Chapter 19
: 8, 12, 14, 16; 20, 22, 24; 30, 32, 33, 38, 40; 44, 48; 52, 54a, 54d, 56, 59, 62, 6466; 72, 74, 76, 78, 80;
82, 84, 91a
19.1
(a)
(b)
Δ
S is positive, because the disorder of the system increases. Each gas has greater motional freedom
as it expands into the second bulb, and there are many more possible arrangements for the mixed
gases.
By
definition,
ideal
gases
experience
no
attractive
or
repulsive
intermolecular
interactions, so
Δ
H for the mixing of ideal gases is zero, assuming heat exchange only between the
two bulbs.
(c)
The process is irreversible. It is inconceivable that the gases would reseparate.
(d)
The entropy change of the surroundings is related to
Δ
H for the system. Since we are mixing ideal
gases and
Δ
H = 0,
Δ
H
surr
is also zero, assuming heat exchange only between the two bulbs.
19.2
(a)
The
process
depicted
is
a
change
of
state
from
a
solid
to
a
gas.
Δ
S
increases
because of the greater motional freedom of the particles.
Δ
H increases because both melting and
boiling are endothermic processes. Since
Δ
G =
Δ
H – T
Δ
S, and both
Δ
H and
Δ
S are positive, the sign
of
Δ
G depends on temperature. This is true for all phase changes. If the temperature of the system
is greater than the boiling point of the substance, the process is spontaneous and
Δ
G is negative. If
the temperature is lower than the boiling point, the process is not spontaneous and
Δ
G is positive.
(b)
If
the
process
is
spontaneous,
the
second
law
states
that
Δ
S
univ
≥
0.
Since
Δ
S
sys
increases,
Δ
S
surr
must decrease. If the change occurs via a reversible pathway,
Δ
S
univ
= 0 and
Δ
S
surr
=
–
Δ
S
sys
. If the pathway is irreversible, the magnitude of
Δ
S
sys
is greater than the magnitude of
Δ
S
surr
,
but the sign of
Δ
S
surr
is still negative.
19.3
In the depicted reaction, both reactants and products are in the gas phase (they are far apart and
randomly placed). There are twice as many molecules (or moles) of gas in the products, so
Δ
S is positive
for this reaction.
19.4
(a)
At 300 K,
Δ
H = T
Δ
S. Since
Δ
G =
Δ
H – T
Δ
S,
Δ
G = 0 at this point. When
Δ
G = 0, the system is at
equilibrium.
(b)
The reaction is spontaneous when
Δ
G is negative. This condition is met when T
Δ
S >
Δ
H. From the
diagram, T
Δ
S >
Δ
H when T > 300 K. The reaction is spontaneous at temperatures above 300 K.
19.5
(a)
Analyze
.
The boxes depict three different mixtures of reactants and products for the reaction A
2
+
B
2
⇌
2AB.
Plan
.
Box 1 is an equilibrium mixture. By definition,
Δ
G = 0 for box 1. Calculate K and
Δ
G
°
for the
reaction from box 1. Boxes 2 and 3 are nonequilibrium mixtures. Calculate Q and
Δ
G for boxes 2
and 3.
.
]
B
[
]
A
[
]
AB
[
K
2
=
Use number of molecules as a measure of concentration.
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Box 1:
.
1
)
3
(
)
3
(
)
3
(
K
2
=
=
Δ
G =
Δ
G
°
+ RT lnK; 0 =
Δ
G
°
– RT ln(1), 0 =
Δ
G
°
– 0;
Δ
G
°
= 0
Box 2:
06
.
0
0625
.
0
16
1
)
4
)(
4
(
)
1
(
Q
2
=
=
=
=
Δ
G =
Δ
G
°
+ RT lnQ = 0 – RT ln(0.0625) = 2.77 RT = 3RT
Box 3:
49
1
46
)
1
)(
1
(
)
7
(
Q
2
=
=
=
Δ
G =
Δ
G
°
+ RT lnQ = 0 – RT ln(49) = –3.89 RT = –4RT
(b)
The magnitudes of
Δ
G (ignoring sign) are: box 1, 0; box 2, 2.8 RT; box 3, 3.9 RT. The order of
increasing
magnitude
of
Δ
G is: box 1 < box 2 < box 3.
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 Spring '06
 larosa
 Thermodynamics, Entropy, NH, G°

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