Version 346 – Exam 4 – mccord – (50970)
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001
10.0 points
Consider the following data at 25.0
◦
C and 1
atm.
Species
Δ
H
0
f
S
0
Δ
G
0
f
kJ/mol
J/mol
·
K
kJ/mol
N
2
(g)
0
.
0
191
.
5
0
.
0
O
2
(g)
0
.
0
205
.
0
0
.
0
N
2
O
5
(g)
11
.
0
?
115
.
0
Calculate the absolute entropy of N
2
O
5
at
25.0
◦
and 1 atm.
1.
+396
.
5 J
·
mol
−
1
·
K
−
1
2.
+355
.
0 J
·
mol
−
1
·
K
−
1
correct
3.

704 J
·
mol
−
1
·
K
−
1
4.
+1053
.
0 J
·
mol
−
1
·
K
−
1
5.
+47
.
5 J
·
mol
−
1
·
K
−
1
6.
+252
.
5 J
·
mol
−
1
·
K
−
1
7.

349
.
0 J
·
mol
−
1
·
K
−
1
Explanation:
Δ
G
rxn
= Δ
H
rxn

T
Δ
S
rxn
so we must find the Δ
H
and Δ
S
of the re
action. To do this, we need to write out the
balanced equation:
2 N
2
+ 5 O
2
→
2 N
2
O
5
Δ
H
rxn
= Δ
H
products

Δ
H
reactants
= (2 mol)(11 kJ
/
mol)

0 = 22 kJ
Δ
S
rxn
= Δ
S
products

Δ
S
reactants
= 2
x

(5 mol)(205 J
/
mol
·
K)

(2 mol)(191
.
5 J
/
mol
·
K)
= 2
x

1408 J
/
K
=
2
x

1408
1000
kJ
/
K
Now substitute into
Δ
G
rxn
= Δ
H
rxn

T
Δ
S
rxn
:
(2 mol)(115 kJ
/
mol)
= 22

(298 K)(2
x

1408) kJ
/
K
1000
Solving for
x
gives
x
= 355
.
007J/mol
·
K.
002
10.0 points
Which of the following has the smallest molar
entropy at 298 K?
1.
He(g)
correct
2.
N
2
(g)
3.
Ne(g)
4.
Cl
2
(g)
5.
F
2
(g)
Explanation:
To find the entropy we can use Boltzmann’s
formula
S
=
k
ln
W
where
W
is the number
of microstates.
At zero K, there is no mo
tion and
W
is given by the number of possible
discernible orientations of the molecule in a
crystal raised to the power of the number of
molecules (1 mole for each in this case).
If
the substance forms a perfect crystal there
should be zero residual entropy.
We would
expect this for these materials since they are
either homonuclear diatomics or monoatomic
species; in the solid, there is only one dis
cernible arrangement.
We now consider the
effect of being above 0 K. The greater the
mass of the species, the closer together are
the vibrational energy levels, so at a given
temperature, more vibrational energy levels
are occupied, thereby increasing the number
of microstates. Thus the substance with the
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Version 346 – Exam 4 – mccord – (50970)
2
lowest standard molar entropy would be He,
the lightest monoatomic species.
003
10.0 points
Which of the following reactions
is
an en
thalpy of formation reaction?
1.
C
diamond
(s) + 2H
2
(g)
←→
CH
4
(g)
2.
2Fe(s) + 3
/
2O
2
(g)
←→
Fe
2
O
3
(s)
cor
rect
3.
CH
4
(g) + 2O
2
(g)
←→
CO
2
(g) + 2H
2
O
4.
Hg(s) + 1
/
2O
2
(g)
←→
HgO(s)
Explanation:
Formation reactions describe production of
exactly one mole of one product from stoichio
metric quantities of elements in their standard
states.
004
10.0 points
Reactions with positive values of Δ
S
◦
r
always
become spontaneous at low temperatures.
1.
True
2.
False
correct
Explanation:
Δ
G
= Δ
H

T
Δ
S
is used to predict spon
taneity.
(Δ
G
is negative for a spontaneous
reaction.)
T
is always positive. When Δ
S
is
positive and
T
is small, then in order for Δ
G
to be negative, Δ
H
must be negative, zero
or have a very small positive value. But this
is not always the case, so not ALL reactions
with a positive Δ
S
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 Fall '08
 wandelt
 Chemistry, Thermodynamics, Enthalpy, Energy, Entropy

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