Chapter 7 Phase Equilibria In A One Component System

系統的内涵性質,另一類勢的測量, ex. 溫度、壓力、化學勢…

1. 溫度: 系統内勢or熱強度一種測量

熱離開系統的傾向, 例如高溫→低熱,有一gradient存在,直到熱流消失,溫度相等,熱平衡為止。

2. 壓力: 系統内質量移動的傾向

ex. 相之間有淨壓力時,質量移動使壓力gradient抵消, 使系統内不同部份,壓力彼此相等。

3. 一相內的成分(component) i的化學勢

測量成分i離開該相的傾向大小；種類i傾向從高化學勢的相移出而轉入較低化學勢的相。”potential gradient” 化學擴散的驅動力,直到化學勢μ_i在不同的相均相等,平衡才成立。

dG=-SdT+VdP+∑μᵢdnᵢ

材料熱力學從能量的觀點,探討物質的變化

§thermodynamic equilibrium, e.g. α, and β phases co-existed at fixed T, P

thermal equil. T_α=T_β , mechanical equil. P_α=P_β, chemical equil. μ_α=μ_β

at equilibrium, G=Gₘᵢₙ ← dG=0 ⸪ G≡H-TS 欲知G(T)=? H(T)=? S(T)=? at P(=1 atm) 進一步了解 G_α(T)=? G_β(T)=?

at T=T₀, if G_α(T₀)<G_β(T₀), reaction: β→α, α stable.

If G_α(T₀)=G_β(T₀), reaction: β↔α, α+β phases co-existed, T₀:相轉換溫度

7-2 研究G-T-P的關係式來討論平衡

若維持在1atm, 0℃時冰水平衡 H₂O₍ₗ₎ ↔ H₂O₍ₛ₎ at 1atm, 273K

T>273K, G_ᴴ₂_ᴼ₍ₛ₎>G_ᴴ₂_ᴼ₍ₗ₎ liquid stable.

T=273K, G_ᴴ₂_ᴼ₍ₛ₎=G_ᴴ₂_ᴼ₍ₗ₎ liquid+solid stable. ←討論點!

T<273K, G_ᴴ₂_ᴼ₍ₛ₎<G_ᴴ₂_ᴼ₍ₗ₎ solid stable.

若含nₛ mole水和nₗ mole冰: G=nₛG_ᴴ₂_ᴼ₍ₛ₎+nₗG_ᴴ₂_ᴼ₍ₗ₎=contant at Tₘ, 其冰水比例無關(i.e. nₛ+nₗ=c)

可以說: 平衡時, H₂O從液相中脫離傾向等於固相中的脫離傾向。

定溫定壓下, G_ᵀ,ᴾ=μₛnₛ+μₗnₗ→比較G=nₛG_ᴴ₂_ᴼ₍ₛ₎+nₗG_ᴴ₂_ᴼ₍ₗ₎ vs. G_ᵀ,ᴾ=μₛnₛ+μₗnₗ

molar Gibbs free energy, μᵢ≡(∂G∕∂nᵢ)_ᴛ,ᴘ,nj=Gᵢ

如何求G_ᴴ₂_ᴼ₍ₛ₎(T), G_ᴴ₂_ᴼ₍ₗ₎(T)? given: cₚ₍ₛ₎=38J/Kmole, cₚ₍ₗ₎=75.44J/Kmole, ∆Hₘ=6008J/mole. At 298K, H₍ₛ₎=-6944, S₍ₛ₎=44.77; H₍ₗ₎=0, S₍ₗ₎=70.08

通式: H(T)=H₂₉₈+∫_₂₉₈^ᵀcₚdT, S(T)=S₂₉₈+∫_₂₉₈^ᵀ(cₚ/T)dT → H₍ₛ₎(T)=..., S₍ₛ₎(T)=..., H₍ₗ₎(T)=..., S₍ₗ₎(T)=...

→ G_ᴴ₂_ᴼ₍ₛ₎(T)=H₍ₛ₎(T)-TS₍ₛ₎(T)=f₍ₛ₎(T), G_ᴴ₂_ᴼ₍ₗ₎(T)=H₍ₗ₎(T)-TS₍ₗ₎(T)=f₍ₗ₎(T)

1. 求ΔH, let H_298(l)=0

H₍ₗ₎_,ᴛ=∫_₂₉₈^ᵀcₚ₍ₗ₎dT=75.44(T-298)=75.44T-22481,

H₍ₛ₎_,ᴛ=∫_₂₉₈^²⁷³cₚ₍ₗ₎dT-ΔHₘ+∫_₂₇₃^ᵀcₚ₍ₛ₎dT=75.44(273-298)-6008+38(T-273)=38T-18268

ΔH_ᴛ=H₍ₗ₎_,ᴛ-H₍ₛ₎_,ᴛ=75.44(T-298)-75.44(273-298)+ΔHₘ-38(T-273)=37.44T+4231.12

2. TΔS

S₍ₗ₎_,ᴛ=S₍ₗ₎_,₂₉₈+∫_₂₉₈^ᵀcₚ₍ₗ₎dlnT=70.08+75.44ln(T/298),

S₍ₛ₎_,ᴛ=S₍ₛ₎_,₂₉₈+∫_₂₉₈^ᵀcₚ₍ₛ₎dlnT=44.77+38ln(T/298)

ΔS=S₍ₗ₎_,ᴛ-S₍ₛ₎_,ᴛ=25.31+37.44ln(T/298)

As ΔH₍ₛ_→ₗ₎= T_m ΔS₍ₛ_→ₗ₎ → 37.44T+4231.12=25.31T+37.44Tln(T/298)→ T_m=?

3. G_ᴴ₂_ᴼ₍ₛ₎(T)=H₍ₛ₎(T)-TS₍ₛ₎(T)=38T-18268-T[44.77+38ln(T/298)]=18268-6.77T+38Tln(T/298)

G_ᴴ₂_ᴼ₍ₗ₎(T)=H₍ₗ₎(T)-TS₍ₗ₎(T)= 75.44T-22481-T[70.08+75.44ln(T/298)]=22481+5.36T+75.44Tln(T/298)

ΔG(T)=G_ᴴ₂_ᴼ₍ₗ₎(T)-G_ᴴ₂_ᴼ₍ₛ₎(T)=4213+12.13T+37.44Tln(T/298)

as T<0℃, G_ᴴ₂_ᴼ₍ₛ₎<G_ᴴ₂_ᴼ₍ₗ₎ solid stable;

T=0℃, liquid+solid stable.(melting point)

consider s → l ∆G=Gₗ-Gₛ

(i) T<Tₘ, ∆G>0

(ii) T=Tₘ, ∆G=0

(iii) T>Tₘ, ∆G<0

重點討論 ∆G? ∆G<0, reaction! (Any transformation)

∆G=∆H-T∆S at Tₘ, ∆H=∆Hₘ>0 and ∆S=∆Sₘ>0 表示溫度上升,∆G會下降(S contribution is more important at high temperature.)

G(T), ∆G(T)的curve shape? 採monotonic change

⸪ dG=-SdT+VdP (∂G∕∂T)ₚ=-S<0代表函數的斜率, (∂²G∕∂T²)ₚ=-(∂S∕∂T)ₚ=-cₚ/T<0代表函數的曲率

e.g. s → l ∆Gₛₗ=∆Gₘ=Gₗ-Gₛ (∂∆Gₘ∕∂T)ₚ=-∆Sₘ<0, (∂²∆Gₘ∕∂T²)ₚ=-(∂∆Sₘ∕∂T)ₚ→0 *Richards rule ∆Sₘ=c₀

G(P), ∆G(P)的curve shape?

At fixed T, (∂G∕∂P)_ᴛ=V>0斜率, (∂²G∕∂P²)_ᴛ =(∂V∕∂P)_ᴛ =-βV<0曲率

e.g. s → l (∂∆G∕∂P)_ᴛ=∆V=Vₗ-Vₛ>0 usual cases, but ∆V=Vₗ-Vₛ<0 for H₂O

(∂²∆G∕∂P²)_ᴛ=(∂∆V∕∂P)_ᴛ=-(βₗVₗ-βₛVₛ)→0

§Phase equilibrium(transformation) T(P)?

e.g. s → l Tₘ(P)=? or l → g T_ᵇ(P)=? consider s ↔ l Gₗ=Gₛ at Tₘ ⸫ dGₗ=dGₛ

dGₛ=-SₛdT+VₛdP, dGₗ=-SₗdT+VₗdP → d(Gₗ-Gₛ)=-(Sₗ-Sₛ)dT+(Vₗ-Vₛ)dP → d∆G=-∆SdT+∆VdP

⸪∆G=0 at Tₘ, -∆SdT+∆VdP=0

⸫ (dP/dT)ₑ_զᵤᵢₗ=∆Sₘ/∆V=∆Hₘ/Tₘ∆V...clapeyron eq. i.e. ∆Gₘ=∆Hₘ-Tₘ∆Sₘ

e.g. s → l for H₂O, ∆Hₛₗ>0, ∆Sₛₗ>0but ∆V=Vₗ-Vₛ<0 ⸫ (dP/dT)<0 P↑ 則Tₘ↓ 例如溜冰時體重施在冰上的壓力會使冰溶化(從理論上冰會因壓力上升而融化,但實際上加壓為瞬間,並未發現冰鞋有水,可能與冰的表面情況有關)

condensed phase transition, α ↔ β T_αβ(P) ?

Clapeyron Eq. (dP/dT)ₑ_զᵤᵢₗ=∆S/∆V=∆H/T∆V, ∆V=V_β-V_α, ∆S=S_β-S_α

∫₁^ᴾdP=∫_ᴛₘ^ᵀ(∆S/∆V)dT≈(∆S/∆V)(T-Tₘ),

e.g. ∆Sₘ>0 and ∆V=Vₗ-Vₛ <0 for H₂O, but ∆V>0 for usual cases.

§solid(liquid) ↔ vapor(gas) phase transition

(dP/dT)ₑ_զᵤᵢₗ=∆S/∆V=∆H/T∆V, ∆V=V_g-V_s≡V_g or V_g-V_l≈ V_g=RT/P

→ (dP/dT)≈∆H/TV_g=P∆H/RT² → dlnP=(∆H/RT²)dT...clausius-clapeyron eq.

積分→ ∫₀^ᴾdlnP=∫_ᴛᵇ^ᵀ(∆H/RT²)dT → lnP=? f(T)

e.g. l → g given: ∆H_ᵇ, T_ᵇ ; ∆H(T)=∆H_ᵇ+∫_ᴛᵇ^ᵀ∆cₚdT, ∆cₚ=cₚ_,g-cₚ_,l=f(T)=? ∆cₚ=∆a+∆bT+∆c/T²

a. ∆H=∆H_ᵇ , if ∆cₚ=0 → lnP=-∆H_ᵇ/R(1/T-1/T_ᵇ)=-∆H_ᵇ/RT+c₀ ⸫ lnP=-∆H_ᵇ/RT

b. ∆cₚ=∆a=c₀ 常數, ∆H=∆H_ᵇ+c₀(T-T_ᵇ) → lnP=∫_ᴛᵇ^ᵀ([∆H_ᵇ+c₀(T-T_ᵇ)]/RT²)dT= (c₀T_ᵇ -∆H_ᵇ)/RT+(c₀/R)lnT+C₀ ⸫ lnP=A₀/T+B₀lnT+C₀, i.e. ∆H=(∆H_ᵇ-c₀T_ᵇ )+c₀T

c. ∆cₚ=∆a+∆bT+∆c/T²

e.g. H₂O, cₚ_,l=75.44, cₚ_,g=30+10.7∙10⁻³T+0.33∙10⁵T⁻², T_ᵇ=373K, ∆H_ᵇ=41090J

求P(T)... vapor pressure of water(l → g)

∆H(T)=∆H_ᵇ+∫_ᴛᵇ^ᵀ∆cₚdT=41090+∫_ᴛᵇ^ᵀ (-45.44+10.7∙10⁻³T+0.33∙10⁵T⁻²)dT=(41090+45.44∙373-5.35∙10⁻³ 373²+0.33∙10⁵/373)-45.44T+5.35∙10⁻³T²-0.33∙10⁵/T 代入∫₀^ᴾdlnP=∫_ᴛᵇ^ᵀ(∆H/RT²)dT

lnP=-52383/RT-45.44lnT/R+5.35∙10⁻³T/R+0.33∙10⁵/2RT²+51.1

ex. cₚ_,H₂O₍ᵥ₎=30+10.7∙10⁻³T+0.33∙10⁵T⁻² , 298~2500K; cₚ_,H₂O₍ₗ₎=75.44, 273~373, Hᵥ=41090

ΔH_ᴛ=ΔH_₃₇₃+∫_₃₇₃^ᵀ∆cₚ₍ₗ_→ᵥ₎dT, dlnP=(ΔH/RT²)dT

→ lnP=-58872/RT-(45.44/R)lnT+(5.35∙10⁻³/R)T+(0.33∙10⁵/2R)T⁻²+C,

B.C.: P(373K)=1 atm 求C=?(51.092)

固液相平衡dP/dT=ΔH_f/TΔV_f , 斜率AOA’ dP/dT=PΔH/RT²*OA’對 gas為介穩態,在相同的 T, P 下,有較高的G值

一元三相圖中如何畫G-T, G-P變化曲線?

G-T curves at fixed P: 

G-P curves at fixed T: 

solid phase transition: e.g. Fe: α(BCC) → γ(FCC) → δ(BCC) → l different crystal structures: allotropes同素異形體, compound ZrO₂: monoclinic → tetragonal → cubic → l, polymorphism多形體

補充: slope of phase change, α ↔ β

e.g. Fe α(BCC) → γ(FCC) ∆V=V_(FCC)-V_(BCC)<0 ⸫ (dP/dT)ₑ_զᵤᵢₗ=∆S/∆V<0

γ(FCC) → δ(BCC) ∆V=V_(BCC)-V_(FCC)>0 ⸫ (dP/dT)ₑ_զᵤᵢₗ=∆S/∆V>0

phase diagram:

for l → g logPₗ=A/T+BlogT+C, s → g logPₛ=A/T+BlogT+C

at triple point, Pₗ=Pₛ → A/T+BlogT+C=A/T+BlogT+C

Gibbs Phase rule f=c-P+2

e.g. one component, c=1⸫ f=3-P (I) one phase region: P=1 → f=2 (ii) two phases curve: P=2 → f=1 (iii) 3 phaes coexist: P=3 → f=0 (triple point)

該系統有P個相, C個chemical species; 前提(1)沒有化學反應發生, (2)每種成分在每一相均存在. 根據(2), 有PC個mole fractions, 再加上T, P → PC+2 intensive variables.

a. 在每相中,例如α相中 x₁^α+x₂^α+x₃^α+x₄^α+...+x_ᶜ^α=1,∵有P個相有P個方程式, 可限制P個變數

b. 在各相間同種的μ_i要平衡時均相同

μ₁^α=μ₁^β=μ₁^γ=..., μ₂^α=μ₂^β=μ₂^γ=..., μ_ᶜ^α=μ_ᶜ^β=μ_ᶜ^γ=...有P-1個等式共有C(P-1) 個方程式

→f=PC+2-P-C(P-1)=C-P+2

在兩相存在區, OA線 or OB, OC線: f=1-2+2=1個自由度(P或T)→ P固定, T即固定。

三相點→ f=1-3+2=0

Ex=1. NaF s → g lnPₛ=-34450/T-2.01lnT+33.74...(1)

l → g lnPₗ=-31090/T-2.52lnT+34.66...(2)

Q1: T_ᵇ=? Boiling at P=1 atm

Pₗ=1 atm, ln1=0=-31090/T-2.52lnT+34.66 T_ᵇ=2006K

Q2: Tₜ=? Pₜ=?

Pₛ=Pₗ, (1)=(2), -34450/T-2.01lnT+33.74=-31090/T-2.52lnT+34.66, -3360/T+0.51lnT-0.92=0 Tₜ=1239K代入(1)或(2), Pₜ=2.29∙10⁻⁴ atm

Q3: ∆H_ᵇ=?

∵ dlnP=(∆H/RT²)dT → ∆H_ᵇ=RT²(dlnPₗ/dT)=RT²[31090/T²-2.52/T]=258500-20.95T=216500J

Q4: when T=Tₜ, ∆Hₛₗ=?

∆Hₛₗ=Hₗ-Hₛ=Hₗ-H_g+H_g-Hₛ=-(H_g-Hₗ)+(H_g-Hₛ)=-∆Hₗ_g+∆Hₛ_g, ∆Hₗ_g=∆H_ᵇ= 258500-20.95T, ∆Hₛ_g=RT²(dlnPₛ/dT)=RT²[34450/T²-2.01/T]=286417-16.71T

∆Hₛₗ=-( 258500-20.95T)+286417-16.71T=27917+4.24T=33170J

Q5: ∆cₚ=cₚ₍ₗ₎-cₚ₍ₛ₎=?

∆H(T)=∆H(T₀)+∫_ᴛ₀^ᵀ∆cₚdT ↔ ∆cₚ=(d∆H/dT)=4.24

Ex-2. cabon(graphite) → diamond at 298K, P≥?

Given: H_ᴅ=1900J/mole, S_g=5.73J/Kmole, S_ᴅ=2.43J/Kmole, ρ_g=2.22g/cm³, ρ_ᴅ=3.515g/cm³

concept: at fixed T, how much P is big enough to make reaction spontaneous(∆G≤0)?

dG=-SdT+VdP, (∂∆G∕∂P)ᴛ=∆V, ∆G=G_ᴅ-G_g=(H_ᴅ-H_g)-T(S_ᴅ-S_g)=(1900-0)-298(2.43-5.73)=2883J

at fixed T, ∫_₂₈₈₃^⁰d∆G=∫_₁^ᴾ∆VdP ∆G₀-2883=∆V(P-1)=(M/ρ_ᴅ-M/ρ_g)(P-1)=-1.99(P-1)∙0.101

∆G₀=-0.201138(P-1)+2883≤0, P≥14334 atm