1. “Chemical Thermodynamics: Enthalpy, Entropy and Free Energy”

2. 2 The Flow of Energy – Heat and Work u OBJECTIVES: Explain how energy, heat, and work are related.

3. 3 The Flow of Energy – Heat and Work u OBJECTIVES: Classify processes as either exothermic or endothermic.

4. 4 The Flow of Energy – Heat and Work u OBJECTIVES: Identify the units used to measure heat transfer.

5. 5 The Flow of Energy – Heat and Work u OBJECTIVES: Distinguish between heat capacity and specific heat capacity (also called simply specific heat).

6. 6 Energy Transformations u “Thermochemistry” - concerned with heat changes that occur during chemical reactions u Energy - capacity for doing work or supplying heat weightless, odorless, tasteless if within the chemical substances- called chemical potential energy

7. 7 Energy Transformations u Gasoline contains a significant amount of chemical potential energy u Heat - represented by “q”, is energy that transfers from one object to another, because of a temperature difference between them. only changes can be detected! flows from warmer  cooler object

8. Exothermic and Endothermic Processes u Essentially all chemical reactions and changes in physical state involve either: a)release of heat, or b)absorption of heat

9. Exothermic and Endothermic Processes u In studying heat changes, think of defining these two parts: the system - the part of the universe on which you focus your attention the surroundings - includes everything else in the universe

10. Exothermic and Endothermic Processes u Together, the system and its surroundings constitute the universe u Thermochemistry is concerned with the flow of heat from the system to its surroundings, and vice-versa.

11. Exothermic and Endothermic Processes u The Law of Conservation of Energy states that in any chemical or physical process, energy is neither created nor destroyed. All the energy is accounted for as work, stored energy, or heat.

12. Exothermic and Endothermic Processes u Heat flowing into a system from its surroundings: defined as positive q has a positive value called endothermic –system gains heat (gets warmer) as the surroundings cool down

13. Exothermic and Endothermic Processes u Heat flowing out of a system into its surroundings: defined as negative q has a negative value called exothermic –system loses heat (gets cooler) as the surroundings heat up

14. Exothermic and Endothermic Exothermic processesEndothermic processes making ice cubesmelting ice cubes formation of snow in cloudsconversion of frost to water vapor condensation of rain from water vaporevaporation of water a candle flameforming a cation from an atom in the gas phase mixing sodium sulfite and bleachbaking bread rusting ironcooking an egg burning sugarproducing sugar by photosynthesis forming ion pairsseparating ion pairs Combining atoms to make a molecule in the gas phase splitting a gas molecule apart

15. 15 Exothermic and Endothermic u Every reaction has an energy change associated with it Gummy Bear Sacrifice u ** Gummy Bear Sacrifice ** u Exothermic reactions release energy, usually in the form of heat. u Endothermic reactions absorb energy u Energy is stored in bonds between atoms

16. 16 Units for Measuring Heat Flow 1)A calorie is defined as the quantity of heat needed to raise the temperature of 1.0 g of pure water 1.0 o C Used except when referring to food a Calorie, (written with a capital C), always refers to the energy in food 1 Calorie = 1 kilocalorie = 1000 cal.

17. 17 Units for Measuring Heat Flow 2)The calorie is also related to the Joule, the SI unit of heat and energy named after James Prescott Joule 4.184 J = 1 cal u Heat Capacity - the amount of heat needed to increase the temperature of an object exactly 1 o C Depends on both the object’s mass and its chemical composition

18. 18 Heat Capacity and Specific Heat u Specific Heat Capacity (abbreviated “c”) - the amount of heat it takes to raise the temperature of 1.0 gram of the substance by 1.0 o C often called simply “Specific Heat” u Water has a HUGE specific heat capacity value, when it is compared to other chemicals

19. Substance Specific Heat Capacity at 25 o C in J/g o C H 2 gas14.267 He gas5.300 H 2 O(liquid)4.184 ethyl alcohol2.460 ethylene glycol2.200 ice @ 0 o C2.090 steam @ 100 o C2.010 vegetable oil2.000 sodium1.23 air1.020 magnesium1.020 aluminum0.900 concrete0.880 glass0.840 potassium0.75 iron0.444 Table of Selected Specific Heat Capacity Values

20. 20 Heat Capacity and Specific Heat u For water, C = 4.184 J/(g o C) in Joules, and C = 1.00 cal/(g o C) in calories. u Thus, for water: it takes a long time to heat up, and it takes a long time to cool off! u Water is used as a coolant!

21. 21 Heat Capacity and Specific Heat u To calculate, use the formula: q = mass (in grams) x  T x c u heat is abbreviated as “q” u  T = change in temperature u c = Specific Heat capacity Units are either: J/(g o C) or cal/(g o C)

22. Problem: It takes 487.5 J to heat 25 grams of copper from 25 °C to 75 °C. What is the specific heat in Joules/g·°C? Use the formula: q = (m) (c) (ΔT) 487.5 J = (25 g)c(75 °C - 25 °C) 487.5 J = (25 g)c(50 °C) Solve for c: c = 487.5 J/(25g)(50 °C) c = 0.39 J/g·°C where q = heat energy m = mass c = specific heat ΔT = change in temperature

23. 23 Measuring and Expressing Enthalpy Changes u OBJECTIVES: Describe how calorimeters are used to measure heat flow.

24. 24 Measuring and Expressing Enthalpy Changes u OBJECTIVES: Construct thermochemical equations.

25. 25 Measuring and Expressing Enthalpy Changes u OBJECTIVES: Solve for enthalpy changes in chemical reactions by using heats of reaction.

26. Calorimetry u Calorimetry - the measurement of the heat into or out of a system for chemical and physical processes. Based on the fact that the heat released = the heat absorbed u The device used to measure the absorption or release of heat in chemical or physical processes is called a “Calorimeter”

27. Calorimetry u Foam cups are excellent heat insulators, and are commonly used as simple calorimeters under constant pressure. u For systems at constant pressure, the “heat content” is the same as a property called Enthalpy (H) of the system (They are good because they are well-insulated.)

28. A styrofoam cup calorimeter - two cups are nestled together for better insulation

29. Calorimetry u Changes in enthalpy =  H u q =  H These terms may be used interchangeably  Thus, q =  H = m ∙ c ∙  T 4  H is negative for an exothermic reaction 4  H is positive for an endothermic reaction

30. Calorimetry u Calorimetry experiments can be performed at a constant volume using a device called a “bomb calorimeter” - a closed system Used by nutritionists to measure energy content of food

31. A Bomb Calorimeter A bomb calorimeter http://www.chm.davidson.edu/ronutt/che115/Bomb/Bomb.htm

32. When a 1.000 g sample of the rocket fuel hydrazine, N 2 H 4, is burned in a bomb calorimeter which contains 1200 g of water, the temperature rises from 24.62°C to 28.16°C. If the C for the bomb is 840 J/°C, calculate q reaction for combustion of a one-gram sample and q reaction for combustion of one mole of hydrazine in the bomb calorimeter

33. For a bomb calorimeter, use this equation: q reaction = -(q water + q bomb ) q reaction = -(4.184 J / g·°C x m water x Δt + c x ΔT) q reaction = -(4.184 J / g·°C x m water + C)Δt where q is heat flow, m is mass in grams, and ΔT is the temperature change. Plugging in the values given in the problem: q reaction = -(4.18 J / g·°C x 1200 g + 840 J/°C)(3.54°C) q reaction = -20,700 J or -20.7 kJ We now know that 20.7 kJ of heat is evolved for every gram of hydrazine that is burned. Using the periodic table to get atomic weights, we can calculate that one mole of hydrazine, N 2 H 4, weight 32.0 g. Therefore, for the combustion of one mole of hydrazine: q reaction = 32.0 g/mol x -20.7 kJ/g = -662 kJ/molperiodic table

34. The following acid-base reaction is performed in a coffee cup calorimeter: H + (aq) + OH - (aq) → H 2 O(l) The temperature of 110 g of water rises from 25.0°C to 26.2°C when 0.010 mol of H + is reacted with 0.010 mol of OH -. Calculate q water ; Calculate ΔH for the reaction ;Calculate ΔH if 1.00 mol OH - reacts with 1.00 mol H + Solution Use this equation: q = c ∙ m ∙ ΔT where q is heat flow, m is mass in grams, and ΔT is the temperature change. Plugging in the values given in the problem: q water = 4.184 (J / g·°C;) x 110 g x (26.6°C - 25.0°C) q water = 550 J ΔH = -(q water ) = - 550 J We know that when 0.010 mol of H + or OH - reacts, ΔH is - 550 J: 0.010 mol H + ~ -550 J Therefore, for 1.00 mol of H + (or OH - ): ΔH = 1.00 mol H + x (-550 J / 0.010 mol H + ) ΔH = -5.5 x 10 4 J or ΔH = -55 kJ

35. 35 C + O 2 → CO 2 Energy ReactantsProducts  C + O 2 CO 2 395kJ given off + 395 kJ

36. 36 Exothermic u The products are lower in energy than the reactants u Thus, energy is released. u ΔH = -395 kJ The negative sign does not mean negative energy, but instead that energy is lost.

37. 37 CaCO 3 → CaO + CO 2 Energy ReactantsProducts  CaCO 3 CaO + CO 2 176 kJ absorbed CaCO 3 + 176 kJ → CaO + CO 2

38. 38 Endothermic u The products are higher in energy than the reactants u Thus, energy is absorbed. u ΔH = +176 kJ The positive sign means energy is absorbed

39. 39 Chemistry Happens in MOLES u An equation that includes energy is called a thermochemical equation  CH 4 + 2O 2  CO 2 + 2H 2 O + 802.2 kJ 1 mole of CH 4 releases 802.2 kJ of energy. When 802.2 kJ of heat energy are produced, so are 2 moles of water

40. 40 Thermochemical Equations u The heat of reaction is the heat change for the equation, exactly as written The physical state of reactants and products must also be given. Standard conditions (SC) for the reaction is 101.3 kPa (1 atm.) and 25 o C (different from STP)

41. 41 CH 4(g) + 2 O 2(g)  CO 2(g) + 2 H 2 O (l) + 802.2 kJ u If 10. 3 grams of CH 4 are burned completely, how much heat will be produced? 10. 3 g CH 4 16.05 g CH 4 1 mol CH 4 802.2 kJ = 514 kJ ΔH = -514 kJ, which means the heat is released for the reaction of 10.3 grams CH 4 Ratio from balanced equation 1 Start with known value Convert to moles Convert moles to desired unit

42. Summary, so far...

43. 43 Enthalpy u The heat content a substance has at a given temperature and pressure Can’t be measured directly because there is no set starting point u The reactants start with a heat content u The products end up with a heat content u So we can measure how much enthalpy changes

44. 44 Enthalpy u Symbol is H  Change in enthalpy is  H (delta H) u If heat is released, the heat content of the products is lower  H is negative (exothermic) u If heat is absorbed, the heat content of the products is higher  H is positive (endothermic)

45. 45 Energy ReactantsProducts  Change is down ΔH is <0 = Exothermic (heat is given off)

46. 46 Energy ReactantsProducts  Change is up ΔH is > 0 = Endothermic (heat is absorbed)

47. 47 Heat of Reaction u The heat that is released or absorbed in a chemical reaction  Equivalent to  H  C + O 2 (g)  CO 2 (g) + 393.5 kJ  C + O 2 (g)  CO 2 (g)  H = -393.5 kJ u In thermochemical equation, it is important to indicate the physical state a) H 2 (g) + 1/2O 2 (g)  H 2 O(g)  H = -241.8 kJ b) H 2 (g) + 1/2O 2 (g)  H 2 O(l)  H = -285.8 kJ

48. 48 Heat of Combustion u The heat from the reaction that completely burns 1 mole of a substance:  C + O 2 (g)  CO 2 (g) + 393.5 kJ  C + O 2 (g)  CO 2 (g)  H = -393.5 kJ

49. 49 Heat in Changes of State u OBJECTIVES: Classify the enthalpy change that occurs when a substance: a) melts, b) freezes, c) boils, d) condenses, or e) dissolves.

50. 50 Heat in Changes of State u OBJECTIVES: Solve for the enthalpy change that occurs when a substance: a) melts, b) freezes, c) boils, d) condenses, or e) dissolves.

51. 51 Heat in Changes of State 1. Molar Heat of Fusion (  H fus. ) = the heat absorbed by one mole of a substance in melting from a solid to a liquid q = mol x  H fus. (no temperature change) 2. Molar Heat of Solidification (  H solid. ) = the heat lost when one mole of liquid solidifies (or freezes) to a solid q = mol x  H solid. (no temperature change)

52. 52 Heat in Changes of State u Note: You may also have the value of these equations as: q = mass x  H This is because some textbooks give the value of  H as kJ/gram, instead of kJ/mol

53. 53 Heat in Changes of State u Heat absorbed by a melting solid is equal to heat lost when a liquid solidifies Thus,  H fus. = -  H solid. u Why is there no value listed for the molar heat of solidification?

54. 54 Heats of Vaporization and Condensation u When liquids absorb heat at their boiling points, they become vapors. 3. Molar Heat of Vaporization (  H vap. ) = the amount of heat necessary to vaporize one mole of a given liquid. q = mol x  H vap. (no temperature change)

55. 55 Heats of Vaporization and Condensation u Condensation is the opposite of vaporization. 4. Molar Heat of Condensation (  H cond. ) = amount of heat released when one mole of vapor condenses to a liquid q = mol x  H cond. (no temperature change) u  H vap. = -  H cond.

56. 56 Heats of Vaporization and Condensation u The large values for water  H vap. and  H cond. is the reason hot vapors such as steam are very dangerous! You can receive a scalding burn from steam when the heat of condensation is released! H 2 0 (g)  H 2 0 (l)  H cond. = - 40.7kJ/mol

57. 57 Heat of Solution NaOH (s)  Na + (aq) + OH - (aq)  H soln. = - 445.1 kJ/mol u The heat is released as the ions separate (by dissolving) and interact with water, releasing 445.1 kJ of heat as  H soln. thus becoming so hot it steams! H 2 O (l)

59. 59 Calculating Heats of Reaction u OBJECTIVES: State Hess’s Law of Heat Summation, and describe how (or why) it is used in chemistry.

60. 60 Calculating Heats of Reaction u OBJECTIVES: Solve for enthalpy changes by using Hess’s law or standard heats of formation.

61. 61 Hess’s Law (developed in 1840) u If you add two or more thermochemical equations to give a final equation, then you can also add the heats of reaction to give the final heat of reaction. Called Hess’s Law of Heat Summation Germain Henri Hess (1802-1850)

62. 62 How Does It Work? 1)If you turn an equation around, you change the sign: If H 2 (g) + 1/2 O 2 (g)  H 2 O(g)  H=-285.5 kJ then the reverse is: H 2 O(g)  H 2 (g) + 1/2 O 2 (g)  H =+285.5 kJ 2)If you multiply the equation by a number, you multiply the heat by that number: 2 H 2 O(g)  H 2 (g) + O 2 (g)  H =+571.0 kJ 3)Or, you can just leave the equation “as is”

63. 63 Hess’s Law - Procedure Options: 1. Use the equation as written 2. Reverse the equation (and change heat sign + to -, etc.) 3. Increase the coefficients in the equation (and increase heat by same amount)

64. 64 Standard Heats of Formation  The  H for a reaction that produces (or forms) 1 mol of a compound from its elements at standard conditions u Standard conditions: 25°C and 1 atm. u Symbol is: u The standard heat of formation of an element in its standard state is arbitrarily set at “ 0” u This includes the diatomic elements

65. 65 Standard Heats of Formation u The heat of a reaction can be calculated by: subtracting the sum of the heats of formation of the reactants from the products HoHo = (Products) -(Reactants) Remember, from balanced equation: Products - Reactants

66. 66 Another Example  CH 4 (g) + 2 O 2 (g)  CO 2 (g) + 2 H 2 O(g) CH 4 (g) = - 74.86 kJ/mol O 2 (g) = 0 kJ/mol CO 2 (g) = - 393.5 kJ/mol H 2 O(g) = - 241.8 kJ/mol  H= [-393.5 + 2(-241.8)] - [-74.86 +2 (0)]   H= - 802.24 kJ (endothermic or exothermic?) (Because it is an element )

67. Entropy Definition: The measure of the disorder of a system, usually denoted by the letter S. A highly ordered system has low entropy. Example: A block of ice will increase in entropy as it melts  Abbreviation for change in enthalpy is ∆S  http://chemistry.about.com/od/chemistryglossary/a/entropydef.htm

68. How does entropy relate to Free Energy and Thermochemistry?  This example problem demonstrates how to use changes in entropy of a system and its surroundings to determine a reaction's spontaneity and whether or not the reaction will be exothermic or endothermic. Problem:  Using the following values for entropy, evaluate three different scenarios to: a) determine if a reaction would be spontaneous. b) determine if the reaction is exothermic or endothermic with respect to the system I) ΔS sys = 30 J/K, ΔS surr = 50 J/K II) ΔS sys = 60 J/K, ΔS surr = -85 J/K III) ΔS sys = 140 J/K, ΔS surr = -85 J/K

69. Entropy problem continued:  Part 1 - Determine if a reaction will be spontaneous. A reaction will be spontaneous if the total entropy change is positive. It will not be spontaneous if the total entropy change is negative. ΔS total = ΔS sys + Δs surr where ΔS total is the total entropy change ΔS sys is the entropy change of the system ΔS surr is the entropy change of the surroundings System I  ΔS sys = 30 J/K, ΔS surr = 50 J/K  Δs total = 30 + 50 = 80 J/K =spontaneous

70. Entropy problem continued: Are these systems endothermic or exothermic? A reaction would be exothermic with respect to the system if it adds entropy to the surroundings and endothermic if it reduces entropy of the surroundings. This means ΔS surr > 0 = exothermic ΔS surr < 0 = endothermic I is exothermic, II and III are endothermic http://chemistry.about.com/od/workedchemistryproblems/a/Entropy-And-Reaction-Spontaneity-Example-Problem.htm System II Δ S total = ΔS sys + ΔS surr ΔS total = 60 J/K + -85 J/K = -25 J/K = not spontaneous System III ΔS total = ΔS sys + ΔS surr ΔS total = 140 J/K + -85 J/K = 55 J/K = spontaneous

71. Gibbs Free Energy (∆G)  Free energy change is the net driving force of a chemical reaction—whether the reaction will be spontaneous or not  If ΔG < 0, the reaction is spontaneous  If ΔG > 0, the reaction is nonspontaneous  If ΔG = 0, the reaction is at equilibrium Generally you will find that most exothermic reactions are spontaneous, even if entropy decreases (becomes more ordered), because enthalpy contributes more to ΔG than does entropy. The exceptions are reactions occurring at high temperatures.  Gibbs Free Energy Equation :  ΔG  = ΔH  - TΔS  OR ΔG = ΣΔG ° products - ΣΔG ° reactants  http://www.saskschools.ca/~chem30_dev/1_energy/energy3_3.htm