(12)
The expression of K in terms of fugacity coefficient is:
The standard state for a gas is the ideal-gas state of thepure gas at the standard-state pressure P0 of 1 bar.
Since the fugacity of an ideal gas is equal to its pressure,fi0 = P0 for each species i .
Thus for gas-phase reactions , and Eq. (12)becomes:
(26)
The equilibrium constant K is a function of temperatureonly.
However, Eq. (26) relates K to fugacities of the reactingspecies 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 thecomposition at equilibrium must change with pressurein such a way that remains constant
The fugacity is related to the fugacity coefficient by
Substitution of this equation into Eq. (26) provides anequilibrium expression displaying the pressure and thecomposition:
Where = ii and P0 i s the standard-state pressure of 1bar, expressed in the same units used for P.
(27)
If the assumption that the equilibrium mixture is an idealsolution is justified, then each becomes i, the fugacitycoefficient of pure species i at T and P.
In this case, Eq. (27) becomes:
(27)
For pressures sufficiently low or temperatures sufficientlyhigh, the equilibrium mixture behaves essentially as anideal gas. In this event, each i = 1, and Eq. (27) reduces to:
(28)
Although Eq. (28) holds only for an ideal-gas reaction,we can base some conclusions on it that are true ingeneral:
According to Eq. (20), the effect of temperature onthe equilibrium constant K is determined by the signof H0:
oH0 > 0 (the reaction is endothermic) T>> K >>.Eq. (28) shows that K >> at constant P >> >>
oH0 < 0 (the reaction is exothermic) T>> K <<.Eq. (28) shows that K << at constant P << e <<
If the total stoichiometric number ( ii) isnegative, Eq. (28) shows that am increase in P atconstant T causes an increase in , implying ashift of the reaction to the right.
If the total stoichiometric number ( ii) ispositive, Eq. (28) shows that am increase in P atconstant T causes a decrease in , implying ashift of the reaction to the left, and a decrease in e.
For a reaction occurring in the liquid phase, we return to
(12)
For the usual standard state for liquids f0i is the fugacityof pure liquid i at the temperature of the system and at 1bar.
The activity coefficient is related to fugacity according to:
(29)
The fugacity ratio can now be expressed
(30)
(31)
Gibbs free energy for pure species i in its standard state atthe same temperature:
(7)
Gibbs free energy for pure species i at P and the sametemperature:
(7.a)
The difference between these two equations is:
Fundamental equation for Gibbs energy:
(32)
For a constant-temperature process:
(32)
For a pure substance undergone a constant-temperatureprocess from P0 to P, the Gibbs free energy change is:
(33)
(34)
Combining eqs. (31) and (34) yields:
(35)
(36)
or
Combining eqs. (31) and (35) yields:
Combining eqs. (36) and (12) yields:
(37)
For low and moderate pressure, the exponential term isclose to unity and may be omitted. Then,
(38)
and the only problem is determination of the activitycoefficients.
An equation such as the Wilson equation or the UNIFACmethod can in principle be applied, and the compositionscan be found from eq. (38) by a complex iterativecomputer program.
However, the relative ease of experimental investigationfor liquid mixtures has worked against the application ofEq. (38).
If the equilibrium mixture is an ideal solution, then i isunity, and Eq. (38) becomes:
(39)
This relation is known as THE LAW OF MASS ACTION.
Since liquids often form non-ideal solutions, Eq. (39) canbe expected in many instances to yield poor results.
Suppose a single reaction occurs in a homogeneoussystem, and suppose the equilibrium constant is known.
In this event, the calculation of the phase compositionat equilibrium is straightforward if the phase isassumed an ideal gas [Eq. (28)] or an ideal solution [Eq.(27) or (39)].
When an assumption of ideality is not reasonable, theproblem is still tractable for gas-phase reactionsthrough application of an equation of state and solutionby computer.
For heterogeneous systems, where more than onephase is present, the problem is more complicated andrequires the superposition of the criterion for phaseequilibrium 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 ofCO. The temperature is 1100 K and the pressure is 1 bar.
Example
Solution
CO
H2O
CO2
H2
– 1
– 1
+ 1
+ 1
A
3,376
3,470
5,547
3,249
B 103
0,557
1,450
1,045
0,422
C 106
– 4,392
0
0
0
D 10-5
– 0,031
0,121
– 1,157
0,083
H0f,298
– 110.525
– 241.818
– 393.509
0
G0f,298
– 137.169
– 228.572
– 394.359
0
20
21
Since the reaction mixture is an ideal gas:
The number of each species at equilibrium is:
While total number of all species at equilibrium is:
Therefore the fraction of the steam that reacts is 0.5
Estimate the maximum conversion of ethylene to ethanolby vapor-phase hydration at 523.15 K and 1.5 bars for aninitial steam-to-ethylene ratio of 5.
Example
Solution
Reaction:
C2H4 (g) + H2O (g) C2H5OH (g)
C2H4
H2O
C2H5OH
– 1
– 1
+ 1
A
1,424
3,470
3,518
B
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
0
H0f,298
52.510
– 241.818
– 235.100
G0f,298
68.460
– 228.572
– 168.490
27
For hign temperature and sufficiently low pressure:
The number of each species at equilibrium is:
While total number of all species at equilibrium is:
The gas-phase oxidation of SO2 to SO3 is carried out at apressure of 1 bar with 20% excess air in an adiabaticreactor. Assuming that the reactants enter at 298.15 K andthat equilibrium is attained at the exit, determine thecomposition and temperature of the product stream fromthe 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) = 0.6
moles of N2 entering = (0.6) (79/21) = 2.257
The amount of each species in the product stream is:
Total amount of all species:
Mole fraction of each species:
Energy balance:
Reactant
T = 298.15 K
SO2 = 1
O2 = 0.6
N2 = 2.257
Product
T = 298.15 K
SO2 = 1 – e
O2 = 0.6 – 0.5 e
N2 = 2.257
Reaction
e
Product
T
SO2 = 1 – e
O2 = 0.6 – 0.5 e
N2 = 2.257
(a)
(b)
SO2
O2
SO3
– 1
– 1
+ 1
A
5.699
3.639
8.060
B
0.801 10-3
0.506 10-3
1.056 10-3
C
0
0
0
D
– 1.015 105
– 0.227 105
– 2.028 105
H0f,298
– 296 830
0
– 395 720
G0f,298
– 300 194
0
– 371 060
38
(c)
(d)
(e)
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 valuejust calculated and the initial value; return to step2.
The scheme converges on the value e = 0.77 and T =855.7 K
The composition of the product is:
