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13.6
H2(g) + CO2(g) = H2O(g) + CO(g)
=
i
=
1 1 1 1
0
=
n0
=
11
=
2
yCO
=
i
By Eq, (13.5),
yH
2
=
yCO
2
1
=
yH2O
2
=
2
With data from Example 13.13, the following vectors represent values for Parts (a) through (d):
1000 1100 T kelvin 1200
3130 150 J G 3190 mol
1300
6170
Combining Eqs. (13.5), (13.11a), and (13.28) gives
2 2 1 1 2 2
=
2
1
G exp R T
2
=
K
exp
=
G R T 0.4531 0.5021 0.5399
1
Ans.
0.5709 13.11 4HCl(g) + O2(g) = 2H2O(g) + 2Cl(g)
=
1
n0
=
H298 114408
6 J mol
T
773.15 kelvin
G298 75948
487
T0 J
mol
298.15 kelvin
The following vectors represent the species of the reaction in the order in which they appear:
4
3.156
1
A
2 2
3.639 3.470
B
4.442
A i
0.151
0.506 10 3 1.450
D
0.344
0.089
B
i
B i
i
D
i
D i
i
B 8 10 T T0
i
i
5
A 0.439 G H298
0.227 105 0.121
i 1 end
end rows ( A)
A
0.623
4
C 0
D 8.23 10
H298 G298
R IDCPH T0 T A B C D R T IDCPS T0 T A B C D J
4
G 1.267 10
K exp
mol
G R T
K 7.18041
By Eq. (13.5)
yO2
=
yHCl
1
=
yH2O
6
=
5 4 6 2 6
0.5
Apply Eq. (13.28);
yCl2
=
2 6
(guess)
4
Given
yHCl
2 6 5 4 1 5 4 6
yHCl 0.3508
yO2 yO2 0.0397
=
2 K
1 6
Find
yH2O
yH2O 0.3048 488
2 6
0.793
yCl2
yCl2 0.3048
2 6 Ans.
13.12 N2(g) + C2H2(g) = 2HCN(g)
=
n0
0
=
2
This is the reaction of Pb. 4.21(x). From the answers for Pbs. 4.21(x), 4.22(x), and 13.7(x), find the following values:
J
H298 42720
G298 39430
mol
3
A 0.060
B 0.173 10
T 923.15 kelvin
G H298
T T0
J mol 5
C 0
D 0.191 10
T0 298.15 kelvin
H298 G298
R IDCPH T0 T A B C D R T IDCPS T0 T A B C D 4
G 3.242 10
J
K exp
mol
G R T
K 0.01464
By Eq. (13.5),
y N2
=
1
yC2H4
2
By Eq. (13.28),
y N2
1 2
y N2 0.4715
1
0.5
2 1
Given
=
2
yHCN
=
2e 2
=
(guess)
2 =
Find
K
yC2H4
1 2
yC2H4 0.4715
0.057
yHCN yHCN 0.057
Ans.
Given the assumption of ideal gases, P has no effect on the equilibrium composition.
489
13.13
CH3CHO(g) + H2(g) = C2H5OH(g)
=
1
n0
=
2.5
This is the reaction of Pb. 4.21(r). From the answers for Pbs. 4.21(r), 4.22(r), and 13.7(r), find the following values:
J
H298 68910
G298 39630
mol
3
G H298
mol
6
A 1.424 B 1.601 10 T 623.15 kelvin
J 5
C 0.156 10
D 0.083 10
T0 298.15 kelvin
T T0
H298 G298
R IDCPH T0 T A B C D R T IDCPS T0 T A B C D
G R T
J
3
G 6.787 10
K exp
mol
By Eq. (13.5), yCH3CHO
=
2.5 1 1.5
yCH3CHO
2.5
=
yCH3CHO 0.108
yH2
=
1.5 2.5
yC2H5OH
=
2.5
(guess)
3 K
1 2.5
yH2
0.5
By Eq. (13.28),
Given
1
K 3.7064
1.5 2.5
yH2 0.4053
Find
yC2H5OH
0.818 2.5
yC2H5OH 0.4867 Ans.
If the pressure is reduced to 1 bar,
Given
2.5 1 1.5
yCH3CHO
=
1 K
1 2.5
yCH3CHO 0.1968
yH2
1.5 2.5
yH2 0.4645 490
Find
yC2H5OH
0.633 2.5
yC2H5OH 0.3387 Ans.
nSO3
By Eq. (4.18),
H753 H298 R IDCPH T 0 T A B C D
H753 98353
13.16
0.1455
By Eq. (13.4),
=
=
J mol
C3H8(g) = C2H4(g) + CH4(g)
=
By Eq. (13.5),
yC3H8
nC3H8
n0 nC3H8 n0
1
=
yC2H4
1
Ans.
mol
1
Basis: 1 mole C3H8 feed. By Eq. (13.4)
Fractional conversion of C3H8 =
J
Q 14314
Q H753
=
=
=
1
1 1 1
1
=
yCH4
=
1
From data in Table C.4,
H298 82670
J
G298 42290
mol
J mol
The following vectors represent the species of the reaction in the order in which they appear:
1 1 1
1.213 A 1.424 1.702
end rows ( A)
A
A i
i 1 end
B
i
i
A 1.913 (a) T
8.824 6 C 4.392 10 2.164
28.785 3 B 14.394 10 9.081
B i
i
C
C
i
i
i
3
B 5.31 10
625 kelvin
i
T0 298.15 kelvin
493
6
C 2.268 10
D 0
G H298
T T0
H298 G298
R IDCPHT0 T A B C D R T IDCPS T0 T A B C D G 2187.9
J
K exp
mol
G R T
0.5
By Eq. (13.28),
K 1.52356
(guess)
2
1 1
Given
0.777
=
Find
K
This value of epsilon IS the fractional conversion. Ans. 2
(b)
0.85
K 1 1
G R T ln( K )
G 4972.3
J mol
K 2.604
Ans.
The problem now is to find the T which generates this value. It is not difficult to find T by trial. This leads to the value: T = 646.8 K Ans.
13.17 C2H6(g) = H2(g) + C2H4(g)
=
1
Basis: 1 mole entering C2H6 + 0.5 mol H2O.
n0
=
1.5
yC2H6
=
By Eq. (13.5),
1
yH
1.5
=
1.5
yC2H4
=
1.5
From data in Table C.4,
H298 136330
J mol
G298 100315
J mol
The following vectors represent the species of the reaction in the order in which they appear: 494
0.0333 y (0.88244)
0.052 0.2496
Ans.
0.6651 T T0
13.27
T 949.23kelvin Ans.
CH4(g) = C(s) + 2H2(g)
=
1
(gases only)
The carbon exists PURE as an individual phase, for which the activity is unity. Thus we leave it out of consideration. From the data of Table C.4,
H298 74520
J
G 298 50460
mol
J mol
The following vectors represent the species of the reaction in the order in which they appear:
1 1 2
1.702 A 1.771 3.249
9.081 3 B 0.771 10 0.422
i 1 3
2.164 6 C 0.0 10 0.0
0.0 5 D 0.867 10 0.083
A
A i
i
B
i
A 6.567
B i
i
C
i
i
i
3
B 7.466 10
T 923.15 kelvin
C 6
509
D
D i
i
i
C 2.164 10
T0 298.15 kelvin
i
4
D 7.01 10
G H298
T T0
H298 G298
R IDCPHT0 T A B C D R T IDCPS T0 T A B C D 4
G 1.109 10
J mol
By Eq. (13.5),
(a)
K exp
n0
=
1
yCH4
4 K
yCH4
0.7173
1
yH2
1
=
K 4.2392
1
yH2
1
2 2 1 1
By Eq. (13.28),
K
G R T
=
4
Given
0.7893 yCH4
1 2
yH2 0.5659
2
K
=
yCH4 0.1646
1
Ans.
yH2 0.8354 n0
.8 =
1
2
1
(b) For a feed of 1 mol CH4 and 1 mol N2,
2 2 2 1
2
(fraction decomposed)
2
By Eq. (13.28),
=
=
2
(guess)
Find
K
(fraction decomposed)
yH2
2 2
yCH4 0.0756
510
y N2 1 yCH4 yH2 y N2 0.3585
Ans.
