1. Lecture Notes Alan D. Earhart Southeast Community College Lincoln, NE Chapter 8 Thermochemistry: Chemical Energy John E. McMurry Robert C. Fay CHEMISTRY Fifth Edition Copyright © 2008 Pearson Prentice Hall, Inc.

2. 8.1 Energy Copyright © 2008 Pearson Prentice Hall, Inc.Chapter 8/2 Units: 1 Cal = 1000 cal = 1 kcal 1 cal = 4.184 J (exactly) Energy: The capacity to supply heat or do work. Kinetic Energy (E K ): The energy of motion. Potential Energy (E P ): Stored energy.

3. 8.2 Energy Changes and Energy Conservation Copyright © 2008 Pearson Prentice Hall, Inc.Chapter 8/3 Law of Conservation of Energy: Energy cannot be created or destroyed; it can only be converted from one form to another.

4. Energy Changes and Energy Conservation Copyright © 2008 Pearson Prentice Hall, Inc.Chapter 8/4 Total Energy = E K + E P

5. Energy Changes and Energy Conservation Copyright © 2008 Pearson Prentice Hall, Inc.Chapter 8/5 Thermal Energy: The kinetic energy of molecular motion and is 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. 8.3 Internal Energy and State Functions Copyright © 2008 Pearson Prentice Hall, Inc.Chapter 8/6 First Law of Thermodynamics: The total internal energy of an isolated system is constant.  E = E final - E initial

7. Internal Energy and State Functions Copyright © 2008 Pearson Prentice Hall, Inc.Chapter 8/7 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.

8. Internal Energy and State Functions Copyright © 2008 Pearson Prentice Hall, Inc.Chapter 8/8 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.

9. 8.4 Expansion Work Copyright © 2008 Pearson Prentice Hall, Inc.Chapter 8/9 w = F x d Work = Force x Distance

10. Expansion Work Copyright © 2008 Pearson Prentice Hall, Inc. Chapter 8/10 Expansion Work: Work done as the result of a volume change in the system. Also called pressure-volume or PV work.

11. 8.5 Energy and Enthalpy Copyright © 2008 Pearson Prentice Hall, Inc. Chapter 8/11 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

12. Energy and Enthalpy Copyright © 2008 Pearson Prentice Hall, Inc. Chapter 8/12 = 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.

13. 8.6The Thermodynamic Standard State Copyright © 2008 Pearson Prentice Hall, Inc. Chapter 8/13 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 = -2219 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. Standard enthalpy of reaction is indicated by the symbol ΔH o 3CO 2 (g) + 4H 2 O(g)C 3 H 8 (g) + 5O 2 (g)  H= -2043 kJ  H° = -2043 kJ

14. 8.7 Enthalpies of Physical and Chemical Change Copyright © 2008 Pearson Prentice Hall, Inc. Chapter 8/14 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.

15. Enthalpies of Physical and Chemical Change Copyright © 2008 Pearson Prentice Hall, Inc. Chapter 8/15

16. Enthalpies of Physical and Chemical Change Copyright © 2008 Pearson Prentice Hall, Inc. Chapter 8/16 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 exothermic

17. Examples Identify each of the following processes as endothermic or exothermic and indicate the sign of ΔH o Sweat evaporating from your skin Water freezing in a freezer Wood burning in fire

18. Example An LP gas tank in a home barbeque contains 13.2 kg of propane, C 3 H 8. Calculate the heat (in kJ) associated with the complete combustion of all of the propane in the tank C 3 H 8 (g) + 5O 2 (g)  3 CO 2 (g) + 4 H 2 O(g) ΔH o = -2044kJ

19. Example How much heat (in kJ) is evolved when 5.00g of aluminum reacts with a stoichiometric amount of Fe 2 O 3 ? 2 Al(s) + Fe 2 O 3 (s)  2 Fe(s) + Al 2 O 3 (s) ΔH o = -852 kJ

20. 8.8Calorimetry and Heat Capacity A temperature change is observed when a system gains or loses energy in the form of heat (ΔH) Amount of heat gained = amount of heat lost q cal = -q rxn q gain = -q lost Copyright © 2008 Pearson Prentice Hall, Inc. Chapter 8/20

21. Calorimetry and Heat Capacity Copyright © 2008 Pearson Prentice Hall, Inc. Chapter 8/21 Measure the heat flow at constant volume (  E).

22. Calorimetry and Heat Capacity Copyright © 2008 Pearson Prentice Hall, Inc. Chapter 8/22 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 m x  T

23. Calorimetry and Heat Capacity Copyright © 2008 Pearson Prentice Hall, Inc. Chapter 8/23

24. Examples Assuming that Coca Cola has the same specific heat as water [4.183 J/g o C], calculate the amount of heat (in kJ) transferred when one can (about 350.0 g) is cooled from 25.0 o C – 3.0 o C

25. Example When 25.0 mL of 1.0 M H 2 SO 4 is added to 50.0ml of 1.0 M NaOH at 25.0 o C in a calorimeter, the temperature of the aqueous solution increases to 33.9 o C. Assuming that the specific heat of the solution is 4.18 J/g o C, that its density is 1.00g/mL, and the calorimeter itself absorbs a negligible amount of heat, calculate ΔH (in kJ/mol) for the reaction H 2 SO 4 (aq) + 2 NaOH(aq)  2 H2O(l) + Na 2 SO 4 (aq)

26. 8.9 Hess’s Law Copyright © 2008 Pearson Prentice Hall, Inc. Chapter 8/26 Haber Process: Multiple-Step Process - Given 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

27. Hess’s Law Copyright © 2008 Pearson Prentice Hall, Inc. Chapter 8/27

28. Example Find ΔH o rxn for the following reaction: 3H 2 (g) + O 3 (g)  3 H 2 O(g) ΔH o rxn = ?? Use the following reactions with known ΔH’s 2H 2 (g) + O 2 (g)  2 H 2 O(g) Δ H o = -483.6 kJ 3O 2 (g)  2 O 3 (g) Δ H o = +285.4 kJ Typo from original note

29. Example Fin ΔH o rxn for the following reaction C(s) + H 2 O(g)  CO(g) + H 2 (g) H o rxn = ? Use the following reactions with known H’s C(s) + O 2 (g)  CO 2 (g) ΔH o = -393.5 kJ 2CO(g) + O 2 (g)  2CO 2 (g) Δ H o = -566.0kJ 2H 2 (g) + O 2 (g)  2H 2 O (g) Δ H o = -483.6 kJ

30. 8.10 Standard Heats of Formation Copyright © 2008 Pearson Prentice Hall, Inc. Chapter 8/30 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

31. Standard Heats of Formation Copyright © 2008 Pearson Prentice Hall, Inc. Chapter 8/31

32. Standard Heats of Formation Copyright © 2008 Pearson Prentice Hall, Inc. Chapter 8/32 cC + dDaA + bB Reactants Products  H° =  H° f (Products) -  H° f (Reactants)  H° = [c  H° f (C) + d  H° f (D)] - [a  H° f (A) + b  H° f (B)]

33. Standard Heats of Formation Copyright © 2008 Pearson Prentice Hall, Inc. Chapter 8/33 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° = ?

34. Example Use the information in Table 8.2 to calculate ΔH o (in kJ) for the reaction of ammonia with oxygen gas to yield nitric oxide (NO) and water vapor, a step in the Ostwald process for the commercial production of nitric acid

35. 8.11 Bond Dissociation Energies Copyright © 2008 Pearson Prentice Hall, Inc. Chapter 8/35 Bond dissociation energies are standard enthalpy changes for the corresponding bond-breaking reactions.

36. Bond Dissociation Energies Copyright © 2008 Pearson Prentice Hall, Inc. Chapter 8/36 2HCl(g)H 2 (g) + Cl 2 (g)  H° = D(Reactant bonds) - D(Product bonds)  H° = (D H-H + D Cl-Cl ) - (2D H-Cl )

37. 8.12 Fossil Fuels, Fuel Efficiency, and Heats of Combustion Copyright © 2008 Pearson Prentice Hall, Inc. Chapter 8/37 CO 2 (g) + 2H 2 O(l)CH 4 (g) + 2O 2 (g)

38. Fossil Fuels, Fuel Efficiency, and Heats of Combustion Copyright © 2008 Pearson Prentice Hall, Inc. Chapter 8/38 CO 2 (g) + 2H 2 O(l)CH 4 (g) + 2O 2 (g) ΔH o c = -890.3 kJ/mol H 2 (g) + ½ O 2 (g)  H 2 O(l) ΔH o c = -285 kJ/mol

39. 8.13 An Introduction to Entropy Copyright © 2008 Pearson Prentice Hall, Inc. Chapter 8/39 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.

40. An Introduction to Entropy

41. Example Predict whether ΔS o is likely to be positive or negative for each of the following reactions H 2 C=CH 2 (g) + Br 2 (g)  BrCH 2 CH 2 Br(l)

42. 8.14 An Introduction to Free Energy Copyright © 2008 Pearson Prentice Hall, Inc. Chapter 8/42 Gibbs Free Energy Change (  G) Entropy change  G = Enthalpy of reaction Temperature (Kelvin) HH SS - T

43. An Introduction to Free Energy Copyright © 2008 Pearson Prentice Hall, Inc. Chapter 8/43 Gibbs Free Energy Change (  G)  G = HH SS - T  G < 0Process is spontaneous  G = 0Process is at equilibrium (neither spontaneous nor nonspontaneous)  G > 0Process is nonspontaneous

44. Example Is the Haber process for the industrial synthesis of ammonia spontaneous or nonspontaneous under standard conditions at 25.0 o C. At what temperature ( o C) does the changeover occur?