Ch. 11 chemical reaction

前面講溶解,只有相變化,無涉及化學反應,這章加上化學反應的氣體產物

A₍_g₎+B₍_g₎=2C₍_g₎ ,∆Gᵢ 當成ideal gas mixture, pᵢ=XᵢP e.g. 2H₂+O₂=2H₂O

引進反應平衡常數,K, 因為用∆G推導時,利用K求nᵢ最快

G_i=n_ᴬG_ᴀ⁰+n_ᴮG_ᴃ⁰= G_ᴀ⁰+G_ᴃ⁰, G_f=n_ᴬG_ᴀ+n_ᴮG_ᴃ+n_ᴄG_ᶜ=n_ᴬ(G_ᴀ+G_ᴃ-2G_ᶜ)+2G_ᶜ ∆G=G_f-G_i

i.e. ∫_Gᵢ⁰^GᵢdGᵢ=∫_₁^pᵢRTdlnpᵢ, Gᵢ-Gᵢ⁰=^ᵢRTlnpᵢ ⸫ Gᵢ=^ᵢGᵢ⁰+RTlnXᵢ+RTlnP代入G_f式中, i=A, B, C,

X_ᴀ=X_ᴃ=n_ᴬ/2, X_ᶜ=(1-n_ᴬ)

→ G_f= n_ᴬ[G_ᴀ⁰+RTlnX_ᴀ+RTlnP+G_ᴃ⁰+RTlnX_ᴃ+RTlnP-2(G_ᶜ⁰+RTlnX_ᶜ+RTlnP)]+2(G_ᶜ⁰+RTlnX_ᶜ+RTlnP)=n_ᴬ(G_ᴀ⁰+G_ᴃ⁰-2G_ᶜ⁰)+2RT[n_ᴬln(n_ᴬ/2)+(1-n_ᴬ)ln(1-n_ᴬ)]+2G_ᶜ⁰

let 2G_ᶜ⁰-(G_ᴀ⁰+G_ᴃ⁰)≡∆G⁰...standard Gibbs free energy change

→ ∆G=(1-n_ᴬ)∆G⁰+2RT[n_ᴬln(n_ᴬ/2)+(1-n_ᴬ)ln(1-n_ᴬ)], ∆G_ᴿ=(1-n_ᴬ)∆G⁰ and ∆G_ᴹ=2RT[n_ᴬln(n_ᴬ/2)+(1-n_ᴬ)ln(1-n_ᴬ)]

equilibrtum: ∆G=∆Gₘᵢₙ → ∂∆G∕∂n_ᴬ=0 → n_ᴬ=? 

⸪G_i= G_ᴀ⁰+G_ᴃ⁰=const, ⸫∂∆G∕∂n_ᴬ=0 → ∂∆G_f∕∂n_ᴬ=0, G_f=n_ᴬG_ᴀ+n_ᴮG_ᴃ+n_ᶜG_ᶜ=n_ᴬ(G_ᴀ+G_ᴃ-2G_ᶜ)+2G_ᶜ

∂∆G_f∕∂n_ᴬ=G_ᴀ+G_ᴃ-2G_ᶜ=G_ᴀ⁰+^ᵢRTlnp_ᴀ+ G_ᴃ⁰+^ᵢRTlnp_ᴃ-2(G_ᶜ⁰+^ᵢRTlnp_ᶜ)=0

→ 2G_ᶜ⁰-(G_ᴀ⁰+G_ᴃ⁰)=-RTln[p_ᶜ²/(p_ᴀp_ᴃ)], Kₚ≡p_ᶜ²/(p_ᴀp_ᴃ) → ∆G⁰=-RTlnKₚ, Kₚ=e^-∆G⁰/RT

def: A₍_g₎+B₍_g₎=2C₍_g₎ , ∆G⁰=2G_ᶜ⁰-(G_ᴀ⁰+G_ᴃ⁰) and Kₚ≡p_ᶜ²/(p_ᴀp_ᴃ)

equil.: ∆G⁰=-RTlnKₚ (∂∆G∕∂n_ᴬ=0得到的結果), given: G_ᴀ⁰, G_ᴃ⁰, G_ᶜ⁰ at T→ ∆G⁰ at T→ Kₚ=e^-∆G⁰/RT

e.g. G_ᴀ⁰=G_ᴃ⁰=0, G_ᶜ⁰=-2500J A₍_g₎+B₍_g₎=2C₍_g₎ A₍_g₎+B₍_g₎=2C₍_g₎ 求 p_ᴀ, p_ᴃ, p_ᶜ

∆G⁰=-5000J,at T=500K, P=1 atm ⸫Kₚ=e^-∆G⁰/RT=3.329=p_ᶜ²/(p_ᴀp_ᴃ)=(X_ᶜP)²/[(X_ᴀP)(X_ᴃP)]=(1-n_ᴬ)²/[(n_ᴬ/2)(n_ᴬ/2)] → 0.671n_ᴬ²-8n_ᴬ+4=0, n_ᴬ=0.523 p_ᴀ=p_ᴃ=0.261, p_ᶜ=0.477

§Effect of temperature on reaction equilibrium(T與Kₚ的關係)

given ∆G⁰=-RTlnKₚ, 通常以反應熱∆H⁰表示 ∆G⁰↔∆H⁰ Gibbs-Helmholtz eq.

[∂(∆G⁰/T)∕∂T]ₚ=-∆H⁰/T² ⸫[∂(-RlnKₚ)∕∂T]ₚ=-∆H⁰/T² → [∂(lnKₚ)∕∂T]ₚ=∆H⁰/RT² vant Hoff eq.

If endothermic, ∆H⁰>0, 溫度增加, ∂(lnKₚ)∕∂T>0 → Kₚ隨之增加

⸪Kₚ≡p_ᶜ²/(p_ᴀp_ᴃ) ⸫p_ᶜ上升 p_ᴀ, p_B下降 A₍_g₎+B₍_g₎=2C₍_g₎反應往右 Le Chatelier principle

aA₍_g₎ + bB₍_g₎ = cC₍_g₎ + dD₍_g₎ ∆G⁰=(cG_ᶜ⁰+dG_ᴅ⁰)-(aG_ᴀ⁰+bG_ᴃ⁰)=f(T), Kₚ≡p_ᶜᶜp_ᴅᵈ/(p_ᴀᵃp_ᴃᵇ)

equil. condition: ∆G⁰=-RTlnKₚ e.g. Given (T,P) → nᵢ=?

Temp. Influence: vant Hoff eq. [∂(lnKₚ)∕∂T]ₚ=∆H⁰/RT²

e.g. Cl₂₍_g₎ → 2Cl₍_g₎ ∆H⁰>0 Kₚ=p_ᴄₗ²/p_ᴄₗ² when T ↑, ∂T>0, ∂lnKₚ>0 → Kₚ ↑ ⸫p_ᴄₗ↑, p_ᴄₗ² ↓ 反應往右

§Influence of total pressure

⸪ Gᵢ⁰ defined at P=1 atm, ⸫∆G⁰與壓力無關

Kₚ=e^-∆G⁰/RT=c₀ if T fixed ⸫Kₚ fixed也與壓力無關; 但改變總壓, pᵢ may change!!

pᵢ=XᵢP → Kₚ= , Kₓ≡X_ᶜᶜX_ᴅᵈ/(X_ᴀᵃX_ᴃᵇ) ⸫Kₚ= Kₓ∙Pᶜ⁺ᵈ⁻ᵃ⁻ᵇ

when a+b=c+d → Kₚ= Kₓ, nᵢ(Xᵢ) no change

when a+b<c+d → c+d-(a+b)>0 ⸫P上升, Kₓ下降 ⸪Kₚ fixed, 因此X_ᶜ, X_ᴅ減少, X_ᴀ, X_ᴃ增加

總壓力增加,反應往左

當平衡時, ∆G⁰=-RTlnKₚ; ∆G=∆H-T∆S at fixed T, ∆Gₘᵢₙ需要∆H↓, ∆S↑兩者妥協

G_f=n_ᴬ(G_ᴀ⁰+G_ᴃ⁰-2G_ᶜ⁰)+2RT[n_ᴬln(n_ᴬ/2)+(1-n_ᴬ)ln(1-n_ᴬ)]+2G_ᶜ⁰

→ ∆G=G_f-2G_ᶜ⁰=-n_ᴬ∆G⁰+2RT[n_ᴬln(n_ᴬ/2)+(1-n_ᴬ)ln(1-n_ᴬ)]=-n_ᴬ∆H⁰+T{n_ᴬ∆S⁰+2R[n_ᴬln(n_ᴬ/2)+(1-n_ᴬ)ln(1-n_ᴬ)]} ⸫∆H=-n_ᴬ∆H⁰, ∆S=-{n_ᴬ∆S⁰+2R[n_ᴬln(n_ᴬ/2)+(1-n_ᴬ)ln(1-n_ᴬ)]}

Ex. SO₂₍_g₎ + ½O₂₍_g₎ = SO₃₍_g₎ given: ∆G⁰=-94600+89.37T, ask nᵢ, pᵢ in equilibrium under T=1000K, P=1 atm, n_SO₂⁰=1 mole, n_O₂⁰=0.5 mole

sol: T=1000K, ∆G⁰=-5230J Kₚ=e^-∆G⁰/RT=1.876≡p_SO₃/(p_SO₂p_O₂^½)=[X_SO₃/(X_SO₂X_O₂^½)]∙P⁻^½

→ Kₚ²=[X_SO₃²/(X_SO₂²X_O₂)]∙P⁻¹ → (1-PKₚ²)x³+(3PKₚ²-3)x²-3PKₚ²x+ PKₚ²=0 代入P=1, Kₚ=1.876

x=0.463

Q2: ask nᵢ, pᵢ in equilibrium at T=900K with the same condition.

T=900K, ∆G⁰=-14167J Kₚ=e^-∆G⁰/RT=6.64=p_SO₃/(p_SO₂p_O₂^½)=[X_SO₃/(X_SO₂X_O₂^½)]∙P⁻^½

→ Kₚ²=[X_SO₃²/(X_SO₂²X_O₂)]∙P⁻¹ → (1-PKₚ²)x³+(3PKₚ²-3)x²-3PKₚ²x+ PKₚ²=0 代入P=1, Kₚ=6.64, x=0.704

Q3: ask nᵢ under equilibrium with P=10 atm at 1000K

(1-PKₚ²)x³+(3PKₚ²-3)x²-3PKₚ²x+ PKₚ²=0 代入P=10, Kₚ=1.876 → x=0.686

n_SO2= 0.314, n_O2 = 0.157, n_SO3= 0.686 → p_SO₃=5.93 atm, p_SO₂=2.71 atm, p_O₂=1.36 atm

check: ∆H⁰=-94600<0 exothermic

T↓ [∂(lnKₚ)∕∂T]ₚ=∆H⁰/RT²<0, ⸪∂T<0 ⸫∂(lnKₚ)>0 → Kₚ=p_SO₃/(p_SO₂p_O₂^½)↑, p_SO₃↑ p_SO₂p_O₂↓反應往右

P↑ Kₚ=[X_SO₃/(X_SO₂X_O₂^½)]∙P⁻^½=c₀, ⸪P⁻^½ ↓ ⸫X_SO₃/(X_SO₂X_O₂^½)↑ → X_SO₃↑ X_SO₂X_O₂↓反應往右

§藉氣體混合來調節氧分壓

SO₂₍_g₎ + ½O₂₍_g₎ = SO₃₍_g₎ at 1000K, 1 atm. Assuming p_O₂=0.1 atm ask n_SO2/n_SO3=a=? under equil.

p_O₂=X_O₂P=x/(2a+2+x)=0.1, a=4.5x-1

Kₚ²=[X_SO₃²/(X_SO₂²X_O₂)]∙P⁻¹=(1-x)²(2a+2+x)/(a+x)²xP=1.876 代入P=1, a=4.5x-1

→ 96.45x²-18.709x-6.481=0, x=0.374 → a=0.683

Q: when p_O₂=max. a=? 氧由SO₃ 分解而來, n_SO2⁰=0 ⸫a=0

H₂/H₂O mixure at T=2000K, P=1 atm given: H₂₍_g₎ + ½O₂₍_g₎ = H₂O₍_g₎ ∆G⁰=-247500+55.85T

keep p_O₂=10⁻¹⁰ atm, ask p_H₂O/p_H₂=?

Kₚ=p_H₂O/(p_H₂ p_O₂^½), Kₚ=e^-∆G⁰/RT=3.521∙10³ → p_H₂O/p_H₂=0.03521, p_H₂=0.966 atm, p_H₂O=0.034 atm

logp_H₂O=-2900/T-4.65logT+19.732, if p_H₂O=0.0352 at T=27℃

CO₍_g₎ + ½O₂₍_g₎ = CO₂₍_g₎ ∆G⁰=-282400+86.81T when T=1000K, P=1 atm and p_O₂=10⁻²⁰ atm, ask p_CO₂/p_CO=?

Sol: Kₚ=p_CO₂/(p_COp_O₂^½), Kₚ=e^-∆G⁰/RT=1.646∙10¹⁰ → p_CO₂/p_CO=1.646

實務上用gas flow(flowmeter or MFC)去控制壓力比值, pf:

before: n_ᴬ=Pa/RT, n_ᴮ=Pb/RT, mixing: a+b=V, p_ᴬ=n_ᴬRT/V=a/(a+b), p_ᴮ=n_ᴮRT/V=b/(a+b)

→ p_ᴬ/p_ᴮ=a/b

Ex-1. P₄₍_g₎ → 2P₂₍_g₎ ∆G⁰=225400+7.90TlnT-209.4T

Q1: What is T for equil. with P=1 atm and X_P₄=X_P₂=0.5

Kₚ=p_P₂²/p_P₄=(X_P₂²/X_P₄)P=0.5, -∆G⁰/RT=lnKₚ, -27109/T-0.95lnT+25.873=0, T=1429K

Q2: What is P for equil. with T=2000K atm and X_P₄=X_P₂=0.5

Kₚ=p_P₂²/p_P₄=(X_P₂²/X_P₄)P, Kₚ=e^-∆G⁰/RT=81.83 → 81.83=0.5P, P=163.66 atm

Ex-2. Decompose of NH₃: 2NH₃₍_g₎ = N₂₍_g₎ + 3H₂₍_g₎ ∆G⁰=87030-25.8TlnT-31.7T J

Q1: final pᵢ=? at T=400℃, P=1 atm

Kₚ≡p_N₂p_H₂³/p_NH₃², Kₚ=e^-∆G⁰/RT=4748

p_N₂=xP/(1+2x), p_H₂=3xP/(1+2x), p_NH₃=(1-2x)P/(1+2x)代入Kₚ, Kₚ=27x⁴P²/[(1+2x)²(1-2x)²]

→ Kₚ^½=5.196x²P/[(1+2x)(1-2x)]代入Kₚ=4748, P=1 → x=4945 → p_N₂, p_H₂, p_NH₃

Q2: final pᵢ=? at T=400℃, and fixed volume

fixed V, P(=1 atm) → P V=RT/P=n_ᴛ RT/P ⸫P=1+2x atm

⸪ Kₚ^½=5.196x²P/[(1+2x)(1-2x)] 代入Kₚ=4748, P=1+2x x=0.4909, P=1.981 atm

p_N₂=0.4954 atm, p_H₂=1.4862 atm, p_NH₃=0.0092 atm

Q3: Synthesis of NH₃ with H₂, N₂; let p_H₂/p_N₂=a, ask a=? when p_NH₃ at max.

解讀:p_NH₃ at max.等同dp_NH₃ /da=0, 求a=?

P=1 atm, P=p_N₂+p_H₂+p_NH₃=(a+1)p_N₂+p_NH₃ → p_N₂=(1-p_NH₃)/(a+1), p_H₂=a(1-p_NH₃)/(a+1)

Kₚ≡p_NH₃²/p_N₂p_H₂³=p_NH₃²(a+1)⁴/[a³(1-p_NH₃)⁴] → Kₚa³(1-p_NH₃)⁴=p_NH₃²(a+1)⁴

取ln → lnKₚ+3lna+4ln(1-p_NH₃)=2lnp_NH₃+4ln(a+1)

∂a → 3da/a-[4/(1-p_NH₃)]dp_NH₃ =(2/p_NH₃)dp_NH₃+4da/(a+1)

→ [4/(1-p_NH₃)+(2/p_NH₃)]dp_NH₃ =[3/a-4/(a+1)]da ⸪dp_NH₃ /da=0 ⸫3/a-4/(a+1)=0, a=3

Ex-3. 1 mole CH₄ + 1 mole CO₂ at T=1000K, P=1atm Final equilibrium composition ? pᵢ=?

Given (1) CH₄_(g)= C_(s)+ 2H₂_(g) ∆G₁⁰=69120-22.25TlnT+65.35 (2) 2C_(s)+ O₂_(g)= 2CO_(g) ∆G₂⁰=-223400-175.3T (3) C_(s)+ O₂_(g) = CO₂_(g) ∆G₃⁰=-394100-0.8T (4) H₂_(g)+½O₂_(g)= H₂O_(g) ∆G₄⁰=-246400+54.8T

sol: CH₄₍_g₎ + CO₂₍_g₎ = 2H₂₍_g₎ + 2CO₍_g₎...(5)

ini     1            1            0             0                Kₚ₅= p_CO²p_H₂²/p_CH₄p_CO₂

f     1-x         1-x          2x            2x

f      1-x       1-x-y       2x+y        2x-y

∆G₅⁰=(1)+(2)-(3)=239820-22.25TlnT-174.5T+65.35=15.95

     H₂₍_g₎ + CO₂₍_g₎ = H₂O₍_g₎ + CO₍_g₎....(6) ∆G₆⁰=(4)-(3)+½(2)=36000-32.05T=0.62

ini  2x       1-x           0           2x              Kₚ₆=p_COp_H₂O/p_H₂p_CO₂

f   2x-y     1-x-y         y         2x+y       →影響(5), (5)(6)兩反應需同時平衡

n_ᴛ=n_CH₄+n_CO₂+n_H₂+n_CO+n_H₂O=(1-x)+(1-x-y)+(2x-y)+(2x+y)+y=2(1+x)

Kₚ₅= p_CO²p_H₂²/p_CH₄p_CO₂=[(2x+y)²(2x-y)²P²/(1-x)(1-x-y)(2+2x)²]=15.95

Kₚ₆=p_COp_H₂O/p_H₂p_CO₂=[(2x+y)y/(1-x-y)(2x-y)]=0.62

聯立解 x=0.785, y=0.0775 → p_CH₄=0.0602, p_CO₂=0.0385, p_H₂=0.4181, p_CO=0.4615, p_H₂O=0.0217

CO₍_g₎ + ½O₂₍_g₎ = CO₂₍_g₎....(7) ∆G₇⁰=(3)-½(2)=-282400+86.85T

Kₚ₇=exp(-∆G₇⁰/RT)=exp(369250/8314.4)=1.64∙10¹⁰=p_CO₂/p_COp_O₂^½ p_O₂^½=0.0385/(0.4615∙1.64∙10¹⁰)

⸫p_O₂=2.5876∙10⁻²³ atm

H₂₍_g₎ + ½O₂₍_g₎ = H₂O₍_g₎....(4) ∆G₄⁰=-246400+54.8T

Kₚ₄=exp(-∆G₄⁰/RT)=exp(191600/8314.4)=1.02∙10¹⁰=p_H₂O/p_H₂p_O₂^½ p_O₂^½=0.0217/(0.4181∙1.02∙10¹⁰)

⸫p_O₂=2.5892∙10⁻²³ atm

整個平衡狀態包括(4),(5),(6),(7)的反應, 最後平衡成分:

p_CH₄=0.0602, p_CO₂=0.0385, p_H₂=0.4181, p_CO=0.4615, p_H₂O=0.0217, p_O₂=2.6∙10⁻²³ atm