Environmental Engineering Reference
In-Depth Information
and the stoichiometric ratios are: 2:1:2 or 1:0.5:1. Thus, if
x
moles of SO
2
react, 1/2
x
moles of O
2
reacts to form
x
moles of SO
3
. So we can write
2SO
2
O
2
2SO
3
Initial moles
1
1
0
Δ
moles
−
x
1
2
x
+
x
−
Moles at equilibrium
(1 −
x
)
1−
x
2
x
For this problem,
x
= 0.85 mol, so the molar concentrations of the species are
ð Þ
V
1
−
x
=
0
:
15
5
L
−1
:
SO½
=
=0
03mol
ð
1
−
ðÞ
x
=
2
Þ
=
0
:
575
5
L
−1
O½
=
=0
:
115mol
V
=
x
V
=
0
:
85
5
L
−1
SO½
=0
:
17mol
Answers
: We can now calculate the equilibrium constants:
2
17
2
SO½
0
:
mol
−1
K
c
=
=
115
= 279 L
2
O½
03
2
×0
0
:
:
SO½
279
ð Þ
Δ
n
=
K
p
=
K
c
R
u
T
082 × 1000
=3
:
4
0
:
Question
: Why is
K
p
dimensionless? Hint: How is
K
p
derived?
Example 5.4 Calculation of the dissociation grade of a reactant
The equilibrium constant
K
c
of the reaction N
2
O
4
!
2NO
2
at 47
Cis0.05mol
L
−1
.
mol
−1
) are present at equilibrium if we fill a
container with a capacity of 1 L with 46 g of N
2
O
4
(MW = 92 g
How many grams of NO
2
(MW = 46 g
mol
−1
)? What is the
dissociation grade of N
2
O
4
at this temperature?
Solution
The stoichiometric ratio for the reaction is 1:2. If
x
are the moles of N
2
O
4
reacted,
we can write
N
2
O
4
2NO
2
Initial number of moles
46/92 = 0.5
0
Δ
moles
−
x
+2
x
Moles at equilibrium
(0.5 −
x
)
2
x
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