1. Chapter 51 Chapter 6 Thermochemistry Jozsef Devenyi Department of Chemistry, UTM

2. Chapter 52 kinetic energy is the energy of motion: The Nature of Energy Recall: force: work: energy: a push or pull on an object the product of force applied to an object over a distance the work done to move an object against a force (energy is the capacity to do work or transfer heat)

3. Chapter 53 potential energy is the energy an object possesses by virtue of its position potential energy can be converted into kinetic energy The Nature of Energy example: a ball is falling from a balcony

4. Chapter 64 We sometimes use the calorie instead of the joule: 1 cal = 4.184 J (exactly) Units of Energy SI Unit for energy is the joule, J: The Nature of Energy for example, an object with a mass of 2 kg that moves at a speed of 1 m/s; its kinetic energy is

5. Chapter 65 system: part of the universe we are interested in Systems and Surroundings surroundings: the rest of the universe The Nature of Energy

6. 6 First Law of Thermodynamics Internal Energy Internal energy: total energy of a system (cannot measure absolute internal energy) Change in internal energy: Chapter 6

7. 7 Relating  E to Heat and Work: internal energy of a system: Energy cannot be created or destroyed. energy of (system + surroundings) is constant Any energy transferred from a system must be transferred to the surroundings (and vice versa). when a system undergoes a physical or chemical change, the change in internal energy is given by the heat released or absorbed by the system plus the work done on or by the system First Law of Thermodynamics first law of thermodynamics: Chapter 6

8. 8 Relating  E to Heat and Work: First Law of Thermodynamics Chapter 6

9. First Law of Thermodynamics Chapter 69

10. 10 Exothermic and Endothermic Processes endothermic: absorbs heat from the surroundings example: (an endothermic reaction feels cold) First Law of Thermodynamics Chapter 6

11. 11 Exothermic and Endothermic Processes exothermic: transfers heat to the surroundings (an exothermic reaction feels warm/hot) example: First Law of Thermodynamics Chapter 6

12. 12 State Functions First Law of Thermodynamics  E is a state function; that is, the value of  E depends only on the initial and final states of system, not on how change occurred Chapter 6

13. 13 we can measure the change in enthalpy: Enthalpy enthalpy is a state function  H = H final – H initial = q p Chapter 6 enthalpy (H): heat transferred between the system and its surroundings while pressure is constant;

14. 14 Enthalpies of Reaction For a reaction: enthalpy is an extensive property (magnitude  H is directly proportional to amount): CH 4 (g) + 2 O 2 (g)  CO 2 (g) + 2 H 2 O (g)  H = - 802 kJ 2 CH 4 (g) + 4 O 2 (g)  2 CO 2 (g) + 4 H 2 O (g)  H = Chapter 6

15. 15 Enthalpies of Reaction When reaction is reversed the sign of  H is reversed: CO 2 (g) + 2 H 2 O (g)  CH 4 (g) + 2 O 2 (g)  H = Change in enthalpy also depends on physical state: H 2 O (g)  H 2 O (l)  H = - 88 kJ Chapter 6

16. 16 Example: Enthalpies of Reaction Chapter 6

17. 17 Example: Enthalpies of Reaction Chapter 6

18. 18 Calorimetry calorimeter = apparatus that measures heat flow by measuring the change in temperature heat capacity (C) = the amount of energy required to raise the temperature of an object by one degree Celsius (J / o C) calorimetry = measurement of heat flow Chapter 6 Heat Capacity and Specific Heat

19. 19 Calorimetry molar heat capacity = heat capacity of 1 mol of a substance [J/(mol. o C)] Chapter 6 Heat Capacity and Specific Heat

20. 20 specific heat (s) = specific heat capacity = heat capacity of 1 g of a substance unit: J/(g. o C) Calorimetry Heat Capacity and Specific Heat Chapter 6 that is, the amount of energy required to raise the temperature of 1 g of substance by one degree Celsius (J /g o C)

21. 21 Calorimetry Heat Capacity and Specific Heat heat released/absorbed: q = (specific heat) x (grams of substance) x  t = = s.h. X m x  t where  t = t final - t initial Chapter 6

22. 22 Examples: A) Calorimetry Chapter 6

23. 23 Examples: Calorimetry B) Chapter 6

24. 24 Constant-Pressure Calorimetry  H = q p at constant atmospheric pressure: in such system, we assume that no heat is lost to surroundings Calorimetry Chapter 6 q rxn + q water = 0 q rxn = - q water therefore

25. 25 Example: Calorimetry Chapter 6 When a student mixes 50.0 mL of 1.0 M NaOH solution and 50.0 mL of 1.0 M HCl solution in a coffee cup calorimeter, the temperature of the resultant solution increases from 21.3 o C to 27.8 o C. Calculate the enthalpy change for this neutralization reaction.

26. 26 Example: Calorimetry Chapter 6 Assume that i. the calorimeter loses only negligible quantity of heat, ii. the total volume of the solution is 100 mL, iii. the density and the specific heat of the solution are the same as those of water, 1.00 g/mL and 4.184 J/g o C, respectively. (these assumptions are usually true, unless stated otherwise)

27. 27 Example: Calorimetry Chapter 6

28. 28 Example: Calorimetry Chapter 6

29. Chapter 529 Constant-Volume Calorimetry (Bomb Calorimetry) rxn carried out under constant volume uses a device called bomb calorimeter usually used to study combustion rxns Note: since pressure is not constant under these conditions, q measured this way is not equal to  H. q water + q cal = - q rxn Calorimetry q rxn + q water + q cal = 0

30. 30 Example: Calorimetry Chapter 6 In a laboratory test 9.20 g of ethanol, C 2 H 5 OH was burned in a bomb calorimeter that contained 500.0 g of water and the heat capacity of the calorimeter is 4.821 kJ/ o C. The temperature increased from 22.9 o C to 24.85 o C. A) Calculate the heat of combustion per gram ethanol.

31. 31 Calorimetry Chapter 6 Example:

32. 32 B) Calorimetry Chapter 6 Example:

33. 33 For example: CH 4 (g) + 2 O 2 (g)  CO 2 (g) + 2 H 2 O (g)  H = - 802 kJ Hess’s Law Hess’s law: if a reaction is carried out in a number of steps,  H for the overall reaction is the sum of  H for each individual step. 2 H 2 O (g)  2 H 2 O (l)  H = - 88 kJ Chapter 6

34. 34  H 1 =  H 2 +  H 3 Hess’s Law Note that: Chapter 6

35. 35 Hess’s Law N 2 (g) + 2 O 2 (g) 2 NO 2 (g)  H 1 = + 67.6 kJ calculate the heat of reaction (enthalpy of reaction,  H rxn ) for the following rxn: N 2 (g) + O 2 (g) 2 NO (g). 2 NO (g) + O 2 (g) 2 NO 2 (g)  H 2 = - 521 kJ From the two reactions: Chapter 6

36. 36 Hess’s Law Chapter 6

37. 37 Hess’s Law Chapter 6

38. 38 If 1 mol of compound is formed from its constituent elements, then the enthalpy change for the reaction is called the enthalpy of formation,  H f. Enthalpies of Formation Standard enthalpy,  H o, is the enthalpy measured when everything is in its standard state (every component). Standard conditions (standard state): 1 atm and 25 o C (298 K). Chapter 6

39. 39 Enthalpies of Formation Molar enthalpy of formation: 1 mol of compound is formed from substances in their standard states. If there is more than one state for a substance under standard conditions, the more stable one is used. Chapter 6

40. 40 Standard enthalpy of formation of the most stable form of an element is zero. Enthalpies of Formation Chapter 6 See Table 6.5

41. 41 Enthalpies of Formation Chapter 6

42. 42 Using a set of  H o f values to calculate  H rxn. We use Hess’ Law to calculate the enthalpy of any reaction using the enthalpies from the table of enthalpies of formation. Enthalpies of Formation where - “  “ means “the sum” - n and m are stoichiometric coefficients for the each product and reactant, respectively

43. Enthalpies of Formation Example: Consider the following combustions reaction of methane: Using Hess’ Law and the relevant standard enthalpies of formation, calculate  H rxn for this reaction. CH 4 (g) + O 2 (g)  CO 2 (g) + H 2 O (g)  H rxn = ?? Chapter 643

44. 44 Enthalpies of Formation Chapter 6 Example:

45. 45 Example: Enthalpies of Formation Chapter 6

46. 46 Example: Enthalpies of Formation Chapter 6

47. 47 Example: Enthalpies of Formation Chapter 6

48. 48 End of Chapter 6 Thermochemistry Chapter 6