Chem
5, 9 Section
Winter,
2005
Exam
1
Solutions
January
27,
1
1.
(10 + 5 points)
In preparation for a study of the equilibrium constant for the reaction
2 NO
2
(
g
)
→
←
N
2
O
4
(
g
)
you take an 8.6 g frozen sample of pure N
2
O
4
(
s
) and place it in an otherwise evacuated
container of volume 2.3 L and allow it to warm up to room temperature, 300 K, where
everything in the container will be a gas.
(a)
If the equilibrium constant for this reaction,
K
P
, was
infinite
, what would the pressure
be when all the sample vaporized?
For the reaction as written above (and if you worked the problem with the reverse reaction
but otherwise drew the correct conclusions, you weren’t penalized), under this assumption
for
K
P
, we can write
K
P
=
P
N
2
O
4
P
NO
2
2
=
∞
which is true if
P
NO
2
is zero, i.e.,
all
the gas is N
2
O
4
.
Thus, the number of moles of gas,
n
, is equal to the number of moles of N
2
O
4
in the 8.6 g sample:
n
=
m
M
=
8.6 g
92.011 g mol
–1
=
0.0935 mol
Thus, the pressure is simply
P
=
nRT
V
=
0.0935 mol
0.08206 L atm mol
–1
K
–1
300 K
2.3 L
=
1.00 atm
(b)
In fact, of course,
K
P
is finite.
How will the actual final pressure compare to the
answer you gave in part (a)?
(circle one)
greater
the same
smaller
Briefly justify your answer.
If
K
P
is finite, then at equilibrium, some of the N
2
O
4
will have dissociated into NO
2
.
But
every one N
2
O
4
that dissociates produces two NO
2
molecules in the gas.
Thus,
n
must
increase, and if
T
and
V
are constant, a larger
n
means a larger
P
.
2.
(10 + 6 + 4 points)
A 0.578 g sample of pure tin (Sn) is reacted with excess fluorine gas
(F
2
) until the resultant solid product has a constant mass of 0.948 g.
(a)
What is the
empirical
formula of the compound formed?
We will need the molar masses of Sn and F: 118.71 g mol
–1
and 18.998 g mol
–1
,
respectively.
The mass of Sn tells us the number of moles of Sn:
n
Sn
=
m
Sn
M
Sn
=
0.578
g
118.71 g mol
=
4.87
×
10
–3
mol
The mass of F in the product is the difference between the product mass and the original Sn
sample mass, and thus the number of moles of F in the product is
n
F
=
m
F
M
F
=
0.948
g
–
g
18.998 g mol
=
1.95
×
10
–2
mol
Thus, the empirical formula, which is simply the ratio of the number of F atoms to the
number of Sn atoms, is