Chapter 6

1. Chapter 6 Thermochemistry JozsefDevenyiDepartment of Chemistry, UTM Chapter 5

2. The Nature of Energy Recall: force: a push or pull on an object work: the product of force applied to an object over a distance energy: the work done to move an object against a force (energy is the capacity to do work or transfer heat) • kinetic energy is the energy of motion: Chapter 5

3. The Nature of Energy • potential energy is the energy an object possesses by • virtue of its position • potential energy can be converted • into kinetic energy example: a ball is falling from a balcony Chapter 5

4. The Nature of Energy Units of Energy SI Unit for energy is the joule, J: for example, an object with a mass of 2 kg that moves at a speed of 1 m/s; its kinetic energy is We sometimes use the calorie instead of the joule: 1 cal = 4.184 J (exactly) Chapter 6

5. The Nature of Energy Systems and Surroundings • system: part of the universe we are interested in • surroundings: the rest of the universe Chapter 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. First Law of Thermodynamics Relating DE to Heat and Work: • Energy cannot be created or destroyed. first law of thermodynamics: energy of (system + surroundings) is constant • Any energy transferred from a system must be • transferred to the surroundings (and vice versa). • internal energy of a system: when a system undergoesa physical orchemical 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 Chapter 6

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

9. First Law of Thermodynamics Chapter 6

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

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

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

13. Enthalpy enthalpy (H): heat transferred between the system and its surroundings while pressure is constant; we can measure the change in enthalpy: DH = Hfinal – Hinitial = qp enthalpy is a state function Chapter 6

14. Enthalpies of Reaction For a reaction: enthalpy is an extensive property (magnitude DH is directly proportional to amount): CH4 (g) + 2 O2 (g) CO2(g) + 2 H2O(g) DH= - 802 kJ 2 CH4 (g) + 4 O2(g) 2 CO2(g) + 4 H2O(g)DH = Chapter 6

15. Enthalpies of Reaction When reaction is reversed the sign of DH is reversed: CO2 (g) + 2 H2O(g) CH4 (g) + 2 O2 (g)DH = Change in enthalpy also depends on physical state: H2O(g)H2O(l)DH = - 88 kJ Chapter 6

16. Enthalpies of Reaction Example: Chapter 6

17. Enthalpies of Reaction Example: Chapter 6

18. Calorimetry Heat Capacity and Specific Heat calorimetry = measurement of heat flow 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 / oC) Chapter 6

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

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

21. Calorimetry Heat Capacity and Specific Heat heat released/absorbed: q = (specific heat) x (grams of substance) xDt= = s.h.X m xDt where Dt = tfinal- tinitial Chapter 6

22. Calorimetry Examples: A) Chapter 6

23. Calorimetry Examples: B) Chapter 6

24. Calorimetry Constant-Pressure Calorimetry • at constant atmospheric • pressure: DH = qp • in such system, we assume that no heat is lost to surroundings qrxn + qwater= 0 therefore qrxn = - qwater Chapter 6

25. Calorimetry Example: When a student mixes 50.0 mL of 1.0 MNaOHsolution and 50.0 mL of 1.0 MHCl solution in a coffee cup calorimeter, the temperature of the resultant solution increases from 21.3 oCto 27.8 oC. Calculate the enthalpy change for this neutralization reaction. Chapter 6

26. Calorimetry Example: 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 oC, respectively. (these assumptions are usually true, unless stated otherwise) Chapter 6

27. Calorimetry Example: Chapter 6

28. Calorimetry Example: Chapter 6

29. Calorimetry Constant-Volume Calorimetry (Bomb Calorimetry) • rxn carried out under constant volume • uses a device calledbomb calorimeter • usually used to study • combustion rxns qrxn+ qwater + qcal= 0 qwater+ qcal = - qrxn Note: since pressure is not constant under these conditions, q measured this way isnotequal to DH. Chapter 5

30. Calorimetry Example: In a laboratory test 9.20 g of ethanol, C2H5OH was burned in a bomb calorimeter that contained 500.0 g of water and the heat capacity of the calorimeter is 4.821 kJ/oC. The temperature increased from 22.9 oC to 24.85 oC. A) Calculate the heat of combustion per gram ethanol. Chapter 6

31. Calorimetry Example: Chapter 6

32. Calorimetry Example: B) Chapter 6

33. 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. For example: CH4 (g) + 2 O2 (g) CO2 (g) + 2 H2O(g) H = - 802 kJ 2 H2O(g) 2 H2O(l)H = - 88 kJ Chapter 6

34. Hess’s Law H1 = H2 + H3 Note that: Chapter 6

35. Hess’s Law From the two reactions: N2(g) + 2 O2(g) 2 NO2(g)DH1 = + 67.6 kJ 2 NO(g) + O2(g)2 NO2(g)DH2 = - 521 kJ calculate the heat of reaction (enthalpy of reaction, DHrxn) for the following rxn: N2(g) + O2(g) 2 NO(g) . Chapter 6

36. Hess’s Law Chapter 6

37. Hess’s Law Chapter 6

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

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

41. Enthalpies of Formation Chapter 6

42. Enthalpies of Formation Using a set of DHof values to calculate DHrxn . We use Hess’ Law to calculate the enthalpy of any reaction using the enthalpies from the table of enthalpies of formation. where - “S“ means “the sum” - n and m are stoichiometric coefficients for the each product and reactant, respectively Chapter 6

43. Enthalpies of Formation Example: Consider the following combustions reaction of methane: CH4 (g) + O2 (g) CO2(g) + H2O(g) DHrxn = ?? Using Hess’ Law and the relevant standard enthalpies of formation, calculate DHrxn for this reaction. Chapter 6

44. Enthalpies of Formation Example: Chapter 6

45. Enthalpies of Formation Example: Chapter 6

46. Enthalpies of Formation Example: Chapter 6

47. Enthalpies of Formation Example: Chapter 6

48. End of Chapter 6 Thermochemistry Chapter 6