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Material equilibrium

ERT 108 PHYSICAL CHEMISTRY. Material equilibrium. NOORULNAJWA DIYANA YAACOB n ajwa_diyana@yahoo.com. Subtopic. Thermodynamic properties of Nonequilibrium system Entropy and Equilibrium The Gibbs and Helmholtz Energies Thermodynamic relations for a system equilibrium

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Material equilibrium

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  1. ERT 108 PHYSICAL CHEMISTRY Materialequilibrium NOORULNAJWA DIYANA YAACOB najwa_diyana@yahoo.com

  2. Subtopic • Thermodynamic properties of Nonequilibrium system • 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 • 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 • Nor 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. Thermodynamics Properties of Nonequilibrium System • Nonequilibrium system: chemical reaction or transport of matter from one phase to another is occurring.

  6. 1. We consider system not in phase equilibrium U= Usol+ UNaCl S =Ssol +SNaCl Partition is removed – removal done by reversibly and adiabatically q= 0 w=0 ΔS = 0 ΔU = 0

  7. 2. We consider system not in reaction equilibrium Catalyst We can measure the ΔU (FIRST LAW) and ΔS if the system is in equilibrium When catalyst was added- system not in reaction equilibrium Changing mixture composition Can ascribe the new value to U and S Water H2 O2

  8. Entropy and Equilibrium 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 to strives to be minimal, entropy strives to be maximal Entropy wants to grow. Energy wants to shrink. Together, they make a compromise.

  9. 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

  10. 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. • Any further process can only decrease entropy –(violate the second law)

  11. Reaction equilibrium is ordinarily studied under two condition. • Reaction involve gases Chemiccal are put in the container of fixed volumed and the system is allowedto reach equilibrium at constant T and V in a constant-temperature batch • Reaction in liquid solution The system is held at atmospheric pressure and allowed to reach equilibrium at constant T and P

  12. To find equilibrium criteria, the system at temperature T is placed in a bath also at T. The system is not in material equilibrium but is in mechanical and thermal equilibrium The surrounding are in material, mechanical and thermal equilibrium System and surroundings can exchanged energy (as heat and work) but not matter Surroundings at T System at T Impermeable wall Rigid, adiabatic, impermeable wall

  13. Since system and surroundings are isolated , we have dqsurr= -dqsyst(Eq1) • Since, the chemical reaction or matter transport within the non equilibrium system is irreversible, dSuniv must be positive: dSuniv= dqsyst + dqsurr > 0 (Eq2)

  14. The surroundings are in thermodynamic equilibrium throughout the process. • Therefore, the heat transfer is reversible, and dsurr= dqsurr/T (Eq3) • The systems is not in thermodynamic equilibrium, and the process involves and irreversible change in the system, therefore Dsyst ≠dqsyst/T (Eq4)

  15. Equation (1) to (3) give dSsyst> -dSsurr = -dqsurr/T = dqsyst/T (Eq5) • Therefore dSsyst > dqsyst/T (Eq6) dSsyst > dqirrev/T (Eq7) closed syst. in them. and mech. Equilib.

  16. 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,a material equilibrium we have, ds ≥ dq/T (Eq8) material change, closed syst. in them. and mech. equilib

  17. The first law for a closed system is dq = dU – dw (Eq 9) • Eq 7 gives dq≤ TdS • Hence for a closed system in mechanical and thermal equilibrium we have dU – dw ≤ TdS or dU ≤ TdS + dw (Eq 10)

  18. dU TdS +SdT – SdT+ dw dU d(TS) – SdT + dw d(U – TS)  – SdT + dw d(U – TS) – SdT - PdV • The Gibbs and Helmholtz Energies dU TdS + dw =: equilibrium P-V work only at constant T and V, dT=0, dV=0 d(U – TS)  0 P-V work only Equality sign holds at material equilibrium

  19. 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

  20. dU TdS + dw d(H – TS)  0 P-V work only Helmholtz free energy A  U - TS Consider for constant T & P, dw = -PdV into dU TdS + SdT – SdT+ PdV + VdP - VdP 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

  21. the state function H-TS, continually decrease (constant T and P) during the spontaneous, irreversible process of chemical reaction and matter transport until material equilibrium is reached

  22. G Constant T, P Equilibrium reached Time Gibbs free energy G  H – TS  U + PV – TS G decreases during the approach at equilibrium dGT,P 0 dAT,V0 Both A and G have units of energy (J or cal)

  23. Gibbs free energy G  H – TS  U + PV – TS In a closed system capable of doing only P-V work, the constant-T-and-V material-equilibrium condition is the minimization of the Helmholtz functionA, and the constant-T-and-P material-equilibrium condition is the minimization of the Gibbs function G. dA = 0 at equilibrium, const. T, V dG = 0 at equilibrium, const. T, P

  24. closed syst., const. T, V, P-V work only Gibbs free energy G  H – TS  U + PV – TS G = G2 – G1 = (H2 – TS2 ) – (H1 – TS1 ) =  H – TS const. T Consider a system in mechanical and thermal equilibrium which undergoes an irreversible chemical reaction or phase change at constant T and P.

  25. Relationship between the minimization-of-G-equilibrium condition (T & P constant) and the maximization-of-S univ closed syst., const. T, V, P-V work only The decrease in G syst as the system proceeds to equilibrium at constant T and P corresponds toa proportional increase in S univ Consider a system in mechanical and thermal equilibrium which undergoes an irreversible chemical reaction or phase change at constant T and P.

  26. const. T, closed syst. 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 change in the Helmholtz energy is equal to the maximum work the system can do:

  27. 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

  28. and G  H – TS H  U + PV When the change is reversible (const. T, P) The maximum non-expansion work from a process at constant P and T is given by the value of G

  29. dU = TdS - PdV H  U + PV A  U – TS G  H - TS • Thermodynamic Reactions for a System in Equilibrium 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

  30. 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 )

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

  32. dU = TdS - PdV dH = TdS + VdP dA = -SdT - PdV dG = -SdT + VdP dA = d(U - TS) = dU - d(TS) = dU - TdS - SdT = (TdS - PdV) - TdS - SdT dG = d(H - TS) = dH - d(TS) = dH - TdS - SdT = (TdS + VdP) - TdS - SdT

  33. dU = TdS - PdV dH = TdS + VdP dA = -SdT - PdV dG = -SdT + VdP The Gibbs Equations First Law closed syst., rev. proc., P-V work only Definitions

  34. 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 The Power of thermodynamics: Difficultly measured properties to be expressed in terms of easily measured properties.

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

  36. dU = TdS - PdV The Maxwell Relations The Gibbs equation (4.33) for dU is dU＝TdS－PdV dV＝0 dS＝0 ↓V is held constant ↓S is held constant =

  37. 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.

  38. 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, andκ

  39. 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 (4.45)

  40. Temperature dependence of U Temperature dependence of H Pressure dependence of H from (4.34) dH＝TdS＋VdP

  41. Temperature dependence of S From (4.31) Pressure dependence of S Temperature and Pressure dependence of G The Gibbs equation (4.36) for dG is dG ＝ -SdT + VdP dT＝0 dP＝0

  42. Reminder: • The equations of this section apply to a closed system of fixed composition and also to a closed system where the composition changes reversibly

  43. Joule-Thomson Coefficient from (2.65)

  44. Heat-Capacity Difference

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

  46. Example:

  47. Internal Pressure 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.

  48. Example

  49. 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

  50. 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 4.5) 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

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