Solutions to CHM2045 Exam #3, Spring 2008, Form Code A
1.
N
2
(g)
+
3H
2
(g)
→
2NH
3
(g)
Using molar ratios:
Consumption of 0.500 atm of N
2
would result in production of 1.00 atm NH
3
.
Consumption of 0.500 atm of H
2
would result in production of 0.33 atm NH
3
.
Therefore, H
2
is the limiting reactant and will be completely consumed.
To consume 0.500 atm of H
2
, 0.170 atm of N
2
is used up, leaving (0.500 – 0.17)
= 0.33 atm N
2
left over.
So the pressure in the vessel is (0.33 atm NH
3
+ 0.33 atm N
2
) =
0.66 atm total
.
2.
As per Graham's Law:
(Rate of S
x
O
y
) / (Rate of O
2
)
=
0.707
= square root of (M of O
2
) / (M of S
x
O
y
).
So,
(0.707)
2
= (32 g/mol) / (? g/mol)
(? g/mol) = 64 g/mol,
and the only listed S
x
O
y
with that molar mass is
SO
2
.
3.
As per the KineticMolecular Theory of Gases, the average kinetic energy is the same
for all gases at the same temperature, and it is directly proportional to the temperature.
So, if the temperature increases from 90.0K to 270.K, the average kinetic energy will
triple (regardless of the gas).
So the answer is
1:3
.
4.
n of Ne = PV/RT = [(1.50 atm)(0.400 L)] / [(0.0821)(513K)] = 0.0142 mol Ne
n of O
2
= PV/RT = [(550/760 atm)(0.150 L)] / [(0.0821)(403K)]=0.00328 mol O
2
Total moles n
tot
= 0.0175 mol Total
P = nRT/V = [(0.0175)(0.0821)(298K)] / (0.500 L)
=
0.856 atm
5.
Attractive interparticular/intermolecular forces between gas particles/molecules in a
real/nonideal gas tend to
decrease the pressure
in the gas relative to that of an ideal gas.
6.
ClausiusClapeyron equation:
ln (P
2
/P
1
) = (
Δ
H
vap
/ R)[(1/T
1
) – (1/T
2
)]
T
1
= (56.2 + 273) = 329 K
P
1
= 760 torr (at boiling point)
T
2
= (25.0 + 273) = 298 K
P
2
= 225 torr
Solving for
Δ
H
vap
results in
32.0 kJ/mol
7.
If Na metal is bodycentered cubic, then there are two atoms of Na per unit cell.
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 Spring '11
 Gower
 Chemistry, Solubility, Vapor pressure

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