1. Chapter 7 Making and Breaking of Bonds Chemical reactions are characterized by the making and breaking of chemical bonds. One possible consequence of a chemical reaction is a transfer of energy from the system to the surroundings.

2. Energy ● Hydrocarbons ● Carbohydrates Reaction with oxygen to release energy.

3. Energy ● Kinetic Energy ● Potential Energy ● Transfer of energy ● Conversion of energy

4. Energy ● Energy is released when making a chemical bond. ● Energy is absorbed when breaking a chemical bond.

5. Energy ● “This chapter is concerned with the energy transfers and conversions associated with chemical reactions. These processes are part of the area of study known as Thermodynamics.” Thermodynamics.”

6. Heat ● "Heat is energy in transit." ● No energy transferred means no heat interaction.

7. Heat ● Heat has units of energy. ● Heat is not the same as temperature. ● In order for a heat interaction to occur between two systems, they must have different temperatures. This heat interaction can form the basis for a definition of temperature!

8. Heat and the Kinetic Molecular Theory ● System ● Surroundings ● Boundary Figure 7.1

9. Specific Heat Consider the following: Figure 7.2

10. Specific Heat ● Water and mercury have different specific heats. ● Specific heat is the energy required to raise the temperature of one gram of material one degree Celsius. Units – cal/°C g – J/°C g

11. Specific Heat ● Molar heat capacity Specific heat × molecular weight Quantity of heat required to raise 1 mole of a substance 1 degree Celsius.

12. Specific Heat Table 7.1

13. State Functions ● State of a system P, T, n, V, … ● Extensive properties of a system depend on the size of the system. ● Intensive properties of a system are independent of the system size. ● State functions are independent of a system’s history.

14. State Functions ● H is a state function So are T, P, V, E,...

15. The First Law of Thermodynamics ● Energy is conserved. ● No exceptions have been observed. Yet. That's why it's called a law.

16. The First Law of Thermodynamics ● Energy can be exchanged between the system and the surroundings. ● Energy cannot appear or disappear. ● Energy entering or leaving a system can do so under two forms: Heat (q) Work (w)

17. Work ● Heat transfers can be made to occur under two separate special conditions: System held at constant volume, or System held at constant pressure.

18. Work ● Energy transfer where the system maintains a constant volume is a heat interaction (q) only. ● Additionally, a work interaction (w) can result if the system changes its volume during the energy transfer.

19. Work ● These special conditions have a profound effect on the form (q or w) in which energy can be transferred. ● Chemistry is typically done at constant pressure. Why?

20. Work ● When energy is exchanged in the form of heat with a system held at constant pressure, the heat energy, q P, is described with a new term, enthalpy (H).

21. Work ● In chemistry, the term enthalpy of reaction (ΔH) is used. It says how much energy is released or consumed in the form of heat if the reaction occurs under the condition of constant pressure.

22. Work Table 7.2

23. The Enthalpy of a System ● Enthalpy, H, is a state function. ● Calorimetry Measures heat interactions associated with chemical and physical changes. Figure 7.6

24. The Enthalpy of a System ● Calorimeters like the one shown measure ΔE. ● A little math is used to determine ΔH once ΔE is measured.

25. Enthalpies of Reaction ● ΔH = heat of reaction at constant P. May be >0endothermic reaction May be <0exothermic reaction

26. Enthalpies of Reaction Enthalpy of reaction (ΔH) is an extensive property. It changes with the stoichiometric coefficients. 2H 2 (g) + O 2 (g) → 2H 2 O(g) ΔH = -483.64 kJ 4H 2 (g) + 2O 2 (g) → 4H 2 O(g) ΔH = -967.28 kJ

27. Enthalpies of Reaction The sign of ΔH changes for a reverse reaction. 2H 2 (g) + O 2 (g) → 2H 2 O(g) ΔH = -483.64 kJ 2H 2 Og) → 2H 2 (g) + O 2 (g) ΔH = +483.64 kJ

28. Enthalpies of Reaction The heat released for any amount of product or reactant can be determined from a single thermochemical equation. 2H 2 (g) + O 2 (g) → 2H 2 O(g) ΔH = -483.64 kJ 2H 2 (g) + O 2 (g) → 2H 2 O(g) ΔH = -483.64 kJ How much heat will be released when 50 grams of H 2 (g) are consumed?

29. Enthalpies of Reaction

30. Enthalpy as a State Function ● H is a state function. ● ΔH = H final - H initial ● Independent of path: a. Convert reactant into atoms b. Make products from same atoms c. No extra or missing atoms!

31. Enthalpy as a State Function ● Reactions do not proceed this way. ● It doesn't matter when calculating ΔH. ● H is a state function.

32. Standard-State Enthalpies of Reaction ● For a reaction carried out at a pressure of 1 bar, ΔH = ΔH°. ● ΔH° is called the standard-state enthalpy of reaction. ● 1 bar is part of the definition of standard conditions. ● Most tabulated enthalpies of reaction are ΔH°.

33. Calculating Enthalpies of Reaction ● Easier to calculate ΔH° than to measure it. ● These are calculations, not estimates. Break all the reactant bonds. Form all the product bonds. ΔH° = the difference in energy between these two processes.

34. Enthalpies of Atom Combination ● N(g) + 3 H(g) → NH 3 (g) ● Notice these reactants are not in their diatomic elemental forms. ● If 1 mole of N atoms combine with 3 moles of H atoms to produce one mole of NH 3 molecules, 1171.76 kJ of energy will be released.

35. Enthalpies of Atom Combination ● N(g) + 3 H(g) → NH 3 (g) ● Reaction is called atom combination. ● Not meant to reflect actual mechanism. ● The enthalpy associated with it is called the enthalpy of atom combination, ΔH° ac. ● In the above example, ΔH° ac = -1171.76 kJ. ● Why is it negative?

36. Enthalpies of Atom Combination The reverse reaction NH 3 (g) → N(g) + 3 H(g) is called atomization. The enthalpy change is called the enthalpy of atomization and in this example equals +1171.76 kJ.

37. Enthalpies of Atom Combination ● Can be used to study physical processes. ● How much heat is required to accomplish the following? CH 3 OH(l) → CH 3 OH(g)

38. Enthalpies of Atom Combination ● Can be used to study physical processes ● How much heat is required to accomplish the following? CH 3 OH(l) → CH 3 OH(g) ΔH° ac = -2037.11 kJ for CH 3 OH(g) ΔH° ac = -2075.11 kJ for CH 3 OH(l)

39. Enthalpies of Atom Combination ● Can be used to study physical processes ● CH 3 OH(l) → CH 3 OH(g) ΔH = -2037.11 kJ - (-2075.11 kJ) = +38.00 kJ

40. Enthalpies of Atom Combination ● Can be used to study chemical processes ● 2H 2 (g) + O 2 (g) → 2H 2 O(g) ΔH° = ?

41. Enthalpies of Atom Combination ● Can be used to study chemical processes ● 2H 2 (g) + O 2 (g) → 2H 2 O(g) ΔH°=2×ΔH° ac (H 2 O(g)) -2×ΔH° ac (H 2 (g))-1×ΔH° ac (O 2 (g) )

42. Enthalpies of Atom Combination ● Can be used to study chemical processes ● 2H 2 (g) + O 2 (g) → 2H 2 O(g) ΔH°=2×ΔH° ac (H 2 O(g)) – 2×ΔH° ac (H 2 (g)) - 1×ΔH° ac (O 2 (g) ) From Appendix B.13 ΔH° = 2×(-926.29 kJ) - 2×(-435.30 kJ) -1×(-498.340 kJ) ΔH° = -483.64 kJ

43. Enthalpies of Atom Combination ● Can be used to study chemical processes. ● 2H 2 (g) + O 2 (g) → 2H 2 O(g) ● Established this as exothermic. ● Determined that 483.64 kJ of energy will be released for each mole of O 2 (g) consumed.

44. Using Enthalpies of Atom Combination to Probe Chemical Reactions Isomers

45. ● Knowing ΔH° ac for each isomer allows for the calculation of the enthalpy change associated with the transformation

46. Using Enthalpies of Atom Combination to Probe Chemical Reactions ● ΔH° ac ● Knowing ΔH° ac gives insight into average bond strengths. ● ΔH° ● These provide a microscopic interpretation of overall ΔH° for a reaction. Endothermic or exothermic character ΔH° Magnitude of ΔH°

47. Using Enthalpies of Atom Combination to Probe Chemical Reactions 4HF(g) + SiO 2 (g) → SiF 4 (g) + 2H 2 O(g) ΔH° = -103.4 kJ 4HCl(g) + SiO 2 (g) → SiCl 4 (g) + 2H 2 O(g) ΔH° = +139.6 kJ Why the difference in sign? Why the difference in sign?

48. Using Enthalpies of Atom Combination to Probe Chemical Reactions 4HF(g) + SiO 2 (g) → SiF 4 (g) + 2H 2 O(g) ΔH° = -103.4 kJ 4HCl(g) + SiO 2 (g) → SiCl 4 (g) + 2H 2 O(g) ΔH° = +139.6 kJ Why the difference in sign? Why the difference in sign? The Si-F bond is stronger than the Si-Cl bond by an amount greater than the difference in bond strength between H-F and H-Cl as shown on the next slide. The Si-F bond is stronger than the Si-Cl bond by an amount greater than the difference in bond strength between H-F and H-Cl as shown on the next slide.

49. Bond Length and the Enthalpy of Atom Combination Table 7.4

50. Bond Length and the Enthalpy of Atom Combination ● Longer bonds tend to be weaker bonds. ● Multiple bonds tend to be stronger than single bonds. This was also covered in section 4.8.

51. Hess's Law ● An alternative method for calculating ΔH°. ● Does not use ΔH° ac. ● Takes advantage of H being a state function.

52. Hess's Law ● Desired reactions are constructed from known reactions. ● ΔH° from known reactions combined in same way to calculate ΔH° for the desired reaction. ● Trial and error method!

53. Hess's Law C(s) + H 2 O(g) → CO(g) + H 2 (g) ΔH° = ? from C(s) + ½O 2 (g) → CO(g) ΔH° = -110.53 kJ H 2 (g) + ½O 2 (g) → H 2 O(g) ΔH° = -241.82 kJ

54. Enthalpies of Formation ● Enthalpy of formation, ΔH° f Combined in same way as ΔH° ac. Used to calculate ΔH° for a reaction. Tabulated in Appendix B.16. ● Don't mix ΔH° f and ΔH° ac in a calculation. Use one set or the other.

55. Enthalpies of Formation ● Defined as enthalpy change associated with the formation of one mole of a substance under standard conditions (1 bar, T,...) from the elements in their thermodynamically stable form at T.

56. Enthalpies of Formation ● At 25 °C and 1 bar: Oxygen is O 2 (g). Carbon is C(solid, graphite). The rest can be determined from B.16 by looking for the entry with ΔH° f = 0. – Why is the entry with ΔH° f = 0 the thermodynamically stable form?