Chapter 17


Published on

1 Like
  • Be the first to comment

No Downloads
Total views
On SlideShare
From Embeds
Number of Embeds
Embeds 0
No embeds

No notes for slide

Chapter 17

  1. 1. Chapter 17The Flow of Energy
  2. 2. Energy TransformationsEnergy is the capacity for doing work or supplying heat.Unlike matter, energy has neither mass nor volumeEnergy is detected only because of its effects – ex: the motion of a race car.Thermochemistry is the study of energy changes that occur during chemical reactions and changes in state.Every substance has a certain amount of energy stored inside it. The energy stored in the chemical bonds of a substance is called chemical potential energy.
  3. 3. Energy TransformationsThe kinds of atoms and their arrangement in the substance determine the amount of energy stored in the substance.During a chemical reaction, a substance is transformed into another substance with a different amount of chemical potential energy.When you buy gasoline, you are actually buying the stored potential energy it contains.The controlled explosions of the gasoline in a car’s engine transform the potential energy into useful work, which can be used to propel the car.
  4. 4. Energy TransformationsAt the same time, heat is also produced, making the car’s engine extremely hot.Energy changes occur as either heat transfer or work, or a combination of both.Heat (q) is energy that transfer from one object to another because of a temperature difference between them.Heat always flows from a warmer object to a cooler object.If two objects remain in contact, heat will flow from the warmer object to the cooler object until the temperature
  5. 5. The UniverseChemical reactions and changes in physical state generally involve either the release or the absorption of heat.System is the part of the universe on which you focus your attention.Surroundings include everything else in the universe.Together the system and its surroundings make up the universe.Thermochemistry examines the flow of heat between the system and its surroundings.
  6. 6. Law of Conservation of EnergyThe law of conservation of energy states that in any chemical or physical process, energy is neither created nor destroyed.If the energy of the system decreases during a process, the energy of the surroundings must increase by the same amount so that the total energy of the universe remains unchanged.In thermochemical calculations, the direction of the heat flow is given from the point of view of the system.
  7. 7. Endothermic ProcessAn endothermic process is one that absorbs heat from the surroundings.In an endothermic process, the system gains heat as the surroundings cool down.Heat flowing into asystem from itssurrounding isdefined as positive(+ q value)
  8. 8. Exothermic ProcessAn exothermic process is one that releases heat into its surroundings.In an exothermic process, the system loses heat as the surroundings heat up.Heat flowing out of a systeminto its surroundings isdefined as negative (- q value)
  9. 9. Units for Measuring Heat FlowHeat flow is measured in two common units, the calorie and the joule.A calorie (cal) is the quantity of heat needed to raise the temperature of 1 g of pure water 1ºC.The word calorie is written with a small c except when referring to the energy contained in food.The dietary Calorie, written with a capital C, always Calorie C refers to the energy in food.1 Calorie (dietary) = 1 kilocalorie = 1000 calories (heat flow)
  10. 10. Units for Measuring Heat FlowThe statement “10 g of sugar has 41 Calories” means that 10 g of sugar releases 41 kilocalories of heat when completely burned.The joule (J) is the SI unit of energy. One joule of heat raises the temperature of 1 g of pure water 0.2390ºC. 1 J = 0.2390 cal 4.164 J = 1 cal
  11. 11. Conceptual ProblemA container of melted paraffin wax is allowed to stand at room temperature until the wax solidifies. What is the direction of heat flow as the liquid wax solidifies. Is the process exothermic or endothermic?Heat flows from the system (paraffin wax) to the surroundings (air). The process is exothermic.When solid barium hydroxide octahydrate is mixed in a beaker with solid ammonium thiocynanate, a reaction occurs. The beaker quickly becomes very cold. Is the reaction exothermic or endothermic?Since the beaker becomes cold, heat is absorbed by the system (chemical within the beaker) from the surroundings (the beaker and surrounding air). The process is endothermic.
  12. 12. Heat CapacityThe amount of heat needed to increase the temperature of an object exactly 1ºC is the heat capacity of that object.The heat capacity of an object depends on both its mass and its chemical composition.The greater the mass of the object, the greater its heat capacity.
  13. 13. Heat CapacityDifferent substances with the same mass may have different heat capacities.On a sunny day, a 20kg puddle of water may be cool, while a nearby 20 kg iron sewer cover may be too hot to touch.
  14. 14. Specific HeatAssuming that both the water and the sewer cover absorb the same amount of radiant energy from the sun, the temperature of the water changes less than the temperature of the cover because the specific heat capacity of water is larger.The specific heat (C) of a substance is the amount of heat it takes to raise the temperature of 1 g of the substance 1ºC.Water has a very high specific heat (it takes more energy to raise the temperature)Metals have low specific heats (it takes less energy to raise the temperature)
  15. 15. Specific HeatHeat affects the temperature of objects with a high specific heat much less than the temperature of those with a low specific heat.It takes a lot of heat to raise the temperature of water, water also releases a lot of heat as it cools.Water in lakes and oceans absorbs heat from the air on hot days and releases it back into the air on cool days.This property of water isresponsible for moderateclimates in coastal areas.
  16. 16. High Specific Heat of WaterTwo other common effects associated with the high specific heat of water.1.Spraying oranges with water to protect the fruit from frost damage during icy weather.  As the water freezes, it releases heat, which helps prevent the fruit from freezing.2.The filling of an apple pie stays hotter than the crust.  The filling that is mostly water, has a higher specific heat than the crust. In order to cool down, the filling must give off a lot of heat.
  17. 17. Calculating Specific Heat C= q = heat (joules or calories) m x ∆T = mass (g) x change in temperature (ºC)∆T = Tfinal - TinitialSpecific heat may be expressed in terms of joules or calories.Therefore, the units of specific heat are either J / g ⋅ ºC or cal / g ⋅ ºCWhat factors do you think affect the specific heat of a substance?Amount of heat and the change in temperature.
  18. 18. QuestionsWhen 435 J of heat is added to 3.4g of olive oil at 21ºC, the temperature increases to 85ºC. What is the specific heat of the olive oil?C= q = 435 J = 2.0 J / g ⋅ ºC m x ∆T (3.4g) (64ºC)How much heat is required to raise the temperature of 250.0g of mercury 52ºC?C x m x ∆T = q (0.14J/g⋅ºC)(250.0g)(52ºC) = 1.8kJ
  19. 19. QuestionsIn what direction does heat flow between two objects?From the object of higher temperature to the object of lower temperature.How do endothermic processes differ from exothermic processes?Endothermic absorbs heat from surroundings; exothermic releases heat to the surroundings.What units are used to measure heat flow?Calories and joules
  20. 20. QuestionsOn what factors does the heat capacity of an object depend?Mass and chemical compositionUsing calories, calculate how much heat 32.0g of water absorbs when it is heated from 25.0ºC to 80.0ºC. How many joules is this?q=C x m x ∆T (1.00cal/g⋅ºC)(55ºC)(32.0g) = 1.76kcal4.164 J = 1 cal (1760cal)(4.164J/cal) = 7.33 kJ
  21. 21. QuestionsA chunk of silver has a heat capacity of 42.8 J/g ⋅ºC and a mass of 181g. Calculate the specific heat of silver.C = 42.8 J/ºC = 2.36 x 10-1 J/g⋅ºC 181gHow many kilojoules of heat are absorbed when 1.00L (1L = 1kg) of water is heated from 18ºC to 85ºC?q=C x m x ∆T (4.18J/g⋅ºC)(67ºC)(1kg) = 2.8 x 102 kJ
  22. 22. End of Section 17.1
  23. 23. CalorimetryHeat that is released or absorbed during many chemical reactions can be measured by a technique called calorimetry.Calorimetry is the precise measurement of the heat flow into or out of a system for chemical and physical processes.In calorimetry, the heat released by the system is equal to the heat absorbed by its surroundingsConversely, the heat absorbed by a system is equal to the heat released by its surroundings.The insulated device used to measure the absorption or release of heat is called a calorimeter.
  24. 24. EnthalpyHeat flows for many chemical reactions can be measured in a constant pressure calorimeter .Because most chemical reactions and physical changes carried out in the laboratory are open to the atmosphere, these changes occur at constant pressure.The heat content of a system at constant pressure is the same as a property called enthalpy (H) of the system.The heat released or absorbed by a reaction at constant pressure is the same as the change in enthalpy (∆H)The terms heat and enthalpy change are used interchangeably when reaction occur under constant pressure. ( q = ∆H)
  25. 25. Measuring Enthalpy (heat absorbed by surroundings) qsurr = ∆H = m x C x ∆TBecause the heat absorbed by the surroundings is equal to (but has the opposite sign of) the heat released by the system, the enthalpy change (∆H) for the reaction can be written as follows. (heat released by the system) qsys = ∆H = -qsurr = - (m x C x ∆T)The sign of ∆H is negative for an exothermic reaction and positive for an endothermic reaction.
  26. 26. Bomb CalorimeterCalorimetry experiments can also be performed at constant volume using a device called a bomb calorimeter.In a bomb calorimeter, a sample of a compound is burned in a constant-volume chamber in the presence of oxygen at high pressure.The heat that is released warms the water surrounding the chamber. By measuring the temperature increase of the water, it is possible to calculate the quantity of heat released.
  27. 27. Foam Cup Calorimeter(Constant pressure)
  28. 28. Bomb Calorimeter(Constant volume)
  29. 29. QuestionsWhen 25.0 mL of water containing 0.025 mol HCl at 25.0ºC is added to 25.0 mL of water containing 0.025 mol NaOH at 25ºC in a foam cup calorimeter, a reaction occurs. Calculate the enthalpy change in kJ during this reaction if the highest temperature observed is 32.0ºC. Assume the densities of the solutions are 1.00g/mL∆H = - (m x C x ∆T) this is an exothermic reactionThe total volume is 25.0 mL + 25.0 mL = 50.0mLYou need the mass of water, so use the densities given to calculate. 50.0mL (1.00 g/mL) = 50.0g
  30. 30. Questions∆H = - (m x C x ∆T)You know the specific heat of water is 4.18 J/g⋅ºC∆T = Tf – Ti = 32.0ºC – 25.0ºC = 7.0ºC∆H = - (50.0g) (4.18 J/g ⋅ºC) (7.0ºC) = -1463J = 1.46 x 103 kJ
  31. 31. QuestionsWhen 50.0 mL of water containing 0.050 mol HCl at 22.5ºC is added to 50.0 mL of water containing 0.50 mol NaOH at 22.5ºC in calorimeter the temperature of the solution increases to 26.0ºC. How much heat in kJ was released by this reaction?q = (m x C x ∆T)The total volume is 50.0 mL + 50.0 mL = 100.0mLYou need the mass of water, so use the densities given to calculate. 100.0mL (1.00 g/mL) = 100.0gYou know the specific heat of water is 4.18 J/g⋅ºC∆T = Tf – Ti = 26.0ºC – 22.5ºC = 3.5ºCq = (100.0g) (4.18 J/g ⋅ºC) (3.5ºC) = 1463J = 1.5 x 103 kJ
  32. 32. QuestionsA small pebble is heated and placed in a foam cup calorimeter containing 25.0 mL of water at 25.0ºC. The water reaches a maximum temperature of 26.4ºC. How many joules of heat were released by the pebble?q = m x C x ∆TYou need the mass of water, so use known 1L = 1kg to calculate. .0250L (1000 g/L) = 25.0gYou know the specific heat of water is 4.18 J/g⋅ºC∆T = Tf – Ti = 26.4ºC – 25.0ºC = 1.4ºCq = (25.0g) (4.18 J/g ⋅ºC) (1.4ºC) = 146J
  33. 33. Thermochemical EquationsWhen you mix calcium oxide with water, 1 mole of calcium hydroxide forms and 65.2 kJ of heat is released.In a chemical equation, the enthalpy change for the reaction can be written as either a reactant or a product.In the equation describing the exothermic reaction of CaO and H2O, the enthalpy change can be considered a product. CaO (s) + H2O (l) → Ca(OH)2 (s) + 65.2 kJA chemical equation that includes the enthalpy change is called a thermochemical equation. equation CaO (s) + H2O (l) → Ca(OH)2 (s) ∆H= -65.2 kJ
  34. 34. Thermochemical EquationsThe heat of reaction is the enthalpy change for the chemical equation exactly as it is written. You will see heats of reaction reported as ∆H, which is equal to the heat flow at constant pressure.The physical state of the reactants and products must also be given.The standard conditions are that the reaction is carried out at 101.3 kPa (1atm) and that the reactants and products are in their usual physical states at 25ºC.The heat or reaction, or ∆H, in the CaO reaction example is -65.2kJ.Each mole of CaO and H2O that react to form Ca(OH)2 produces 65.2 kJ of heat.
  35. 35. Thermochemical EquationsOther reactions absorb heat from the surroundings. Baking soda decomposes when heated. The carbon dioxide released in the reaction causes a cake to rise while baking. This process in endothermic. 2NaHCO3 (s) + 129kJ → Na2CO3 (s) + H2O (g) + CO2 (g)Remember that ∆H is positive for endothermic reactions. Therefore, you can write the reactions as follows:2NaHCO3 (s) → Na2CO3 (s) + H2O (g) + CO2 (g) ∆H=129kJ
  36. 36. Thermochemical EquationsChemistry problems involving enthalpy changes are similar to stoichiometry problems.The amount of heat released or absorbed during a reaction depends on the number of moles of the reactants involved.The decomposition of 2 mol of sodium bicarbonate requires 129kJ of heat.2NaHCO3 (s) → Na2CO3 (s) + H2O (g) + CO2 (g) ∆ H=129kJTherefore, the decomposition of 4 mol of the same substance would require twice as much heat or 258 kJ.
  37. 37. Thermochemical EquationsIn endothermic processes, the potential energy of the products(s) is higher than the potential energy of the reactants.The physical state of the reactants and products must also be given.H2O (l) → H2 (g) + 1/2O2 (g) ∆H = 285.8 kJH2O (g) → H2 (g) + 1/2O2 (g) ∆H = 241.8 kJAlthough the two equations are very similar, the different physical states of H2O result in different ∆H values.
  38. 38. QuestionsCalculate the amount of heat (in kJ) required to decompose 2.24 mol NaHCO3 (s)2NaHCO3(s) → Na2CO3(s) + H2O(g) + CO2(g) ∆H=129kJThe thermochemical equation indicates that 129 kJ of heat are needed to decompose 2 mole of NaHCO3 (s)∆H = 2.24 mole NaHCO3 (s) | 129 kJ = 144kJ | 2 mol NaHCO3 (s)Think Logically: Because the decomposition of 2 mol of NaHCO3 requires 129kJ, then the decomposition of 2.24 mol should absorb about 10% more heat than 129kJ.
  39. 39. QuestionsWhen carbon disulfide is formed from its elements, heat is absorbed. Calculate the amount of heat (in kJ) absorbed when 5.66 g of carbon disulfide is formed. C(s) + 2S(s) → CS2(l) ∆H= 89.3kJThe thermochemical equation indicates that 89.3 kJ of heat are needed to form 1 mole of CS2 (l)∆H = 5.66 g CS2 (l) | 1 mol CS2 (l) | 89.3 kJ = 6.64kJ | 76.1g CS2 (l) | 1mol CS2 (l)
  40. 40. QuestionsThe production of iron and carbon dioxide from Iron(III) oxide and carbon monoxide is an exothermic reaction. How many kJ of heat are produced when 3.40 mol Fe2O3 reacts with an excess of CO? Fe2O3(s) + 3CO(g) → 2Fe(s) +3CO2(g) + 26.3kJ∆H = 3.40 mol Fe2O3(s) | 26.3 kJ = -89.4kJ | mol Fe2O3(s)
  41. 41. QuestionsExplain the difference between H and ∆H.H is enthalpy or heat content; ∆H represents a change in heat content.Calorimetry is based on what basic concepts?The heat released by the system is equal to the heat absorbed by the surroundings. Vice versa.How are enthalpy changes treated in chemical equationsThe enthalpy change in a chemical reaction can be written as either a reactant or a product.
  42. 42. QuestionsWhen 2 mol of solid magnesium combines with 1 mole of oxygen gas, 2 mol of solid magnesium oxide is formed and 1204kJ of heat is released. Write the thermochemical equation for this combustion reaction. 2Mg (s) + O2 (g) → 2MgO + 1204kJ 2Mg (s) + O2 (g) → 2MgO ∆H = -1204kJ
  43. 43. QuestionsGasohol contains ethanol (C2H5OH) (l) which when burned reacts with oxygen to produce CO2(g) and H2O (g). How much heat is released when 12.5 g of ethanol burns?C2H5OH (l) + 3O2 (g) → 2CO3 (g) + 3H2O (l) + 1368kJ∆H = 12.5g C2H5OH(l) | 1 mol C2H5OH(l) |1368kJ | 46g C2H5OH(l) | mol = -372 kJExplain the term heat of combustionThe heat of reaction for the complete burning of one mole
  44. 44. End of Section 17.2
  45. 45. Heats of Fusion and SolidificationAll solids absorb heat as they melt to become liquids.The gain of heat causes a change of state instead of a change in temperature.Whenever a change of state occurs by a gain or loss of heat, the temperature of the substance remains constant.The heat absorbed by one mole of a solid substance as it melts to a liquid at constant temperature is the molar heat of fusion. (∆Hfus)The molar heat of solidification (∆Hsolid) is the heat lost when one mole of a liquid solidifies at constant temperature.
  46. 46. Heats of FusionThe quantity of heat absorbed by a melting solid is exactly the same as the quantity of heat released when a liquid solidifies. ∆Hfus = -∆HsolidWhen a liquid solidifies, it loses heat, thus the negative sign.
  47. 47. Heats of Fusion and SolidificationMelting 1 mol of ice at 0ºC to 1 mol of water at 0ºC requires the absorption of 6.01kJ of heat. (this quantity of heat is the molar fusion of water)The conversion of 1 mol of water at 0ºC to 1 mol of ice at 0ºC releases 6.01kJ of heat. (this quantity of heat is the molar heat of solidification of water) H2O (s) → H2O (l) (∆Hfus)= 6.01 kJ/mol H2O (l) → H2O (s) (∆Hsolid)= 6.01 kJ/mol
  48. 48. Sample ProblemHow many grams of ice at 0ºC will melt is 2.25kJ of heat are added?2.25 kJ 1 mol ice 18.0 g ice = 6.74 g ice 6.01 kJ 1 mol iceUse your common sense to check. 6.01 kJ of heat is required to melt 1 mol of ice. You are only adding about 1/3 of that heat, so only about 1/3 of the ice should melt.
  49. 49. Sample ProblemHow many kJ of heat are required to melt a 10.0g popsicle at 0ºC? Assume the popsicle has the same molar mass and heat of fusion as water.10.0 g pop 1 mol pop 6.01 kJ = 3.34 kJ 18 g pop 1 mol popHow many grams of ice at 0ºC could be melted by the addition of 0.400 kJ of heat?0.400 kJ 1 mol ice 18 g ice = 1.20 g ice 6.01 kJ 1 mol ice
  50. 50. Heats of Vaporization and CondensationWhen liquids absorb heat at their boiling points, they become vapors. The amount of heat necessary to vaporize one mole of a given liquid is called its molar heat of vaporization (∆Hvap)The molar heat of vaporization of water is 40.7 kJ /mol It takes 40.7 of energy to convert 1 mol of water to 1 mole of water vapor at the normal boiling point of water. H2O (l) → H2O (g) ∆Hvap = 40.7 kJ/molCondensation is the exact opposite of vaporization
  51. 51. Heats of Vaporization and CondensationWhen a vapor condenses, heat is released. The amount of heat released when 1 mol of vapor condenses at the normal boiling point is called its molar heat of condensation. (∆Hcond)The value is numerically the same as the molar heat of vaporization, however, the value has the opposite sign. ∆Hvap = -∆HcondHeat is released during condensation, thus the negative sign.Condensation is the exact opposite of vaporization
  52. 52. (∆Hcond) (∆Hvap)∆Hfus∆Hsolid
  53. 53. Changes of State
  54. 54. Sample ProblemHow much heat (in kJ) is absorbed when 24.8 g H20 (l) at 100ºC and 101.3 kPa is coverted to steam at 100ºC?24.8g H2O 1 mol H2O 40.7 kJ = 56.1 kJ 18 g H2O 1 mol H2OHow much heat is absorbed when 63.7 g H2O at 100ºC and 101.3 kPa is converted to steam at 100ºC?63.78g H2O 1 mol H2O 40.7 kJ = 144 kJ 18 g H2O 1 mol H2O
  55. 55. Heat of SolutionDuring the formation of a solution, heat is either released or absorbed.The enthalpy change caused by dissolution of one mole of substance is the molar heat of solution (∆Hsoln)Hot packs are an example. When CaCl2 and H2O are mixed, heat is produced. (solution releases heat and the reaction is exothermic)A cold pack is an example of an endothermic reaction, where the solution absorbs heat.
  56. 56. Sample ProblemHow much heat is released when 0.677 mol NaOH is dissolved in water.0.677 mol NaOH -445.1 kJ = -301 kJ 1 mol NaOHHow many moles of NH4NO3 must be dissolved in water so that 88.0 kJ of heat is absorbed from the water? (∆Hsoln for NH4NO3 = 25.4 kJ/mol)88.0 kJ 1 mol NH4NO3 = 3.42 mol NH4NO3 25.4 kJ
  57. 57. Calculating Heats of ReactionHess’s law of heat summation states that 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.Use Hess’s law to find the enthalpy change for the conversion of diamond to graphite as follows:C(s,graphite) + O2 → CO2(g) ∆H = -393.5 kJC(s, diamond) + O2 → CO2(g) ∆H = -395.4 kJWrite the first equation in reverse because you want graphite on the product side. When you reverse the equation, the sign of ∆H is also reversed.
  58. 58. End of Section 17.3
  59. 59. Calculating Heats of ReactionCO2(g) → C(s,graphite) + O2(g) ∆H = 393.5 kJ (in reverse)Add both equations to get:CO2(g) → C(s,graphite) + O2 ∆H = 393.5 kJC(s, diamond) + O2 → CO2(g) ∆H = -395.4 kJC(s, diamond) → C(s,graphite) ∆H = -1.9 kJThe conversion of diamond to graphite is an exothermic process, so its heat of reaction has a negative sign.Conversely, the change of graphite to diamond is an endothermic process.
  60. 60. Calculating Heats of ReactionFind the enthalpy change of the change of graphite to COC(s,graphite) + O2(g) → CO2(g) ∆H = -393.5 kJCO(g) + 1/2O2(g) → CO2(g) ∆H = -283.0 kJWrite the second equation in reverse to get CO on the product side. (don’t forget to change the sign)C(s,graphite) + O2(g) → CO2(g) ∆H = -393.5 kJCO2(g) → CO(g) + 1/2O2(g) ∆H = 283.0 kJC(s,graphite) + 1/2O2(g) → CO(g) ∆H = -110.5 kJ
  61. 61. Standard Heats of FormationEnthalpy changes generally depend on conditions of the process. In order to compare enthalpy changes, scientists specify a common set of conditions as a reference point.These conditions, called the standard state, refer to the stable form of a substance at 25ºC and 101.3 kPa.The standard heat of formation (∆Hf0) of a compound is the change in enthalpy that accompanies the formation of one mole of a compound from its elements with all substances in their standard states.The ∆Hf0 of a free element is arbitrarily set at zero.
  62. 62. Standard Heats of FormationStandard heats of formation provide an alternative to Hess’s law in determining heats of reaction indirectly.For a reaction that occurs at standard conditions, you can calculate the heat of reaction by suing standard heats of formation.This enthalpy change is called the standard heats of reaction (∆H0)The standard heat of reaction is the difference between the standard heats of formation of all the reactants and products. ∆H0 = ∆Hf0 (products) - ∆Hf0 (reactants)
  63. 63. Sample ProblemWhat is the standard heat of reaction for the reaction of CO(g) with O2 (g) to form CO2 (g) ∆Hf0 O2 = 0kJ/mol (free element) ∆Hf0 CO2 = -393.5kJ/mol ∆Hf0 CO = -110.5kJ/molFirst write a balanced equation: 2CO (g) + O2 (g) → 2CO2 (g)Next find and add the ∆Hf0 of all of the reactants, taking into account the number of moles of each.∆Hf0 (reactants) = (2 mol CO)(-110.5kJ/mol) + 0kJ = -221.0kJ
  64. 64. Sample Problem∆Hf0 (products) = (2 mol CO2) (-393.5kJ/mol) = -787 kJLastly, plug your values calculated for ∆Hf0 (products) and ∆Hf0 (reactants) into the equation. ∆H0 = ∆Hf0 (products) - ∆Hf0 (reactants)∆H0 = (-787.0kJ) – (-221.0kJ) = -566.0 kJ
  65. 65. Sample ProblemWhat is the standard heat of reaction for the reaction Br2(g) → Br2 (l) ∆Hf0 Br2 (g) = 0kJ/mol (free element) ∆Hf0 Br2 (l) = -393.5kJ/mol ∆The equation is already balancedNext find and add the ∆Hf0 of all of the reactants, taking into account the number of moles of each.∆Hf0 (reactants) = (1 mol Br2 (g))(30.91kJ/mol) = 30.91 kJ ∆Hf0 (products) = 0∆H0 = (0kJ) – (30.91kJ) = -30.91kJ
  66. 66. Sample ProblemWhat is the standard heat of reaction for the reaction CaCO3(s) → CaO(s) + CO2(g) ∆Hf0 CaCO3(s) ) = -1207.0 kJ/mol ∆Hf0 CaO(s) = -635.1 kJ/mol ∆Hf0 CO2(g) = -393.5 kJ/molThe equation is already balancedNext find and add the ∆Hf0 of all of the reactants, taking into account the number of moles of each.∆Hf0 (products) = (1 mol CaO(s) )(-635.1 kJ/mol ) + (1 mol CO2(g) )(-393.5 kJ/mol ) = -1028.6 kJ∆Hf0 (reactants) = -1207.0kJ∆H0 = (-1028.6 kJ) – (-1207.9 kJ) = 179.3 kJ
  67. 67. End of Chapter 17