Esami svolti chimica fisica (I) seconda parte

18/06/12

T=298K

ΔH_f Be3N2(s) = ? a T=298K

Reazioni

• Be3N2(s)
• (298,4)
• Be(s) + O2(g) → BeO(s)

ΔH⁰ _r (KJ/mol)

ΔH_f P Be3N2(s, 180K) = 1265,0 KJ/mol

Cp N2(g) = 25,58 + 0,01197

Cp Be3N2(s) = -11,912 + 0,25017

1. 3 Be + N2 → Be3N2 + 3/2 O2
2. 2 O2 → 2 O3

3 BECl2 + N2 + 3Be + 3Cl2 → Be3N2 + 3BeCl2

N2 + 2O3 → 2N2 + 2O

ΔH⁰ = (-584,9 + -496,4):3 = -585,3 KJ/mol

1. 3 Be + 2 NH3 → Be3N2 + 3/2 O2
2. 3 Be + 2O2 → 2NH2 + Be3N2

3Be + 2NH2 + 3 H₂ → 2NH3 + 3/2O2 + 3HCl

3Be + N2 → Be3N2

ΔH_r⁰ = -405(124) + (-426,11) = -584,62 KJ/mol

ΔH_f Be3H2 = -585,3 + (-581,62):2 = -586,06 KJ/mol

ΔH=cpΔT

d(ΔH_fT)=∫Δr_pdT

• Cp = Cp
• (-11,912 + 0,25017)(Δ_pBe₃O₂) - Cp= ε

T=298K Trovo ΔH_f o 298K

ΔH_f = -0,874 = -86,812 + (0,25017x + 298,15):2

T=180K Trovo ΔH_f o 180K

ΔH_f o 180K= -58,766.812 + 86.812

• -583,128 x51/mol

T(°C) 97,6 88 94 100 94

• P_i^St(bar) 4,285 0,258 0,616 1,852 4
• P_t^ot(bar) 4 5166 5186 5981 5166

a) composizione a Pcr? ...

b) X=0,5 La composizione?

c) X=0,183 ...

PA=PCα; PB=xB*PB

XB=PA=PCα; YB=PBPCα

xB=α;

YB=0,633; YB=0,633:1,285=0,493

PA=PCα; PB

XA=YA; YB=PBPCα

xB=α:0,633;

YA=0,633; yB=0,6330,661⇒0,402

(TCC) 70,14 88 94 400 440

• XB=0,15⇒dal grafico si ottiene che Teb=93°C
• YB=0,42

ΔStccc=7; ampliessione isoterma; T=353,2 K

n(C6H6=3mol; Pi=1,0532·10⁵PaPz=1,0133·10⁵Pa

condensazione e raffredamento; T2=333,2 K

Cp(C6H6)=(e) =4,18 J/gK

ΔHap,C6H6=30844,92 J/mol

• (C6H6 (g)) → n(C6H6) (g)
• T=353,2 K
• P=1,0532·10⁵Pa

Raffredamento a P=Cose

• T=333,2 K
• T=353,2 K
• P=1,0133·10⁵Pa
• P=ΔvapnP2P1
• =4,618,3·8,18 n 1,0133·10⁵ Pa
• ΔS₂=ΔHap,n/np;
• =30844,92,9.4/353,2
• =84,423 J/K

ΔS₂ = Φp; M⌐⌐⌐C6H6= en tf

• =4,18J · 8,18 J en 333,2 8,618;8
• - 8118,967 S/T KI/K

A). V=ydm³ PN2≈0,233 mol P(ALS=0,361mol T=250°C

• PCαeq=29,38 Keg Keq=?
• PC(3g)==PC(3g)+2C(g+n )
• ∆Hp,T = T 250CP per vietreior- modinamica

ΔHp conostante rispeta a T.

da Pcαeq compenda che la pressione dal gas

inerta (N2)= bisogna ogsåiera;

^PV/rt=prt

PN2=Nortr2=0,186 n 0,3198=0,186

0.101.131⋅bar

PCαeq = 29,33-101,33^x=1,19534 bar

• PAUS x = x
• ς = ς

√/√^{X+o x}

PC¹⁸³=29,089 0 X=(PA)

VNRT

K_p = ^Pco₂/_Po

(Δ = 0,169)

= 0,1690

ΔG_500.15K = -RTlnK_p

R = 8,316·500,15·ln(0,169)

= 17,0 kJ/mol

• K_p non varia

PCl₂, eq = 0,2 bar

P_NCl = 0,2 bar

= 0,2 bar

= 0,7 bar

P(Cl₂)

P_NCl + P_NO + PCl₂ = 0,7 bar

ΔG_298K = YP_APGP-VRΔ_GR

1 = 2ΔNO + ΔG = 0 + 2ΔG°col

1 = 51,31·10³·298.15 (-29,11)

- ΔH_298K = YP_APSSR

= 47,05 KJ/mol

Δ = ΔH° - RTlnK_R

lnK(12) = lnK(7)

K_p,500.15K = e^lnK = 0,017

ΔG°_500.15K = -RTlnK

1 = 8,316·500,15

= 17,5455 KJ/mol

E° = 0,0885 V a T = 25°C

catodo:

anodo:

= eè_{RT lnQR}

E°'> = E° - E°_cata - E°_anodo

(-log(H⁺))

= 6,13

Legge Henry generalizzata!

Pt = aiKi = ∑ Xi · Ki

Q₁, C_{S 2} → P_S2 = 232 X_CS2 = 0.182

Q₁, Ac → P_Ac = 281 X_{C Ac} = 0.28

Q₁, Ac → 0.15 · 1320 = (Q) calcolato per tutte e Q P (door) v

E° = 7 T = 25°C gas reali

Ptl H₂(g), 2 atm, 1M, Cl (aq), 0.1 mol/kg (L₂, G₁, 2 atm) Pt

• catodo: 2H₂ (g) + 2e^- → 2H⁺ (aq) =
• anodo: Cl^- (aq) + 2e^- → 1/2 H₂ (g)
• 1/2 Cl₂(g) → 1/2 H₂(g) + H⁺ (aq) + Cl^- (aq)

ε = RT / b Check

ε° = ε° (valore) + ε = 4.36 - 0 = 4.36 V

Q = 2 H⁺ Cl^-

ai = X γ = Cl (P_H2 X X P_H2)₅ P_H2 = 40 atm

C_H Cl = Cl H₂ → H Cl^-

γ = X/Y M = Cl = γH₂ → YCl^- Cl₂

i = Y (4^2.01 + 4^-) = Y = 0.4

log bt = - 0.51 + Y/2 J/Y 0.3

log y = 0.5 | Y | 0.01 0.3 | 0.1 | = Y |,107

y t = | 0.3 + 0.034

ε = 1 + 831 - 0.311 + ε n - 4 = 4.56 J

21/06/13

• ΔH°_CCu (g) ΔS°_CCu = ?
• c_pCu = 83.3 J/Kmol T = 298K

G. di Hess:

(C₁ + 2Cl² (g) → C₂ Cl₂ (g))

1. 1 CCu (g), 350 K
2. Cu(g): 350 K
3. Cu(g), 298.15K
4. ΔH₃, ¹/₂ Cu (e)

Cu

ΔH₃ & ΔH_C = Cu_(G) n n n ΔH₁ & ∫ 131,75, 5 (350 - 298.15)

Cu(e)

= 68.3 105 J/mol

ΔH = ΔH₃ hop = ΔH tZe

Cu(g)

= 131.105 5/mol

CCHC = -10,283 KJ/mol

n S Cu(g) 298K

• 5Cu_e En Tf

  = 131.75 ∏en 850 298.15

  = 211.20 51 Km o5

• ΔS₃= 1.2 300₀
• f tps 350
• ΔS₄= C_pCu(g) En Tf
• = 83.3 En, 13.356 51 Kmol
• S Cu = S Cu, T = 298K = 308.188 51 Kmol