厦门大学：《物理化学 Physical Chemistry》课程电子教案（PPT教学课件，英文版）chapter04-1 Material Equilibrium

Physical Chemistry Material Equilibrium Helmholtz free energy A=U-S (4.15)* Consider for constant T& p, dw =-Pav into(4.9) dU≤TdS+dhw (4.9) dU≤T+SdT-ST+PdI+P-P du s d(Tsy-SdT-d(Pn+ vdP d(+Pk-tss-SdT+ vdp d(H-tss-sdii-yakp at constant t and p dt=0. dP=0 (H-TS) 0 P-v work only (4.16)

A  U - TS Helmholtz free energy (4.15)* 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 P-V work only (4.16) Consider for constant T & P, dw = -PdV into (4.9) Physical Chemistry Material Equilibrium dU  TdS + dw (4.9) dU  TdS + SdT – SdT + PdV + VdP - VdP

Physical Chemistry Material Equilibrium Gibbs free energy G≡H-Ts≡U+P-TS (4.17) G Constant T p Fig. 4.3 Equilibrium reached Ime dGrp≤0 d4ru≤0

G  H – TS  U + PV – TS Gibbs free energy (4.17)* dAT,V  0 dGT,P  0 Physical Chemistry Material Equilibrium Equilibrium reached Constant T, P Time G Fig. 4.3

Physical Chemistry Material Equilibrium Gibbs free energy G≡H-Ts≡U+P-TS (4.17) 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 function a, and the constant T-and-P material-equilibrium condition is the minimization of the gibbs function g da=o at equilibrium, const T,v(4. 18)* dg =o at equilibrium, const T, P (4.19)

G  H – TS  U + PV – TS Gibbs free energy (4.17)* 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 function A, and the constant￾T-and-P material-equilibrium condition is the minimization of the Gibbs function G. dA = 0 at equilibrium, const. T, V (4.18)* dG = 0 at equilibrium, const. T, P (4.19)* Physical Chemistry Material Equilibrium

Physical Chemistry Material Equilibrium Gibbs free energy G≡H-Ts≡U+P-TS (4.17) AG=G2-G1=(H2-TS2)-(H1-TS1 AH-TAS △G=△H-TS const. T (420) e Consider a system in mechanical and thermal . equilibrium which undergoes an irreversible chemical reaction or phase change at constant T and p ASam=△Sm+ASst=△AHg7+△S syst (△Hast-7ASs)/7 △G/T △S uniy AGsyst /T closed syst, const T,I P-V work only(4.2

G  H – TS  U + PV – TS Gibbs free energy (4.17)* Consider a system in mechanical and thermal equilibrium which undergoes an irreversible chemical reaction or phase change at constant T and P. Physical Chemistry Material Equilibrium Suniv = Ssurr + Ssyst = Hsyst T + Ssyst / = −(Hsyst −TSsyst)/T = −Gsyst /T Suniv = −Gsyst /T closed syst., const. T, V, P-V work only (4.21) G = G2 – G1 = (H2 – TS2 ) – (H1 – TS1 ) =  H – TS G = H −TS const. T (4.20)

Physical Chemistry Material Equilibrium d(U-TS)≤-ST+ah (4.12) da<-Sar+aw da s dv const. T △4< W wby<-AA const T, closed syst. (4.22) It turns out that a carries a greater significance than e being simply a signpost of spontaneous change The change in the Helmholtz energy is equal to the maximum work the system can do =△4 max

Physical Chemistry Material Equilibrium const. T d(U −TS)  −SdT + dw (4.12) dA −SdT +dw dA dw A w wby= −w wby  −A const. T, closed syst. (4.22) 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: wmax = A