Environmental Engineering Reference
In-Depth Information
and hydrogen are produced at high temperatures using high temperature steam. Of
course, a tradeoff exists between temperature and pressure for a given fuel.
Example
Calculate the equilibrium conversion for the reaction:
C
(
s
)
+ H
2
O
(
g
)
↔
CO
(
g
)
+ H
2(
g
)
at 1, 10, and 34 atm for each of these temperature: 800, 1000, and 1500 K.
Solution
The following constants are given:
K
800
= 0.04406,
K
1000
= 2.617,
K
1500
= 608.1
G
0
from the following table. Also given are free energy (∆
kcal), and heat of reaction
H
0
(∆
kcal).
∆
H
T
0
, kcal
T
, K
∆
G
T
0
, kcal
log
10
K
K
298.16
21.827
-15.998
1.005 × 10
-16
31.382
400
18.510
-10.113
7.709 × 10
-11
31.723
500
15.176
-6.633
2.328 × 10
-7
31.978
600
11.796
-4.296
5.058 × 10
-5
32.164
800
4.966
-1.356
4.406 × 10
-2
32.371
1000
-1.912
+0.418
2.617 × 10
-0
32.445
1500
-19.107
+2.784
6.081 × 10
+2
32.295
Basis: 1 mole of steam feed
Pressure: P atm
Converted moles of steam =
x
= fractional conversion
At equilibrium:
C + H
2
O ↔
CO + H
2
Steam: 1 -
x
mol, CO:
x
mol, H
2
:
x
, Total: 1 +
x
mol
Activities of gaseous components are considered equal to partial pressure, and
the activity of solid graphite is unity. Thus,
(
)
xP
x
xP
x
1
1
xP
x
a
=
,
a
=
,
a
=
CO
H
H O
2
1
+
1
+
+
2
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