1. Chem. 412 – Phys. Chem. I

2. Sign Convention System Surroundings  ΔU = Internal Energy.  q = heat flow; transfer of energy between two objects.  w = work; product of force applied to an object over a distance.

3. Work – Basic Formulation Work Done = Force x (Distance Moved) Differentially => dw = F dS

4. Work – Gas Expansions and Compressions – I (F14)

5. Work – Gas Expansions and Compressions – II (F14)

6. Work – Gas Expansions and Compressions – III (F14)

7. Work – Gas Expansions and Compressions – IV (F14)

8. Work – Gas Expansions and Compressions – V (F14)

9. Work – Gas Expansions and Compressions – I (F13)  Work is area under curve of P-V Diagram.  dw = F dS = - P dV

10. Work – Gas Expansions and Compressions – II (F13)  Work is area under curve of P-V Diagram.  dw = F dS = - P dV

11. Work – Gas Expansions and Compressions – III (F13)  Work is area under curve of P-V Diagram.  Reversible versus Irreversible Work

12. Work – Gas Expansions and Compressions – IV (F13)  Work is area under curve of P-V Diagram.  Reversible versus Irreversible Work

13. Work – Gas Expansions and Compressions – V (F13)  Reversible versus Irreversible Work

14. Tutorial Problem on ‘Work’ Consider one mole of an ideal gas kept in a right-circular cylinder with a movable piston cap at constant temperature. The cap is attached to an external engine that is capable of moving the piston in either directions. (a)Initially, the cylinder has a volume of 1.0 m 3 at a pressure of 10. Pa. The cap was moved to give a final pressure of 1.0 Pa. In this case, the piston cap was moved at an infinitely slow rate to achieve this final pressure, thereby following the ideal gas law. Calculate the work involved in this process. This work is referred to as reversible work, w rev. [ 23 J ] (b)The initial and final volume/pressure was kept the same as in part (a) but the piston cap was suddenly released to get to the final pressure value. Calculate the work involved in this process. This work is referred to as irreversible work, w irrev. [ 9 J ] (c)Draw one P-V diagram for parts (a) and (b). (d)Repeat parts (a), (b), and (c) by reversing the initial and final volume/pressure conditions; that is, compression instead of expansion. [ 23 J, 90 J ] (e)Compare/Contrast/Discuss the above results.

16. First Law of Thermodynamics

17.  U = Internal Energy = Heat Flow under Constant Volume  H = Enthalpy = Heat Flow under Constant Pressure  H = U + PV

18. Two Heat Capacities from  q

19. Calorimetry Reaction carried out under constant volume. Use a bomb calorimeter. Usually study combustion. Bomb Calorimetry Bomb Calorimetry (Constant Volume Calorimetry)

20. Constant Pressure (Solution) Calorimetry Atmospheric pressure is constant! Calorimetry

21. Calorimetry Constant Pressure Calorimetry

22. Calorimetry Examples 1.In an experiment similar to the procedure set out for Part (A) of the Calorimetry experiment, 1.500 g of Mg(s) was combined with 125.0 mL of 1.0 M HCl. The initial temperature was 25.0 o C and the final temperature was 72.3 o C. Calculate: (a) the heat involved in the reaction and (b) the enthalpy of reaction in terms of the number of moles of Mg(s) used. Ans: (a) –25.0 kJ (b) –406 kJ/mol 2.50.0 mL of 1.0 M HCl at 25.0 o C were mixed with 50.0 mL of 1.0 M NaOH also at 25.0 o C in a styrofoam cup calorimeter. After the mixing process, the thermometer reading was at 31.9 o C. Calculate the energy involved in the reaction and the enthalpy per moles of hydrogen ions used. Ans: -2.9 kJ, -58 kJ/mol [heat of neutralization for strong acid/base reactions]Ans CyberChem video

23. Calorimetry ExamplesCalorimetry Examples: Hints 50.0 mL of 1.0 M HCl at 25.0 o C were mixed with 50.0 mL of 1.0 M NaOH also at 25.0 o C in a styrofoam cup calorimeter. After the mixing process, the thermometer reading was at 31.9 o C. Calculate the energy involved in the reaction and the enthalpy per moles of hydrogen ions used. Ans: -2.9 kJ, -58 kJ/mol [heat of neutralization for strong acid/base reactions]Ans

25. Comparison of Heat Capacities: C P vs. C V

28. Comparison of Heat Capacities: C P vs. C V (F08)

29. Example on State and Path Functions 20.0 grams of Argon gas was heated from 20.0 o C to 80.0 o C. Considering Argon behaving as an ideal gas, calculate q, w,  U, and  H for the following processes: (a)under Constant Volume, and (b)under Constant Pressure.

30. Constant Volume Constant Pressure q 374 J 623 J q w 0 -249 J w  U 374 J  U  H 623 J  H C Vm 12.5 J mol -1 K -1 20.8 J mol -1 K -1 C Pm

31. Enthalpy of Phase Transitions Constant P:s     g s  g

32. Enthalpy of Phase Transitions

34. Relating  U rxn and  H rxn

35. 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: CH 4 (g) + 2O 2 (g)  CO 2 (g) + 2H 2 O(g)  H = -802 kJ 2H 2 O(g)  2H 2 O(l)  H = -88 kJ CH 4 (g) + 2O 2 (g)  CO 2 (g) + 2H 2 O(l)  H = -890 kJ Hess’s Law

36. Given:(i)Fe 2 O 3 (s) + 3CO(g)  2Fe(s) + 3CO 2 (g)∆H = -26.7 kJ/mol (ii)CO(g) + ½O 2 (g)  CO 2 (g)∆H = -283.0 kJ/mol Calculate the heat of reaction for:2Fe(s) + 3/2 O 2 (g)  Fe 2 O 3 (s) Ans: -822.3 kJ/mol

37. If 1 mol of compound is formed from its constituent elements (standard state), then the enthalpy change for the reaction is called the enthalpy of formation,  H o f. Standard conditions (standard state): Most stable form of the substance at 1 atm and 25.00 o C (298.15 K). Standard enthalpy,  H o, is the enthalpy measured when everything is in its standard state. Standard enthalpy of formation: 1 mol of compound is formed from substances in their standard states. Standard enthalpy of formation of the most stable form of an element is zero. Enthalpies of Formation

39. Enthalpies of Formation: Example Example: Write the balanced reaction equation for the standard enthalpy of formation of solid ammonium carbonate.

40. Enthalpies of Formation Substance  o f (kJ/mol) C(s, graphite)0 O(g)247.5 O 2 (g)0 N 2 (g)0

42. Bucky Ball drawn by HyperChemHyperChem

43. Using Enthalpies of Formation to Calculate Enthalpies of Reaction For a reaction Calculate/Compare heat of reactions for the combustion of methanol gas and ethanol gas giving carbon dioxide and water. Enthalpies of Formation

46. Temperature Dependence of C P and  H rxn

47. Example: Find the heat of reaction for the following reaction at 1000. K (  H o rxn,1000 K ? ) NaCl(s)  Na(g) + ½ Cl 2 (g) [ Given:  H o rxn,298K = 519.23 kJ ] SpeciesC P / J K -1 mol -1 Na(g)20.80 NaCl(s)49.70 Cl 2 (g)31.70 + 10.14x10 -3 T – 2.72x10 -7 T 2