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CHAPTER 4 Material equilibrium

PHYSICAL CHEMISTRY. CHAPTER 4 Material equilibrium. ANIS ATIKAH BINTI AHMAD anisatikah@unimap.edu.my. Subtopic. Introduction to Material Equilibrium Entropy and Equilibrium The Gibbs and Helmholtz Energies Thermodynamic Relations for a System Equilibrium

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CHAPTER 4 Material equilibrium

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  1. PHYSICAL CHEMISTRY CHAPTER 4Material equilibrium ANIS ATIKAH BINTI AHMAD anisatikah@unimap.edu.my

  2. Subtopic • 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

  3. What 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

  4. Reaction equilibrium Material equilibrium Phase equilibrium

  5. 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. Entropy and Equilibrium

  6. Entropy and Equilibrium • Example: In isolated system (not in material equilibrium) • 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)

  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

  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 Consider a system at T; • Since system and surroundings are isolated , we have dqsurr= -dqsyst(1) • Since, the chemical reaction or matter transport within the non equilibrium system is irreversible, dSuniv must be positive: dSuniv= dSsyst+ dSsurr > 0 (2)

  9. The surroundings are in thermodynamic equilibrium throughout the process. • Therefore, the heat transfer is reversible, and dSsurr= dqsurr/T (3) • The systems is not in thermodynamic equilibrium, and the process involves an irreversible change in the system, therefore DSsyst ≠dqsyst/T (4)

  10. Equation (1) to (3) give dSsyst> -dSsurr = -dqsurr/T = dqsyst/T (5) • Therefore dSsyst > dqsyst/T dS > dqirrev/T (6)closed syst. in them. and mech. equilib. • dqsurr= -dqsyst(1) dSuniv= dSsyst+ dSsurr >0 (2) dSsurr= dqsurr/T (3)

  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 = dqrev/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)

  13. THE GIBSS & HELMHOLTZ ENERGIES 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

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

  15. Helmholtz free energy • 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

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

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

  18. Gibbs free energy • 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. d(H – TS)  0 GIBBS FREE ENERGY, G=H-TS G = H – TS = U + PV - TS

  19. G Constant T, P Equilibrium reached Time • Gibbs free energy G  H – TS  U + PV – TS G decreases during the approach to equilibrium, reaching minimum at equilibrium dGT,P 0

  20. closed syst., const. T, V, P-V work only • Gibbs free energy As G of the system decrease at constant T & P, Sunivincreases. WHY? Consider a system in mechanical and thermal equilibrium which undergoes an irreversible chemical reaction or phase change at constant T and P. The decrease in Gsyst as the system proceeds to equilibrium at constant T and P corresponds to a proportional increase in S univ

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

  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)

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

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

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

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

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

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

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

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

  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.

  32. Dependence of State Functions 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

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

  34. Temperature dependence of U Temperature dependence of H Pressure dependence of H From Basic Equations from Gibbs equations, dH＝TdS＋VdP From Maxwell Relations

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

  36. Joule-Thomson Coefficient (easily measured quantities) from Chapter 2 From pressure dependence of H

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

  38. Heat-Capacity Difference • As T 0, CP CV • CP  CV (since  > 0) • CP= CV (if  = 0)

  39. EXAMPLE 1 ÷ n

  40. Internal Pressure Ideal gases Solids, Liquids, & Non-ideal Gases Solids 300 J/cm3 (25 oC, 1 atm) Liquids 300 J/cm3 (25 oC, 1 atm) Strong intermolecular forces in solids and liquids.

  41. Calculation of Changes in State Function Calculation of ΔS Suppose a closed system of constant composition goes from state (P1,T1) to state (P2,T2), the system’s entropy is a function of T and P

  42. 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 P1 and change T from T1 to T2. Then T is held constant at T2, and P is changed from P1 to P2. For step (a), dP=0 and gives For step (b), dT=0 and gives

  43. EXAMPLE 2

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

  45. 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: from slide 28 A special case: [Since ]

  46. Phase Equilibrium A phase equilibrium involves the same chemical species present in different phase. [ eg:C6H12O6(s) C6H12O6(g) ] - - Phase equilib, in closed syst, P-V work only

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

  48. Suppose that substance j is initially absent in phase . If initially > , then j flows from phase β to phase δ until the equilibrium is reached. However, if Then, j cannot flow out of δ (since it is absent from δ ). The system will therefore unchanged with time and hence in equilibrium. So, when a substance is absent from a phase, the equilibrium condition becomes: Phase equilib, j absent from

  49. Reaction Equilibrium A reaction equilibrium involves different chemical species present in the same phase.

  50. During a chemical reaction, the change Δn in the no. of moles of each substance is proportional to its stoichometric coefficient v, where the proportionality constant is the same for all species. This proportionality constant is called the extent of reaction For general chemical reaction undergoing a definite amount of reaction, the change in moles of species i, , , equals multiplied by the proportionality constant :

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