1. © 2015 Pearson Education, Inc. Chapter 5 Thermochemistry James F. Kirby Quinnipiac University Hamden, CT Lecture Presentation

2. Thermochemistry © 2015 Pearson Education, Inc. Energy Energy is the ability to do work or transfer heat. –Energy used to cause an object that has mass to move is called work. Work = force x distance W = f x d –Energy used to cause the temperature of an object to rise is called heat. This chapter is about thermodynamics, which is the study of energy transformations, and thermochemistry, which applies the field to chemical reactions, specifically.

3. Thermochemistry © 2015 Pearson Education, Inc. Kinetic Energy Kinetic energy is energy an object possesses by virtue of its motion:

4. Thermochemistry © 2015 Pearson Education, Inc. Potential Energy Potential energy is energy an object possesses by virtue of its position or chemical composition. The most important form of potential energy in molecules is electrostatic potential energy, E el : Q = charge D = distance

5. Thermochemistry © 2015 Pearson Education, Inc. Units of Energy The SI unit of energy is the joule (J): An older, non-SI unit is still in widespread use, the calorie (cal): 1 cal = 4.184 J (Note: this is not the same as the calorie of foods; the food calorie is 1 kcal!) 1 J = 1 kg·m 2 /s 2

6. Thermochemistry © 2015 Pearson Education, Inc. Definitions: System and Surroundings The system includes the molecules we want to study (here, the hydrogen and oxygen molecules). The surroundings are everything else (here, the cylinder and piston).

7. Thermochemistry © 2015 Pearson Education, Inc. Definitions: Work Energy used to move an object over some distance is work: w = F  d where w is work, F is the force, and d is the distance over which the force is exerted.

8. Thermochemistry © 2015 Pearson Education, Inc. Heat Energy can also be transferred as heat. Heat flows from warmer objects to cooler objects.

9. Thermochemistry © 2015 Pearson Education, Inc. Conversion of Energy Energy can be converted from one type to another. The cyclist has potential energy as she sits on top of the hill. As she coasts down the hill, her potential energy is converted to kinetic energy until the bottom, where the energy is converted to kinetic energy.

10. Thermochemistry © 2015 Pearson Education, Inc. A bowler lifts a 5.4-kg (12-lb) bowling ball from ground level to a height of 1.6 m (5.2 ft) and then drops it. (a)What happens to the potential energy of the ball as it is raised? (b)What quantity of work, in J, is used to raise the ball? (c)After the ball is dropped, it gains kinetic energy. If all the work done in part (b) has been converted to kinetic energy by the time the ball strikes the ground, what is the ball’s speed just before it hits the ground? (Note: The force due to gravity is F = m × g, where m is the mass of the object and g is the gravitational constant; g = 9.8 m ⁄ s 2.)

11. Thermochemistry © 2015 Pearson Education, Inc. Solve (a) Because the ball is raised above the ground, its potential energy relative to the ground increases. (b) The ball has a mass of 5.4 kg and is lifted 1.6 m. To calculate the work performed to raise the ball, we use Equation 5.3 and F = m × g for the force that is due to gravity: W = F × d = m × g × d = (5.4 kg)(9.8 m ⁄ s 2 )(1.6 m) = 85 kg-m 2 ⁄ s 2 = 85 J Thus, the bowler has done 85 J of work to lift the ball to a height of 1.6 m.

12. Thermochemistry © 2015 Pearson Education, Inc. (c) When the ball is dropped, its potential energy is converted to kinetic energy. We assume that the kinetic energy just before the ball hits the ground is equal to the work done in part (b), 85 J: E k = mv 2 = 85 J = 85 kg-m 2 ⁄ s 2 We can now solve this equation for v:

13. Thermochemistry © 2015 Pearson Education, Inc. First Law of Thermodynamics Energy is neither created nor destroyed. In other words, the total energy of the universe is a constant; if the system loses energy, it must be gained by the surroundings, and vice versa.

14. Thermochemistry © 2015 Pearson Education, Inc. Internal Energy The internal energy of a system is the sum of all kinetic and potential energies of all components of the system; we call it E.

15. Thermochemistry © 2015 Pearson Education, Inc. Internal Energy By definition, the change in internal energy,  E, is the final energy of the system minus the initial energy of the system:  E = E final − E initial

16. Thermochemistry © 2015 Pearson Education, Inc. Changes in Internal Energy If  E > 0, E final > E initial –Therefore, the system absorbed energy from the surroundings. –This energy change is called endergonic or endothermic.

17. Thermochemistry © 2015 Pearson Education, Inc. Changes in Internal Energy If  E < 0, E final < E initial –Therefore, the system released energy to the surroundings. –This energy change is called exergonic or exothermic.

18. Thermochemistry © 2015 Pearson Education, Inc. Changes in Internal Energy When energy is exchanged between the system and the surroundings, it is exchanged as either heat (q) or work (w). That is,  E = q + w.

19. Thermochemistry © 2015 Pearson Education, Inc.  E, q, w, and Their Signs

20. Thermochemistry © 2015 Pearson Education, Inc. Example: Gases A(g) and B(g) are confined in a cylinder-and-piston arrangement like that in the Figure and react to form a solid product C(s): A(g) + B(g) → C(s). As the reaction occurs, the system loses 1150 J of heat to the surroundings. The piston moves downward as the gases react to form a solid. As the volume of the gas decreases under the constant pressure of the atmosphere, the surroundings do 480 J of work on the system. What is the change in the internal energy of the system?

21. Thermochemistry © 2015 Pearson Education, Inc. Solution Solve Heat is transferred from the system to the surroundings, and work is done on the system by the surroundings, so q is negative and w is positive: q = −1150 J and w = 480 kJ. Thus, ΔE = q + w = (−1150 J) + (480 J) = −670 J The negative value of ΔE tells us that a net quantity of 670 J of energy has been transferred from the system to the surroundings.

22. Thermochemistry © 2015 Pearson Education, Inc. Exchange of Heat between System and Surroundings When heat is absorbed by the system from the surroundings, the process is endothermic.

23. Thermochemistry © 2015 Pearson Education, Inc. Exchange of Heat between System and Surroundings When heat is released by the system into the surroundings, the process is exothermic.

24. Thermochemistry © 2015 Pearson Education, Inc. State Functions Usually we have no way of knowing the internal energy of a system; finding that value is simply too complex a problem. However, we do know that the internal energy of a system is independent of the path by which the system achieved that state. –In the system below, the water could have reached room temperature from either direction.

25. Thermochemistry © 2015 Pearson Education, Inc. State Functions Therefore, internal energy is a state function. It depends only on the present state of the system, not on the path by which the system arrived at that state. And so,  E depends only on E initial and E final.

26. Thermochemistry © 2015 Pearson Education, Inc. State Functions However, q and w are not state functions. Whether the battery is shorted out or is discharged by running the fan, its  E is the same. –But q and w are different in the two cases.

27. Thermochemistry © 2015 Pearson Education, Inc. Work Usually in an open container the only work done is by a gas pushing on the surroundings (or by the surroundings pushing on the gas).

28. Thermochemistry © 2015 Pearson Education, Inc. Work We can measure the work done by the gas if the reaction is done in a vessel that has been fitted with a piston: w = −P  V 5.3

29. Thermochemistry © 2015 Pearson Education, Inc. Enthalpy If a process takes place at constant pressure (as the majority of processes we study do) and the only work done is this pressure–volume work, we can account for heat flow during the process by measuring the enthalpy of the system. Enthalpy is the internal energy plus the product of pressure and volume: H = E + PV

30. Thermochemistry © 2015 Pearson Education, Inc. Enthalpy When the system changes at constant pressure, the change in enthalpy,  H, is  H =  (E + PV) This can be written  H =  E + P  V

31. Thermochemistry © 2015 Pearson Education, Inc. Enthalpy Since  E = q + w and w = −P  V, we can substitute these into the enthalpy expression:  H =  E + P  V  H = (q + w) − w  H = q So, at constant pressure, the change in enthalpy is the heat gained or lost.

32. Thermochemistry © 2015 Pearson Education, Inc. Endothermic and Exothermic A process is endothermic when  H is positive. A process is exothermic when  H is negative.

33. Thermochemistry © 2015 Pearson Education, Inc. Example: A fuel is burned in a cylinder equipped with a piston. The initial volume of the cylinder is 0.250 L, and the final volume is 0.980 L. If the piston expands against a constant pressure of 1.35 atm, how much work (in J) is done? (1 L-atm = 101.3 J)

34. Thermochemistry © 2015 Pearson Education, Inc. Solve The volume change is ΔV = V final − V initial = 0.980 L − 0.250 L = 0.730 L Thus, the quantity of work is w = −PΔV = −(1.35 atm)(0.730 L) = −0.9855 L-atm Converting L-atm to J, we have

35. Thermochemistry © 2015 Pearson Education, Inc. Enthalpy of Reaction The change in enthalpy,  H, is the enthalpy of the products minus the enthalpy of the reactants:  H = H products − H reactants 5.4

36. Thermochemistry © 2015 Pearson Education, Inc. Enthalpy of Reaction This quantity,  H, is called the enthalpy of reaction, or the heat of reaction.

37. Thermochemistry © 2015 Pearson Education, Inc. The Truth about Enthalpy 1.Enthalpy is an extensive property. 2.  H for a reaction in the forward direction is equal in size, but opposite in sign, to  H for the reverse reaction. 3.  H for a reaction depends on the state of the products and the state of the reactants.

38. Thermochemistry © 2015 Pearson Education, Inc. H 2 O (s) H 2 O (l)  H = 6.01 kJ The stoichiometric coefficients always refer to the number of moles of a substance Thermochemical Equations If you reverse a reaction, the sign of  H changes H 2 O (l) H 2 O (s)  H = - 6.01 kJ If you multiply both sides of the equation by a factor n, then  H must change by the same factor n. 2H 2 O (s) 2H 2 O (l)  H = 2 x 6.01 = 12.0 kJ 6.4

39. Thermochemistry © 2015 Pearson Education, Inc. H 2 O (s) H 2 O (l)  H = 6.01 kJ The physical states of all reactants and products must be specified in thermochemical equations. Thermochemical Equations 6.4 H 2 O (l) H 2 O (g)  H = 44.0 kJ How much heat is evolved when 266 g of white phosphorus (P 4 ) burn in air? P 4 (s) + 5O 2 (g) P 4 O 10 (s)  H = -3013 kJ 266 g P 4 1 mol P 4 123.9 g P 4 x 3013 kJ 1 mol P 4 x = 6470 kJ

40. Thermochemistry © 2015 Pearson Education, Inc. Calorimetry Since we cannot know the exact enthalpy of the reactants and products, we measure  H through calorimetry, the measurement of heat flow. The instrument used to measure heat flow is called a calorimeter. 5.5

41. Thermochemistry © 2015 Pearson Education, Inc. Heat Capacity and Specific Heat The amount of energy required to raise the temperature of a substance by 1 K (1  C) is its heat capacity, usually given for one mole of the substance.

42. Thermochemistry © 2015 Pearson Education, Inc. Heat Capacity and Specific Heat We define specific heat capacity (or simply specific heat) as the amount of energy required to raise the temperature of 1 g of a substance by 1 K (or 1  C).

43. Thermochemistry © 2015 Pearson Education, Inc. Heat Capacity and Specific Heat Specific heat, then, is

44. Thermochemistry © 2015 Pearson Education, Inc. Constant Pressure Calorimetry By carrying out a reaction in aqueous solution in a simple calorimeter, the heat change for the system can be found by measuring the heat change for the water in the calorimeter. The specific heat for water is well known (4.184 J/g ∙ K). We can calculate  H for the reaction with this equation: q = m  C s   T

45. Thermochemistry © 2015 Pearson Education, Inc. Bomb Calorimetry Reactions can be carried out in a sealed “bomb” such as this one. The heat absorbed (or released) by the water is a very good approximation of the enthalpy change for the reaction. q rxn = – C cal × ∆T

46. Thermochemistry © 2015 Pearson Education, Inc. Bomb Calorimetry Because the volume in the bomb calorimeter is constant, what is measured is really the change in internal energy,  E, not  H. For most reactions, the difference is very small.

47. Thermochemistry © 2015 Pearson Education, Inc. Hess’s Law  H is well known for many reactions, and it is inconvenient to measure  H for every reaction in which we are interested. However, we can estimate  H using published  H values and the properties of enthalpy. 5.6

48. Thermochemistry © 2015 Pearson Education, Inc. Hess’s Law Hess’s law: If a reaction is carried out in a series of steps,  H for the overall reaction will be equal to the sum of the enthalpy changes for the individual steps. Because  H is a state function, the total enthalpy change depends only on the initial state (reactants) and the final state (products) of the reaction.

49. Thermochemistry © 2015 Pearson Education, Inc. Enthalpies of Formation An enthalpy of formation,  H f, is defined as the enthalpy change for the reaction in which a compound is made from its constituent elements in their elemental forms. 5.7

50. Thermochemistry © 2015 Pearson Education, Inc. Standard Enthalpies of Formation Standard enthalpies of formation, ∆H f °, are measured under standard conditions (25 ºC and 1.00 atm pressure).

51. Thermochemistry © 2015 Pearson Education, Inc. Calculation of  H Imagine this as occurring in three steps: C 3 H 8 (g) + 5 O 2 (g)  3 CO 2 (g) + 4 H 2 O(l) 1) Decomposition of propane to the elements: C 3 H 8 (g)  3 C (graphite) + 4 H 2 (g)

52. Thermochemistry © 2015 Pearson Education, Inc. Calculation of  H Imagine this as occurring in three steps: 2) Formation of CO 2 : 3 C (graphite) + 3 O 2 (g)  3 CO 2 (g) C 3 H 8 (g) + 5 O 2 (g)  3 CO 2 (g) + 4 H 2 O(l)

53. Thermochemistry © 2015 Pearson Education, Inc. Calculation of  H Imagine this as occurring in three steps: 3) Formation of H 2 O: 4 H 2 (g) + 2 O 2 (g)  4 H 2 O(l) C 3 H 8 (g) + 5 O 2 (g)  3 CO 2 (g) + 4 H 2 O(l)

54. Thermochemistry © 2015 Pearson Education, Inc. Calculation of  H So, all steps look like this: C 3 H 8 (g)  3 C (graphite) + 4 H 2 (g) 3 C (graphite) + 3 O 2 (g)  3 CO 2 (g) 4 H 2 (g) + 2 O 2 (g)  4 H 2 O(l) C 3 H 8 (g) + 5 O 2 (g)  3 CO 2 (g) + 4 H 2 O(l)

55. Thermochemistry © 2015 Pearson Education, Inc. C 3 H 8 (g)  3 C (graphite) + 4 H 2 (g) 3 C (graphite) + 3 O 2 (g)  3 CO 2 (g) 4 H 2 (g) + 2 O 2 (g)  4 H 2 O(l) C 3 H 8 (g) + 5 O 2 (g)  3 CO 2 (g) + 4 H 2 O(l) Calculation of  H The sum of these equations is the overall equation! C 3 H 8 (g) + 5 O 2 (g)  3 CO 2 (g) + 4 H 2 O(l)

56. Thermochemistry © 2015 Pearson Education, Inc. Calculation of  H We can use Hess’s law in this way:  H =  n  H f,products –  m  H f °,reactants where n and m are the stoichiometric coefficients.

57. Thermochemistry © 2015 Pearson Education, Inc.  H= [3(−393.5 kJ) + 4(−285.8 kJ)] – [1(−103.85 kJ) + 5(0 kJ)] = [(−1180.5 kJ) + (−1143.2 kJ)] – [(−103.85 kJ) + (0 kJ)] = (−2323.7 kJ) – (−103.85 kJ) = −2219.9 kJ Calculation of  H using Values from the Standard Enthalpy Table C 3 H 8 (g) + 5 O 2 (g)  3 CO 2 (g) + 4 H 2 O(l)

58. Thermochemistry © 2015 Pearson Education, Inc. Energy in Foods Most of the fuel in the food we eat comes from carbohydrates and fats.

59. Thermochemistry © 2015 Pearson Education, Inc. Energy in Fuels The vast majority of the energy consumed in this country comes from fossil fuels.

60. Thermochemistry © 2015 Pearson Education, Inc. Other Energy Sources Nuclear fission produces 8.5% of the U.S. energy needs. Renewable energy sources, like solar, wind, geothermal, hydroelectric, and biomass sources produce 7.4% of the U.S. energy needs.