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7.(c) (5 points)
You can solve the questions in parts (a) and (b)
also by defining the
fraction that is dissociated as
α
so that the
pressure of N
2
O
5
at equilibrium is
n
(1
α
),
where
n
is the initial
pressure.
Derive
an equation for K
p
that expresses it only as a
function of
α
and the total pressure P
t
at equilibrium.
N
2
O
5
(g)
↔
NO
3
(g) + NO
2
(g)
Initial pp:
n
0
0
Equilibrium:
n
(1
α
)
α
n
α
n
P
t
=
n
(1
α
) +
2
α
n
=
n
+
α
n =
n
(1+
α
)
>
n
= P
t
/ (1+
α
)
22
2
()
(
)
(1
)
)
1
)
)
t
p
t
P
n
K
P
n
n
αα
ααα
α
==
=
=
+
−−−
−
32
25
NO
NO
p
NO
PP
K
P
=
L17
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View Full Document (a)
In the experiment
n
atm of N
2
O
5
(g) were used to prepare NO
3
.
Calculate the partial pressure of NO
3
,
p
NO3
, in a mixture that has a
total pressure at equilibrium of
P
t
= 1.00 atm
.
K
p
= 4.96
(b)
What was the initial pressure,
n
?
K
p
= P
t
α
2
/(1 
α
2
);
n
= P
t
/ (1+
α
);
N
2
O
5
(g)
↔
NO
3
(g) + NO
2
(g)
K
p
and
P
t
are given; can find
α
K
p
(1 
α
2
) = P
t
α
2
K
p
K
p
α
2
= P
t
α
2
K
p
=
α
2
(
K
p
+
P
t
)
>
α
2
=
K
p
/
(
K
p
+
P
t
)
α
= {K
p
/ (K
p
+
P
t
)}
1/2
= {4.96 / (4.96
+
1.00}
1/2
=
0.912 =
α
n
= P
t
/ (1+
α
)
= 1.00/1.912 =
0.523 atm; initial pressure of N
2
O
5
α
n
= 0.523x0.912 =
0.477 atm = pressure of NO
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This note was uploaded on 07/22/2008 for the course CHEM 115b taught by Professor Reisler during the Spring '08 term at USC.
 Spring '08
 Reisler
 Chemistry, Equilibrium

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