1. CHAPTER 4 M ATERIAL EQUILIBRIUM ANIS ATIKAH BINTI AHMAD hakitasina@gmail.com PHYSICAL CHEMISTRY 1

2. S UBTOPIC  Introduction to Material Equilibrium  Entropy and Equilibrium  The Gibbs and Helmholtz Energies  Thermodynamic Relations for a System Equilibrium  Calculation of Changes in State Function  Phase Equilibrium  Reaction Equilibrium 2

3. W HAT IS MATERIAL EQUILIBRIUM ? In each phase of the closed system, the number of moles of each substances present remains constant in time No net chemical reactions are occurring in the system No net transfer of matter from one part of the system to another Concentration of chemical species in the various part of the system are constant 3

4. Material equilibrium Reaction equilibrium Phase equilibrium 4

5.  E NTROPY AND E QUILIBRIUM Entropy, is a measure of the "disorder" of a system. What "disorder refers to is really the number of different microscopic states a system can be in, given that the system has a particular fixed composition, volume, energy, pressure, and temperature. While energy strives to be minimal, entropy strives to be maximal Entropy wants to grow. Energy wants to shrink. Together, they make a compromise. 5

6. E NTROPY AND E QUILIBRIUM Example: In isolated system (not in material equilibrium)isolated system The spontaneous chemical reaction or transport of matter are irreversible process that increase the ENTROPY The process was continued until the system’s entropy is maximized. Once it is maximized, any further process can only decrease entropy –(violate the second law) 6

7. isolated systems : is one with rigid walls that has no communication (i.e., no heat, mass, or work transfer) with its surroundings. An example of an isolated system would be an insulated container, such as an insulated gas cylinder isolated (Insulated) System : U = constant Q = 0 7

8. The system is not in material equilibrium but is in mechanical and thermal equilibrium The surroundings are in material, mechanical and thermal equilibrium System and surroundings can exchange energy (as heat and work) but not matter Since system and surroundings are isolated, we have dq surr = -dq syst ( 1) Since, the chemical reaction or matter transport within the non equilibrium system is irreversible, dS univ must be positive: dS univ = dS syst + dS surr > 0 ( 2) Consider a system at T; 8

9. The surroundings are in thermodynamic equilibrium throughout the process. Therefore, the heat transfer is reversible, and dS surr = dq surr /T (3) The systems is not in thermodynamic equilibrium, and the process involves an irreversible change in the system, therefore DS syst ≠dq syst /T (4) 9

10. Equation (1) to (3) give dS syst > -dS surr = -dq surr /T = dq syst /T (5) Therefore dS syst > dq syst /T dS > dq irrev /T (6) closed syst. in them. and mech. equilib. dq surr = -dq syst ( 1) dS surr = dq surr /T (3) dS univ = dS syst + dS surr >0 (2) 10

11. 11 When the system has reached material equilibrium, any infinitesimal process is a change from a system at equilibrium to one infinitesimally close to equilibrium and hence is a reversible process. Thus, at material equilibrium we have, ds = dq rev /T (7) Combining ( 6) and (7): ds ≥dq/T (8) material change, closed syst. in them & mech. Equilib

12. The first law for a closed system is dq = dU – dw (9) Eq 8 gives dq≤ TdS Hence for a closed system in mechanical and thermal equilibrium we have dU – dw ≤ TdS Or dU ≤ TdS + dw (10) ds ≥ dq/T (8) 12

13. A spontaneous process at constant -T-and-V is accompanied by a decrease in the Helmholtz energy, A. A spontaneous process at constant -T-and-P is accompanied by a decrease in the Gibbs energy, G. dA = 0 at equilibrium, const. T, V dG = 0 at equilibrium, const. T, P  THE GIBSS & HELMHOLTZ ENERGIES 13

14. dU  TdS + SdT – SdT + dw dU  d(TS) – SdT + dw d(U – TS)  – SdT + dw d(U – TS)  – SdT - PdV at constant T and V, dT=0, dV=0 d(U – TS)  0 dU  TdS + dw Equality sign holds at material equilibrium  H ELMHOLTZ FREE ENERGY A  U - TS Consider material equilibrium at constant T and V dw = -P dV for P-V work only 14

15. For a closed system (T & V constant), the state function U-TS, continually decrease during the spontaneous, irreversible process of chemical reaction and matter transport until material equilibrium is reached d(U-TS)=0 at equilibrium  H ELMHOLTZ FREE ENERGY 15

16. dU  d(TS) – SdT – d(PV) + VdP d(H – TS)  – SdT + VdP d(U + PV – TS)  – SdT + VdP at constant T and P, dT=0, dP=0 d(H – TS)  0 Consider material equilibrium for constant T & P, into with dw = -P dV dU  T dS + dw dU  T dS + S dT – S dT + P dV + V dP – V dP  G IBBS FREE ENERGY G  H – TS  U + PV – TS 16

17. the state function H-TS, continually decrease during material changes (constant T and P), until material equilibrium is reached. This is the minimisation of Gibbs free energy. GIBBS FREE ENERGY, G=H-TS GIBBS FREE ENERGY, G=H-TS d(H – TS)  0 G = H – TS = U + PV - TS  G IBBS FREE ENERGY 17

18. G  H – TS  U + PV – TS dG T,P  0 Equilibrium reached Constant T, P Time G G decreases during the approach to equilibrium, reaching minimum at equilibrium  G IBBS FREE ENERGY 18

19. As G of the system decrease at constant T & P, S univ increases. WHY? Consider a system in mechanical and thermal equilibrium which undergoes an irreversible chemical reaction or phase change at constant T and P. closed syst., const. T, V, P-V work only  G IBBS FREE ENERGY The decrease in G syst as the system proceeds to equilibrium at constant T and P corresponds to a proportional increase in S univ 19

20. const. T const. T, closed syst. It turns out that A carries a greater significance than being simply a signpost of spontaneous change: The change in the Helmholtz energy is equal to the maximum work the system can do: Closed system, in thermal &mechanic. equilibrium 20

21. G  H – TS  U + PV – TS G  U– TS + PV  A + PV const. T and P, closed syst. If the P - V work is done in a mechanically reversible manner, then const. T and P, closed syst. or 21

22. For a reversible change The maximum non-expansion work from a process at constant P and T is given by the value of -  G (const. T, P ) 22

23.  Thermodynamic Reactions for a System in Equilibrium 6 Basic Equations: dU = TdS - PdV H  U + PV A  U – TS G  H - TS closed syst., rev. proc., P-V work only closed syst., in equilib., P-V work only 23

24. The rates of change of U, H, and S with respect to T can be determined from the heat capacities C P and C V. Key properties closed syst., in equilib. Basic Equations Heat capacities (C P C V ) 24

25. dG = -SdT + VdP dA = -SdT - PdV dH = TdS + VdP dU = TdS - PdV The Gibbs Equations closed syst., rev. proc., P-V work only How to derive dH, dA and dG ? 25

26. The Gibbs Equations dH = d(U + PV) dH = TdS + VdP = dU + d(PV) = dU + PdV + VdP = (TdS - PdV) + PdV + VdP H  U + PV dH = ? dU = TdS - PdV 26

27. dA = d(U - TS) dG = d(H - TS) dG = -SdT + VdPdA = -SdT - PdV = dU - d(TS) = dU - TdS - SdT = (TdS - PdV) - TdS - SdT = dH - d(TS) = dH - TdS - SdT = (TdS + VdP) - TdS - SdT dA = ? dU = TdS - PdV dH = TdS+VdP dG = ? A  U - TS G  H - TS 27

28. The Power of thermodynamics: Difficultly measured properties to be expressed in terms of easily measured properties. The Gibbs equation dU= T dS – P dV implies that U is being considered a function of the variables S and V. From U= U (S,V) we have 28 (dG = -SdT + VdP)

29. The Euler Reciprocity Relations If Z ＝ f(x ， y) ， and Z has continuous second partial derivatives, then That is 29

30. The Gibbs equation (4.33) for dU is dU＝TdS－PdVdU＝TdS－PdV dS＝0dS＝0 dV＝0dV＝0 Applying Euler Reciprocity, dU = TdS - PdV The Maxwell Relations (Application of Euler relation to Gibss equations) 30

31. These are the Maxwell Relations The first two are little used. The last two are extremely valuable. The equations relate the isothermal pressure and volume variations of entropy to measurable properties. 31

32. D EPENDENCE OF S TATE F UNCTIONS ON T, P, AND V We now find the dependence of U, H, S and G on the variables of the system. The most common independent variables are T and P. We can relate the temperature and pressure variations of H, S, and G to the measurable Cp,α, and κ 32

33. Volume dependence of U The Gibbs equation gives dU ＝ TdS － PdV From Maxwell Relations Divided above equation by dV T, the infinitesimal volume change at constant T, to give For an isothermal process dU T ＝ TdS T － PdV T T subscripts indicate that the infinitesimal changes dU, dS, and dV are for a constant-T process 33

34. from Gibbs equations, d H ＝ T d S ＋ V d P Pressure dependence of H Temperature dependence of U Temperature dependence of H 34 From Basic Equations From Maxwell Relations

35. Temperature and Pressure dependence of G Temperature dependence of S Pressure dependence of S From Basic Equations The Gibbs equation (4.36) for dG is d G ＝ - S d T + V d P dT＝0dT＝0dP＝0dP＝0 35 From Maxwell Relations The equations of this section apply to a closed system of fixed composition and also to a closed system where the composition changes reversibly

36. Joule-Thomson Coefficient (easily measured quantities) from (2.65) From pressure dependence of H 36

37. Heat-Capacity Difference (easily measured quantities) From volume dependence of U 37

38. 1.As T  0, C P  C V Heat-Capacity Difference 2.C P  C V (since  > 0) 3.C P = C V (if  = 0) 38

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40. Ideal gases Solids 300 J/cm 3 (25 o C, 1 atm) Internal Pressure Liquids 300 J/cm 3 (25 o C, 1 atm) Strong intermolecular forces in solids and liquids. 40 Solids, Liquids, & Non-ideal Gases

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42.  C ALCULATION OF C HANGES IN S TATE F UNCTION 1. Calculation of ΔS Suppose a closed system of constant composition goes from state (P 1,T 1 ) to state (P 2,T 2 ), the system’s entropy is a function of T and P 42

43. Integration gives: Since S is a state function, ΔS is independent of the path used to connect states 1 and 2. A convenient path (Figure 4.3) is first to hold P constant at P 1 and change T from T 1 to T 2. Then T is held constant at T 2, and P is changed from P1 to P2. For step (a), dP=0 and gives For step (b), dT=0 and gives 43

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45. ΔU can be easily found from ΔH using : ΔU = ΔH – Δ (PV) Alternatively we can write down the equation for ΔU similar to: 2. Calculation of ΔH 45

46. 3. Calculation of ΔG For isothermal process: Alternatively, ΔG for an isothermal process that does not involve an irreversible composition change can be found as: A special case: [Since ] 46 from slide 35

47.  Phase Equilibrium A phase equilibrium involves the same chemical species present in different phase. [ eg:C 6 H 12 O 6 (s) C 6 H 12 O 6 (g) ] - - Phase equilib, in closed syst, P-V work only 47

48. For the spontaneous flow of moles of j from phase to phase - Closed syst that has not yet reached phase equilibrium 48

49. One EXCEPTION to the phase equilibrium, Then, j cannot flow out of (since it is absent from ). The system will therefore unchanged with time and hence in equilibrium. So the equilibrium condition becomes: Phase equilib, j absent from 49

50.  Reaction Equilibrium A reaction equilibrium involves different chemical species present in the same phase. Let the reaction be: reactants products a, b,…..e, f….. Are the coefficients 50

51. Adopt the convention of of transporting the reactant to the right side of equation: are negative for reactant and positive for products During a chemical reaction, the change Δn in the no. of moles of each substance is proportional to its stoichometric coefficient v. This proportionality constant is called the extent of reaction (xi) For general chemical reaction undergoing a definite amount of reaction, the change in moles of species i,, equals multiplied by the proportionality constant : 51

52. The condition for chemical-reaction equilibrium in a closed system is Reaction equilib, in closed system., P-V work only 52

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