CHAPTER 18
THERMODYNAMICS
18.6
The probability (
P
) of finding all the molecules in the same bulb becomes progressively smaller as the
number of molecules increases.
Based on the information given in the problem, we can come up with an
equation relating the probability to the number of molecules present.
1
2
⎛⎞
=
⎜⎟
⎝⎠
N
P
where,
N
is the total number of molecules present.
Using the above equation, we calculate the probability of finding 100 molecules in the same bulb to be:
100
1
2
==
31
81
0
P
−
×
Extending the calculation to Avogadro’s number of molecules would result in such a small number that for
all practical purposes, there is zero probability that all molecules would be found in the same bulb.
18.9
(a)
This is easy.
The liquid form of any substance always has greater entropy (more microstates).
(b)
This is hard.
At first glance there may seem to be no apparent difference between the two substances
that might affect the entropy (molecular formulas identical).
However, the first has the
−
O
−
H structural
feature which allows it to participate in hydrogen bonding with other molecules.
This allows a more
ordered arrangement of molecules in the liquid state.
The standard entropy of CH
3
OCH
3
is larger.
(c)
This is also difficult.
Both are monatomic species.
However, the Xe atom has a greater molar mass
than Ar.
Xenon has the higher standard entropy.
(d)
Same argument as part (c).
Carbon dioxide gas has the higher standard entropy (see Appendix 2).
(e)
O
3
has a greater molar mass than O
2
and thus has the higher standard entropy.
(f)
Using the same argument as part (c), one mole of N
2
O
4
has a larger standard entropy than one mole of
NO
2
.
Compare values in Appendix 2.
Use the data in Appendix 2 to compare the standard entropy of one mole of N
2
O
4
with that of two
moles of NO
2
.
In this situation the number of atoms is the same for both.
Which is higher and why?
18.10
In order of increasing entropy per mole at 25
°
C:
(c)
<
(d)
<
(e)
<
(a)
<
(b)
(c) Na(
s
): ordered, crystalline material.
(d) NaCl(
s
): ordered crystalline material, but with more particles per mole than Na(s).
(e) H
2
: a diatomic gas, hence of higher entropy than a solid.
(a) Ne(
g
): a monatomic gas of higher molar mass than H
2
.
(b) SO
2
(
g
): a polyatomic gas of higher molar mass than Ne.
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THERMODYNAMICS
424
18.11
Using Equation (18.7) of the text to calculate
rxn
Δ
S
o
(a)
rxn
2
2
(
SO) [ (O)
(
S
)
]
Δ=
°
−
°
+
°
SS
S
S
o
rxn
(1)(248.5 J/K mol)
(1)(205.0 J/K mol)
(1)(31.88 J/K mol)
Δ
=
⋅−
⋅
=
⋅
11.6 J/K mol
S
o
(b)
rxn
2
3
(MgO)
(CO )
(MgCO )
°
+
°
−
°
S
S
o
rxn
(1)(26.78 J/K mol)
(1)(213.6 J/K mol)
(1)(65.69 J/K mol)
Δ
=
⋅+
⋅
=
⋅
174.7 J/K mol
S
o
18.12
Strategy:
To calculate the standard entropy change of a reaction, we look up the standard entropies of
reactants and products in Appendix 2 of the text and apply Equation (18.7). As in the calculation of enthalpy
of reaction, the stoichiometric coefficients have no units, so
rxn
Δ
S
o
is expressed in units of J/K
⋅
mol.
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 Spring '07
 BELLEW
 Chemistry, Thermodynamics, Mole, Entropy, kJ/mol, G°

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