1 / 57

Unit 13: Thermochemistry

Unit 13: Thermochemistry. Chapter 17 By: Jennie Borders. Section 17.1 – The Flow of Energy. Energy is the capacity to do work or supply heat . Energy has no mass or volume . Chemical potential energy is energy stored in chemicals .

hisa
Download Presentation

Unit 13: Thermochemistry

An Image/Link below is provided (as is) to download presentation Download Policy: Content on the Website is provided to you AS IS for your information and personal use and may not be sold / licensed / shared on other websites without getting consent from its author. Content is provided to you AS IS for your information and personal use only. Download presentation by click this link. While downloading, if for some reason you are not able to download a presentation, the publisher may have deleted the file from their server. During download, if you can't get a presentation, the file might be deleted by the publisher.

E N D

Presentation Transcript

  1. Unit 13: Thermochemistry Chapter 17 By: Jennie Borders

  2. Section 17.1 – The Flow of Energy • Energy is the capacity to do work or supply heat. • Energy has no mass or volume. • Chemical potential energy is energy stored in chemicals. • The kinds of atoms and the arrangement of the atoms in a substance determine the amount of energy stored in the substance.

  3. Heat • Heat is a form of energy that always flows from a warmer object to a cooler object. • Heat is represented by q.

  4. Thermochemistry • Thermochemistry is the study of the heat changes that occur during chemical reactions and physical changes of state. • The law of conservation of energy states that in any chemical or physical process, energy is neither created nor destroyed.

  5. The Great Debate 1. Exothermic reactions lose heat. 2. Endothermic reactions absorb heat.

  6. Exothermic and Endothermic • Thermochemistry is concerned with the flow of heat between a chemical system (reaction) and its surroundings. • A system is the specific part of the universe on which you focus your attention. • The surroundings include everything outside the system. • The system and the surroundings constitute the universe.

  7. Exothermic and Endothermic • In thermochemical calculations the direction of the heat flow is given from the point of view of the system. • A process that absorbs heat from the surroundings is called an endothermic process. • A process that loses heat to the surroundings is called an exothermic process.

  8. Exothermic and Endothermic

  9. Units of Heat • A calorie is the quantity of heat that raises the temperature of 1 gram of pure water 1oC. • A Calorie, or dietary Calorie, is equal to 1000 calories. • A Joule is the SI unit of heat and energy. 1 Calorie = 1000 cal = 1 kcal = 4184 J 1 cal = 4.184 J

  10. Practice Problems Make the following conversions. • 444 calories to Joules • 850 Joules to calories 444 cal x 4.184 J = 1857.7 J 1 cal 850 J x 1 cal= 203.2 cal 4.184 J

  11. Heat Capacity • The heat capacity of an object is the amount of heat it takes to change an object’s temperature by exactly 1oC. • The greater the mass of an object, the greater the heat capacity. • The heat capacity of an object also depends on its chemical composition.

  12. Specific Heat • The specific heat capacity of a substance is the amount of heat it takes to raise the temperature of 1 gram of the substance 1oC. • Specific heat is represented by C. • The units of specific heat are J/goC. • Water has a higher specific heat than most substances.

  13. Heat Heat = mass x specific heat x change in temp q = m.C.DT Mass is in grams Specific heat is in J/goC Change in temp is in oC

  14. Sample Problem • The temperature of a 95.4g piece of copper increases form 25oC to 48oC when the copper absorbs 849J of heat. What is the specific heat of copper? q = m.c.DT c = q DT = 48oC – 25oC = 23oC m.DT c = 849J = 0.39 J/goC 95.4g.23oC

  15. Practice Problems 1. When 435J 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? q = m.c.DT c = q DT = 85oC – 21oC = 64oC m.DT c = 435J = 1.99 J/goC 3.4g.64oC

  16. Practice Problems 2. How much heat is required to raise the temperature of 250g of mercury 52oC? (specific heat of mercury = 0.14 J/goC) q = m.c.DT q = 250g (0.14J/goC) (52oC) = 1820J

  17. Section 17.1 Review • In what direction does heat flow between two objects? • How do endothermic processes differ from exothermic processes? • On what factors does the heat capacity of an object depend? • How many kilojoules of heat are absorbed when 1000g of water is heated from 18oC to 85oC? (specific heat of water = 4.184 J/goC) q = 1000g (4.184J/goC) (67oC) = 280328J 280328J x 1kJ = 280.328kJ 1000 J

  18. Section 17.1 Review • Using calories, calculate how much heat 32.0g of water absorbs when it is heated from 25oC to 80oC. How many joules is this? (specific heat of water = 4.184 J/goC) q = 32g (4.184 J/goC) (55oC) = 7363.8J 7363.8J x 1 cal = 1759.9 cal 4.184 J

  19. Section 17.2 – Measuring and Expressing Enthalpy Changes • Calorimetry is the accurate and precise measurement of heat change for chemical and physical processes. • Calorimeters are devices used to measure the amount of heat absorbed or released during chemical and physical processes. • Enthalpy is the heat content of a system at constant pressure. • Enthalpy is represented by H.

  20. Calorimeter q = DH = m . C . DT

  21. Heat Change Sign Convention

  22. Thermochemical Equations • An equation that included the heat change is a thermochemical equation. • A heat of reaction is the heat change for the equation exactly as written. • Ex: CaO(s) + H2O(l) Ca(OH)2(s) DH = -65.2 KJ 2NaHCO3(s) Na2CO3(s) +H2O(g) +CO2(g) DH = +129 KJ

  23. Sample Problem • Calculate the amount of heat (in kJ) required to decompose 2.24 moles of NaHCO3. 2NaHCO3 Na2CO3 + H2O + CO2 DH = 129kJ 2.24 mol NaHCO3 x 129kJ = 144.48kJ 2 mol NaHCO3

  24. Practice Problems • Calculate the amount of heat (in kJ) absorbed when 5.66g of carbon disulfide is formed. C + 2S  CS2DH = 89.3kJ 5.66g CS2 x 1 mol CS2 x 89.3kJ = 6.65kJ 76g CS2 1 mol CS2

  25. Practice Problems 2. How many kilojoules of heat are produced when 3.40 mole Fe2O3 reacts with an excess of CO? Fe2O3 + 3CO  2Fe + 3CO2 DH = -26.33 kJ 3.40 mol Fe2O3 x -26.3kJ = -89.42kJ 1 mol Fe2O3

  26. Section 17.2 Review • When 2 mol of solid magnesium combines with 1 mol of oxygen gas, 2 mol of solid magnesium oxide is formed and 1204kJ of heat is release. Write the thermochemical equation for this combustion reaction. 2Mg(s) + O2(g) 2MgO DH = -1204kJ

  27. Section 17.2 Review 2. How much heat is released when 12.5g of ethanol burns? C2H5OH + 3O2 2CO2 + 3H2O DH = -1368kJ 12.5g x 1mol x -1368kJ = -371.74kJ 46g 1 mol

  28. Section 17.3 – Heat in Changes of State • The heat of combustion is the heat of reaction for the complete burning of one mole of a substance.

  29. DH (fusion and solidification) • The heat absorbed by one mole of a substance melting from a solid to a liquid at constant temperature is the molar heat of fusion. • The heat lost when one mole of a liquid changes to a solid at a constant temperature is the molar heat of solidification. DHfus = -DHsolid

  30. DH (vaporization and condensation) • The heat absorbed by one mole of a substance changing from a liquid to a vapor is the molar heat of vaporization. • The heat released by one mole of a substance changing from a vapor to a liquid is the molar heat of condensation. DHvap = -DHcond

  31. Practice Problems What is the DHvap of acetone? What is the DHcond of water? What is the DHfus of rubbing alcohol? What is the DHsolid of diethyl ether? 29 kJ/mol -41 kJ/mol 6 kJ/mol -7 kJ/mol

  32. DH (solution) • The heat change caused by dissolution of one mole of a substance is the molar heat of solution. • Ex. NaOH(s) Na+(aq) + OH-(aq) DHsoln = -445.1 KJ

  33. Heating Curve for Water

  34. Sample Problem • How much heat (in kJ) is absorbed when 24.8g H2O(l) at 100oC and 101.3kPa is converted to steam at 100oC. 24.8g H2O x 1 mol H2O x 40.7kJ = 56.08kJ 18g H2O 1 mol

  35. Practice Problems 1. How much heat is absorbed when 63.7g H2O(l) at 100oC and 101.3kPa is converted to steam at 100oC? 63.7g H2O x 1 mol H2O x 40.7kJ = 144.03kJ 18g H2O 1 mol H2O

  36. Practice Problems 2. How many kilojoules of heat are absorbed when 0.46g of chloroethane (C2H5Cl) vaporizes at its boiling point? (The molar heat of vaporization for chloroethane is 26.4 kJ/mol.) 0.46g C2H5Cl x 1 mol C2H5Cl x 26.4kJ = 0.19kJ 64g C2H5Cl 1 mol

  37. Practice Problems 3. How much heat (in kJ) is released when 2.5 mol of NaOH is dissolved in water? (The molar heat of solution is -445.1 kJ/mol.) 2.5 mol NaOH x -445.1kJ = -1112.75kJ 1 mol

  38. Practice Problems 4. How many moles of NH4NO3 must be dissolved in water so that 88kJ of heat is absorbed from the water? (The molar heat of solution 25.7 kJ/mol.) 88kJ x 1 mol = 3.42mol NH4NO3 25.7kJ

  39. Heating Curve Problems • Sometimes a question might ask you about a physical change that involves a temperature change and a phase change. • You will have to do these problems in multiple steps. • You have to work your way through the heating curve.

  40. Heating Curve

  41. Sample Problem • Calculate the amount of heat required to change 50g of water at 75oC to steam. q = m.c.DT q = 50g (4.184 J/goC) (25oC) = 5230J 50g H2O x 1 mol H2O x 40.7kJ = 113.06kJ 18g H2O 1 mol 113.06kJ = 113060J 113060J + 5230J 118290J

  42. Practice Problems 1. Calculate the amount of heat needed to change 175g of ice at -20oC to water at 15oC. q = 175g (2.1 J/goC) (20oC) = 7350J 175g H2O x 1 mol H2O x 6.01kJ = 58.43kJ 18g H2O 1 mol q = 175g (4.184 J/goC) (15oC) = 10983J 58.43kJ = 58430J 10983J 58430J +7350J 76763J

  43. Practice Problems 2. Calculate the amount of heat released when 100g of steam at 130oC is cooled to water at 60oC. q = 100g (1.7 J/goC) (-30oC) = -5100J 100g H2O x 1 mol H2O x -40.7kJ = -226.11kJ 18g H2O 1 mol q = 100g (4.184 J/goC) (-40oC) = -16736J -226.11kJ = -226110J -226110J -16736J -5100J -247946J

  44. Section 17.3 Review • How does the molar heat of fusion of a substance compare to its molar heat of solidification? • How does the molar heat of vaporization of a substance compare to its molar heat of condensation?

  45. Section 17.3 Review 3. Identify each enthalpy change by name and classify each change as exothermic or endothermic. • 1 mol C3H8(l) 1 mol C3H8(g) • 1 mol Hg(l) 1 mol Hg(s) • 1 mol NH3(g)  1 mol NH3(l) • 1 molNaCl(s) + 3.88kJ/mol  1 molNaCl(aq) • 1 molNaCl(s)  1 molNaCl(l) Endothermic Exothermic Exothermic Endothermic Endothermic

  46. Section 17.4 – Calculating Heats of Reaction • Hess’ 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 heat changes to give the final heat change.

  47. Sample Problem • Calculate the enthalpy change for the following reaction given the following information: PbCl2 + Cl2 PbCl4DH = ? Pb + 2Cl2  PbCl4DH = -329.2kJ Pb + Cl2  PbCl2DH = -359.4kJ 1st reaction: keep the same DH = -329.2kJ 2nd reaction: flip the reaction (change sign) +DH = 359.4kJ DH = 30.2kJ

  48. Practice Problems • Find the enthalpy change for the reaction using the following information. 2P + 5Cl2 2PCl5DH = ? PCl5  PCl3 + Cl2DH = 87.9kJ 2P + 3Cl2  2PCl3DH = -574kJ 1st reaction: flip reaction (change sign) and x2 DH = -175.8kJ 2nd reaction: keep the same +DH = -574kJ DH = -749.8kJ

  49. Practice Problems 2. Calculate the enthalpy change for the reaction using the following information: N2 + O2 2NO DH = ? 4NH3 + 3O2  2N2 + 6H2O DH = -1530kJ 4NH3 + 5O2  4NO + 6H2O DH = -1170kJ 1st reaction: flip reaction (change sign) /2 DH = 765kJ 2nd reaction: /2 +DH = -585kJ DH = 180kJ

  50. Practice Problems 3. Calculate the enthalpy change for the reaction using the following information. C2H4 + H2 C2H6DH = ? 2H2 + O2  2H2O DH = -572kJ C2H4 + 3O2  2H2O + 2CO2DH = -1401kJ 2C2H6 + 7O2  6H2O + 4CO2DH = -3100kJ 1st reaction: /2 DH = -286kJ 2nd reaction: keep the same DH = -1401kJ 3rd reaction: flip reaction (change sign) /2 +DH = 1550kJ DH = -137kJ

More Related