Transcript The intensive state of a PVT system containing N chemical species and  phases in equilibrium is characterized by the intensive variables,

The intensive state of a PVT system containing N
chemical species and  phases in equilibrium is
characterized by the intensive variables, temperature T,
pressure P, and N - 1 mole fractions for each phase.
These are the phase-rule variables, and their number is
2 + (N – 1)().
The masses of the phases are not phase-rule variables,
because they have no influence on the intensive state of
the system.
An independent phase-equilibrium equation may be
written connecting intensive variables for each of the N
species for each pair of phases present.
Thus, the number of independent phase-equilibrium
equations is ( – 1 )(N).
The difference between the number of phase-rule
variables and the number of independent equations
connecting them is the number of variables that may be
independently fixed.
Called the degrees of freedom of the system F, the
number is:
F  2  N  1       1N
or
F 2N
(42)
For reacting system, the phase-rule variables are
unchanged: temperature, pressure, and N – 1 mole
fractions in each phase.
The total number of these variables is 2 + (N – 1)().
The same phase-equilibrium equations apply as before,
and they number ( – 1)(N).
However, Eq. (12) provides for each independent reaction
an additional relation that must be satisfied at
equilibrium.
Since the pi's are functions of temperature, pressure, and
the phase compositions, Eq. (12) represents a relation
connecting phase-rule variables.
If there are r independent chemical reactions at equilibrium within the system, then there is a total of ( – 1)(N) + r
independent equations relating the phase-rule variables.
Taking the difference between the number of variables
and the number of equations gives:
F  2  N  1       1N  r 
or
F  2    N r
This is the phase rule for reacting systems
(43)
Example
Determine the number of degrees of freedom F for a
system of two miscible nonreacting species which exists in
vapor/liquid equilibrium
Solution
F  2    N r
 2220  2
Example
Determine the number of degrees of freedom F for a
system prepared by partially decomposing CaCO3 into an
evacuated space.
Solution
A single chemical reaction occurs:
CaCO3(s)  CaO(s) + CO2(g)
r=1
N=3
 = 3 (solid CaCO3, solid CaO, and gaseous CO2)
F = 2 –  + N – r =2 – 3 + 3 – 1 = 1
When liquid and gas phases are both present in an
equilibrium mixture of reacting species,
ˆfiV  ˆfiL
a criterion of vapor-liquid equilibrium, must be satisfied
along with the equation of chemical-reaction equilibrium.
Consider, for example, the reaction of gas A with liquid
water B to form an aqueous solution C.
Several choices in the method of treatment exist.
Method 1
• The reaction may be assumed to occur entirely in the
gas phase with simultaneous transfer of material
between phases to maintain phase equilibrium.
• In this case, the equilibrium constant is evaluated
from G0 data based on standard states for the
species as gases, i.e., the ideal-gas states at 1 bar and
the reaction temperature.
Method 2
• The reaction may be assumed to occur in the liquid
phase.
• In this case, the equilibrium constant is evaluated
from G0 data based on standard states for the
species as liquids.
Method 3
Alternatively, the reaction may be written:
A(g) + B(l)  C(aq)
in which case the G0 value is for mixed standard states: C
as a solute in an ideal 1-molal aqueous solution, B as a
pure liquid at 1 bar, and A as a pure ideal gas at 1 bar.
For this choice of standard states, the equilibrium
constant as given by Eq. (12) becomes:
Henry’s Law
ˆfC fC0
mC

K
0
0
0
ˆfB fB ˆfA fA   B xB ˆfA fA 
(44)
Jika keadaan keseimbangan dalam suatu sistem reaksi
tergantung pada dua atau lebih reaksi kimia independen,
maka langkah-langkah untuk menentukan komposisi
keseimbangan adalah:
• Tentukan reaksi kimia yang terjadi.
• Komposisi keseimbangan dihitung seperti yang telah
dibahas.
• Konstanta keseimbangan untuk setiap reaksi dievaluasi
dengan cara seperti yang telah dibahas.
Untuk reaksi tunggal:
i
ˆ
 fi 
 0   K
i
 fi 
(12)
Untuk reaksi multi:
i ,j
ˆ
 fi 
 0   K j
i
 fi 
(45)
dengan j adalah nomor reaksi kimia.
Untuk reaksi fasa gas, persamaan (45) menjadi:
i ,j
ˆ
 fi 
 0   K j
i
P 
(46)
Jika campuran keseimbangan berupa gas ideal:
 y i 
i ,j
i
 j
P

  0  Kj
P 
(47)
CONTOH
Terhadap senyawa n-butana dilakukan reaksi cracking
pada 750 K dan 1,2 bar sehingga dihasilkan olefin.
Hanya dua reaksi yang memiliki konversi keseimbangan
yang signifikan, yaitu:
C4H10  C2H4 + C2H6
(I)
C4H10  C3H6 + CH4
(II)
Jika kedua reaksi ini mencapai keseimbangan,
bagaimana komposisi produk?
Data: K(I) = 3,856
K(II) = 268,4
Penyelesaian:
j
C4H10
–1
–1
I
II
C2H4
+1
0
i,j
C2H6
+1
0
Basis: 1 mol umpan C4H10
n   1

n  n
C 4H10 0
0
C 4H10 0
1
C3H6
0
+1
CH4
0
+1
j
+1
+1
nC4H10  1  I  II
nC2H4  I
nC2H6  I
nC3H6  II
nCH4  II
n0  1  I  II
y C4H10
1  I  II

1  I  II
y C2H4
I

1  I  II
y C2H6
I

1  I  II
y C3H6
II

1  I  II
y CH4
II

1  I  II
Keseimbangan kimia:
 y i 
i ,j
i
P

  0  Kj
P 
y C2H4 y C2H6
y C 4H10
y C3H6 y CH4
y C 4H10
 j
1
P

  0  KI
P 

2
I
1
P

  0  KI
1  I  II 1  I  II   P 
(A)
1
P

  0  KII
P 

2
II
1
P

  0  KII
1  I  II 1  I  II   P 
(B)
 A  I2 KI 3,856
 2 
 0 ,0144
B  II KII 268,4
I
 0 ,0144  0 ,12
II
I  0 ,12 II
Per. (C) dimasukkan ke pers. (A):

1
2
I
P

  0  KI
1  I  II 1  I  II   P 
0,12II 
1,2 


 3,856
1  0,12II  II 1  0,12II  II   1 
2
1
(C)
0,12II 
1,2 


 3,856
1  0,12II  II 1  0,12II  II   1 
1
2
0,12II 
 3,2133
1  1,12II 1  1,12II 
2
0,0144II2  3,21331  1,2544II2 
0,0144II2  3,2133  4,0308II2
4,0452II2  3,2133
3,2133
II 
 0,8913
4,0452
I  0 ,107
y C4H10
1  I  II

 0,0009
1  I  II
y C2H4
I

 0,0535
1  I  II
y C2H6
I

 0,0535
1  I  II
y C3H6
II

 0,4460
1  I  II
y CH4
II

 0,4460
1  I  II
Contoh
Setumpukan batubara (dianggap terdiri dari karbon
murni) dalam sebuah gasifier dialiri steam dan udara
sehingga terjadi reaksi yang menghasilkan gas yang
terdiri dari H2, CO, O2, CO2, dan N2. Jika gas yang
diumpankan ke dalam gasifier terdiri dari 1 mol steam
dan 2,38 mol udara, hitung komposisi keseimbangan
dari aliran gas yang keluar dari gasifier pada P = 20 bar
dan temperatur 1500 K. Diketahui nilai G0f,1500 untuk:
H2O = – 164.310 J/mol
CO = – 243.740 J/mol
CO2 = – 396.160 J/mol
Karena temperatur cukup tinggi, maka campuran gas
dapat dianggap sebagai gas ideal.
Penyelesaian
Kemungkinan reaksi yang terjadi:
C + O2  CO2
(a)
C + CO2  2 CO
(b)
H2O + C  H2 + CO
(c)
Harga K untuk masing-masing reaksi pada 1500K adalah:
0
  G1500

K a  exp

 RT 
  G0f ,1500 CO 
    396.160 
2
 exp
  exp




RT
8
,
314
1500




 6,25  1013
Dengan cara yang sama diperoleh:
Kb  1514,12
K c  583,58
Harga Ka sangat besar, yang berarti bahwa reaksi (a)
merupakan reaksi irreversibel dan semua O2 habis
bereaksi dengan C menjadi CO2 [reaksi (a)].
Jumlah mol gas mula-mula:
H2O = 1 mol
N2
= 0,79  2,38 = 1,88 mol
O2
= 0,21  2,38 = 0,5 mol
n0 =3,38
Jumlah mol gas setelah reaksi:
nH2O = 1 – c
nN2 = 1,88
nCO2 = 0,5 – b
n  3,38  b   c
nCO = 2 b + c
H2
= c
Fraksi mol masing-masing komponen setelah reaksi:
1  c
y H2O 
3,38  b   c
1,88
y N2 
3,38  b   c
0 ,5  b

3,38  b   c
2 b   c

3,38  b   c
y CO 2
c
y H2 
3,38  b   c
y CO
Untuk reaksi (b):
 b     i   1  2  1
 i b
 y i 
i
i
y CO 2
y CO2
P

 0
P 
b
Kb
1
20 

   1514,12
1
2b   c 2
 75,706
0,5   c 3,38  b  b 
2b  c 2  75,7060,5  b 3,38  b  b   0
(A)
Untuk reaksi (c):
 c     i   1  1  1  1
 i c
 y i 
i
i
P

 0
P 
c
Kc
y CO yH2  20  1
   583,58
yH2O
1
2b   c  c
 29,179
1   c 3,38  b   c 
2b  c c  29,1791  c 3,38  b  c   0
(B)
Persamaan (A) dan (B) diselesaikan secara simultan
dengan menggunakan Excel Solver dengan batasan:
 0 ,5  b  0 ,5
dan
0  c  1
Hasil perhitungan:
b = 0,4895 dan c = 0,9863
Jika nilai b dan c ini dimasukkan ke persamaan untuk
fraksi mol masing-masing komponen maka akan
diperoleh:
yH2  0,203
y CO2  0,002
y CO  0,405
yN2  0,387
yH2O  0,003