1. AP CHEMISTRY NOTES Ch 6 Thermochemistry Ch 8.8 Covalent bond energies
Ch. 16 Spontaneity, Entropy, and Free Energy

2. 6.1 The Nature of Energy Energy? Law conservation of energy
The capacity to do work to produce heat
Law conservation of energy
Energy can be converted from one form to another but can be either created nor destroyed.

3. 6.1 The Nature of Energy Kinetic energy Temperature Heat 1/2mv2
Property that reflects the random motions of the particles in a particular substance.
Heat
Transfer of energy b/w two objects due to a temperature difference.

4. 6.1 The Nature of Energy Work Transfer of energy
Force acting over a distance
Transfer of energy
Through work
Through heat
In other words, the energy transferred into two ways.

5. 6.1 The Nature of Energy
- Energy change is independent of the pathway; however, work and heat are dependent on the pathway.

6. 6.1 The Nature of Energy
Sign
- Or +
*Chemical energy Released or Absorbed through chemical reaction *System *Surroundings (anything other than reactants and products)

7. Exothermic Process Energy flows out of the system
Delta PE is the change in PE

8. In any exothermic reaction, some of the potential energy stored in the chemical bonds converted to thermal energy via heat

9. Endothermic Process
Energy flow is into a system

10. Thermodynamics First law of thermodynamics
The law of conservation of energy
(The energy of the universe is constant)

11. Internal energy The internal energy E of a system
= the sum of the kinetic and potential energies in the system
Delta E = q + w
(q = heat, w = work)
Consist of a number (magnitude) and sign (direction of the flow)

12. q = -x (out of the system) q = +x (into the system)
Work from the system’s point of view

13. Figure 6.4 The Piston, Moving a Distance Against a Pressure P, Does Work On the Surroundings
P=F/A
Work = F x ∆ h
= PA x ∆ h
= P x ∆ V
Sign?
Work and ∆V must have opposite sign
W= - P x ∆ V

14. 6.2 Enthalpy and Calorimetry
Enthalpy (H)
H=E + PV
E is internal energy (KE+PE)
PV is pressure-volume P
Enthalpy is heat at constant pressure
∆H=∆E + ∆(PV)
∆H=qP (heat or energy flow)

15. Ex 6.4

16. Calorimetry Calorimeter Calorimetry Heat capacity C
(heat absorbed /increase in temp)
Specific heat capacity s
(per gram or per mol)
Cwater=4.18 J/oC∙g
4.18J is required to increase 1oC of 1g water
Coffee cup calorimeter

17. Extensive or intensive properties?
Heat=an extensive prop., depends on amount of substance
Temp=an intensive prop., not related to the amount of substance

18. Ex 6.5 Constant pressure calorimetry
q=mc∆t
Assume solution specific heat capacity is same as water
∆t= final t – initial t

19. Example
Consider a calorimeter at constant pressure ml of a 4.0 M HCl solution is mixed with 50.0 ml of a 2.0 M NaOH solution at 25˚C. After mixing, the temperature of the solution is 31.9˚C.
Calculate the energy released.

20. Calculate the energy released per mole of H+ neutralized
Consider a calorimeter at constant pressure ml of a 4.0 M HCl solution is mixed with 50.0 ml of a 2.0 M NaOH solution at 25˚C. After mixing, the temperature of the solution is 31.9˚C.
Calculate the energy released per mole of H+ neutralized

21. Constant volume calorimetry
Coffee cup and Bomb calorimeter 1

22. Constant volume calorimetry
∆E=q+w (w=0, constant V)
∆E=q
Ex 6.6
E released / mass

23. Constant volume
A calorimeter has a heat capacity of 195 J/K. It contains 100. grams of water at 25˚C grams of aluminum (heat capacity of J/gK) at ˚C was added to it. What is the final temperature of the solution?

24. 30.1oC

25. Constant volume
A calorimeter has a heat capacity of 50. cal/K grams of cold water at 15˚C is added to it. Then some hot water at 80˚C is added to it. The final temperature of the water is 30. ˚C. What is the mass of hot water added?

26. 45g

27. Constant pressure
A calorimeter contains water at 25˚C. Describe the heat flow when a metal at 50.0˚C is placed into the water.

28. Constant pressure
A 15.0g sample of nickel metal is heated to 100.0˚C and dropped into 55.0g of water, initially at 23.0˚C. Assuming that all the heat lost by the nickel is absorbed by the water, calculate the final temperature of the nickel and water. Nickel has a specific heat of J/˚Cg.

29. In a coffee cup calorimeter, 1
In a coffee cup calorimeter, 1.60g of ammonium nitrate is mixed with 75.0g of water at an initial temperature of 25.00˚C. After dissolution of the salt, the final temperature of the calorimeter contents is 23.34˚C. Assuming the solution has a heat capacity of 4.18 J/˚Cg and assuming no heat loss to the calorimeter, calculate the enthalpy change for the dissolution of ammonium nitrate in units of kJ/mol.

30. 6.3 Hess’s law ∆H is not dependent on the reaction pathway.
Is a state of function
Initial
Final

31. 6.3 Hess’s law
The change in enthalpy is the same whether the reaction takes place in one step or in a series of steps.

32. 6.3 Hess’s law
The change in enthalpy is the same whether the reaction takes place in one step or in a series of steps.

33. Rules of Hess’s law
If a sub-reaction is multiplied by a coefficient, ∆Ho value is also multiplied by the same coefficient.
If a sub-reaction is reversed (flipped), ∆Ho value is switched the sign.
Hint!
Work backward
Reverse any reactions
Multiply reactions

34. Example of Hess’s law
H2 (g) + Cl2 (g)  2HCl (g) ΔH˚ = ? Given: NH3 (g) + HCl (g)  NH4Cl (s) ΔH˚ = kJ N2 (g) + 3H2 (g)  2NH3 (g) ΔH˚ = kJ N2 (g) + 4H2 (g) + Cl2 (g)  2NH4Cl (s) ΔH˚ = kJ

35. ΔH˚ = kJ

36. Example of Hess’s law
C2H2(g) + 2H2(g)  C2H6(g) ΔH˚ = ? Given: 2C2H2(g) + 5O2(g)  4CO2(g) + 2H2O(l) ΔH˚ = kJ 2C2H6(g) + 7O2(g)  4CO2(g) + 6H2O(l) ΔH˚ = kJ H2(g) + 1/2 O2(g)  H20(l) ΔH˚ = kJ

37. ΔH˚ = -312 kJ

39. Example of Hess’s law Given: 2H2(g) + C(s)  CH4(g) ΔH˚ = 74.81 kJ
CH4(g) + 2O2(g)  CO2(g) + 2H2O(l) ΔH˚ = ?
Given:
2H2(g) + C(s)  CH4(g) ΔH˚ = kJ
2H2(g) + O2(g)  2H2O(l) ΔH˚ = kJ
C(s) + O2(g)  CO2(g) ΔH˚ = kJ

40. ΔH˚ = kJ

41. Example of Hess’s law
2NH3(g) + 3N2O(g)  4N2(g) + 3H2O(l) ΔH˚ = ? Given: 2NH3(g) + 3/2 O2(g)  N2(g) + 3H2O(l) ΔH˚ = kJ/ mole 3N2O(g) + 3H2(g)  3N2(g) + 3H2O(l) ΔH˚ = kJ/ mole 3H2O(l)  3H2(g) + 3/2O2(g) ΔH˚ = kJ/ mole

42. ΔH˚ = kJ/mol

43. 6.4 Standard Enthalpies of Formation
∆Hof (under standard condition)
The change is enthalpy that accompanies the formation of one mole of a compound from its elements with all substances in their states.
We can’t determine absolute values of enthalpy

44. Reference P
Conc.=1M
25oC

45. ∆Hof = per mole of product (kJ/mol)
See appendix 4
Tabulated ∆Hof value – using this for calculation

46. Ex. CH4(g)+2O2(g)CO2(g)+2H2O(l)
*Add all up A: CH4 CH4 C + 2H2 ∆H=-75kJ/mol Flip! C + 2H2  CH4 ∆H=75kJ/mol B: 2O2 ∆H=0kJ (element) C: CO2 C + O2  CO2 ∆H=-394 kJ D: 2H2O H2 + ½ O2 H2O ∆H=2x(-286) kJ (75kJ)+(-394kJ)+(-572kJ)= -891kJ

47. Ex. CH4(g)+2O2(g)CO2(g)+2H2O(l)
*Sum of produced – sum of required A: CH4 CH4 C + 2H2 ∆H=-75kJ/mol You don’t have to flip! B: 2O2 ∆H=0kJ (element) C: CO2 C + O2  CO2 ∆H=-394 kJ D: 2H2O H2 + ½ O2 H2O ∆H=2x(-286) kJ (-394kJ)+(-572kJ)-[0+(-75kJ)]= -891kJ

50. Ex. 6.9 4NH3(g)+7O2(g)4NO2(g)+6H2O(l)

51. Table 6.2 Standard Enthalpies of Formation for Several Compounds at 25°C
Copyright © Houghton Mifflin Company. All rights reserved.

52. Ex. 6.9 4NH3(g)+7O2(g)4NO2(g)+6H2O(l)
-1396kJ

53. Ex. 6.10 2Al(s)+Fe2O3(s)Al2O3(s)+2Fe(s)

54. Ex. 6.10 2Al(s)+Fe2O3(s)Al2O3(s)+2Fe(s)
-850.kJ

55. 6.11 CH3OH or C8H18? Need two equations
_CH3OH (l) + _O2(g)_CO2(g)+_H2O(l)
_C8H18(l) + _O2(g)_CO2(g)+_H2O(l)
Finally divide by molar mass for both

56. 6.11 CH3OH or C8H18? 2CH3OH (l) + 3O2(g)2CO2(g)+4H2O(l)
-1454kJ/2mol*32.0g = kJ/g
2C8H18(l) + 25O2(g)16CO2(g)+18H2O(l)
-10914kJ/2mol*114.2g = kJ/g

57. 6.5 Present sources of energy

58. 6.6 New energy sources