1. Chapter 5 Written by JoAnne Swanson University of Central Florida Thermodynamics

2. Topics P Types of energy and units of energy P Exothermic vs. endothermic reactions P Heat capacity and specific heat P Energy transfers in changes of state P Calorimetry P Enthalpy and entropy P Hess’s Law

3. Energy is defined as the capacity to do work. M There are two categories of energy: M Kinetic energy. M Potential energy M Kinetic Energy is the energy of motion M Potential energy is energy of position,

4. Chemical energy is a form of potential energy stored in the structure of a chemical substance due to its composition.

5. L L Consider the potential energy of a rock up on a cliff. If one rock is up on a 1000 ft. cliff and another is up on a 10 ft. hill, the rock that is 1000ft. up will hit the ground with much more energy than the other if it falls.

6. L L Chemical energy has similar differences in that the structure of one substance may contain much more potential energy than another substance.

7. Kinetic Energy of an object depends on its mass and velocity: E k = ½ (mv 2 ) or mv 2 /2 mass is in Kg, and Kg. m 2 /s 2 = 1 Joule, the unit used to designate energy. From this equation you should see that E k increases with mass and with velocity

8. Potential Energy depends on gravity, mass, and position: E p = mgh = (mass, gravity, height) (gravitational constant = 9.8 m/s 2 ) But E p of submicroscopic substances (molecules, ions, etc.) electrical charges between particles.

9. Electrical charge is designated by the symbol ‘Q’ and has the value of an electron charge = 1.60 x 10 -19 C The unit for electrical charge is the coulomb ( C). E el = kQ 1 Q 2 / d where k=8.99x10 9 Jm E el = kQ 1 Q 2 / d where k=8.99x10 9 Jm C 2 C 2 This equation shows the electrostatic forces of attraction between two particles, 1 and 2, are inversely proportional to the distance between the particles.

10. Another energy unit we will use is the calorie. 1 c = 1 cal = 4.184 J This differs from the nutritional calorie which is 1C = 1000c = 1Kcal

11. Thermochemistry is the study of heat transfers occurring in chemical and physical changes of substances. L Thermal energy Lrandom motion of atoms and molecules.

12. The Universe is made up of the and the The Universe is made up of the system and the surroundings The system is the part of the universe that is of interest to us. The surroundings is everything outside of the system.

13. Energy is transferred as heat and work. Work = Force x Distance = pressure x volume = pressure x volume It usually pertains to either the electrical work (for instance in a battery) or mechanical work (like the pressure exerted on or by a gas in a reaction).

14. The change in Internal Energy of a system =  E = q + w  E is the change in internal energy q = heat w = work Heat can flow into the system or out of the system. sign on qsign on q Work can be done on the system or by the system. sign on wsign on w

15. Exothermic vs. Endothermic Processes   Any process that gives off heat to the surroundings is an Exothermic process.   bonds are made

16. L L When a process absorbs heat from the surroundings it is an Endothermic process. When ice melts it absorbs heat from the surroundings, thus it is an endothermic process.

17. To measure the change in internal energy of a chemical reaction, the reaction is often carried out in a “bomb calorimeter”. There is no change in volume in this sealed container, therefore there is no work involved and  E = q reaction (at constant volume)  E = q reaction (at constant volume)

18. The symbol  H is used to represent the change in heat into or out of the system. It is defined as the change in enthalpy.

19. Enthalpy and Internal Energy are nearly equal. Enthalpy takes into account reactions at constant pressure where an expanding gas does work on the surroundings. Remember w = PV, but if w is done by the system, w = -PV H = E +  PV and  H =  E +  PV w = -  PV so,  H = (q + w) – w w = -  PV so,  H = (q + w) – w  H = q (at constant pressure)  H = q (at constant pressure)

20. 20 1 mole gas at 25 o C and 1atm has a volume of 24.5L =( 24.5 L. atm / mol )1 mole gas at 25 o C and 1atm has a volume of 24.5L =( 24.5 L. atm / mol ) 1 L. atm = 0.1013 kJ1 L. atm = 0.1013 kJ  PV makes little difference in the  H value, and even less difference for liquids or solids.  PV makes little difference in the  H value, and even less difference for liquids or solids.why?

21. If 1 L. atm = 0.1013 kJ how many kJ are in the molar volume of a gas at 25 o C? 24.5 L. atm x 0.1013 kJ = 2.5 kJ 1 L. atm see next example

22. Examples: given the values for PV, calculate the change in enthalpy for the combustion of methane where  E = -885 kJ PV at 25 o C CH 4 (g)2.5 KJ/ mol O 2 (g)2.5 KJ/ mol CO 2 (g)2.5 KJ/ mol H 2 0(l)0.0018 KJ/ mol  H =  E +  PV  H =  E +  PV (  PV = PV products – PV reactants) continued………

23. CH 4 (g) + 2 O 2 (g)  CO 2 (g) + 2 H 2 O (l)  PV = PV CO 2 + 2PV H 2 O –[PV CH 4 + 2PV O 2  PV = [2.5 KJ + 2(0.0018 KJ)] – [2.5 KJ + 2(2.5 KJ) ]  PV = -5.0 KJ  H = -885 KJ - 5.0 KJ = -890 KJ Carried out at constant pressure,  H = -885 KJ Not a big difference when  PV is added

24. EXAMPLE 2: a.Calculate the kinetic energy in joules, of a 45 g golf ball moving at 61 m/s. b.Convert this energy to calories. c.What happens to the energy when the ball strikes a tree? Since 1J = 1kg m 2 /s 2 change g to kg. 45 g = 0.045 kg

25. E k = mv 2 / 2 = 0.045 kg x (61m/s) 2 2 = 84 kg m 2 = 84 J s 2 b.84 J x 1cal = 20 cal 4.184 J c. When the ball hits the tree, velocity = 0 and so does Kinetic energy. Kinetic Energy is transferred as potential energy to the tree and the deformed golf ball.

26. Example 4: Calculate  E when a balloon is inflated completely by heating it. The volume changes from 4.00 x 10 6 L to 4.50 x 10 6 L, with the addition of 1.3 x 10 8 J of heat. The balloon expands against an internal pressure of 1.0 atm.  E = q+w w = -  PV because ________________ q is positive because _________________

27. 27  V = 4.50 x 10 6 L - 4.00 x 10 6 L = 5.00 x 10 5 L -P  V = 5.00 x 10 5 L x (-1.0 atm) =5.00 x 10 5 Latm = -5.07 x 10 7 J  E = 1.3 x 10 8 J – 5.07 x 10 7 J = 7.9 x 10 7 J

28. Example 3: Calculate  E when a gas is compressed from 6.0 x 10 4 L to 5.3 x 10 4 L under an external pressure of 1.3 atm. The amount of heat transferred to the surroundings was 4.3 x 10 2 J.  E = w + q, where w = +  PV why?  V = 5.3 x 10 4 L - 6.0 x 10 4 L = - 7.0 x 10 3 L  PV = (1.3 atm)(- 7.0 x 10 3 L) = -9.1 x 10 3 Latm

29. -9.1 x 10 3 Latm x 0.1013 kJ = -9.2 x 10 2 KJ Latm Latm In this case since the gas was compressed, w = + and since heat was transferred to the surroundings, q = -.  E = w – q, = -9.2 x 10 5 J – 4.3 x 10 2 J = = - 9.2043 x 10 5 J = - 9.2 x 10 5 J

30. The enthalpy of reaction is the difference between the enthalpies of the products and reactants.  H (rxn) =  H f (products) –  f (reactants)  H (rxn) =  H f (products) –  f (reactants)

31. reactants productsreactants products reaction progression Energy Energy energy added energy expelled EaEaEaEa EaEaEaEa EndothermicExothermic

32. The diagram illustrates energy given off when bonds are made between H 2 and O 2, (a). Energy is absorbed when bonds are broken in HgO, (b).

33. The units of energy are Kilojoules and Calories. Enthalpy of reaction will be expressed as Kilojoules (Kj).

34. It is important to note the following when calculating the enthalpy of reaction: 6 6The enthalpy of a substance in its standard state is equal to zero. 6 6Enthalpy is dependent on the quantity of the substance therefore the enthalpy of a substance must be multiplied by its coefficient in the chemical equation.

35. 35  When a rxn is reversed, the sign on the enthalpy of rxn is changed.  If a rxn is divided through or multiplied through by a number, so is the enthalpy of rxn.  the state of matter is important in enthalpy calculations.

36. Enthalpy of formation values of compounds are found in tables of thermodynamic data. The enthalpy of formation is defined as the energy associated with the formation of 1 mole of a substance from its elements in their standard states. Na (s) + 1/2 Cl 2 (g)  NaCl The formation equation for sodium chloride

37. Ex. 1 Determine the enthalpy for the following chemical reaction: CO 2 (g) + 2 H 2 O (l) -----> 2 O 2 (g) + CH 4 (g)

38. Use appendix C pg. 1041 in your text to find the individual enthalpy values. CO 2 (g) = -393.509, H 2 O (l) = -285.83 CH 4 (g) = -74.81, O 2 (g) = 0  H =   H f (products) –  f  reactants)  H = (0 + -74.81)-(-393.509 + 2(- 285.83)) = -74.81 + 965.17 = 890.36 Kj CO 2 + 2 H 2 O  2 O 2 + CH 4

39. HESS’S LAW ENTHALPY IS A STATE FUNCTION. ENTHALPY OF A REACTION WILL BE THE SAME NO MATTER WHAT PATH IS TAKEN TO ARRIVE AT THE PRODUCTS.

40. This means that if we need to calculate the enthalpy of a reaction for which we do not know the enthalpies of formation, we can algebraically manipulate other reactions to arrive at the enthalpy for the desired reaction.

41. Ex.Determine the  H rxn for Sn (s) + 2 Cl 2 (g)  SnCl 4(l) Given the following: Sn (s) + Cl 2 (g)  SnCl 2(s)  H 1 = -349.8 kJ SnCl 2(s) + Cl 2 (g)  SnCl 4(l)  H 2 = -195.4 kJ Sn (s) + 2 Cl 2 (g)  SnCl 4(l)  H 3 = ?  H 1 +  H 2 =  H 3 = -545.2 kJ

42. Ex. Calculate the  H rxn for S (s) + O 2 (g)  SO 2(g) Given S (s) + 3/2 O 2 (g)  SO 3(g)  H= -395.2 kJ S (s) + 3/2 O 2 (g)  SO 3(g)  H= -395.2 kJ SO 2 (g) + O 2 (g)  2 SO 3(g)  H= -198.2 kJ SO 2 (g) + O 2 (g)  2 SO 3(g)  H= -198.2 kJ NOTICE THAT WE NEED SO 2 (g) TO BE ON THE PRODUCT SIDE. THEREFORE WE MUST REVERSE THE SECOND RXN CONTINUED

43. S (s) + O 2 (g)  SO 2(g) S (s) + 3/2 O 2 (g)  SO 3(g)  H= -395.2 kJ S (s) + 3/2 O 2 (g)  SO 3(g)  H= -395.2 kJ 2 SO 3 (g)  O 2 (g) + 2 SO 2(g)  H= +198.2 kJ 2 SO 3 (g)  O 2 (g) + 2 SO 2(g)  H= +198.2 kJ IT IS ALSO NECESSARY TO DIVIDE EQUATION 2 BY 2, SO THAT WE HAVE 1 SO 2(g) IN THE PRODUCT. WE MUST ALSO DIVIDE THE  H +198.2 kJ / 2 = 99.10 kJ

44. Now add the reactions and the enthalpies. S (s) + 3/2 O 2 (g)  SO 3(g)  H= -395.2 kJ SO 3 (g)  1/2 O 2 (g) + SO 2(g)  H= +99.1 kJ S (s) + O 2 (g)  SO 2(g) S (s) + O 2 (g)  SO 2(g)  H= -296.1 kJ

45. The enthalpy of formation of a substance in a chemical reaction can also be calculated if the enthalpy of reaction is known and the enthalpies of the other substances in the reaction are known. Simply call the unknown enthalpy ‘X’ and plug in all known values to solve for ‘X’. example on transparency

46. Specific heat and heat Capacity Specific heat and heat Capacity Heat capacity is the amount of heat required to raise the temperature of a given quantity of a substance by 1 o C. Specific Heat is the amount of heat required to raise the temperature of 1 gram of a substance by 1 o C. J/g o C Molar heat capacity is J / mol o C

47. a substance’s ability to absorb heat and to store heat.a substance’s ability to absorb heat and to store heat. For example, a metal _______________ ________________________________ A liquid like water (which contains strong intermolecular forces, hydrogen bonds), _________________________________ _________________________________

48. The metal has a ______________ and the water has a ________________. The specific heat capacity ( C ) of water is an important and easy number to remember ; C = 1 cal / g o C, (or 4.184 J / g o C ) Since o C and Kelvin have the same size increments, they are interchangeable in these equations.

49. Calorimetry is a technique used in the lab to measure the change in enthalpy of a reaction. One apparatus used is the Bomb Calorimeter. It is a Heavy walled, steel container which has a known specific heat.

50. Types of Calorimetry problems we will cover: 1.simple change in water temperature 2.change in state of water 3.a rxn in a coffee cup calorimeter 4.a rxn in a bomb calorimeter 5.a mixture of hot substance with cold substance in a calorimeter.

51. Equations needed: 1.q = C x m x  T 2.q = q ice +  f + q water +  H v + q steam 3.q rxn =  H rxn = -q water  H rxn = -[q water + q bomb ], or = -(C cal x  T) 5.q cold = -q hot

52. Some needed values: specific heat of water(l) = 4.184 J/gK specific heat of water(s) = 2.05 J/gK specific heat of water(g) = 2.01 J/gK  H f = enthalpy of fusion = energy involved in a change of state from liquid to solid or vice versa = 333 J/g  H v = enthalpy of vaporization = energy involved in a change of state from liquid to vapor or vice versa = 2444 J/g

53. The heat of the reaction (q) is equal to the negative of the heat absorbed by the (bomb plus the water surrounding the bomb). q rxn =  H rxn = - (q bomb + q water )

54. q is used to represent the quantity and direction of heat transferred, using a calorimeter.

55. Necessary Equations: q = C x m x  T q = (specific heat)(mass)(change in Temp.) q = (J/gK)(g)(  T) q(rxn) = - (q of the water + q of the bomb)

56. Subliminal message..... Wake up !!!

57. 1. Calorimetry A 466g sample of water is heated from 8.50 o C to 74.60 o C. Calculate the amount of heat absorbed by the water. q = (sp. heat)(mass)(  q = ( 4.184 J/g o C)(466g)(74.60-8.50 o C) q = 1.29 x 10 5 J = 129 KJ

58. 58 change of state of water example on transparency. coffee cup calorimeter on transparency

59. Ex. 3 1.435 g of Naphthalene (molar mass =128.2) was burned in a bomb calorimeter. The temp. rose from 20.17 o C to 25.84 o C. The mass of the water surrounding the calorimeter was 2000.g and the heat capacity of the bomb was 1.80 KJ/ o C. Calculate the heat of combustion of Naphthalene.

60. q(rxn) = -(q water + q bomb) q(water) = (2000g)(4.184 J/g o C)(5.67 o C) = 4.74 x 10 4 q(bomb) = (1.80 x 10 3 J / o C)(5.67 o C) = 1.02 x 10 4 J q(rxn)= -(4.74 x 10 4 J + 1.02 x 10 4 J) = -5.76 x 10 4 J

61. 61 MORE EXAMPLES WILL BE GIVEN IN CLASS. SINCE I HAD TO RUSH TO GET THESE NOTES ON LINE, THEY WERE NOT COMPLETELY EDITED. THERE IS SOME REPETITION, and they may need some corrections.