Chem 338 Homework Set 5 Solutions Workseet
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5Chem 338
Homework Set #5 solutions
October 10, 2001
From Atkins:
5.2, 5.9, 5.12, 5.13, 5.15, 5.17, 5.21
-3
5.2) The density of rhombic sulfur is 2.070 g cm
and that of monoclinic sulfur is
-3
1.957 g cm
. Can the application of pressure be expected to make monoclinic
sulfur more stable than rhombic sulfur? (See Exercise 5.1.)
From 5.1, ∆
=
∆
=
G
(
S
)
0 0
.
kJ / mol ;
G
(
S
)
0 33
.
kJ / mol
f
rho
f
mono
At constant T, G depends on p via:
dG
= V
dp , hence as p increases, so
m
m
does G in proportion to V
. The molar volume is related to the density by:
m
MW
1
=
∝
V
, therefore, V
m
m
ρ
ρ
ρ(rhom) > ρ (mono), hence V
(rhom) < V
(mono) so increasing the pressure
m
m
will make the monclinic form less stable relative to the rhombic form rather than
more stable. (i.e., G of the monoclinic form will increase faster than G of the
rhombic form)
-1
-1
5.9) The standard molar entropy of rhombic sulfur is 31.80 J K
mol
and that of
-1
-1
mooclinic sulfur is 32.6 J K
mol
. (a) Can an increase in temperature be
expected to make monoclinic sulfur more stable than rhombic sulfur? (b) If so, at
what temperature will the transition occur at 1 bar? (See Exercise 5.1)
a) G
depends on temperature via:
dG
= –S
dT . Hence as T increases, G
m
m
m
m
decreases in proportion to S
. Since S
(mono) > S
(rhom), G
(mono) will
m
m
m
m
decrease faster than that of the rhombic form, so eventually a suitable increase in
temperature will make monoclinic sulfur more stable than the rhombic form.
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