Hall, Inc.Chapter 8/36 Standard Heats of Formation 1.Multiply the coefficients by a factor and multiply the standard enthalpy change by the same factor. 6C(s) + 6O 2 (g)6CO 2 (g) Why does the calculation “work”? ∆H° = 6(393.5 kJ) = 2361.0 kJ Becomes C(s) + O 2 (g)CO 2 (g) ∆H° = -393.5 kJ C 6 H 12 O 6 (s) + 6O 2 (g)6CO 2 (g) + 6H 2 O(l) ∆H° = 2802.5 kJ (3)(1)(2)

37. Copyright © 2010 Pearson Prentice Hall, Inc.Chapter 8/37 Standard Heats of Formation 2.Reverse the “reaction” and reverse the sign on the standard enthalpy change. C 6 H 12 O 6 (s) + 6O 2 (g)6CO 2 (g) + 6H 2 O(l) ∆H° = 2802.5 kJ Why does the calculation “work”? ∆H° = 285.8 kJ Becomes ∆H° = -285.8 kJ (3)(1)(2) H 2 (g) + O 2 (g)H 2 O(l) 2 1 H 2 (g) + O 2 (g) 2 1

38. Copyright © 2010 Pearson Prentice Hall, Inc.Chapter 8/38 Standard Heats of Formation 2.Multiply the coefficients by a factor and multiply the standard enthalpy change by the same factor. 6H 2 (g) + 3O 2 (g)6H 2 O(l) Why does the calculation “work”? ∆H° = 6(285.8 kJ) = 1714.8 kJ Becomes C 6 H 12 O 6 (s) + 6O 2 (g)6CO 2 (g) + 6H 2 O(l) ∆H° = 2802.5 kJ (3)(1)(2) ∆H° = 285.8 kJ H 2 (g) + O 2 (g)H 2 O(l) 2 1

39. Copyright © 2010 Pearson Prentice Hall, Inc.Chapter 8/39 Standard Heats of Formation C 6 H 12 O 6 (s) + 6O 2 (g)6CO 2 (g) + 6H 2 O(l) ∆H° = 1714.8 kJ Why does the calculation “work”? 6H 2 (g) + 3O 2 (g)6H 2 O(l) C 6 H 12 O 6 (s)6C(s) + 6H 2 (g) + 3O 2 (g) 6C(s) + 6O 2 (g)6CO 2 (g) ∆H° = 2802.5 kJ ∆H° = -1273.3 kJ ∆H° = 2361.0 kJ C 6 H 12 O 6 (s) + 6O 2 (g)6CO 2 (g) + 6H 2 O(l) ∆H° = 2802.5 kJ (3)(1)(2)

40. Copyright © 2010 Pearson Prentice Hall, Inc.Chapter 8/40 Bond Dissociation Energies The amount of energy that must be supplied to break a chemical bond in an isolated molecule in the gaseous state and is thus the amount of energy released when the bond forms. or Standard enthalpy changes for the corresponding bond-breaking reactions. Bond Dissociation Energy:

41. Bond Dissociation Energies

42. Copyright © 2010 Pearson Prentice Hall, Inc.Chapter 8/42 Bond Dissociation Energies 2HCl(g)H 2 (g) + Cl 2 (g) (2mol)(432 kJ/mol) ∆H° = D(Reactant bonds) - D(Product bonds) ∆H° = (D H-H + D Cl-Cl ) - (2D H-Cl ) ∆H° = [(1 mol)(436 kJ/mol) + (1mol)(243 kJ/mol)] - = -185 kJ

43. Copyright © 2010 Pearson Prentice Hall, Inc.Chapter 8/43 Fossil Fuels, Fuel Efficiency, and Heats of Combustion CO 2 (g) + 2H 2 O(l)CH 4 (g) + 2O 2 (g) = -890.3 kJ ∆H° c = [∆H° f (CO 2 (g)) + 2 ∆H° f (H 2 O(l))] - [∆H° f (CH 4 (g))] [(1 mol)(-74.8 kJ/mol)] = [(1 mol)(-393.5 kJ/mol) + (2 mol)(-285.8 kJ/mol)] -

44. Fossil Fuels, Fuel Efficiency, and Heats of Combustion CO 2 (g) + 2H 2 O(l)CH 4 (g) + 2O 2 (g)

45. Copyright © 2010 Pearson Prentice Hall, Inc.Chapter 8/45 An Introduction to Entropy Spontaneous Process: A process that, once started, proceeds on its own without a continuous external influence. Spontaneous processes are favored by a decrease in H (negative ∆H). favored by an increase in S (positive ∆S). Nonspontaneous processes are favored by an increase in H (positive ∆H). favored by a decrease in S (negative ∆S). Entropy (S): The amount of molecular randomness in a system.

47. Copyright © 2010 Pearson Prentice Hall, Inc.Chapter 8/47 An Introduction to Free Energy Gibbs Free Energy Change (∆G) Entropy change ∆G = Enthalpy of reaction Temperature (Kelvin) ∆H∆H∆S∆S-T

48. Copyright © 2010 Pearson Prentice Hall, Inc.Chapter 8/48 An Introduction to Free Energy ∆G < 0Process is spontaneous ∆G = 0Process is at equilibrium (neither spontaneous nor nonspontaneous) ∆G > 0Process is nonspontaneous Gibbs Free Energy Change (∆G) ∆G =∆H∆H∆S∆S-T

49. An Introduction to Free Energy