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 is the capacity to do work or supply heat. Energy has no mass or volume. Energy has no mass or volume. Chemical potential energy is energy stored in chemicals. 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. 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 a form of energy that always flows from a warmer object to a cooler object. Heat is represented by q. 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. 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. 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. 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. A system is the specific part of the universe on which you focus your attention. The surroundings include everything outside the system. The surroundings include everything outside the system. The system and the surroundings constitute the universe. 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. 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 absorbs heat from the surroundings is called an endothermic process. A process that loses heat to the surroundings is called an exothermic 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 1 o C. A calorie is the quantity of heat that raises the temperature of 1 gram of pure water 1 o C. A Calorie, or dietary Calorie, is equal to 1000 calories. A Calorie, or dietary Calorie, is equal to 1000 calories. A Joule is the SI unit of heat and energy. 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. 1. 444 calories to Joules 2. 850 Joules to calories 444 cal x 4.184 J = 1857.7 J 1 cal 1 cal 850 J x 1 cal = 203.2 cal 4.184 J 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 1 o C. The heat capacity of an object is the amount of heat it takes to change an object’s temperature by exactly 1 o C. The greater the mass of an object, the greater the heat capacity. The greater the mass of an object, the greater the heat capacity. The heat capacity of an object also depends on its chemical composition. 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 1 o C. The specific heat capacity of a substance is the amount of heat it takes to raise the temperature of 1 gram of the substance 1 o C. Specific heat is represented by C. Specific heat is represented by C. The units of specific heat are J/g o C. The units of specific heat are J/g o C. Water has a higher specific heat than most substances. Water has a higher specific heat than most substances.

13. Heat Heat = mass x specific heat x change in temp q = m. C.  T Mass is in grams Specific heat is in J/g o C Change in temp is in o C

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

15. Practice Problems 1. When 435J of heat is added to 3.4g of olive oil at 21 o C, the temperature increases to 85 o C. What is the specific heat of the olive oil? q = m. c.  T c = q  T = 85 o C – 21 o C = 64 o C m.  T c = 435J = 1.99 J/g o C 3.4g. 64 o C

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

17. Section 17.1 Review 1. In what direction does heat flow between two objects? 2. How do endothermic processes differ from exothermic processes? 3. On what factors does the heat capacity of an object depend? 4. How many kilojoules of heat are absorbed when 1000g of water is heated from 18 o C to 85 o C? (specific heat of water = 4.184 J/g o C) q = 1000g (4.184J/g o C) (67 o C) = 280328J 280328J x 1kJ = 280.328kJ 1000 J

18. Section 17.1 Review 5. Using calories, calculate how much heat 32.0g of water absorbs when it is heated from 25 o C to 80 o C. How many joules is this? (specific heat of water = 4.184 J/g o C) q = 32g (4.184 J/g o C) (55 o C) = 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. 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. 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 the heat content of a system at constant pressure. Enthalpy is represented by H. Enthalpy is represented by H.

20. Calorimeter q =  H = m. C.  T

21. Heat Change Sign Convention Direction of Heat Flow Sign Reaction Type Heat Flows Out of the System -H-H-H-HExothermic Heat Flows Into the System +H+H+H+HEndothermic

22. Thermochemical Equations An equation that included the heat change is a thermochemical equation. 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. A heat of reaction is the heat change for the equation exactly as written. Ex: Ex: CaO (s) + H 2 O (l)  Ca(OH) 2(s)  H = -65.2 KJ  H = -65.2 KJ 2NaHCO 3(s)  Na 2 CO 3(s) +H 2 O (g) +CO 2(g)  H = +129 KJ

23. Sample Problem Calculate the amount of heat (in kJ) required to decompose 2.24 moles of NaHCO 3. Calculate the amount of heat (in kJ) required to decompose 2.24 moles of NaHCO 3. 2NaHCO 3  Na 2 CO 3 + H 2 O + CO 2  H = 129kJ 2.24 mol NaHCO 3 x 129kJ = 144.48kJ 2 mol NaHCO 3

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

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

26. Section 17.2 Review 1. 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) + O 2(g)  2MgO  H = -1204kJ

27. Section 17.2 Review 2. How much heat is released when 12.5g of ethanol burns? C 2 H 5 OH + 3O 2  2CO 2 + 3H 2 O  H = -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. The heat of combustion is the heat of reaction for the complete burning of one mole of a substance.

29.  H (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 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. The heat lost when one mole of a liquid changes to a solid at a constant temperature is the molar heat of solidification.  H fus = -  H solid

30.  H (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 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. The heat released by one mole of a substance changing from a vapor to a liquid is the molar heat of condensation.  H vap = -  H cond

31. Practice Problems 1.What is the  H vap of acetone? 2.What is the  H cond of water? 3.What is the  H fus of rubbing alcohol? 4.What is the  H solid of diethyl ether? 29 kJ/mol - 41 kJ/mol 6 kJ/mol -7 kJ/mol

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

33. Heating Curve for Water

34. Sample Problem How much heat (in kJ) is absorbed when 24.8g H 2 O (l) at 100 o C and 101.3kPa is converted to steam at 100 o C. How much heat (in kJ) is absorbed when 24.8g H 2 O (l) at 100 o C and 101.3kPa is converted to steam at 100 o C. 24.8g H 2 O x 1 mol H 2 O x 40.7kJ = 56.08kJ 18g H 2 O 1 mol

35. Practice Problems 1. How much heat is absorbed when 63.7g H 2 O (l) at 100 o C and 101.3kPa is converted to steam at 100 o C? 63.7g H 2 O x 1 mol H 2 O x 40.7kJ = 144.03kJ 18g H 2 O 1 mol H 2 O

36. Practice Problems 2. How many kilojoules of heat are absorbed when 0.46g of chloroethane (C 2 H 5 Cl) vaporizes at its boiling point? (The molar heat of vaporization for chloroethane is 26.4 kJ/mol.) 0.46g C 2 H 5 Cl x 1 mol C 2 H 5 Cl x 26.4kJ = 0.19kJ 64g C 2 H 5 Cl 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 NH 4 NO 3 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 NH 4 NO 3 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. 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 will have to do these problems in multiple steps. You have to work your way through the heating curve. 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 75 o C to steam. Calculate the amount of heat required to change 50g of water at 75 o C to steam. q = m. c.  Tq = 50g (4.184 J/g o C) (25 o C) = 5230J 50g H 2 O x 1 mol H 2 O x 40.7kJ = 113.06kJ 18g H 2 O 1 mol 113.06kJ = 113060J 113060J + 5230J 118290J

42. Practice Problems 1. Calculate the amount of heat needed to change 175g of ice at -20 o C to water at 15 o C. q = 175g (2.1 J/g o C) (20 o C) = 7350J 175g H 2 O x 1 mol H 2 O x 6.01kJ = 58.43kJ 18g H 2 O 1 mol q = 175g (4.184 J/g o C) (15 o C) = 10983J 58.43kJ = 58430J 10983J 58430J +7350J 76763J

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

44. Section 17.3 Review 1. How does the molar heat of fusion of a substance compare to its molar heat of solidification? 2. 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. a. 1 mol C 3 H 8(l)  1 mol C 3 H 8(g) b. 1 mol Hg (l)  1 mol Hg (s) c. 1 mol NH 3(g)  1 mol NH 3(l) d. 1 mol NaCl (s) + 3.88kJ/mol  1 mol NaCl (aq) e. 1 mol NaCl (s)  1 mol NaCl (l) Endothermic Exothermic 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. 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: Calculate the enthalpy change for the following reaction given the following information: PbCl 2 + Cl 2  PbCl 4  H = ? Pb + 2Cl 2  PbCl 4  H = -329.2kJ Pb + Cl 2  PbCl 2  H = -359.4kJ 1 st reaction: keep the same  H = -329.2kJ 2 nd reaction: flip the reaction (change sign) +  H = 359.4kJ  H = 30.2kJ

48. Practice Problems 1. Find the enthalpy change for the reaction using the following information. 2P + 5Cl 2  2PCl 5  H = ? PCl 5  PCl 3 + Cl 2  H = 87.9kJ 2P + 3Cl 2  2PCl 3  H = -574kJ 1 st reaction: flip reaction (change sign) and x2  H = -175.8kJ 2 nd reaction: keep the same +  H = -574kJ  H = -749.8kJ

49. Practice Problems 2. Calculate the enthalpy change for the reaction using the following information: N 2 + O 2  2NO  H = ? 4NH 3 + 3O 2  2N 2 + 6H 2 O  H = -1530kJ 4NH 3 + 5O 2  4NO + 6H 2 O  H = -1170kJ 1 st reaction: flip reaction (change sign) /2  H = 765kJ 2 nd reaction: /2 +  H = -585kJ  H = 180kJ

50. Practice Problems 3. Calculate the enthalpy change for the reaction using the following information. C 2 H 4 + H 2  C 2 H 6  H = ? 2H 2 + O 2  2H 2 O  H = -572kJ C 2 H 4 + 3O 2  2H 2 O + 2CO 2  H = -1401kJ 2C 2 H 6 + 7O 2  6H 2 O + 4CO 2  H = -3100kJ 1 st reaction: /2  H = -286kJ 2 nd reaction: keep the same  H = -1401kJ 3 rd reaction: flip reaction (change sign) /2 +  H = 1550kJ  H = -137kJ

51. Standard Heat of Formation The standard heat of formation of a compound is the change in enthalpy that accompanies the formation of one mole of the compound from its element with all substances in their standard states at 25 o C. The standard heat of formation of a compound is the change in enthalpy that accompanies the formation of one mole of the compound from its element with all substances in their standard states at 25 o C. The  H f o of a free element in its standard state is zero. The  H f o of a free element in its standard state is zero.  H o =  H f o (products) –  H f o (reactants)

52. Sample Problem Calculate the standard heat of formation for the following reaction. Calculate the standard heat of formation for the following reaction. 2CO (g) + O 2(g)  2CO 2(g)  H o =  H f o (products) –  H f o (reactants)  H o = (-787kJ) – (-221kJ + 0kJ) = -566kJ CO 2(g) = 2mol x -110.5 kJ/mol = -221kJ O 2(g) = 1 mol x 0 kJ/mol = 0kJ CO 2(g) = 2 mol x -393.5 kJ/mol = -787kJ

53. Practice Problems 1. Calculate the standard heat of formation for the following reaction. CaCO 3(s)  CaO (s) + CO 2(g) CaCO 3(s) = 1 mol x -1207 kJ/mol = -1207kJ CaO (s) = 1 mol x -635.1 kJ/mol = -635.1kJ CO 2(g) = 1 mol x -393.5 kJ/mol = -393.5kJ  H o =  H f o (products) –  H f o (reactants)  H o = (-635.1kJ + -393.5kJ) – (-1207kJ) = 178.4kJ

54. Practice Problems 2. Calculate the standard heat of formation for the following reaction. 2NO (g) + O 2(g)  2NO 2(g) NO (g) = 2 mol x 90.37 kJ/mol = 180.74kJ O 2(g) = 1 mol x 0 kJ/mol = 0kJ NO 2(g) = 2 mol x 33.85 kJ/mol = 67.7kJ  H o =  H f o (products) –  H f o (reactants)  H o = (67.7kJ) – (180.74kJ + 0kJ) = -113.04kJ

55. Section 17.4 Review 1. Calculate the enthalpy change in kJ for the following reaction. 2Al + Fe 2 O 3  2Fe + Al 2 O 3 Use the enthalpy changes for the combustion of aluminum and iron: 2Al + 3/2O 2  Al 2 O 3  H = -1669.8kJ 2Fe + 3/2O 2  Fe 2 O 3  H = -824.2kJ 1 st reaction: keep the same  H = -1669.8kJ 2 nd reaction: flip reaction (change sign)+  H = 824.2kJ  H = -845.6kJ

56. Section 17.4 Review 2. What is the standard heat of reaction for the decomposition of hydrogen peroxide? 2H 2 O 2(l)  2H 2 O (l) + O 2(g) H 2 O 2(l) = 2 mol x -187.8kJ/mol = -375.6kJ H 2 O (l) = 2 mol x -285.8kJ/mol = -571.6kJ O 2(g) = 1 mol x 0kJ/mol = 0kJ  H o =  H f o (products) –  H f o (reactants)  H o = (-571.6kJ + 0kJ) – (-375.6kJ) = -196kJ

57. THE END