武汉大学：《物理化学 Physical Chemistry》课程教学资源（PPT课件讲稿）05 Helmholtz and Gibbs energies

热力学定律— Helmholtz and Gibbs Energies Consider a system in thermal equilibrium with surroundings, at temp. T For change in the system with transfer of heat, the Clausius inequality is s当。0变不等式到状态函数 Heat transfer at constant v Heat transfer at constant p do 0 Tds> U ( no non-expansion work) T dS Helmholtz Energy, A Gibbs Energy, G S H- S du- Tds at constant T dh- TdS 7F≤0

热力学定律 ⎯ Helmholtz and Gibbs Energies 变不等式 到 状态函数

热力学定律— Helmholtz and Gibbs Energies Helmholtz Energy a(亥姆霍兹能) U- S 恒温恒容 du- Tds da 7,F≤ 0 最大功函( Maximum Work function) 物理意义 wmax da dQ≤TdS dU≤Tds+dW→dW≥aU-TdS △U=Q+W

热力学定律 ⎯ Helmholtz and Gibbs Energies Helmholtz Energy A (亥姆霍兹能) 恒温 恒容 最大功函 (Maximum Work Function) dwmax = dA Why? dQ  TdS U = Q + W } dU  TdS + dW  dW  dU - TdS < 0 物理意义

热力学定律— Helmholtz and Gibbs Energies Gibbs Free Energy G(吉布斯自由能) G H- TS 恒温恒压 Chemical reactions are G dh- Tds spontaneous in the direction of decreasing Gibbs energy 最大非体积功( Maximum Non-Expansion Work) C dG △G e. max e. max 标准 Gibbs能 △G°=△H°-T△S° ∑v ∑v△G。=∑v△1G products reactants 标准生成 Gibbs能

Gibbs Free Energy G (吉布斯自由能) 热力学定律 ⎯ Helmholtz and Gibbs Energies 恒温 恒压 最大非体积功(Maximum Non-Expansion Work) Chemical reactions are spontaneous in the direction of decreasing Gibbs energy. 标准Gibbs能 标准生成Gibbs能

Since H=U+ pv for a general change, then +a+(D If the change is reversible, dw drey and dq= dorey t ds,thus Tds+ dwr+ d(pn-Tds Work consists of expansion work, Wrey=-p dv, and some other kind of work(e.g, eletrical work pushing electrons through a circuit, work raising a column of fluid, etc. ) The latter work is non-expansion work, dw G=(-pdr+dwe rey)+pdv+va dwe. rev 丿dh And, if work occurs at constant p, then dG= dwe rev because the process is reversible work has its maximum value here

Combining First and Second Laws First Law dU=dQ +dW dW=PdⅤdQ=TdS(可逆&只有体积功) Tds- pdr Fundamental Equation Since H=U+pV: dH= dU+ pdv+vdp dh Tas VdP dG dh- tds- sdT dG= vdp- sdT Because we control p and T, g is a very important quantity in chemistry G carries the combined consequences of the 1st and 2nd laws!

Combining First and Second Laws First Law dU = dQ + dW dWrev = -PdV dQrev = TdS (可逆 & 只有体积功) dH = TdS + VdP

dG= vdp- sdr G S aG aT Liquid Solid Liquid Solid Temperature, T P

Temperature Dependence ofG 温度变化关系 OG G-H OG G H aT aT T T 0(G 1(G d(1 +g OT(T TLaT dT(t 1/aG G 自由能 G 可测 0(G H Gibbs- Helmholz OT(T T Equation P

Temperature Dependence of G T G H S T G P −  = − =        T H T G T G P  − = −               +        =                 dT T d G T G T T G T P P 1 1 2 1 T G T G T P  −        =        −        = T G T G T P 1 2 T H T G T P = −                 Gibbs – Helmholz Equation 温度变化关系 自由能 焓 可测

△G求算 (1)恒温过程 dG= vap- sdT P2 恒温dT=0AG=VaP Perfect Gases AG=nRT hn -2=nrT In T*△S (2)相变过程 可逆相变 △G=0 不可逆相变需设计可逆过程 (3)化学反应 恒温恒压△G=△H-TS

G求算 (1) 恒温过程 恒温 dT = 0   = 2 1 P P G VdP Perfect Gases 2 1 1 2 ln ln V V nRT P P G = nRT = (2) 相变过程 可逆相变 G = 0 不可逆相变 需设计可逆过程 (3) 化学反应 恒温恒压  rG =  rH − T rS - T * S

具体计算 课下先看例子!(P68,69)

具体计算 课下先看例子! (P68,69)