Let us examine practice example 5.6, but from a different “angle.”
The reaction is:
2
2
3
2
)
(
2
O
SO
g
SO
(I)
Let us suppose we were told that we start off with 12.5 atm of (pure)
sulfur trioxide
and, at equilibrium, we find (measure) that we have 2.5 atm of molecular
oxygen
.
Now calculate K.
We set up the rxn table as before:
2SO
3
=
2SO
2
+
O
2
I
12.5
0
0
a
+b
+c
F/Eq
d
e
2.5
This simply “puts into place” all of the given data; ae denote entries that we do not have as
given.
Note that I have made the signs explicit in the delta line.
(1) Now we take advantage of the fact that for each column, I +
= F.
Thus for the
O
2
column:
0 + c = 2.5.
Thus, c = 2.5 atm.
(2) Once we have any single entry in the delta row, we have all the others by relative
stoichiometry.
For example:
b/c = 2/1.
Thus,
b = 2.5*(2/1) = 5.0 atm
.
And: a/c = 2/1 so
a = 5.0 atm
.
(3) We now obtain ‘d’ and ‘e’ by:
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 Fall '07
 Warren
 Chemistry, Equilibrium, Sulfur, Oxide, Sulfur dioxide

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