Chemistry 311 Physical Chemistry (Fall 2000)
Examination 2
1. Acetic acid (ac) is purified by fractional freezing at its melting point of 16.6◦ C. A flask containing acetic
acid at 16.6◦ C is briefly lowered into a very large ice bath holding several kilograms of ice at 0◦C. When
it is removed, it is found that exactly 1 mol of acetic acid has frozen. (a) What is the entropy change of
the acetic acid that has frozen? (b) What is the entropy change for the water bath during this process? (c)
How many grams of ice will freeze or melt as a result of the acetic acid freezing? (d) If the acetic acid
is the system and the ice bath is the only part of the surroundings affected, is the freezing of the acetic
◦ = 11.45 kJ mol−1 ,
◦
acid described above a spontaneous process? Why/Why not? f us ac
f us H2 O =
−1
5.98 kJ mol .
∆ H
∆ H
Equations given:
For a phase change,
∆pcS = ∆TpcpcH .
Answer
(a) Entropy change for the acetic acid (system) [ note the negative sign, since freezing is the reverse of
fusion]:
∆S= −∆fTus Hac
−11450 J mol−1 = −39.52 J K−1mol−1.
= (273
.15 + 16.6) K
◦
(b) Entropy change of the water bath (surroundings) [the water bath gains
J mol
∆Ssurr = 11450
273.15 K
=41.92 J K−1mol−1.
11, 450 J of heat at 273.15 K]:
−1
11 450
(c) The amount of heat gained by the water bath as a result of acetic acid freezing is ,
J. This heat
J of heat to melt one mole of ice. The amount of ice melted with
melts some of the ice. It takes ,
,
J:
11 450
5 980
11450 J × 18.016 g mol−1
5980 J mol−1
=34.946 g.
wice =
(d) The entropy change for the Universe is
∆Suniv = ∆S + ∆Ssurr
= −39.52 + 41.92 J K−1mol−1.
= 2.40 J K−1mol−1.
Since this is a positive number, we conclude that the process is spontaneous.
1
2. One mole of water at 0◦C and 1 atm pressure is turned into steam at 100◦ C and 1 atm pressure. From
the following information, calculate the entropy change for the entire process. CP,m;H2 O(l) = 75.48 J
K−1 mol−1 ; vap H◦ 2 O = 40.67 kJ mol−1.
∆ H
Equations given:
∆pcS
For a phase change,
=
H
T .
∆pc
pc
For a constant pressure process,
S C
∆
=
P,m ln
T 2
T1 .
Answer
The total entropy change is the sum of the entropy change for heating liquid water from 0 to 100◦C,
and for vaporizing it at 100◦ C:
∆S
.
= 75 48
373.15 J K−1mol−1 × ln
.
273 15
+
J mol−1 .
373.15 K
J K−1 mol−1 .
40670
.
= 132 54
3. At 100◦ C and 2 bar pressure, the degree of dissociation of phosgene is 6.30 × 10−5. The equilibrium is
written as COCl2 (g) CO(g) + Cl2 (g ). (a) Calculate KP and Kc for the equilibrium. (b) What is the
free energy change for the reaction if the standard state is 1 bar pressure? (c) What is the free energy
change if the standard state is 1 mol dm−3 (1 mol L−1)?
Equations given:
= Kc(RT/P ◦)∆ng
∆G −RT K , where K
KP
=
ln
=
KP , or Kc , depending on the standard state.
Answer
(a) Since the degree of dissociation is very small, it is a good assumption that the total pressure is nearly
equal to the initial pressure of COCl2. Then,
COCl2(g)
(1 − α)P
CO(g ) + Cl2 (g )
αP
αP
α2 P
(6.30 × 10−5)2 × 2
=
1 − α 1 − 6.30 × 10−5
=7.94 × 10−9.
KP =
Kc =KP /(
)
RT/P ◦ (2−1)
.
=2 56
×
= 7.94 × 10 ×
−9
.
0 083145
−10.
L bar K−1 mol−1 × 373.15 K
1 bar
10
G = −RT ln KP :
−1 mol−1 × 373.15 K × ln(7.94 × 10−9)
= −8.3145 J K
−1.
= 57.9 kJ mol
(b) For a standard state of 1 bar pressure,
∆
2
−1
(c) For a standard state of 1 mol dm−3 (1 mol L−1 ),
G = −RT ln Kc :
∆
.
J K−1 mol−1 × 373.15 K × ln(2.56 × 10−10 )
−1 .
=68.5 kJ mol
−
=
8 3145
4. At 25.0◦C, the equilibrium constant for the reaction CO(g) + H2O(g) CO2 (g) + H2(g) is 1.00 × 10−5 .
If 2 mol of CO and 2 mol of H2O are introduced into an evacuated 10.0 L vessel at 25◦ C, what are the
concentrations of CO, H2O, CO2 and H2 at equilibrium?
Equations given: None
Answer
The initial concentrations of CO and H2 O are 0.2 mol L−1 each. Let x mol of each react to reach
equilibrium. This gives:
CO(g ) +H2O(g)
CO2 (g)+H2 (g )
0.2 − x+0.2 − x
x
x
Kc
2
= (0.2x− x)2 = 1.00 × 10−5.
Taking square root of both sides and solving for x, we get
x
. × 10−5 =
0.2 − x
x=6.30 × 10−4 mol L−1
1 00
Therefore, [CO] = [H2 O] =
.
0 199
mol L−1 ; [CO2 ] = [H2 ] =
. × 10−4 mol L−1 .
6 30
5. The variation of pressure with temperature of liquid and solid chlorine in the vicinity of the triple point
is given by
ln
ln
Pl = −2661 + 22.76,
Ps =
T
−
3755
T
. .
+ 26 88
where the unit of temperature is K and that of pressure is Pa. Calculate the triple point temperature and
pressure.
Equations given:none
Answer
At the triple point, the two equations must be equal. Therefore, equating the right-hand sides, we get
(3755
−
2661)
1
.
−
. , or
T = 265.5 K.
= 26 88
T
22 76
Substituting this temperature into one of the equations above, we get the triple-point pressure:
−
2661
Pl =
+ 22.76
265.5
Pl = 3.4024 × 105 Pa.
ln
3
6. (a) Given the parent equation dA = −SdT − P dV , derive the corresponding Maxwell’s relation showing
all necessary steps. (b) For a gas obeying the van der Waal’s equation, with a = 5.46 bar L2 mol−2, and
b = 0.064 L mol−1 , what is the value of (∂S/∂V )T at a molar volume of 1.0 L?
Equations given:
a
For a van der Waal’s gas, P + 2
Vm
(Vm − b) = RT.
Answer
(a) Expressing
A as a function of V and T (as suggested by the parent equation), we get
dA =
∂A ∂T V
dT +
∂A ∂V T
dV.
Comparing this with the parent equation, we get
∂A ∂T
V
∂A
∂V T
Now, since
= −S,
= −P.
A is a state property (or exact differential),
∂ ∂A ∂ ∂A = ∂T ∂V
∂V ∂T V T
T V
∂S ∂P which yields
∂V T
which is the desired Maxwell relation.
(b) For a van der Waal’s gas,
P
∂S Therefore,
∂V T
∂V
∂T V
,
= V RT− b − Va2 .
=
m
∂P ∂T V
At a molar volume of 1.0 L mol−1, we get
∂S =
m
= V R− b .
m
L bar K mol
= 0.083145
(1.0 − 0 064) L mol−1
T
8 3145 J K−1mol−1 = 8 9 J K−1L−1
=0 089 bar K−1 × 0 083145
L bar K−1 mol−1
−1
−1
.
.
.
.
.
4
.