1. Dr. Paul Charlesworth Michigan Technological University Dr. Paul Charlesworth Michigan Technological University C h a p t e rC h a p t e r C h a p t e rC h a p t e r 8 8 Thermochemistry: Chemical Energy Chemistry 4th Edition McMurry/Fay Chemistry 4th Edition McMurry/Fay

2. Prentice Hall ©2004 Chapter 08Slide 2 Thermodynamics01 Energy: is the capacity to do work, or supply heat. Energy = Work + Heat Kinetic Energy: is the energy of motion. E K = 1 / 2 mv 2 (1 Joule = 1 kg  m 2 /s 2 ) (1 calorie = 4.184 J) Potential Energy: is stored energy.

3. Prentice Hall ©2004 Chapter 08Slide 3 Thermodynamics 02 Thermal Energy is the kinetic energy of molecular motion (translational, rotational, and vibrational). Thermal energy is proportional to the temperature in degrees Kelvin. E thermal  T(K) Heat is the amount of thermal energy transferred between two objects at different temperatures.

4. Prentice Hall ©2004 Chapter 08Slide 4 Thermodynamics03 In an experiment: Reactants and products are the system; everything else is the surroundings. Energy flow from the system to the surroundings has a negative sign (loss of energy). Energy flow from the surroundings to the system has a positive sign (gain of energy).

5. Prentice Hall ©2004 Chapter 08Slide 5 Thermodynamics04 Closed System: Only energy can be lost or gained. Isolated System: No matter or energy is exchanged.

6. Prentice Hall ©2004 Chapter 08Slide 6 Thermodynamics05 The law of the conservation of energy: Energy cannot be created or destroyed. The energy of an isolated system must be constant. The energy change in a system equals the work done on the system + the heat added.  E = E final – E initial = E 2 – E 1 = q + w q = heat, w = work

7. Prentice Hall ©2004 Chapter 08Slide 7 Thermodynamics06 Pressure is the force per unit area. (1 N/m 2 = 1 Pa) (1 atm = 101,325 Pa) Work is a force (F) that produces an object’s movement, times the distance moved (d): Work = Force x Distance

8. Prentice Hall ©2004 Chapter 08Slide 8 Thermodynamics07 Click to Play w = –P  V

9. Prentice Hall ©2004 Chapter 08Slide 9 Thermodynamics08 How much work is done (in kilojoules), and in which direction, as a result of the following reaction?

10. Prentice Hall ©2004 Chapter 08Slide 10 Thermodynamics09 The amount of heat exchanged between the system and the surroundings is given the symbol q. q =  E + P  V At constant volume (  V = 0): q v =  E At constant pressure: q p =  E + P  V =  H Enthalpy change:  H = H products – H reactants

11. Prentice Hall ©2004 Chapter 08Slide 11 Thermodynamics10 The following reaction has  E = –186 kJ/mol. Is the sign of P  V positive or negative? What is the sign and approximate magnitude of  H?

12. Prentice Hall ©2004 Chapter 08Slide 12 Thermodynamics11 The reaction between hydrogen and oxygen to yield water vapor has  H° = –484 kJ. How much PV work is done, and what is the value of  E (in kilojoules) for the reaction of 0.50 mol of H 2 with 0.25 mol of O 2 at atmospheric pressure if the volume change is –5.6 L? The explosion of 2.00 mol of solid TNT with a volume of approximately 0.274 L produces gases with a volume of 489 L at room temperature. How much PV (in kilojoules) work is done during the explosion? Assume P = 1 atm, T = 25°C. 2 C 7 H 5 N 3 O 6 (s)  12 CO(g) + 5 H 2 (g) + 3 N 2 (g) + 2 C(s)

13. Prentice Hall ©2004 Chapter 08Slide 13 State Functions01 State Function: A function or property whose value depends only on the present state (condition) of the system. The change in a state function is zero when the system returns to its original condition. For nonstate functions, the change is not zero if the path returns to the original condition.

14. Prentice Hall ©2004 Chapter 08Slide 14 State Functions02 State and Nonstate Properties: The two paths below give the same final state: N 2 H 4 (g) + H 2 (g)  2 NH 3 (g) + heat (188 kJ) N 2 (g) + 3 H 2 (g)  2 NH 3 (g) + heat (92 kJ) State properties include temperature, total energy, pressure, density, and [NH 3 ]. Nonstate properties include the heat.

15. Prentice Hall ©2004 Chapter 08Slide 15 Enthalpy Changes01 Enthalpies of Physical Change:

16. Prentice Hall ©2004 Chapter 08Slide 16 Enthalpy Changes02 Enthalpies of Chemical Change: Often called heats of reaction (  H reaction ). Endothermic: Heat flows into the system from the surroundings and  H has a positive sign. Exothermic: Heat flows out of the system into the surroundings and  H has a negative sign.

17. Prentice Hall ©2004 Chapter 08Slide 17 Enthalpy Changes03 Reversing a reaction changes the sign of  H for a reaction. C 3 H 8 (g) + 5 O 2 (g)  3 CO 2 (g) + 4 H 2 O(l)  H = –2219 kJ 3 CO 2 (g) + 4 H 2 O(l)  C 3 H 8 (g) + 5 O 2 (g)  H = +2219 kJ Multiplying a reaction increases  H by the same factor. 3 C 3 H 8 (g) + 15 O 2 (g)  9 CO 2 (g) + 12 H 2 O(l)  H = –6657 kJ

18. Prentice Hall ©2004 Chapter 08Slide 18 Enthalpy Changes04 How much heat (in kilojoules) is evolved or absorbed in each of the following reactions? Burning of 15.5 g of propane: C 3 H 8 (g) + 5 O 2 (g)  3 CO 2 (g) + 4 H 2 O(l)  H = –2219 kJ Reaction of 4.88 g of barium hydroxide octahydrate with ammonium chloride: Ba(OH) 2 ·8 H 2 O(s) + 2 NH 4 Cl(s)  BaCl 2 (aq) + 2 NH 3 (aq) + 10 H 2 O(l)  H = +80.3 kJ

19. Prentice Hall ©2004 Chapter 08Slide 19 Enthalpy Changes05 Thermodynamic Standard State: Most stable form of a substance at 1 atm pressure and 25°C; 1 M concentration for all substances in solution. These are indicated by a superscript ° to the symbol of the quantity reported. Standard enthalpy change is indicated by the symbol  H°.

20. Prentice Hall ©2004 Chapter 08Slide 20 Hess’s Law01 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. 3 H 2 (g) + N 2 (g)  2 NH 3 (g)  H° = –92.2 kJ

21. Prentice Hall ©2004 Chapter 08Slide 21 Hess’s Law02 Reactants and products in individual steps can be added and subtracted to determine the overall equation. (a) 2 H 2 (g) + N 2 (g) N 2 H 4 (g)  H° 1 = ? (b) N 2 H 4 (g) + H 2 (g) 2 NH 3 (g)  H° 2 = –187.6 kJ (c) 3 H 2 (g) + N 2 (g) 2 NH 3 (g)  H° 3 = –92.2 kJ  H° 1 =  H° 3 –  H° 2 = (–92.2 kJ) – (–187.6 kJ) = +95.4 kJ

22. Prentice Hall ©2004 Chapter 08Slide 22 Hess’s Law03 The industrial degreasing solvent methylene chloride (CH 2 Cl 2, dichloromethane) is prepared from methane by reaction with chlorine: CH 4 (g) + 2 Cl 2 (g) CH 2 Cl 2 (g) + 2 HCl(g) Use the following data to calculate  H° (in kilojoules) for the above reaction : CH 4 (g) + Cl 2 (g) CH 3 Cl(g) + HCl(g)  H° = –98.3 kJ CH 3 Cl(g) + Cl 2 (g) CH 2 Cl 2 (g) + HCl(g)  H° = –104 kJ

23. Prentice Hall ©2004 Chapter 08Slide 23 Standard Heats of Formation 01 Standard Heats of Formation (  H° f ): The enthalpy change for the formation of 1 mole of substance in its standard state from its constituent elements in their standard states. The standard heat of formation for any element in its standard state is defined as being ZERO.  H° f = 0 for an element in its standard state

24. Prentice Hall ©2004 Chapter 08Slide 24 Standard Heats of Formation 02 H 2 (g) + 1 / 2 O 2 (g)  H 2 O(l)  H ° f = –286 kJ/mol 3 / 2 H 2 (g) + 1 / 2 N 2 (g)  NH 3 (g)  H ° f = –46 kJ/mol 2 C(s) + H 2 (g)  C 2 H 2 (g)  H ° f = +227 kJ/mol 2 C(s) + 3 H 2 (g) + 1 / 2 O 2 (g)  C 2 H 5 OH(g)  H ° f = –235 kJ/mol

25. Prentice Hall ©2004 Chapter 08Slide 25 Standard Heats of Formation 03 Calculating  H° for a reaction:  H° =  H° f (Products) –  H° f (Reactants) For a balanced equation, each heat of formation must be multiplied by the stoichiometric coefficient. aA + bB cC + dD  H° = [c  H° f (C) + d  H° f (D)] – [a  H° f (A) + b  H° f (B)]

26. Prentice Hall ©2004 Chapter 08Slide 26 Standard Heats of Formation04 -1131Na 2 CO 3 (s)49C6H6(l)C6H6(l)-92HCl(g) -127AgCl(s)-235C 2 H 5 OH(g)95.4N2H4(g)N2H4(g) -167Cl - (aq)-201CH 3 OH(g)-46NH 3 (g) -207NO 3 - (aq)-85C2H6(g)C2H6(g)-286H 2 O(l) -240Na + (aq)52C2H4(g)C2H4(g)-394CO 2 (g) 106Ag + (aq)227C2H2(g)C2H2(g)-111CO(g) Some Heats of Formation,  H f ° (kJ/mol)

27. Prentice Hall ©2004 Chapter 08Slide 27 Standard Heats of Formation 05 Calculate  H° (in kilojoules) for the reaction of ammonia with O 2 to yield nitric oxide (NO) and H 2 O(g), a step in the Ostwald process for the commercial production of nitric acid. Calculate  H° (in kilojoules) for the photosynthesis of glucose from CO 2 and liquid water, a reaction carried out by all green plants.

28. Prentice Hall ©2004 Chapter 08Slide 28 Bond Dissociation Energy01 Bond Dissociation Energy: Can be used to determine an approximate value for  H° f.  H = D (Bonds Broken) – D (Bonds Formed) For the reaction between H 2 and Cl 2 to form HCl:  H = D(H–Cl) – ∑ { D(H–H) + D(O=O) }

29. Prentice Hall ©2004 Chapter 08Slide 29 Bond Dissociation Energy02

30. Prentice Hall ©2004 Chapter 08Slide 30 Bond Dissociation Energy03 Calculate an approximate  H° (in kilojoules) for the synthesis of ethyl alcohol from ethylene: C 2 H 4 (g) + H 2 O(g)  C 2 H 5 OH(g) Calculate an approximate  H° (in kilojoules) for the synthesis of hydrazine from ammonia: 2 NH 3 (g) + Cl 2 (g)  N 2 H 4 (g) + 2 HCl(g)

31. Prentice Hall ©2004 Chapter 08Slide 31 Calorimetry and Heat Capacity01 Calorimetry is the science of measuring heat changes (q) for chemical reactions. There are two types of calorimeters: Bomb Calorimetry: A bomb calorimeter measures the heat change at constant volume such that q =  E. Constant Pressure Calorimetry: A constant pressure calorimeter measures the heat change at constant pressure such that q =  H.

32. Prentice Hall ©2004 Chapter 08Slide 32 Calorimetry and Heat Capacity02 Constant Pressure Bomb

33. Prentice Hall ©2004 Chapter 08Slide 33 Calorimetry and Heat Capacity03 Heat capacity (C) is the amount of heat required to raise the temperature of an object or substance a given amount. Specific Heat: The amount of heat required to raise the temperature of 1.00 g of substance by 1.00°C. Molar Heat: The amount of heat required to raise the temperature of 1.00 mole of substance by 1.00°C.  C= q  T

34. Prentice Hall ©2004 Chapter 08Slide 34 Calorimetry and Heat Capacity04 What is the specific heat of lead if it takes 96 J to raise the temperature of a 75 g block by 10.0°C? When 25.0 mL of 1.0 M H 2 SO 4 is added to 50.0 mL of 1.0 M NaOH at 25.0°C in a calorimeter, the temperature of the solution increases to 33.9°C. Assume specific heat of solution is 4.184 J/(g –1 ·°C –1 ), and the density is 1.00 g/mL –1, calculate  H for the reaction.

35. Prentice Hall ©2004 Chapter 08Slide 35 Calorimetry and Heat Capacity05

36. Prentice Hall ©2004 Chapter 08Slide 36 Introduction to Entropy01 Second Law of Thermodynamics: Reactions proceed in the direction that increases the entropy of the system plus surroundings. A spontaneous process is one that proceeds on its own without any continuous external influence. A nonspontaneous process takes place only in the presence of a continuous external influence.

37. Prentice Hall ©2004 Chapter 08Slide 37 Introduction to Entropy02 The measure of molecular disorder in a system is called the system’s entropy; this is denoted S. Entropy has units of J/K (Joules per Kelvin).  S = S final – S initial Positive value of  S indicates increased disorder. Negative value of  S indicates decreased disorder.

38. Prentice Hall ©2004 Chapter 08Slide 38 Introduction to Entropy03

39. Prentice Hall ©2004 Chapter 08Slide 39 Introduction to Entropy04 To decide whether a process is spontaneous, both enthalpy and entropy changes must be considered: Spontaneous process:Decrease in enthalpy (–  H). Increase in entropy (+  S). Nonspontaneous process:Increase in enthalpy (+  H). Decrease in entropy (–  S).

40. Prentice Hall ©2004 Chapter 08Slide 40 Introduction to Entropy05 Predict whether  S° is likely to be positive or negative for each of the following reactions. Using tabulated values, calculate  S° for each: a. 2 CO(g) + O 2 (g)  2 CO 2 (g) b. 2 NaHCO 3 (s)  Na 2 CO 3 (s) + H 2 O(l) + CO 2 (g) c. C 2 H 4 (g) + Br 2 (g)  CH 2 BrCH 2 Br(l) d. 2 C 2 H 6 (g) + 7 O 2 (g)  4 CO 2 (g) + 6 H 2 O(g)

41. Prentice Hall ©2004 Chapter 08Slide 41 Introduction to Free Energy01 Gibbs Free Energy Change (  G): Weighs the relative contributions of enthalpy and entropy to the overall spontaneity of a process.  G =  H – T  S  G < 0Process is spontaneous  G = 0Process is at equilibrium  G > 0Process is nonspontaneous

42. Prentice Hall ©2004 Chapter 08Slide 42 Introduction to Free Energy02 Situations leading to  G < 0:  H is negative and T  S is positive  H is very negative and T  S is slightly negative  H is slightly positive and T  S is very positive Situations leading to  G = 0:  H and T  S are equally negative  H and T  S are equally positive Situations leading to  G > 0:  H is positive and T  S is negative  H is slightly negative and T  S is very negative  H is very positive and T  S is slightly positive

43. Prentice Hall ©2004 Chapter 08Slide 43 Introduction to Free Energy03 Which of the following reactions are spontaneous under standard conditions at 25°C? a.AgNO 3 (aq) + NaCl(aq)  AgCl(s) + NaNO 3 (aq)  G° = –55.7 kJ b.2 C(s) + 2 H 2 (g)  C 2 H 4 (g)  G° = 68.1 kJ c.N 2 (g) + 3 H 2 (g)  2 NH 3 (g)  H° = –92 kJ;  S° = –199 J/K

44. Prentice Hall ©2004 Chapter 08Slide 44 Introduction to Free Energy04 Equilibrium (  G° = 0): Estimate the temperature at which the following reaction will be at equilibrium. Is the reaction spontaneous at room temperature? N 2 (g) + 3 H 2 (g)  2 NH 3 (g)  H° = –92.0 kJ  S° = –199 J/K Equilibrium is the point where  G° =  H° – T  S° = 0

45. Prentice Hall ©2004 Chapter 08Slide 45 Introduction to Free Energy05 Benzene, C 6 H 6, has an enthalpy of vaporization,  H vap, equal to 30.8 kJ/mol and boils at 80.1°C. What is the entropy of vaporization,  S vap, for benzene?