1. 1 Unit 2: Energy Icons are used to prioritize notes in this section. Make some notes: There are SOME important items on this page that should be copied into your notebook or highlighted in your printed notes. Copy as we go: There are sample problems on this page. I EXPECT YOU to copy the solutions into a notebook— Even if you have downloaded or printed the notes!!! Look at This: There are diagrams or charts on this page. Look at them and make sure you understand them. (but you don’t need to copy them) Extra Information: This page contains background information that you should read, but you don’t need to copy it. R Review: There is review material on this page. It is up to you to decide if you want to make notes or highlight it, depending on how well you remember it.

2. 2 Chapter 3 Energy Transfer

3. 3 Heat, Work & Temperature Overview: Heat, work and temperature are frequently confused by students. The terms are closely related to each other, but they are not the same. Heat and work are a forms of energy transfer, temperature is a measure of molecular agitation. Heat and temperature both involve random movement of particles, while work is an orderly movement. Heat and work measurements depend on the mass of a material, while temperature does not. 3.1

4. 4 Energy is sometimes defined as the ability to do work or to produce heat. This means that energy, work and heat can all be measured with the same unit: the joule (J) Work is a transfer of kinetic energy that results in an orderly movement of particles, such as the motion of an object, or the expansion of a balloon. Work and

5. 5 Heat is a transfer of thermal energy that occurs when two systems of different temperatures come into contact. Heat is transferred as a result of molecular agitation, that is random motion of particles. Click on the boxes below to link to more on heatheat

6. 6 Heat Transfer Hot ObjectCold Object Equal Temperature Thermal Energy When two systems of different temperatures are placed in contact… Thermal energy (or heat) flows from the hotter one into the colder one… Until their temperature is the same. This concept will be developed fully in a later lesson. PREVIEW

7. 7 Temperature You remember this from the Gas Laws. We often round this to 3 significant digits: -273°C

8. 8 Assignments Exercises page 130 # 1 to 4

9. 9 Law of Conservation of Energy 3.2 *It’s a bit more complicated with nuclear changes, since matter can change into energy and vice-versa at the nuclear level, as Einstein calculated E=mc 2

10. 10 Thermodynamic Systems Nothing is lost, Nothing is created, Everything is transformed. Antoine Lavoisier Promoter of the Laws of Conservation

11. 11 Three Types of System An Open System Both matter and energy can easily enter or leave the system, for example, a beaker. A Closed System Energy can enter or leave the system, but matter cannot, for example, a sealed balloon. An Isolated System Neither energy nor matter can easily enter or leave the system. For example, an insulated container or calorimeter.

12. 12 Imperfection of Isolated Systems In the real world it is impossible to create a perfectly isolated system. Even the best Dewar flask or Thermos™ bottle will eventually let some heat in or out. However, if experiments are conducted quickly, an insulated container is close enough to an isolated system to give us acceptable results. A Dewar Flask

13. 13 Calorimeters Simplified calorimeter Typical calorimeter

14. 14 Homemade Calorimeters For quick experiments a simple, but effective calorimeter can be made out of Styrofoam cups and a good thermometer. The chemicals are mixed in the cup, and the Styrofoam insulates long enough to measure temperature differences. Measurements must be made quickly with this type of calorimeter, before heat loss can occur. When using a homemade calorimeter promptness is important. Every minute you waste will reduce the accuracy of your measurements. Make sure you gather your data quickly but carefully!

15. 15 Energy Relationships 3.3

16. 16 Specific Heat Capacity (c) Specific heat capacity is the amount of energy required to raise temperature of one gram of a substance by 1°C. Specific heat is a characteristic property, it is different for each substance, if measured at a standard temperature. SubstanceSpecific Heat Capacity Liquid Water4.184 J/(g∙°C) Water vapour1.41 J/(g∙°C) Ice (solid water)2.05 J/(g∙°C) Ethylene glycol2.20 J/(g∙°C) Aluminum0.90 J/(g∙°C) Copper0.39 J/(g∙°C) Glass0.84J/(g∙°C) Air (dry)1.02 J/(g∙°C)  Remember the specific heat capacity of water. Textbooks sometimes use 4.19 or 4.18 if the problem requires no more than 3 significant figures.

17. 17 Heat Gained or Lost by a Substance The heat gained or lost by a substance can be measured by the calorimeter formulas: Where:Q = thermal energy (heat), in joules m=mass of a substance, in grams c = specific heat capacity of a substance, in J/(g∙°C) ΔT = temperature change (T f -T i ), in °C Where:T f = the final temperature of the substance, in °C T i = the initial temperature of the substance, in °C

18. 18 Sample Question Calculate the energy that is absorbed by a 3.00 kg block of aluminum when its temperature changes from 17.1 °C to 35.5 °C. (From the internet, the specific heat capacity of aluminum is found to be 0.900 J/g°C) Data: m= 3.00kg = 3000g c =0.900 J/(g∙°C) T i =17.1 °C T f =35.5 °C To find: ΔT= Q= Step 1: Calculate ΔT: ΔT= T f – T i = 35.5 °C – 17.1 °C = 18.4 °C 18.4 °C Step 2: Calculate Q: Q= mcΔT = 3000 g 0.900 J/(g∙ °C) 18.4 °C = 49 680 J Answer: The aluminum block must absorb 49.7 kilojoules of thermal energy. (3 significant figures were used to match m and c precision) 49 680 J Always show equation Note: large amounts of energy should be converted to kilojoules and answers rounded to a sensible number of significant digits. = 49.68 kJ

19. 19 Read pages 134-136 Examine sample questions on page 136. Do questions 1 to 16 on page 137

20. 20 Calculating Energy Transfer Overview: Energy transfer occurs when energy moves from one body to another. When two systems at different temperatures come in contact, the thermal energy from the hotter system is transferred into the cooler system until both systems reach the same temperature. The amount of heat lost by the hotter system should exactly equal the amount of heat gained by the cooler system, unless heat is lost to the surroundings. 3.4

21. 21 Heat Content of an Object The heat content of an object or system that is available for transfer is calculated by the same formula used in calorimetry, which was described in the previous lesson: Now we are going to combine two systems, so we will be calculating

22. 22 Heat Transfer Hot ObjectCold Object Equal Temperature Thermal Energy When two systems of different temperatures are placed in contact… Thermal energy (or heat) flows from the hotter one into the colder one… Until their temperature is the same. Where:Q 1 = Heat lost by the 1 st system Q 2 = Heat gained by the 2 nd system The heat lost by the warmer object will equal the heat gained by the cooler one.

23. 23 Heat Transfer Calculations  Combining  Gives us: Where:m 1 = mass of system 1, in grams c 1 = specific heat capacity of system 1, in J/(g.°C) ΔT 1 = temperature change in system 1 (T f -T i1 ), in °C m 2 = mass of system 1, in grams c 2 = specific heat capacity of system 1, in J/(g.°C) ΔT 1 = temperature change in system 2 (T f -T i2 ), in °C Special note:T f is the same for both systems, but T i is different

24. 24 Sample Question Calculate the mass of cold water at 10 °C that it would take to cool 10.0 g of hot (95°C) glass to 30°C. Data: m 1 = 10g c 1 =0.84 J/(g∙°C) T i1 =95 °C T f =30 °C c 2 = 4.184 J/(g∙°C) Ti 2 = 10 °C To find: ΔT 1 ΔT 2 m 2 Step 1: Calculate ΔT 1 : ΔT 1 = T f – T i1 = 30 °C – 95 °C = -65 °C = -65 °C Step 2: Calculate ΔT 2 : ΔT 2 = T f –T i2 = 30 °C – 10 °C = 20 °C = +20 °C Answer: It would take 6.5 g of water to cool the glass. = 6.5 g Formula: ̶ m 1 c 1 ΔT 1 =m 2 c 2 ΔT 2 = 6.5248565965 g  From Table Round to a sensible number of significant digits. Step 3: calculate the mass of water (m 2 )

25. 25 Finding the Final Temperature Quite often questions will ask for the final temperature of the two systems. This rearrangement of the previous formula is sometimes useful: Where: T i1 = initial temperature of system 1 T i2 = initial temperature of system 2 T f = final temperature of both systems Everything else means the same as before. The derivation of this formula is shown on page 139 of your textbook, so I won’t repeat it here. You can actually use formulas given before to solve this type of problem, but this form is easier to input if you are working with a TI83 calculator or the equivalent.

26. 26 Sample Question A 500 g package of frozen strawberries at -4.0 °C has a specific heat capacity of 3.50 J/(g.°C). It is placed in 2.00 kg of warm (40.0°C) water in an insulated cooler to thaw. What is the final temperature of the strawberries and water? Data: m 1 = 500 g c 1 = 3.5 J/(g∙°C) T i1 =-4.0 °C m 2 = 2 kg = 2000 g c 2 = 4.19 J/(g∙°C) T i2 = 40 °C To find: T f = 32.4 °C Answer: The final temperature is 32.4 degrees celsius. Round to a sensible number of significant digits. I did this in two steps to show the cancellation of units more clearly. You can do it in a single step if you have a scientific calculator, as shown on the following calculator display.

27. 27 Personal Observation Personally, I have never bothered to memorize the formula on the previous slide. I just use the three simple formulas earlier in the chapter: Q=mcΔT, –Q 1 =Q 2 and ΔT=T f -T i. Then I break my calculations into many small steps. It takes me a bit longer, but I don’t have to commit a really complex formula to memory, and I find I make fewer mistakes.

28. 28 Read Section 3.4, pp. 138-140 Do Questions 1 to 11 on page 141 Also: Chapter End Questions on page 145 to 146

29. 29 Chapter 4 Enthalpy Change

30. 30 Enthalpy (H) Overview: Enthalpy is defined as the total energy of a system. In theory, enthalpy is the sum of the kinetic and potential energy that a system contains at a given pressure. In practice, it is impossible to measure the total enthalpy of a system, since there are too many variables involved. Instead, we concern ourselves with the energy changes that occur in a system. This gives us a way of finding how much potential energy changes to kinetic, or vice versa, during a chemical change. 4.1 Page 148

31. 31 Enthalpy (H) Enthalpy is the total energy of a system It includes all the potential energy and all the kinetic energy of the system. In practice, it is impossible to find these values exactly, so we never talk about total enthalpy. Where: H = total enthalpy of a system E k = the kinetic energy of the system E p = the potential energy of the system

32. 32 Enthalpy Change (ΔH) AKA: Heat of Reaction Enthalpy Change is the energy exchanged between a system and its surroundings during a physical change or chemical change. In theory, the enthalpy change should equal the difference between the enthalpy of the products and the enthalpy of the reactants: Where:Δ H = the heat of reaction (enthalpy change) H p = enthalpy of the products H r = enthalpy of the reactants

33. 33 Enthalpy Change in Practice Although finding the actual total enthalpy (H) of a system is nearly impossible, finding the enthalpy change (∆H) is quite easy to do by experiment. Since nearly all the energy change of most reactions is in the form of heat (thermal energy), a calorimeter can find the amount of heat absorbed or released during a change. This will be your heat of reaction, from which we get the enthalpy change (∆H)

34. 34 Endothermic and Exothermic 4.2

35. 35 EndothermicExothermic Endothermic and Exothermic Definitions

36. 36 Endothermic and Exothermic Physical Changes Physical changes, like freezing, melting, boiling vaporization, condensation and dissolution can absorb or release energy. Endothermic ChangesEndothermic or ExothermicExothermic Changes Vaporization (evaporation)DissolutionCondensation (deposition) Vaporization (boiling)(depends on the solute)Condensation (liquid) Fusion (melting)Solidification (freezing) Sublimation Heat Absorbed Heat Released

37. 37 Liquid Solid Gas Melting (fusion) Freezing (solidification) Vaporization Liquid Liquid Condensation Sublimation Solid Condensation / Deposition Terminology associated with Change of Phase And the energy exchanges Rapid vaporization is called “boiling”, Slow vaporization is “evaporation” Sublimation occurs when a material “evaporates” from a solid straight to a gas, like dry ice or iodine. Exothermic Process Endothermic Process 37

38. 38 The Staircase Analogy another way to remember physical changes Gas Most energy Liquid More energy Solid Low energy Imagine the states of matter to be like a staircase. Solids are lowest in energy, so they are on the bottom step. Gases are highest, so they are the top step. To go from solid to liquid or gas means “climbing” the stairs. Using up energy to go up. These changes are endothermic. Going from gas to liquid, or from liquid to solid gives you back energy, as you bounce down the stairs. These changes are exothermic

39. 39 Heat Curve of a Pure Substance 150 140 130 120 110 100 90 80 70 60 50 40 30 20 10 0 -10 -20 -30 -40 -50 100 200 300 400 500 600 700 800 900 1000 1100 1200 1300 1400 1500 1600 Temperature (°C) Energy absorbed by substance (joules / gram) Boiling Point Melting Point solid liquid gas liquid & gas solid & liquid Heat of fusion Heat of vaporization ΔH fus ΔH (l) ΔT (l) ΔH vap If a pure, cold solid is slowly heated, its temperature will increase, until it starts to melt. Then, even though you are still heating it, the temperature stays the same until all the solid is melted. The energy absorbed to melt it is called the heat of fusion Specific heat capacity of the liquid can be determined from the inverse slope of the curve

40. 40 Reversing a Heat Curve 150 140 130 120 110 100 90 80 70 60 50 40 30 20 10 0 -10 -20 -30 -40 -50 100 200 300 400 500 600 700 800 900 1000 1100 1200 1300 1400 1500 1600 Temperature (°C) Energy (joules / gram) released solid liquid gas liquid & gas solid & liquid Heat of solidification Heat of condensation ΔHsΔHs ΔH cond Condensation point Freezing Point If instead of heating a cold, solid substance, we cool a hot gaseous substance, we get a similar heat curve, but reversed in appearance. Some of the terminology changes when we reverse a heat curve. Instead of a boiling point, we have a condensation point. Instead of heat of vaporization we have heat of condensation. Instead of a melting point we have a freezing point, and the heat change is called heat of solidification.

41. 41 Things You Must be Able to Find From the Heat Curve of a Pure Substance The melting/freezing point temperature From the vertical axis (1 st plateau level) The vaporization/condensation temperature From the vertical axis (2 nd plateau level) The heat of fusion (+) or solidification (–) From the horizontal axis (difference of 2 measurements) The heat of vaporization (+) or condensation (–) From the horizontal axis (difference of 2 measurements) The specific heat capacity of the solid or liquid substance. Inverse slope of a section of the curve.

42. 42 Heat Curve of a Mixture 150 140 130 120 110 100 90 80 70 60 50 40 30 20 10 0 -10 -20 -30 -40 -50 100 200 300 400 500 600 700 800 900 1000 1100 1200 1300 1400 1500 1600 Temperature (°C) Energy absorbed by substance (joules / gram) Fractional Distillation #3 Melting Range solid liquid gas Fractional Distillation #2 Fractional Distillation #1 slushy Boiling range We can use the different boiling temperatures of the mixtures components to separate them by capturing the vapours separately—a process called fractional distillation. The heat curve of a mixture of substances is less precise than that of a pure substance. The mixture “melts” over a range of temperatures, and each component “boils” at a different temperature. The mixture melts more gradually than pure matter.

43. 43 Sample Problem Heat curve of “Imaginarium” 150 140 130 120 110 100 90 80 70 60 50 40 30 20 10 0 -10 -20 -30 -40 -50 100 200 300 400 500 600 700 800 900 1000 1100 1200 1300 1400 1500 1600 Temperature (°C) Energy absorbed by substance (joules / gram) Heat of vaporization Use the heat curve to determine: a)The melting point of imaginarium. b)The boiling point of imaginarium c)The specific heat capacity of liquid imaginarium d)the heat of fusion of imaginarium e)the heat of vaporization of imaginarium ΔH (l) = (900-500) = 400J/g ΔT (l) = (70-(-20)) = 90℃ Heat of fusion ΔH fus = (500–300)= 200 J/g Answers: a)–20°C b)+70°C c)4.44 J/g°C d)200 J/g* e)400 J/g* ΔH vap =(1300-900)=400 J/g *Heat of fusion and vaporization are given in J/g. To convert to the more standard form ( j/mol) we could multiply this answer by the molar mass of imaginarium (if we knew it).

44. 44 Endothermic and Exothermic Chemical Reactions During a chemical reaction, energy is absorbed as the old bonds in the reactants are broken. This phase of the reaction is endothermic Energy is released as new bonds form in the products. This phase is exothermic. If more energy is absorbed than is released, the overall reaction is endothermic. If more energy is released than is absorbed, then the overall reaction is exothermic.

45. 45 Energy Released Energy Absorbed H H H Visualization of a Chemical Change O O H H H H Reactants 2H 2 + O 2 atoms 4H + 2O O O H H H H Products 2H 2 O O O H H O O H H H H O O H H H In the chemical reaction visualized here, hydrogen burns with oxygen to form water. This happens in two stages. ① The collisions of hydrogen (H 2 ) and oxygen (O 2 ) molecules break the molecules into atoms, absorbing a little energy. ② The oxygen and hydrogen atoms then form new bonds, making water molecules and releasing large amounts of energy. The overall reaction is exothermic, but the initial stage is endothermic. ① ② Bond breaking Bond forming

46. 46 In reality, the process of reaction is a bit more complicated. Molecules do not instantly jump apart into atoms… they tend to clump into intermediate complexes, as we shall see in the next chapter. However, the mathematics of energy is the same for the simple visualization shown on the last slide as it is for the actual reaction pathway, so we’ll use the simple version for now.

47. 47 Simple Enthalpy Diagrams Simple enthalpy diagrams show the difference between the enthalpy of the reactants and the products. In an endothermic reaction, the enthalpy increases (ΔH is positive) In an exothermic reaction, the enthalpy decreases (ΔH is negative)

48. 48 Assignments Read sections 4.1 and 4.2 Do questions 1 to 6 on page 155

49. 49 Energy Balance Overview: Energy Balance is the sum of the energy released when old bonds are broken (a positive number) and the energy absorbed as new bonds are formed (a negative number). The energy balance can tell us what the enthalpy change is during a reaction. Energy balance can also be calculated from bond energies, many of which have been found experimentally and recorded in charts. 4.3

50. 50 Bond Energies The text book contains tables of bond energies, such as the one on page 419, or the small one here (from page 156). Each bond energy can represent the energy required to break a bond, or the energy released if a bond is formed (per mole). The energy absorbed while breaking bonds is represented by a positive number, and the energy released by a negative number, so breaking the bond between two hydrogen atoms would require +436 kJ of energy per mole of atoms. If you formed new H—H bonds, you would release energy. (–436 kJ/mol) p. 156 Also see p. 419

51. 51 Error in Textbook Table on page 156 The textbook gives the value of 418 kJ/mol for the N  N triple bond. This is incorrect. That is the value for the N=N double bond. Please make a note in the margin of your textbook, showing N=N The table on page 419 is correct for this bond. N=N

52. 52 Energy Balance Formula Your textbook gives the following formula for energy balance: This is technically correct, but since the second number is always negative, isn’t it easier to treat this as a subtraction? Okay, purists won’t like that, since I’m not using the standards of enthalpy notation, but I think its easier to remember.

53. 53 Energy Balance Diagrams Energy balance diagrams are a graphical way of representing the calculations of energy balance. They show the energy absorbed breaking bonds, and the energy released forming new bonds. 1 2 3 4 5 6

54. 54 Energy Balance diagrams Energy balance diagrams are, by definition, diagrammatic. They do NOT need to be drawn to exact scale, although the arrows should approximately represent the energy proportions. Actual numbers are usually not shown on the axes. The y-axis is labelled Enthalpy, and represents total energy of the system, presumably in kilojoules or kilojoules/mole. The x-axis is labelled reaction Progress. It is in arbitrary time units, representing the beginning, mid-point and end of the reaction. This means that we don’t actually record how long the reaction takes, just that it proceeds from reactants to products.

55. 55 Energy Balance Example negative positive 4x 2x 4x 2x

56. 56 Draw an Energy Balance Diagram for the previous sample problem Energy balance diagram should show: Enthalpy on y-axis (kJ/mol, but usually not numbered) Reaction progress on x axis (in arbitrary time units) Formulas of reactants Energy from broken bonds (+ ΔH) Separate atoms Energy from bonds formed(-ΔH) Formulas of products The heat of reaction (ΔH) Enthalpy (kJ/mol) Reaction Progress CH 4 + 2 O 2 +ΔH broken = 2648 kJ C + 4 H + 4 O -ΔH formed = -3330 kJ CO 2 + 2 H 2 O ΔH= -682 kJ 1 2 3 4 5 6

57. 57 Thermochemical Equations When we add energy information to a chemical equation, it is called a thermo-chemical equation There are two main techniques for doing this: We can treat the energy like a reactant or product and put it directly into the equation: Or, we can record the ΔH value separately at the end of the equation:

58. 58 Disagreement with the Textbook’s method. Okay, here we come to a place where I disagree with the textbook’s method of writing thermo-chemical equations. The textbook wrote the thermo-chemical equation on the previous slide this way: ΔH=-824 kJ/mol Exothermic reaction:4 Fe (s) + 3 O 2(g)  2Fe 2 O 3(s) ΔH=-824 kJ/mol Not what I wrote: ΔH=-1648 kJ Exothermic reaction:4 Fe (s) + 3 O 2(g)  2Fe2O 3(s) ΔH=-1648 kJ IF of Fe 2 O 3 Now, there is nothing wrong with what they put in the textbook. They have divided the total enthalpy by the number of moles, but they should only do this IF they state which substance they are referring to.. Its ΔH=-824 kJ/mol of Fe 2 O 3 ! Without stating which substance, how are we to know they aren’t giving it with reference to the number of moles of iron (in which case it would be ΔH=-412 kJ/mol of Fe), or of oxygen (ΔH=-549.3 kJ/mol of O 2 ) kJ kJ/mol Overall, I prefer to write my thermo-chemical equations with the total enthalpy change (just kJ) rather than the molar enthalpy change (kJ/mol), because otherwise its just too confusing.

59. 59 Assignments Read section 4.3, pp. 156 to 160 Do Questions on page 160, #1 to 6

60. 60 Energy Change and Stoichiometry Overview: Calculating enthalpy change using stoichiometry allows us to find the energy change that accompanies a chemical reaction when we know the mass of the reactants or products. Although normally we write equations with whole number coefficients, sometimes during calculations with thermoequations it is more convenient to use fractional coefficients. 4.4

61. 61 Stoichiometry and Thermochemical Equations

62. 62 Sample Problem Carrying this idea one step further, can we calculate how much energy we would get from burning 20g of H 2 ? 2H 2(g) + O 2(g)  2 H 2 O (g) + 488 kJ Data: m H2 = 20g M H2 = 2g/mol n H2 = To Find: Energy= 10 mol 2440 kJ Answer: Burning 20g of hydrogen gas produces 2440 kJ of energy

63. 63 Chapter 5 Graphical Representation of Enthalpy

64. 64 Activated Complexes Overview The molecular collisions that cause chemical reactions happen very fast, but in those fractions of a second that the particles are colliding many things happen. As reactant particles strike, some of their kinetic energy is momentarily converted to potential energy. The particles temporarily form an unstable cluster or “complex”. Then the complex springs apart into product molecules as some of the potential energy turns back into kinetic energy. 5.1

65. 65 Activated Complex Formation and Breakup When two reactant molecules collide They can form an unstable cluster, called an activated complex. The activated complex will then break apart into the products. Activated Complex H Cl H

66. 66 Activation Energy Reactants Activation Energy Reaction

67. 67 Energy Diagrams 5.2 Overview An energy diagram is a visualization of the energy relationship between the reactants, the activated complex and the products. An energy diagram is a graph of Potential Energy An energy diagram resembles an energy balance diagram, but there are some important differences, as we shall see on the next slide.

68. 68 Energy Diagrams vs. Energy Balance Diagrams

69. 69 Energy Diagram for an endothermic reaction Progress of the Reaction Potential Energy Activation energy Heat of Reaction EaEa ∆H∆H Products Activated Complex Reactants

70. 70 Energy Diagram for an exothermic reaction Progress of the Reaction Potential Energy Products Activated Complex Reactants EaEa Heat of Reaction ∆H∆H

71. 71 The “Over the Hill” Analogy Getting a reaction started is a bit like pushing a boulder over a hill. It takes some energy (E a ) to get it to the top. ie. start the reaction Once it’s at the top it rolls down the other side easily. ie. the reaction gets going Diagram from http://www2.ucdsb.on.ca/tiss/stretton/CHEM2/rate03.htm Sisyphus

72. 72 Slow, Faster and Spontaneous Reactions Progress of the Reaction Potential Energy Actually reactions with zero activation energy don’t exist, but some have activation energies that are so low that ambient temperature can cause them to begin.

73. 73 Direct and Reverse Reactions Some chemical reactions are reversible. For example: Hydrogen & oxygen burn to form water 2 H 2 + O 2  2 H 2 O Water can be decomposed by electrolysis 2 H 2 O  2 H 2 + O 2 These two reactions are opposites of each other. The first is exothermic (it gives off energy as heat and flames) its opposite is endothermic (it absorbs energy, in the form of electricity)

74. 74 Energy Diagram for a reversible reaction Progress of the Reaction Potential Energy Products of the direct reaction Activated Complex Reactants of the direct reaction E a(direct) –∆ H (direct) E a(reverse) +∆ H (reverse) Direct Reaction (Exothermic in this example) Reverse Reaction (Endothermic in this example)

75. 75 Read Chapter 5, pp. 171 - 181 Do the Chapter end questions on pages 183- 184

76. 76 Chapter 6 Molar Heat of Reaction

77. 77 Molar Heat of Reaction and Heat of Dissolution Overview: The molar heat of a reaction is an enthalpy change involving one mole of a substance. When a solute dissolves in a solvent, its particles disperse themselves between the particles of solvent. Water in particular is a good solvent because it is a polar molecule, having a slight residual electric charge at each end. During dissolution, heat may be released or absorbed. The amount of this heat is called the heat of dissolution. 6.1

78. 78 Molar Heat of Reaction (ΔH) Molar heat of reaction is the enthalpy change involved in the transformation of one mole of a substance. In most cases, the symbol ΔH is used for the molar enthalpy, and Q for the heat measured in a calorimeter. Since the calorimeter formula usually gives Q in joules, and ΔH is usually measured in kilojoules, you may have to convert the joules into kilojoules first. Where:ΔH = molar heat of reaction, in kilojoules Q = total heat exchanged, converted to kilojoules n = number of moles of substance transformed. Q ΔH n

79. 79 Dissolution

80. 80 Dissolution in Detail A substance dissolves in two steps. 1.The solute particles are separated from each other. For ionic compounds, they may actually dissociate. This phase is endothermic. Energy is absorbed from the solvent to break up solute molecules. 2.The solute particles disperse and rearrange themselves in the solvent. This step is exothermic. H2OH2O H2OH2O H2OH2O H2OH2O H2OH2O H2OH2O H2OH2O H2OH2O S S S S

81. 81 Molar Heat of Dissolution ( ΔH d ) (a specific example of heat of reaction)

82. 82 Dissociation of Ions When ionic compounds dissolve, many of them will also dissociate into individual ions: Examples: NaCl (s) + H 2 O  Na + (aq) + Cl – (aq) + H 2 OΔH d = +4.3 kJ / mol Since the water doesn’t change, we can leave it out of the equation. CaBr 2(s)  Ca 2+ (aq) + 2Br - (aq) ΔH d = -81 kJ / mol H 2 SO 4(s)  2H + (aq) + SO 4 2- (aq) ΔH d = -74 “ NaNO 3(l)  Na + (aq) + NO 3 – (aq) ΔH d = +21 “ exothermic endothermic

83. 83 Calculating Heat of Dissolution typical steps and useful formulas Dissolve a certain number of moles of solute in sufficient solvent in a calorimeter (if solute is in grams, change it to moles) Measure the temperature change Use the calorimeter formula to find the total heat exchange (Q in joules ). Now m w = the mass of the solvent (ie water) in the calorimeter. Convert joules to kilojoules, Divide the heat (Q in kilojoules ) by the number of moles of solute.

84. 84 Sample Problem (easy) 112.2 grams of potassium hydroxide, KOH (s) are dissolved in 500mL (5.00x10 2 g) of water in an insulated cup. The temperature of the solution rises from 12.1°C to 64.7°C. Calculate the molar heat of dissolution of potassium hydroxide. Data m KOH = 112.2 g M KOH = 56.1 g/mol m water = 500 g c water = 4.184 J/g ° C T f = 64.7 ° C T i = 12.1 ° C ΔT= 52.6 ° C To Find: n KOH Q ΔH Preliminary Step: M KOH = 39.1 + 16.0 + 1.0 = 56.1 g/mol = 110.0 kJ = 55.0 kJ/mol = 2.00 mol Step 2: Calculate Q = mc ΔT note: use m water = 500 g x 4.184 J/g ° C x52.6 ° C = 110 039 joules Step 3: Convert to kilojoules: 110039 J ≈ 110.0 kJ Answer: The molar heat of dissolution of KOH is 55.0 kJ/mol. Step 2: Find ΔT = T f – T i = 64.7–12.1 = 52.6 ° C

85. 85 Sample Problem (tough) In an insulated cup, 4.25 g of sodium nitrate (NaNO 3 ) were dissolved in 100 mL of water at 23.7C. The molar heat of dissolution of sodium nitrate is known to be 21.0 kJ/mol. What will the final temperature of the water be? Data: m water = 100g  100mL@1g/mL c water = 4.184 J/(g∙°C) T i = 23.7°C ΔH d = +21.0 kJ/mol m NaNO 3 = 4.25g To find: M NaNO 3 n Q T f Step 1: Calculate the number of moles of NaNO 3 = 85.0 g/mol Step 2: Calculate the quantity of heat (Q) Answer: The final Temperature is 21.2°C = 0.0500 mol = –1.05 kJ = –1050 J But, since ΔH d is positive, the reaction is endothermic, which means the water loses heat. Make Q negative! =21.2°C

86. 86 Page 195, Questions #1 to 10

87. 87 Heat of Neutralization Overview: When an acid and base are mixed, some of the H + ions from the acid will combine with some OH - ions from the base to produce H 2 O, water. This process will be discussed in more detail in future chapters. Heat is often released during this neutralization, and can be measured by calorimetry. 6.2

88. 88 Neutralization A neutralization reaction occurs when an acidic solution and a basic solution are mixed. The general reaction is: A more specific example: Nitric Acid Potassium hydroxide Water Potassium nitrate

89. 89 Molar Heat of Neutralization (ΔH n ) (another specific example of heat of reaction) The molar heat of neutralization is the quantity of energy that is absorbed or released in the neutralization of one mole of an acid or base.

90. 90 Calorimeters and Neutralization One way to find the molar heat of neutralization is to mix the two solutions inside a calorimeter – or a Styrofoam cup calorimeter. If the two solutions (acid and base) start out at the same temperature, any increase in temperature must come from the heat of neutralization.

91. 91 Simplifying the Problem Simplification #1: For dilute solutions, assume the solution has the same properties as water Density = 1 g / mLSpecific heat = 4.184 J/g°C While this is not strictly true, it is close enough to give us a good answer, so long as the amount of water in each solution is much greater than the mass of solute. While this is not strictly true, it is close enough to give us a good answer, so long as the amount of water in each solution is much greater than the mass of solute. Simplification #2: Assume that the mass of water in the calorimeter is the sum of the two individual solutions: Mass in calorimeter= mass of acid plus mass of base m w = m A + m B

92. 92 Molar Concentration Formula Another useful formula that we learned long ago is sometimes needed in molar problems: This may also be rearranged: C= molar concentration n s = moles of solute V=volume (in litres) n C V R

93. 93 Sample Problem In a calorimeter 100 mL (1.00x10 2 ) of a 0.500 mol/L solution of NaOH is neutralized by an equal amount of of 0.500 mol/L solution of HCl, both at 22.5°C. The temperature after they are mixed is 25.9°C. Calculate the molar heat of neutralization of Sodium Hydroxide. Data: m w = 100g + 100g = 200g c w = 4.184 J/(g∙°C) T f = 25.9°C T i = 22.5°C ΔT = 25.9-22.5 = 3.4 °C C NaOH = 0.5 mol/L V NaOH = 0.100 L To find: Q n ΔH n Step 1: Calculate Q, the quantity of heat transferred Q=m w c w ΔT = 200 g ∙ 4.184 J/(g∙°C) ∙ 3.4°C = 2845.1 J or 2.8451 kJ = -2.8451 kJ Reaction is exothermic, so Q is negative. Q= –2.8451 kJ Step 2: Find the number of moles of NaOH. n NaOH = C NaOH ∙ V NaOH = 0.5 mol/L ∙ 0.100 L = 0.05 mol Answer: the molar heat of neutralization of NaOH is -56.9 kJ/mol = 0.05 mol = -56.9 kJ/mol n C V

94. 94 Trickier Problems What do you do if the temperature of the two initial solutions is not the same? Find the combined inital temperature of the two solutions, using the formula: and then assume they both have this combined initial temperatre

95. 95 Trickier Problems What if you are asked to find the final temperature instead of the ΔH? Use the ΔH value given (or looked up in a table) to find Q, then solve for ΔT and use that to find the final temperature.

96. 96 Other Heats of Reaction Any type of chemical reaction or physical change that absorbs or releases energy can have a heat of reaction (ΔH). Some common examples are… NameSymbolExplanation Heat of vaporizationΔH vap, or ΔH v Heat absorbed when a material evaporates Heat of fusionΔH fus Heat absorbed when a material melts Heat of solidificationΔH sol Heat released by freezing (-ΔH fus ) Heat of condensationΔH cond Head released by condensation(-ΔH vap ) Heat of sublimationΔH sub Heat absorbed by sublimation of Heat of combustionΔH comb Heat released by burning of a fuel Heat of dissolutionΔHdΔHd Heat aborbed/released during dissolving Standard Heat of Formation Special heat of reaction, representing the enthalpy change in creating one mole of a substance from its elements.

97. 97 Read Chapter 6 Page 196, questions 11 to 14

98. 98 Chapter 7 Hess’s Law

99. 99 Reaction Mechanisms Overview: Sometimes a reaction occurs in a series of steps, rather than all in one instant. A reaction mechanism is a series of simple reactions that combine into a more complex reaction. Germain Henri Hess discovered that the total enthalpy change of a complex reaction is equal to the sum of the enthalpy changes of the simple steps that make up the more complex mechanism. 7.1

100. 100 Reaction Mechanism

101. 101 Mechanism Energy Graphs When a reaction has a multi- step mechanism, its energy graph will show an activation energy for each step. Because the activation energy of several small steps is usually lower than a single activation energy would be, it is easier for a reaction to occur by means of a mechanism.

102. 102 Comparison of Overall Reaction and Reaction mechanism Overall Reaction 2NO + 2 H 2  N 2 + 2H 2 O Reaction Mechanism 2NO + 2 H 2  N 2 O 2 + 2H 2  N 2 O + H 2 O + 2H 2 O  N 2 + 2H 2 O Photsynthesis,An extreme example 6CO 2 + 6H 2 O  C 6 H 12 O 6 + 6O 2 Overall Reaction Mechanism  2x 2x2x 1 2 3

103. 103 Photosyntheis: An extreme example of reaction mechanisms Overall Reaction: 6CO 2 + 6H 2 O  C 6 H 12 O 6 + 6O 2 Mechanism:

104. 104 Energy Graph for a Complex Reaction  Step 1  Step 2  Step 3  ΔH1ΔH1 ΔH2ΔH2 ΔH3ΔH3 ΔH Total

105. 105 Summation of Enthalpies Overview: According to Hess’ Law, also called the law of heat summation, if a reaction can be broken down into several simple reactions, its enthalpy change is equal to the sum of the enthalpy changes of each of the simple reactions. 7.2

106. 106 Hess’ Law The total enthalpy change of a reaction is the sum of the individual enthalpy changes that make up its mechanism. Where: ΔH = Total enthalpy change of the reaction ΔH 1 = Enthalpy change of 1 st simple reaction ΔH 2 = Enthalpy change of 2 nd simple reaction etc… Germain Henri Hess 1802-1850 Born: August 8, 1802 Geneva, Switzerland Died: November 30, 1850 St. Petersburg, Russia Fields: Chemistry, Medicine Known for: Law of constant heat summation (Hess’ Law) Special Note: When working with Hess’ Law you are sometimes allowed to break the rule about having only whole number coefficients in chemical equations. Using fraction coefficients may be necessary!

107. 107 The most common use of Hess’ Law is to calculate the enthalpy change of a reaction using data from a table of standard heats of formation (AKA. Standard Molar Enthalpy of Formation) These tables list the enthalpy associated with the formation of one mole of a substance from its elements, and these values have been worked out for many substances. Your textbook has a list of these on page 418, unfortunately it does not show the formation equations, which renders it awkward to use.

108. 108 Sample Table of Standard Heats of Formation Substance NameFormulaFormation Equation (balanced for 1 mole) ΔH° f (kJ/mol) WaterH 2 O (l) H 2(g) + ½ O 2(g)  H 2 O (l) -285.8 kJ/mol Water vapourH 2 O (g) H 2(g) + ½ O 2(g)  H 2 O (g) -242.8 kJ/mol Carbon dioxideCO 2(g) C (s) + O 2(g)  CO 2(g) -393.5 kJ/mol Carbon monoxideCO (g) C (s) + ½ O 2(g)  CO (g) -110.5 kJ/mol ButaneC 4 H 10(g) 4C (s) + 5 H 2(g)  C 4 H 10(g) -125.6 kJ/mol PropaneC 3 H 8(g) 3C (s) + 4 H 2(g)  C 3 H 8(g) -104.7 kJ/mol MethaneCH 4(g) C (s) + 2 H 2(g)  CH 4(g) -74.4 kJ/mol OzoneO 3(g) 3 / 2 O 2(g)  O 3(g) +142.7 kJ/mol Nitrogen dioxideNO 2(g) ½ N 2(g) + O 2(g)  NO 2(g) +33.2 kJ/mol Oxygen gas*O 2(g) It’s the common element form at 25 ℃ 0 kJ/mol *The ΔH° f value for the most common form of any element, in its most common state at 25°C is always zero. Therefore, all diatomic gases ( H 2, N 2, O 2, Cl 2, F 2 ) have a heat of formation of zero. Same with the diatomic liquid ( Br 2 ) and solid ( I 2 ).

109. 109 Working Out Standard Formation Equations Since the table on page 418 only gives the Standard Enthalpy of Formation values, and not the entire formation equation, we may have to work out the formation equations ourselves. Method: Write out a balanced equation showing the formation of the required compound from its components. If the balanced equation produces more than one mole of product, divide the whole equation to reduce the product to one mole. Add the given ΔH° f value, from the table on p. 418, to the end of the equation.

110. 110 Sample Problem Find the standard enthalpy of formation equation for barium oxide: Formula: BaO (s) Elements: Ba (s) O 2(g) Step 1: Find the balanced equation. Ba (s) + O 2(g)  BaO ΔH= ? 22 Step 2: Divide all coefficients by 2 to get a single mole of BaO Ba (s) + ½ O 2(g)  BaO ΔH= ? Step 3: Add ΔH° f info from the table on p. 418. Ba (s) + ½ O 2(g)  BaO ΔH= –944.7 kJ/mol The equation is: Ba (s) + ½ O 2(g)  BaO ΔH= –944.7 kJ/mol

111. 111 Hess’s Law If two or more thermo-chemical equations are added together to give a final equation, then the enthalpy changes can be added to give the enthalpy change for the final equation. ΔH T = ΔH 1 + ΔH 2 +...

112. 112 Adding Equations ReactantsProducts Enthalpy ( Δ H) N2 +O2N2 +O2N2 +O2N2 +O2  2 NO +180.6kJ 2 NO+O 2  2 NO 2 - 122.2 kJ N 2 +2NO+2O 2  2 NO + 2NO 2 + 66.4 kJ N 2 + 2O 2  2 NO 2 + 66.4 kJ

113. 113 In the example, we cancelled the like substances at the end of the problem. You may also cancel them earlier, as long as you are careful to only cancel a substance in the reactant column with an identical substance in the product column. You may multiply an entire equation, including the ΔH value by a simple number to make the coefficients match. You can only cancel if coefficients they match exactly! In the final step, you may subtract the same amount of a reactant from both sides (like solving an algebra equation) You may switch the substances in the reactant column with the substances in the product column, but you then have to change the sign of the Δ H value.

114. 114 Example 2: ( see study guide page 47 question 2) Find the heat of combustion of methanol (CH 3 OH) and write its thermo-chemical equation using the following heat of formation data from p.418: CO 2(g) ΔH° f = - 394 kJ/mol H 2 O (l) ΔH° f = - 286 kJ/mol CH 3 OH (l) ΔH° f = - 239 kJ/mol Note: sometimes the “phase markers” ie. (s) (l) (g) (aq) are left out, but be aware that they are important, especially if a material changes state before or after it reacts. Note also, I have rounded all ΔH values to 3 sig.figs. for simplicity. Remember, combustion is a form of rapid oxidation!

115. 115 The Reactions we know: C (s) + O 2(g)  CO 2(g) ΔH= -394 kJ H 2(g) + ½ O 2(g)  H 2 O (l) + ΔH= -286 kJ C (s) + 2H 2(g) + ½ O 2(g)  CH 3 OH (l) ΔH= -239 kJ The reaction we are trying to get will contain the following: ___CH 3 OH (l) + ___O 2(g)  ___CO 2 + ___H 2 O We can balance it to find the coefficients, but I’m not going to do that yet, since all I need to know for now is what side of the arrow each substance is on.

116. 116 Solution, as you should show it in your notebooks. flip x2 ReactantsProductsΔHΔH C (s) + O 2(g)  CO 2(g) -394 kJ H 2(g) + ½ O 2(g)  H 2 O (l) -286 kJ C (s) + 2H 2(g) + ½ O 2(g)  CH 3 OH (l) -239 kJ 2 2H 2(g) + O 2(g) 2 H 2 O (l) C (s) + 2H 2(g) + ½ O 2(g) CH 3 OH (l) -572 kJ CH 3 OH (l) + __O 2  __CO 2 + __H 2 O CH 3 OH (l) + 2 O 2(g)  CO 2 + 2 H 2 O + ½O 2 1½ +239 kJ Target: -727 kJ ΔH=-727 kJ Leave Blank Lines

117. 117 Hess’ Law Assignment Sheet do the sheet and check your answers before starting the textbook questions. Textbook Page 208, Questions #1 to 6 plus 9 and 11 Question 1 has the additional complication of not telling you what the “target” equation is. You must find it as you add the equations together.

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