Biomedical Engineering Reference
In-Depth Information
Example 11.1
The production of methanol from syngas is given by the reaction:
CO
2H
2
-
CH
3
OH
(11.i)
1
The reaction heat at 25
Cis
90.7 kJ/mol. Using Gibb's equation, calculate the
equilibrium constant, K. Using the constant, K, find the fraction of the hydrogen in
the syngas that will be converted into methanol at 1 atm at that temperature.
2
Solution
Let us assume that the reaction started with 1 mol of CO and 2 mol of H
2
.Ifin
the equilibrium state only x moles of CO have been converted, it will have con-
sumed 2x moles of H
2
and produced x moles of CH
3
OH (as per
Eq. (11.i)
), leav-
ing (1
x) moles of H
2
. The total number
of moles will comprise unreacted moles and the methanol produced. Hence,
the total moles will be 1
x) moles of unreacted CO and 2(1
2
2
2x.
Noting that partial pressure is proportional to mole fraction, the equilibrium
constant is defined as:
x
2(1
x)
x
3
2
1
2
1
5
2
2
P
CH
3
OH
P
CO
xP
Þ
3
ð
3
2x
Þ
3
2x
2
2
K
(ii)
5
5
P
3
2
ð
3
2x
ð
1
x
Þ
2
ð
1
x
Þ
P
ð
P
H
2
Þ
2
2
2
The equilibrium constant, K, is calculated from the Gibbs free energy using
Eq. (7.56).
2
Δ
G
T
RT
K
exp
(iii)
5
G
T
for methanol from
So, for T
5
25
1
273
5
298K, we take the value of
Δ
Table 7.5:
G
298
52
mol
The universal gas constant, R, is known to be 0.008314 kJ/mol K.
Substituting these values in
Eq. (11.iii)
we get:
Δ
161
:
6kJ
=
10
28
Using this value in
Eq. (11.i)
, we get a quadratic equation of x. Now, solving
x, we get the following:
K
exp
ð
161
:
6
=ð
0
:
008314
298
ÞÞ
5
2
:
12
5
3
3
0
So the equilibrium concentration of the product is:
CO
x
1
:
5
1 mol
At 25
C, the reaction will produce 2 mol of hydrogen and 1 mol of methanol.
1
1
0
H
2
2
ð
1
1
Þ
5
0
and CH
3
OH
5
5
;
5
2
;
5
11.4.2 Fischer
Tropsch Synthesis
Fischer-Tropsch or FTS synthesis is a highly successful method for the pro-
duction of liquid hydrocarbons from syngas. The FTS process can produce
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