Chapter 9 Energy, Enthalpy and Thermochemistry

1. Chapter 9Energy, Enthalpy and Thermochemistry • The study of energy and its interconversions is called thermodynamics. • Kinetic Energy: energy due to the motion of the object (1/2 mv2) • Potential Energy: energy due to position or composition • Heat: the transfer of energy between two objects due to a temperature difference • Work: a force acting over a distance

2. Figure 9.1 (a): Initial position of balls

3. Figure 9.1 (b): Final position of balls

4. Energy of Matters

5. State Function • A property of the system depends only on its present state. A state function does not depend in any way on the system’s past. • Energy is a state function, but work and heat are not state function

6. Combustion of methane

7. Nitrogen/oxygen

8. First Law of Thermodynamics The energy of the universe is constant In closed system ∆E= ∆U=q+w (內能=熱能+功) q: the heat added to the system during the process w: the work done on the system during the process q>0 heat flows into the system from the surroundings q<0 an outflow of heat from the system to the surroundings w>0 work is done on the system by the surroundings w<0 the system does work on the surroundings

9. P-V Work

10. P-V Work

11. Enthalpy (焓) The heat qp absorbed in a constant-pressure process equals the system’s enthalpy change.

12. For a chemical reaction ΔH=ΔHproducts-ΔHreactants If ΔHreactants<ΔHproducts (endothermic) If ΔHreactants>ΔHproducts (exothermic) Consider a constant-volume process dw=-PdV=0 (體積固定不做功) ∆U=q+w=qv ∆U=qv

13. Heat Capacity (熱容量)

14. Thermodynamics of Ideal Gases

16. Heat Capacity of Heating an Ideal Monatomic Gas • Under constant volume, the energy flowing into the gas is used to increase the translational energy of the gas molecules.

17. Heat Capacity of Ideal Monatomic Gases

18. Heat Capacity of DiatomicGases

19. Heat Capacity of PolyatomicGases

20. Heat Capacity of Heating a Polyatomic Gas • Polyatomic gases have observed values for Cv that are significantly greater than 3/2 R. • This larger value for Cv results because polyatomic molecules absorb energy to excite rotational and vibrational motions in addition to translational motions.

21. Cv and Cp of molecules

22. Cv and Cp of Monatomic Gas

23. Cv and Cp of H2O at 373K

25. 2 mol of monatomic ideal gas Calculate q, w, ∆U and ∆H for both pathway

26. TA=122K, TC=366K, TB=183K, TD=61K Cv=3/2R, Cp=5/2R Path 1(A→C) w1=-P∆V=-2atm×(30-10)L×101.3J/Latm=-4.05×103J q1=qp=nCp∆T=2×5/2(R)×(366-122)=1.01×104J=∆H1 ∆U1=nCv∆T= 2×3/2(R)×(366-122)=6.08×103J Path 2(C→B) q2=qv=nCv∆T=2×3/2(R)×(183-366)=-4.56×103J=∆U2 ∆H2=nCp∆T= 2×5/2(R)×(183-366)=-7.6×103J ∆V=0 w2=-P∆V =0

27. Path 3(A→D) q3=qv=nCv∆T=2×3/2(R)×(61-122)=-1.52×103J=∆U3 ∆H3=nCp∆T= 2×5/2(R)×(61-122)=-2.53×103J ∆V=0 w2=-P∆V =0 Path 4(D→B) w1=-P∆V=-1atm×(30-10)L×101.3J/Latm=-2.03×103J q4=qp=nCp∆T=2×5/2(R)×(183-61)=5.08×104J=∆H4 ∆U4=nCv∆T= 2×3/2(R)×(183-61)=3.05×103J

28. Path 1 qpath1=q1+q2=5.5 ×103J wpath1=w1+w2= -4.05×103J ∆Hpath1= ∆H1 +∆H2= 2.55×103J ∆Upath1= ∆U1 +∆U2= 1.52×103J Path 2 qpath2=q3+q4=3.56×103J wpath2=w3+w4= -2.03×103J ∆Hpath2= ∆H3 +∆H4= 2.55×103J ∆Upath2= ∆U3 +∆U4= 1.52×103J Summary

30. Calorimetry Specific heat capacity with unit JK-1g-1 Molar heat capacity with unit JK-1mol-1

31. A constant-pressure calorimetry is used in determining the change in enthalpy equals the heat. ∆H=qp=nCp∆T Coffee Cup Calorimeter

32. ∆V=0 ∆U=qv+0=qv ∆U=qv=nCv ∆ T Bomb Calorimeter

33. 2SO2(g)+O2(g)→2SO3(g) ∆H=-198 KJ 2 mol. 1 mol. 2 mol. Calculate ∆H and ∆U

34. P is constant, ∆H=qp=-198 KJ (energy flow out of system) ∆U= qp +w w=-P∆V and ∆V=∆n(RT/P) T and P are constant, ∆n=nfinal-ninitial=-1 mol So w=-P∆V=-P×∆n× (RT/P) =- ∆nRT=-(-1)(8.314)(298)=2.48 kJ ∆U= qp +w=-198 kJ+2.48 kJ=-196 kJ

35. Hess’s Law • If a reaction is carried out in a series of steps, ∆H for the reaction will be equal to the sum of the enthalpy changes for the individual steps • The overall enthalpy change for the process is independent of the number of steps or the particular nature of the path by which the reaction is carried out.

36. Consider the combustion reaction of methane to form CO2 and liquid H2O CH4(g) + 2O2(g)→ CO2(g) + 2H2O(l) ∆H1 =-890KJ/mol This reaction can be thought of as occurring in two steps: CH4(g) + 2O2(g)→ CO2(g) + 2H2O(g) ∆H2 = -802 kJ/mol 2H2O(g)→2H2O(l) ∆H3 =-88KJ/mol ∆H1 = ∆H2 + ∆H3

38. Standard Enthalpies of Formation • The change in enthalpy that accompanies the formation of 1 mole of a compound from its elements with all substances in their standard states. • The superscript zero indicates that the corresponding process has been carried out under standard conditions.

42. Present Sources of Energy Petroleum and Natural Gas Coal New Energy Sources Coal Conversion Hydrogen as a fuel

46. CO2 capture