Problems
37
36
The CIosed-SystemConstraint and ChemicaJ Stoichiometry
PROBLEMS
The element-abundance vector b*, corresponding to A* is given by
2.1
(a)
Write equation
2.2-3
in full
f-or
the system {(H
3
P0
4
,
H
2
PO;,
A*n
o
=
b*,
(2.4-26)
HPOr-,
pol-,
H+
,
OH-
, H
2
0),
(H,
0,
P,
p)},
if the system
where n° is any allowable composition of the system, such as the starting results from dissolving
2
moles of
H
3
P0
4
in
I
mole of H
2
0.
composi
tion_
Thus the element-abundance constraints are
(b)
From
the result in part a, write equations
2.2-}
for the system, that
is, by multiplying out equation
2.2-3.
A*n
=
b*.
2.2
Balance each of the following
by
Gauss-Jordan reduction,
and
in
so
Smith
(1976)
has also treated Example
2.5
by this implicit approach. doing show that only one chemical equation
i5
required in each case:
(a)
Na
2
0
2
+
CrCI
3
+
NaOH
=F
Na
2
Cr0
4
+
NaCI
+
H
2
0
Example 2.7
Consider the system {(C
2
H
6
(C),
C
3
H
6
(e),
C
3
H
8
(e),
C
2
H
6
(g),
C
3
H
6
(g), C
3
H
ll
(g», (C, H)},
in
which only mass transfer
of
the
substances
(b)
K 2
Cr
20
7
+
H
2
S0
4
+
H
2
S0
3
=F
CriS04h
+
H
2
0
+
K
2
S0
4
between the two phases
is
allowed.
Find
A*
and
b*
50
that the problem may be
(c)
KCI0
3
+
NaN0
2
=1=
KCI
+
NaN0
3
treated in standard
formo
(d)
KMn0
4
+
H
20
+
Na2Sn02
=1=
MnO
z
+
KOH
+
Na
z
Sn0
3
2.3
For
each of the folJowing 5ystems, determine the number C and
a
Solution
The
N
matrix is generated at the outset from the chemical
permissible set of components and the maximum number
R
and
a
equations
permissible set of independent chemicaI equations:
C
2
H
6
(g)
=
C
2
H
6
(r),
(a)
{(CO, CO2, H,
H
2
,
H
2
0,
0,
02'
OH,
N
2
,
NO),
(C, H, O. N)}
C
3
H
6
(g)
=
C
3
H
6
(P),
(b)
{(CH
4
,
C
2
H
2
,
C
2
H4, C
2
1l
6
,
C
6
H
ó
'
H
2
,
H20), (C,
H,
H
2
0)}
(c)
{(CH
4
,
CH
3
D,
CH
z
D
2
,
CHD
3
•
CD
4
),
(C,
H,
O)}
(Apse
and
Missen.
and
1967)
C
3
H
g
(g)
=
C
3
H
8
(C),
(d)
((C(gr), CO(g), CO
2
(g), Zn(g), Zn(
e).
ZnO(s»,
(C,
O.
Zn)}
(e)
((Fe(C
2
0
4
)+
.
Fe(C
2
0
4
)2"
.Fe(C204)~-
.Fe
3
+
,SOJ-
,HS04-
.H+,
as
HC
2
0
4
- ,
H
2C204'
C
2
0,;-
}.
(C,
Fe.
H,
0,
S,
p)}
(Swinnerton and
1
O
Miller,
1959)
O
1
~\
(f)
{(H
2
0,
H
2
0
2
•
H+
,
K
+ •
Mn0
4
- ,
Mn2+
,
02'
SOr-
).
(H,
K,
Mn,
O
O
I
N=I
0,
S.
p)}
,
-}
O
O
(g)
{(C
6
H
6
(f).
C(,Hó(g),
C
7
H!(l'), C
7
H x
(g),
o-CgH1oU).
o-CKHIO(g)·
O
-]
O
m-CxHIO(I'),
m-CgHIO(g),
p-Cl\HlOU). p-Cy,HIO(g».
(C.
H)}
O
O
-}
(h)
{(0Z<g).
H
2
0(g),
CHig),
CO(g), CO
2
(g). H
2
{g),
N
2
(g). CHO(g),
This is in the same form as equation
2.4-24,
with
~
=
3.
A compatible formula
CH
20(g), OH(g), Fe(s), FeO(s). Fé
3
0
4
(s), C(gr), CaO(s).
CaC0
3
(s»,
matrix
A*,
from equation
2.4-25,
is (O. H,
C.
Fe, Ca, N2
)}
(Madeley and Toguri, 1973b)
2.4
Ethylene can be made by the dehydrogenation
of
ethane. Methane
is
a
1 O O 1 O O)
A*
=
O 1 O O 1 O
possible by-product. and it
is
undesirable for the system
to
approach
.
(
O O
100
1
equilihrium with respect to all these species at lhe outlet of the reactor.
as the following figures show. For a feed that contains
0.4
mole of steam
Fromequation
2.4-26,
(inert)permoleofC
2
H
ó
'
andforanoutlettemperatureof
1100
Kand
pressure
of
}.6
atm, it can be calculated that
if
equilibrium obtained at
,
n~
+
n~
)
the oudet, there wouid be
0.515
mole
of
ethylene per mole
of
ethane
in
lhe
feed
and
0.950
mole of methane. Calculate the mole fraction
of
each
b*
=
A*n
c
= .
n~
+
n~
..
I
species in the oudet mixture
011
a steam-free basis.
n~
+
nâ