This preview shows pages 1–4. Sign up to view the full content.
280
CHAPTER TEN
SPONTANEITY, ENTROPY, AND FREE ENERG
Y
Spontaneity and Entropy
11.
A spontaneous process is one that occurs without any outside intervention.
12.
a.
Entropy is a measure of disorder or randomness.
b. The quantity T
∆
S has units of energy.
13.
We draw all of the possible arrangements of the two particles in the three levels.
2 kJ
x
x
xx
1 kJ
x
xx
x
0 kJ
xx
x
x
Total E =
0 kJ
1 kJ
2 kJ
2 kJ
3 kJ
4 kJ
The most likely total energy is 2 kJ.
14.
2 kJ
AB
B
A
B
A
1 kJ
AB
B
A
A
B
0 kJ
AB
A
B
A
B
E
T
=
0 kJ
2 kJ
4 kJ
1 kJ
1 kJ
2 kJ
2 kJ
3 kJ
3 kJ
The most likely total energy is 2 kJ.
15.
Processes a, b, d, and g are spontaneous.
Processes c, e, and f require an external source of energy
in order to occur since they are nonspontaneous.
16.
Of the three phases, solids are most ordered and gases are most disordered.
Thus, a, b, c, e, and g
involve an increase in entropy.
All increase disorder.
17.
a.
Entropy increases; there is a greater volume accessible to the randomly moving gas molecules
which increases disorder.
This preview has intentionally blurred sections. Sign up to view the full version.
View Full Document CHAPTER 10
SPONTANEITY, ENTROPY, AND FREE ENERGY
281
b. The positional entropy doesn’t change.
There is no change in volume and thus, no change in
the numbers of positions of the molecules.
The total entropy (
∆
S
univ
) increases because the
increase in temperature increases the energy disorder (
∆
S
surr
).
c.
Entropy decreases because the volume decreases (P and V are inversely related).
18.
a.
N
2
O;
It is the more complex molecule, i.e., more parts, more disorder.
b. H
2
at 100
°
C and 0.5 atm;
Higher temperature and lower pressure means greater volume and
hence, greater positional entropy.
c.
N
2
;
N
2
at STP has the greater volume.
d. H
2
O(l) is more disordered than H
2
O(s).
19.
There are more ways to roll a seven.
We can consider all of the possible throws by constructing a
table.
There are six ways to get a seven, more than any other number.
The seven is not favored by
energy; rather it is favored by probability.
To change the probability we would have to expend
energy (do work).
20.
Arrangement I:
S = k ln
Ω
;
Ω
= 1;
S = k ln 1 = 0
Arrangement II:
Ω
= 4;
S = k ln 4 = 1.38
×
10
23
J/K ln 4,
S = 1.91
×
10
23
J/K
Arrangement III:
Ω
= 6;
S = k ln 6 = 2.47
×
10
23
J/K
one die
1
2
3
4
5
6
1
2
3
4
5
6
7
2
3
4
5
6
7
8
3
4
5
6
7
8
9
sum of the two dice
4
5
6
7
8
9
10
5
6
7
8
9
10
11
6
7
8
9
10
11
12
CHAPTER 10
SPONTANEITY, ENTROPY, AND FREE ENERGY
282
Energy, Enthalpy, and Entropy Changes Involving Ideal Gases and Physical Changes
21.
15.0 g He
×
g
4.003
mol
1
= 3.75 mol He
q
v
=
∆
E = nC
v
∆
T = 3.75 mol (12.47 J K
1
mol
1
) (56.5 K) = 2640 J = 2.64 kJ
22.
1.00
×
10
3
g C
2
H
6
×
g
30.07
mol
1
= 33.3 mol
q
v
=
∆
E = nC
v
∆
T = 33.3 mol
(44.60 J mol
1
K
1
) (48.4 K) = 7.19
×
10
4
J = 71.9 kJ
At constant volume, 71.9 kJ of energy are required and
∆
E = 71.9 kJ.
At constant pressure (assuming ethane acts as an ideal gas):
C
p
= C
v
+ R = 44.60 + 8.31 = 52.91 J mol
1
K
1
Energy required = q
p
=
∆
H = nC
p
∆
T = 33.3 mol (52.91 J mol
1
K
1
) (48.4 K) = 8.53
×
10
4
J = 85.3 kJ
For the constant pressure process,
∆
E = 71.9 kJ as calculated previously (
∆
E is unchanged).
This preview has intentionally blurred sections. Sign up to view the full version.
View Full Document
This is the end of the preview. Sign up
to
access the rest of the document.
This homework help was uploaded on 04/06/2008 for the course CHEM 142 taught by Professor Zoller,williamh during the Fall '07 term at University of Washington.
 Fall '07
 ZOLLER,WILLIAMH

Click to edit the document details