(12) Kff i0ii
i
The expression of K in terms of fugacity coefficient is:
The standard state for a gas is the ideal-gas state of the pure gas at the standard-state pressure P0 of 1 bar.
Since the fugacity of an ideal gas is equal to its pressure, fi
0 = P0 for each species i .
Thus for gas-phase reactions , and Eq. (12) becomes:
0i
0ii Pfff
KPf i
0i
i
(26)
The equilibrium constant K is a function of temperature only.
However, Eq. (26) relates K to fugacities of the reacting species as they exist in the real equilibrium mixture.
These fugacities reflect the nonidealities of the equili-brium mixture and are functions of temperature, pressure, and composition.
This means that for a fixed temperature the composition at equilibrium must change with pressure in such a way that remains constant i0
ii
Pf
The fugacity is related to the fugacity coefficient by
Substitution of this equation into Eq. (26) provides an equilibrium expression displaying the pressure and the composition:
Pyˆf iii
KPPˆy 0ii
i
i
Where = ii and P0 i s the standard-state pressure of 1 bar, expressed in the same units used for P.
(27)
If the assumption that the equilibrium mixture is an ideal solution is justified, then each becomes i, the fugacity coefficient of pure species i at T and P.
In this case, Eq. (27) becomes:
i
KPP
y 0iii
i
(27)
For pressures sufficiently low or temperatures sufficiently high, the equilibrium mixture behaves essentially as an ideal gas. In this event, each i = 1, and Eq. (27) reduces to:
KPP
y 0ii
i
(28)
Although Eq. (28) holds only for an ideal-gas reaction, we can base some conclusions on it that are true in general:According to Eq. (20), the effect of temperature on the
equilibrium constant K is determined by the sign of H0:o H0 > 0 (the reaction is endothermic) T>> K >>. Eq.
(28) shows that K >> at constant P >> >>
o H0 < 0 (the reaction is exothermic) T>> K <<. Eq. (28) shows that K << at constant P << e <<
i
ii
y
i
ii
y
If the total stoichiometric number ( ii) is negative, Eq. (28) shows that am increase in P at constant T causes an increase in , implying a shift of the reaction to the right.
If the total stoichiometric number ( ii) is positive, Eq. (28) shows that am increase in P at constant T causes a decrease in , implying a shift of the reaction to the left, and a decrease in e.
i
ii
y
i
ii
y
For a reaction occurring in the liquid phase, we return to
LIQUID PHASE REACTION
(12) Kff i0ii
i
For the usual standard state for liquids f0i is the fugacity
of pure liquid i at the temperature of the system and at 1 bar.
The activity coefficient is related to fugacity according to:
iiii fxf (29)
The fugacity ratio can now be expressed
0
i
iii0
i
iii0
i
i
ff
xf
fxff
(30)
(31)
Gibbs free energy for pure species i in its standard state at the same temperature:
0ii
0i flnRTTG (7)
Gibbs free energy for pure species i at P and the same temperature:
iii flnRTTG (7.a)
The difference between these two equations is:
0i
i0ii f
flnRTGG
Fundamental equation for Gibbs energy:
dTSdPVdG iii (32)
For a constant-temperature process:
dPVdG ii (32)
For a pure substance undergone a constant-temperature process from P0 to P, the Gibbs free energy change is:
P
Pi
G
G 0
i
0i
dPVdG (33)
(34) 0i
0ii PPVGG
Combining eqs. (31) and (34) yields:
(35)
RT
PPVff
ln0
i0
i
i
RT
PPVexpx
ff 0
iii0
i
i
(36)
RT
PPVexp
ff 0
i0
i
ior
Combining eqs. (31) and (35) yields:
K
RTPPV
expx0
iiii
Combining eqs. (36) and (12) yields:
K
RTPPV
expxu
i
0i
iiii
u
i
RTPPV
expKx0
i
iiii
iii
0
iiiV
RTPP
expKx i (37)
For low and moderate pressure, the exponential term is close to unity and may be omitted. Then,
Kx i
iii
(38)
and the only problem is determination of the activity coefficients.
An equation such as the Wilson equation or the UNIFAC method can in principle be applied, and the compositions can be found from eq. (38) by a complex iterative computer program.
However, the relative ease of experimental investigation for liquid mixtures has worked against the application of Eq. (38).
If the equilibrium mixture is an ideal solution, then i is unity, and Eq. (38) becomes:
Kx i
ii
(39)
This relation is known as THE LAW OF MASS ACTION.
Since liquids often form non-ideal solutions, Eq. (39) can be expected in many instances to yield poor results.
Suppose a single reaction occurs in a homogeneous system, and suppose the equilibrium constant is known.
In this event, the calculation of the phase composition at equilibrium is straightforward if the phase is assumed an ideal gas [Eq. (28)] or an ideal solution [Eq. (27) or (39)].
When an assumption of ideality is not reasonable, the problem is still tractable for gas-phase reactions through application of an equation of state and solution by computer.
For heterogeneous systems, where more than one phase is present, the problem is more complicated and requires the superposition of the criterion for phase equilibrium developed in Sec. 11.6
Single-Phase Reactions
The water-gas shift reaction,
CO (g) + H2O (g) CO2 (g) + H2 (g)
Is carried out under the different set of conditions below. Calculate the fraction of steam reacted in each case. Assume the mixture behaves as an ideal gas.
The reactants consist of 1 mol of H2O vapor and 1 mol of CO. The temperature is 1100 K and the pressure is 1 bar.
Example
Solution
CO H2O CO2 H2
– 1 – 1 + 1 + 1A 3,376 3,470 5,547 3,249
B 103 0,557 1,450 1,045 0,422C 106 – 4,392 0 0 0D 10-5 – 0,031 0,121 – 1,157 0,083H0
f,298 – 110.525 – 241.818 – 393.509 0
G0f,298 – 137.169 – 228.572 – 394.359 0
950,1A
310540,0B
610392,4C
510164,1D
10298 molJ166.41H
10298 molJ618.28G
260.103
15,298314,8618.28
expRT
GexpK
0
00
0
20
TT
1RT
HexpK 0
0
00
1
610527,5
110015,298
115,298314,8
166.41exp
6894,315,298
1100TT
0
189,2K2
189,28354,310527,5261.103KKKK 6210
21
01111i
i
Since the reaction mixture is an ideal gas:
Ky i
ii
189,2Kyyyy
OHCO
HCO
2
22
eiii 0nn
e0nn
The number of each species at equilibrium is:
While total number of all species at equilibrium is:
189,2Kyyyy
OHCO
HCO
2
22
189,2
1 2
2
222 21189,21189,2
0189,2378,4189,1 2
5436,0e
Therefore the fraction of the steam that reacts is 0.5
Estimate the maximum conversion of ethylene to ethanol by vapor-phase hydration at 523.15 K and 1.5 bars for an initial steam-to-ethylene ratio of 5.
Example
Solution
Reaction:C2H4 (g) + H2O (g) C2H5OH (g)
C2H4 H2O C2H5OH – 1 – 1 + 1A 1,424 3,470 3,518B 14,394 10-3 1,450 10-3 20,001 10-3
C – 4,392 10-6 0 – 6,002 10-6
D 0 0,121 105 0H0
f,298 52.510 – 241.818 – 235.100G0
f,298 68.460 – 228.572 – 168.490
376,1A
310157,4B
610610,1C
510121,0D
10298 molJ792.45H
10298 molJ378.8G
366,29
15,298314,8378.8
expRT
GexpK
0
00
0
27
TT
1RT
HexpK 0
0
00
1
4105,3
15,52315,298
115,298314,8
792.45exp
7547,115,29815,523
TT
0
9778,0K2
34210 1005.109778,0105,3366,29KKKK
1111i
i
KPP
y 0ii
i
KPP
yyy
0OHHC
OHHC
242
52
KPP
y 0ii
i
For hign temperature and sufficiently low pressure:
eiii 0nn
ee0 6nn
The number of each species at equilibrium is:
While total number of all species at equilibrium is:
eHC 5n42
eOH 1n2
eOHHC 52n
e
eHC 6
5y
42
e
eOH 6
1y
2
e
eOHHC 6
y52
KPP
yyy
0OHHC
OHHC
242
52
33
ee
ee 1007.151005.105.115
6
ee3
ee 151007.156
00754.009045.601507.1 e2e
0135.0e
The gas-phase oxidation of SO2 to SO3 is carried out at a pressure of 1 bar with 20% excess air in an adiabatic reactor. Assuming that the reactants enter at 298.15 K and that equilibrium is attained at the exit, determine the composition and temperature of the product stream from the reactor.
Example
Solution
Reaction:SO2 (g) + ½ O2 (g) SO3 (g)
Basis: 1 mole of SO2 entering the reactor:moles of O2 entering = (0.5) (1.2) = 0.6moles of N2 entering = (0.6) (79/21) = 2.257
The amount of each species in the product stream is:eiii 0
nn
eSO 1n2
eO 5.06.0n2
eSO3n
257.2n3N
e5.0857.3n
Total amount of all species:
Mole fraction of each species:
e
eSO 5.0857.3
1y
2
e
eO 5.0857.3
5.06.0y
2
e
eSO 5.0857.3
y3
eN 5.0857.3
257.2y
2
Energy balance:
ReactantT = 298.15 KSO2 = 1O2 = 0.6N2 = 2.257
ProductT = 298.15 KSO2 = 1 – e O2 = 0.6 – 0.5 e
N2 = 2.257
Reactione
ProductTSO2 = 1 – e O2 = 0.6 – 0.5 e
N2 = 2.257
0HHH 0Pe
0298
i
T
T
Pi
0P
0
i dTRC
RnH
i
T
T
2iiii
0
dTTDTBARn
T
T
2
iii
iii
iii
0P
0
dTTDnTBnAnRH
0iii
20
2iii
0i
ii T1
T1
DnTT2
BnTTAnR
(a)
(b)
SO2 O2 SO3
– 1 – 1 + 1A 5.699 3.639 8.060B 0.801 10-3 0.506 10-3 1.056 10-3
C 0 0 0D – 1.015 105 – 0.227 105 – 2.028 105
H0f,298 – 296 830 0 – 395 720
G0f,298 – 300 194 0 – 371 060
278.1A
310251,0B
0C
510786.0D
10298 molJ98890H
10298 molJ70866G
12
0
00
0 106054,215,298314,8
70866exp
RTG
expK
38
5.0
e
e
e
e05
OSO
SO
5.06.05.0857.3
1yyy
K22
3
KKPP
y 0ii
i
For hign temperature and pressure of 1 bar:
(f)
Algorithm:
1. Assume a starting value of T
2. Evaluate K1 [eq. (c)], K2 [eq. (d)], and K [eq. (e)]
3. Solve for e [eq. (f)]
4. Evaluate T [eq. (a)]
5. Find a new value of T as the arithmatic mean value just calculated and the initial value; return to step 2.
The scheme converges on the value e = 0.77 and T = 855.7 K
The composition of the product is:
0662.0
472.323.0
77.05.0857.377.01
5.0857.31
ye
eSO2
0619.0472.3215.0
77.05.0857.377.05.06.0
y2O
2218.0472.377.0
y3SO
6501.0472.3257.2
y2N
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