Ans.
4.2
(a)
T
0
473.15 K
⋅
:=
n
10 mol
⋅
:=
Q
800 kJ
⋅
:=
For ethylene:
A
1.424
:=
B
14.394 10
3
−
⋅
K
:=
C
4.392
−
10
6
−
⋅
K
2
:=
τ
2
:=
(guess)
Given
Q
n R
⋅
A T
0
⋅
τ
1
−
(
)
⋅
B
2
T
0
2
⋅
τ
2
1
−
(
)
⋅
+
⎡
⎢
⎣
⎤
⎥
⎦
C
3
T
0
3
⋅
τ
3
1
−
(
)
⋅
+
⎡
⎢
⎣
⎤
⎥
⎦
⋅
=
τ
Find
τ
( )
:=
τ
2.905
=
T
τ
T
0
⋅
:=
T
1374.5K
=
Ans.
(b)
T
0
533.15 K
⋅
:=
n
15 mol
⋅
:=
Q
2500 kJ
⋅
:=
For 1-butene:
A
1.967
:=
B
31.630 10
3
−
⋅
K
:=
C
9.873
−
10
6
−
⋅
K
2
:=
Chapter 4 - Section A - Mathcad Solutions
4.1
(a)
T
0
473.15 K
⋅
:=
T
1373.15 K
⋅
:=
n
10 mol
⋅
:=
For SO2:
A
5.699
:=
B
0.801 10
3
−
⋅
:=
C
0.0
:=
D
1.015
−
10
5
⋅
:=
∆
H
R ICPH T
0
T
,
A
,
B
,
C
,
D
,
(
)
⋅
:=
∆
H
47.007
kJ
mol
=
Q
n
∆
H
⋅
:=
Q
470.073kJ
=
Ans.
(b)
T
0
523.15 K
⋅
:=
T
1473.15 K
⋅
:=
n
12 mol
⋅
:=
For propane:
A
1.213
:=
B
28.785 10
3
−
⋅
:=
C
8.824
−
10
6
−
⋅
:=
D
0
:=
∆
H
R ICPH T
0
T
,
A
,
B
,
C
,
0.0
,
(
)
⋅
:=
∆
H
161.834
kJ
mol
=
Q
n
∆
H
⋅
:=
Q
1.942
10
3
×
kJ
=
76

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