1. Thermochemistry Chapter 5 Thermochemistry John D. Bookstaver St. Charles Community College St. Peters, MO  2006, Prentice Hall, Inc. Chemistry, The Central Science, 10th edition Theodore L. Brown; H. Eugene LeMay, Jr.; and Bruce E. Bursten

2. Thermochemistry December 16 HOMEWORK Do questions 1,2,9,11,17,19,20 First law of thermodynamics 23,25,29 Enthalpy 31 to 45 odd Lab Notebook due by next Monday

3. Thermochemistry Energy The ability to do work or transfer heat.  Work: Energy used to cause an object that has mass to move.  Heat: Energy used to cause the temperature of an object to rise.

4. Thermochemistry Potential Energy Energy an object possesses by virtue of its position or chemical composition.

5. Thermochemistry Electrostatic PE Arises from the interaction between charged particles. E el = k Q 1 Q 2 k = 8.99x10 9 Jm/C 2 d Q charge of particles d distance If charges are same sign they repel and E el is positive. If they have opposite sign they attract and E el is negative. The lower the energy of system the more stable it is. Thus the more strongly opposite charged particles attract each other.

6. Thermochemistry Kinetic Energy Energy an object possesses by virtue of its motion. 1 2 KE =  mv 2

7. Thermochemistry 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 1 J = 1  kg m 2 s2s2

8. Thermochemistry Joule A mass of 2 kg moving at a speed of 1 m/s possesses a kinetic energy of 1 J. calorie: the amount of heat needed to increase the temperature of 1 g of water from 14.5 0 C to 15.5 0 C. 1 Cal = 1000 cal = 1 kcal

9. Thermochemistry System and Surroundings Consider the reaction inside the piston 2H 2 (g) +O 2 ->2H 2 O(g)+heat The system includes the molecules we want to study (here, the hydrogen and oxygen molecules). Reactants an products. The surroundings are everything else (here, the cylinder and piston).

10. Thermochemistry Work Energy is used to move an object over some distance. w = F  d, where w is work, F is the force, and d is the distance over which the force is exerted.

11. Thermochemistry Heat Energy can also be transferred as heat. Heat flows from warmer objects to cooler objects. HEAT THE ENERGY TRANSFERRED BETWEEN A SYSTEM AND ITS SURROUNDINGS AS A RESULT OF THEIR DIFFERENCE IN TEMPERATURE

12. Thermochemistry December 17 First law of thermodynamics Internal energy Exchange of heat – endo and exo processes State functions Enthalpy – Heat of a reaction Calorimetry

13. Thermochemistry Transferal of Energy a)The potential energy of this ball of clay is increased when it is moved from the ground to the top of the wall.

14. Thermochemistry Transferal of Energy a)The potential energy of this ball of clay is increased when it is moved from the ground to the top of the wall. b)As the ball falls, its potential energy is converted to kinetic energy.

15. Thermochemistry Transferal of Energy a)The potential energy of this ball of clay is increased when it is moved from the ground to the top of the wall. b)As the ball falls, its potential energy is converted to kinetic energy. c)When it hits the ground, its kinetic energy falls to zero (since it is no longer moving); some of the energy does work on the ball, the rest is dissipated as heat.

16. Thermochemistry 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. Use Fig. 5.5

17. Thermochemistry 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. Use Fig. 5.5

18. Thermochemistry 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 Use Fig. 5.5

19. Thermochemistry 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.

20. Thermochemistry 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.

21. Thermochemistry 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.

22. Thermochemistry  E, q, w, and Their Signs

23. Thermochemistry Exchange of Heat between System and Surroundings When heat is absorbed by the system from the surroundings, the process is endothermic.

24. Thermochemistry Exchange of Heat between System and Surroundings When heat is absorbed by the system from the surroundings, the process is endothermic. When heat is released by the system to the surroundings, the process is exothermic.

25. Thermochemistry State Functions Usually we have no way of knowing the internal energy of a system; finding that value is simply too complex a problem.

26. Thermochemistry State Functions 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.

27. Thermochemistry 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.

28. Thermochemistry 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.

29. Thermochemistry Work PV work is the work associated with a change of volume (  V) that occurs against a resisting pressure.

30. Thermochemistry 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

31. Thermochemistry 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

32. Thermochemistry 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

33. Thermochemistry 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.

34. Thermochemistry Endothermicity and Exothermicity A process is endothermic, then, when  H is positive.

35. Thermochemistry Endothermicity and Exothermicity A process is endothermic when  H is positive. A process is exothermic when  H is negative.

36. Thermochemistry Enthalpies of Reaction The change in enthalpy,  H, is the enthalpy of the products minus the enthalpy of the reactants:  H = H products − H reactants

37. Thermochemistry Enthalpies of Reaction This quantity,  H, is called the enthalpy of reaction, or the heat of reaction.

38. Thermochemistry 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.

39. Thermochemistry Practice problem– Change in enthalpy – Heat of reaction 3CuCl 2 + 2Al  3Cu + 2AlCl 3 ΔH = -793.5 kJ 1.What is ΔH for the following reaction? 6CuCl 2 + 4Al  6Cu + 4AlCl 3 2.What is ΔH for the following reaction? 9 Cu + 6 AlCl 3  9 CuCl 2 + 6 Al 3.Given the equation at the top of the page, how much energy will be given off by the reaction of 2.07 g of Al with excess CuCl 2 ?

40. Thermochemistry December 20 Enthalpy practice Calorimetry

41. Thermochemistry Constant Pressure Calorimetry By carrying out a reaction in aqueous solution in a simple calorimeter such as this one, one can indirectly measure the heat change for the system by measuring the heat change for the water in the calorimeter.

42. Thermochemistry Heat Capacity and Specific Heat Calorimeter = apparatus that measures heat flow. Heat capacity = the amount of energy required to raise the temperature of an object (by one degree). (J °C -1 ) Molar heat capacity = heat capacity of 1 mol of a substance. (J mol -1 °C -1 ) Specific heat = specific heat capacity = heat capacity of 1 g of a substance. (J g -1 °C -1 )

43. Thermochemistry Remember a change in T in K and in 0 C are equal in magnitude  T in K =  T in 0 C

44. Thermochemistry Molar heat capacity The heat capacity of one mole of substance. The amount of heat required to increase the temperature of 1 mole of substance 1 K.

45. Thermochemistry Constant Pressure Calorimetry Atmospheric pressure is constant! HEAT LOST = HEAT GAINED Object 1Object 2

46. Thermochemistry Constant Pressure Calorimetry Because the specific heat for water is well known (4.184 J/mol-K), we can measure  H for the reaction with this equation: q = m  s   T

47. Thermochemistry Heat Capacity and Specific Heat Specific heat, then, is Specific heat = heat transferred mass  temperature change s = q m  Tm  T

48. Thermochemistry Sample problem 1 Find the specific heat of water if 209 J is required to increased the temperature of 50.0 g of water by 1.00K s=q / m  T

49. Thermochemistry Sample Problem 2 How much heat is needed to warm 250g of water from 25 0 C to 75 0 C. Specific heat of water 4.18 J/gK Find the molar heat capacity of water.

50. Thermochemistry Sample problem 3 25.0 mL of 0.100 M NaOH is mixed with 25.0 mL of 0.100 M HNO 3, and the temperature of the mixture increases from 23.1°C to 49.8°C. Calculate ΔH for this reaction in kJ mol -1. Assume that the specific heat of each solution is equal to the specific heat of water, 4.184 J g -1 °C -1.

51. Thermochemistry Sample problem 4 An unknown metal is to be analyzed for specific heat. A 12.84-g piece of the metal is placed in boiling water at 101.5°C until it attained that temperature. The metal is then removed from the boiling water and placed into a styrofoam cup containing 54.92 g of water at 23.2°C. The final temperature of the water was 26.1°C. What is the specific heat of the metal sample, in J g -1 °C -1 ?

52. Thermochemistry Bomb Calorimetry A more sophisticated model is the bomb calorimeter, it has a chamber where a chemical reaction takes place and a device to start the reaction.

53. Thermochemistry HEAT CAPACITY OF THE CALORIMETER Heat is released when combustion occurs. The heat is absorbed by the calorimeter and the T of the water rises. THE HEAT CAPACITY OF THE CALORIMETER HAS TO BE CALCULATED. It is done by measuring  T in a reaction that releases a known amount of heat. The heat capacity of the calorimeter C cal = amount of heat absorbed /  T

54. Thermochemistry Reaction carried out under constant volume. Use a bomb calorimeter. Usually study combustion. Bomb Calorimetry (Constant Volume Calorimetry)

55. Thermochemistry 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.

56. Thermochemistry Sample problem 5 A 0.5269 g sample of octane is placed in a bomb calorimeter with a heat capacity of 11.3 kJ °C -1. The original temperature of the water in the calorimeter is 21.93°C. After combustion, the temperature increases to 24.21°C. Determine the enthalpy of combustion per mole and per gram of octane.

57. Thermochemistry Sample problem 6 After 50 mL of 1.0 M HCl and 50mL of 1.0M NaOH are mixed in a coffe cup calorimeter, the temperature of the solution increases form 21.0 0 C to 27.5 0 C. Calculate the enthalpy change for the rx in kJ/mol of HCl assuming that the calorimeter loses only a negligible quantity of heat, total volume of solution is 100mL, that its density is 1.0 g/mL and the specific heat is 4.18J/gk

58. Thermochemistry December 21 MORE HESS LAW ENTHALPIES OF FORMATION

59. Thermochemistry 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  H values that are published and the properties of enthalpy.

60. Thermochemistry Hess’s Law Hess’s law states that “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.”

61. Thermochemistry Hess’s Law Because  H is a state function, the total enthalpy change depends only on the initial state of the reactants and the final state of the products.

62. Thermochemistry Example The bombardier beetle uses an explosive discharge as a defensive measure. The chemical reaction involved is the oxidation of hydroquinone by hydrogen peroxide to produce quinone and water: C 6 H 4 (OH) 2 (aq) + H 2 O 2 (aq)  C 6 H 4 O 2 (aq) + 2 H 2 O (l)

63. Thermochemistry C 6 H 4 (OH) 2 (aq)  C 6 H 4 O 2 (aq) + H 2 (g)ΔH = +177.4 kJ H 2 (g) + O 2 (g)  H 2 O 2 (aq)ΔH = -191.2 kJ H 2 (g) + ½ O 2 (g)  H 2 O (g)ΔH = -241.8 kJ H 2 O (g)  H 2 O (l)ΔH = -43.8 kJ C 6 H 4 (OH) 2 (aq) + H 2 O 2 (aq)  C 6 H 4 O 2 (aq) + 2 H 2 O (l) ΔH = ???

64. Thermochemistry 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.

65. Thermochemistry  H o f. Standard conditions (standard state): 1 atm and 25 o C (298 K). Standard enthalpy,  H o, is the enthalpy measured when everything is in its standard state. Standard enthalpy of formation: 1 mol of compound is formed from substances in their standard states. If there is more than one state for a substance under standard conditions, the more stable one is used. Standard enthalpy of formation of the most stable form of an element is zero.

66. Thermochemistry Enthalpies of Formation

67. Thermochemistry CONVENTIONAL DEFINITIONS OF STANDARD STATES FOR ELEMENTS Is the form in which the element exist under conditions of 1 atm and 25 C. For O 2 is gas. For Na is solid For Hg is liquid For C is graphite By definition the  H of formation of the most stable form of an element is zero.

68. Thermochemistry Examples – write formation reactions for each: 1.H 2 SO 4 2.C 2 H 5 OH 3.NH 4 NO 3

69. Thermochemistry Examples Use the tables in the back of your book to calculate ΔH for the following reactions: 1.4 CuO (s)  2 Cu 2 O (s) + O 2 (g) 2.C 3 H 8 (g) + 5 O 2 (g)  3 CO 2 (g) + 4 H 2 O (l) 3.NH 3 (g) + HCl (g)  NH 4 Cl (s)

70. Thermochemistry Using Enthalpies of Formation of Calculate Enthalpies of Reaction We use Hess’ Law to calculate enthalpies of a reaction from enthalpies of formation.

71. Thermochemistry Using Enthalpies of Formation of Calculate Enthalpies of Reaction For a reaction

72. Thermochemistry 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. 

73. Thermochemistry Calculation of  H Imagine this as occurring in 3 steps: C 3 H 8 (g) + 5 O 2 (g)  3 CO 2 (g) + 4 H 2 O (l) 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)

74. Thermochemistry Calculation of  H Imagine this as occurring in 3 steps: C 3 H 8 (g) + 5 O 2 (g)  3 CO 2 (g) + 4 H 2 O (l) 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)

75. Thermochemistry Calculation of  H Imagine this as occurring in 3 steps: C 3 H 8 (g) + 5 O 2 (g)  3 CO 2 (g) + 4 H 2 O (l) 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)

76. Thermochemistry C 3 H 8 (g) + 5 O 2 (g)  3 CO 2 (g) + 4 H 2 O (l) 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:

77. Thermochemistry C 3 H 8 (g) + 5 O 2 (g)  3 CO 2 (g) + 4 H 2 O (l) Calculation of  H  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

78. Thermochemistry Energy in Foods Most of the fuel in the food we eat comes from carbohydrates and fats.

79. Thermochemistry Fuels The vast majority of the energy consumed in this country comes from fossil fuels.

80. Thermochemistry HW PRELAB FOR HESS LAW LAB DUE TOMORROW (Lab 6 – page 25) Remember test on monday Hw for Thursday up to 5.77 odd

81. Thermochemistry Example Calculate the total energy produced from the combustion of one gram of gasoline (assume it is pure octane, C 8 H 18 ). Compare that to the energy produced from the combustion of one gram of ethanol (a renewable fuel – C 2 H 5 OH)

82. Thermochemistry Hess’s law: if a reaction is carried out in a number of steps,  H for the overall reaction is the sum of  H for each individual step. For example: CH 4 (g) + 2O 2 (g)  CO 2 (g) + 2H 2 O(g)  H = -802 kJ 2H 2 O(g)  2H 2 O(l)  H = -88 kJ CH 4 (g) + 2O 2 (g)  CO 2 (g) + 2H 2 O(l)  H = -890 kJ

83. Thermochemistry Note that:  H 1 =  H 2 +  H 3

84. Thermochemistry JANUARY 8 Prelab Foods and fuels. Read from txbook Hess law problems

85. Thermochemistry Foods Fuel value = energy released when 1 g of substance is burned. 1 nutritional Calorie, 1 Cal = 1000 cal = 1 kcal. Energy in our bodies comes from carbohydrates and fats (mostly). Intestines: carbohydrates converted into glucose: C 6 H 12 O 6 + 6O 2  6CO 2 + 6H 2 O,  H = -2816 kJ Fats break down as follows: 2C 57 H 110 O 6 + 163O 2  114CO 2 + 110H 2 O,  H = -75,520 kJ Fats: contain more energy; are not water soluble, so are good for energy storage.

86. Thermochemistry Fuels In 2000 the United States consumed 1.03  10 17 kJ of fuel. Most from petroleum and natural gas. Remainder from coal, nuclear, and hydroelectric. Fossil fuels are not renewable.

87. Thermochemistry Sources of energy consumed in USA