Biomedical Engineering Reference
In-Depth Information
Now, to find the value at 1100K, we use
Eq. (8.19)
:
X
ð
1100
298
ð
H
1100
5
Δ
H
298
1
Δ
C
p;CH
4
1
C
p;CO
2
Þ
dT
product
(8.21)
X
ð
1100
2
C
p;H
2
O
dT
298
reactants
The specific heats of gases are taken from Table C.4 (Appendix C) as:
C
p;CH
4
5
22
:
35
1
0
:
0481T kJ
=
kmol
T
2
kJ
C
p;CO
2
43
:
28
0
:
0114T
818363
=
=
mol
5
1
2
0000056 T
2
kJ
C
p;H
2
O
34
:
4
0
:
00062T
0
:
=
mol
5
1
1
The integrations of respective gas components are:
2
4
3
5
1100
298
5
0
:
0481T
2
2
0
:
0481
2
C
p;CH
4
5
22
:
35T
22
:
35
3
ð
1100
298
Þ
1
1
2
1100
2
298
2
3
ð
Þ
5
44
:
8895 kJ
=
mol
2
2
4
3
5
1100
0114T
2
2
0
:
818
;
363
0
:
0114
2
C
p;CO
2
43
:
28T
43
:
28
3
ð
1100
298
Þ
1
5
1
1
298
5
2
T
818
;
363
1100
2
298
2
3
ð
Þ
1
2
1100
2
298
42
:
1218 kJ
=
mol
5
2
4
3
5
1100
298
5
000628T
2
2
0000056T
3
3
0
:
0
:
C
p;H
2
O
34
:
4T
34
:
4
3
ð
1100
298
Þ
5
1
1
2
0
:
000628
2
1100
2
298
2
3
ð
Þ
1
2
0
:
0000056
3
1100
3
298
3
ð
Þ
1
2
mol
Substituting these values and integrating the above expression, we get:
30
:
376 kJ
=
5
H
1100
5
Δ
7
:
65
1
104
:
58
2
33
:
578
5
78
:
65 kJ
=
mol
Thus, this reaction is endothermic and it is written as:
1
2
CH
4
1
2
CO
2
C
H
2
O
ð
gas
Þ-
78
:
65 kJ
=
mol
1
1
1
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