Chemistry 360
Dr. Jean M. Standard
Problem Set 10 Solutions
1.
Sketch (roughly to scale) a phase diagram for molecular oxygen given the following information: the
triple point occurs at 54.3 K and 1.14 torr; the critical point occurs at 154.6 K and 37828 torr; the normal
melting point is 54.75 K; and the normal boiling point is 90.25 K. Does solid molecular oxygen melt
under applied pressure?
The data points that are given for the phase diagram are plotted below with pressure on the y-axis and
temperature on the x-axis.
Note that critical point occurs at such a high pressure compared to the other points that it is difficult to see the
other regions (these regions will be focused on later).
We know that the triple point, normal boiling point, and critical point must lie along the liquid-vapor
coexistence curve, with the triple point and critical point corresponding to the terminators of the curve. This is
indicated in the plot above by dashed lines connecting these three points. On the lower temperature side of the
L-V coexistence curve lies the liquid phase, and on the higher temperature side lies the vapor phase.
We also know that the triple point and the normal melting point lie along the solid-liquid coexistence curve.
This curve most likely extends to pressures above that of the normal melting point, but we have no data to
indicate the pressure and temperature values above 1 atm=760 torr. Thus, the solid phase must lie somewhere
on the left side of the plot, at temperatures lower than that of the normal melting point, as indicated.
The other coexistence curve, for the solid-vapor phase equilibrium, must also be present on the phase diagram.
However, we are given no data to characterize this portion of the phase diagram.
2
1.
continued
To get a better view of the coexistence curves at lower pressures, the scale of the y-axis is adjusted to show
pressures of 800 torr and lower; therefore, the position of the critical point is not shown in the figure below.
In this zoomed-in version of the O2 phase diagram, the liquid-vapor coexistence curve connecting the triple
point and the normal boiling point (and eventually the critical point) is clearly seen and the liquid and vapor
regions noted. Similarly, the solid-liquid coexistence curve connecting the triple point and the normal melting
point is more easily seen and the solid (and liquid) regions defined. Once again, however, no data is given to
locate the solid-vapor coexistence curve. It must occur at very low pressure, however, since it must lie below
the triple point at 1.14 torr.
To answer the question of whether or not solid O2 melts when the pressure is increased, we need to estimate the
slope of the coexistence curve. For a graph of pressure vs. temperature, the slope is
slope =
ΔP
P −P
= 2 1 .
ΔT
T 2 − T1
Using the two known points along the solid-liquid coexistence curve, the normal melting point ( T1 = 54.75 K ,
P1 = 760 torr ) and the triple point ( T 2 = 54.3 K , P2 = 1.14 torr ), the slope is
€
ΔP
P −P
= 2 1
ΔT
T 2 − T1
€
1.14 torr − 760 torr
=
54.3 K − 54.75 K
slope = 1686 torr/K .
slope =
€
€
€
Since the slope of the solid-liquid coexistence curve is positive, the solid will not melt as the pressure increases.
€
3
2.
Calculate the melting point of ice under a pressure of 50 bar given that the melting point at 1 bar is 0°C.
Assume that the density of ice under these conditions is 0.92 g/mL and the density of liquid water is 1.00
g/mL. The molar enthalpy of fusion of water is 6.01 kJ/mol.
The Clapeyron equation for solid-liquid phase equilibrium is
ΔH fus,m
dP
=
.
dT
T fusΔVm
We can approximate the left side of the equation as
€
dP
ΔP
≈
.
dT
ΔT
Substituting,
€
ΔH fus,m
ΔP
=
.
ΔT
T fus ΔVm
To get the melting point of ice at a pressure of 50 bar, we can solve for ΔT and use the fact that the melting
point of ice at 1 bar pressure is 0˚C (273.15 K). Solving for ΔT yields,
€
ΔT =
€
or T 2 − T1 =
T fus ΔVm ΔP
€
,
ΔH fus,m
T fus ΔVm ( P2 − P1 )
ΔH fus,m
.
Solving for T 2 ,
€
T 2 = T1 +
€
T fus ΔVm ( P2 − P1)
ΔH fus,m
.
The molar volumes of the liquid and solid phases can be calculated from the molecular weight and the densities.
For ice,
€
M # 1L &
⋅%
(
D $ 1000 mL '
# 18.016 g mol−1 & # 1L &
(⋅
= %%
(
−1 ( %
$ 0.92 g mL ' $ 1000 mL '
Vs,m =
Vs,m = 0.01958 L/mol.
For liquid water, the molar volume is
€
M # 1L &
⋅%
(
D $ 1000 mL '
# 18.016 g mol−1 & # 1L &
(⋅
= %%
−1 ( % 1000 mL (
'
$ 1.00 g mL
' $
Vl,m =
Vl,m = 0.01802 L/mol.
€
4
2.
Continued
Substituting,
T 2 = T1 +
T fus ΔVm ( P2 − P1 )
ΔH fus,m
= 273.15 K +
( 273.15 K)( 0.01802 − 0.01958 L/mol)( 50 − 1 bar )
( 6010 J/mol)
$ 100 J '
= 273.15 K − 0.00348 K L bar J -1 &
)
% 1 L bar (
= 273.15 K − 0.35 K
T 2 = 272.80 K or − 0.35! C .
€
3.
If it takes an increase of 1.334 megabars of pressure to change the melting point of a substance from
222°C to 122°C for a change in molar volume of –3.22 cm3/mol, what is the molar enthalpy of fusion of
the substance in J/mol?
The Clapeyron equation for solid-liquid phase equilibrium is
ΔH fus,m
dP
=
.
dT
T fusΔVm
Using an approximation for the left side in terms of finite changes leads to the relation
€
ΔH fus,m
ΔP
≈
.
ΔT
T fus ΔVm
Solving for the molar entropy of fusion, we have
€
# ΔP &
ΔH fus,m = %
( T fus ΔVm .
$ ΔT '
−3
Substituting ΔP = 1.334 × 10 6 bar, ΔT = −100 K , T fus = 495.15 K , and ΔVm = −3.22 × 10 L/mol yields
€
# ΔP &
ΔH fus,m = %
( T fus ΔVm
$ ΔT
€'
€
€
# 1.334 × 10 6 bar &
(( ( 495.15 K) −3.22 × 10−3 L/mol
= %%
−100
K
$
'
€
(
# 100 J &
= 21270 L bar/mol %
(
$ 1 L bar '
ΔH fus,m = 2.127 × 10 6 J/mol or 2127 kJ/mol.
€
)
5
4.
The vapor pressure of liquid benzene obeys the relation
ln P = 29.411 −
5893.5
,
T
where the pressure is in torr and temperature is in Kelvin. Calculate the molar enthalpy of vaporization
of benzene and the normal boiling point.
€
The liquid-vapor equilibrium coexistence curve is given by the Clausius-Clapeyron equation as
ln P = −
ΔH vap,m
RT
+ C .
Since the equation above is in the same form, we have that
€
−
ΔH vap,m
R
= − 5893.5K ,
or ΔH vap,m = R ( 5893.5K)
(
)
= 8.314 J mol−1K−1 ( 5893.5K)
ΔH vap,m = 49000 J/mol or 49.0 kJ/mol.
The normal boiling point can be determined by solving the equation given for T and substitution of P = 760 torr,
€
5893.5
T
5893.5
ln P − 29.411 = −
T
−5893.5
T =
.
ln P − 29.411
ln P = 29.411 −
Then, for P = 760 torr,
€
T =
−5893.5
ln 760 − 29.411
T = 258.7 K or − 14.45! C .
€
6
5.
At what pressure does the boiling point of water become 300°C? If oceanic pressure increases by 1 atm
for every 10 m, to what ocean depth does this pressure correspond?
Since we are dealing with a liquid/vapor equilibrium, the Clausius-Clapeyron equation applies. The integrated
form is
"P %
ΔH vap,m " 1
1 %
ln $ 2 ' =
−
$
'.
R
T2 &
# P1 &
# T1
In order to use this equation to determine P2 , we need the enthalpy of vaporization of water. This can be
obtained from values given in€the appendix ( ΔH vap,m = 44.03 kJ/mol ). Substituting,
ΔH vap,m $ 1
1 '
ln P2 = €ln P1 +
−
&
)
R
T2 (
% T1
€
= ln (1 atm) +
(44.03 × 10
)
3
J/mol $
'
1
1
−
&
)
(8.314 J/molK) % 373.15 K 573.15 K (
ln P2 = 4.9524 .
Taking the exponential gives the pressure,
€
P2 = e 4.9524
P2 = 142 atm.
This would correspond to an ocean depth of about 1420 meters.
€
6.
The sublimation pressures of solid Cl2 are 352 Pa at –112˚C and 35 Pa at –126.5˚C. The vapor pressures
of liquid Cl2 are 1590 Pa at –100˚C and 7830 Pa at –80˚C. Calculate the molar enthalpies of sublimation,
vaporization, and fusion.
Since the problem deals with S-V and L-V equilibria, the Clausius-Clapeyron equation applies. The equation is
"P %
ΔH m
ln $ 2 ' =
P
R
# 1&
"1
1 %
−
$
'.
T2 &
# T1
Solving for the enthalpy change,
€
ΔH m =
€
R ln ( P2 / P1)
.
$1
1'
−
&
)
T2 (
% T1
7
6.
continued
For the S-V equilibrium, the equation is
R ln ( P2 / P1)
$1
1 '
−
&
)
T2 (
% T1
ΔH sub,m =
(8.314 J mol
−1 −1
=
ΔH sub,m
K
$
'
35 Pa
)
) ln &% 352
Pa (
$
'
1
1
−
&
)
146.65 K (
% 161.15 K
= 31280 J/mol or 31.28 kJ/mol.
For the L-V equilibrium, the equation becomes
€
ΔH vap,m =
R ln ( P2 / P1)
$1
1 '
−
&
)
T2 (
% T1
(8.314 J mol
−1 −1
=
ΔH vap,m
K
$
'
7830 Pa
)
) ln &% 1590
Pa (
$
'
1
1
−
&
)
193.15K (
% 173.15K
= 22160 J/mol or 22.16 kJ/mol.
Since enthalpy is a state function, the enthalpy of fusion can be calculated from the following relation
€
ΔH sub,m = ΔH fus,m + ΔH vap,m .
Solving for the enthalpy of fusion,
€
€
ΔH fus,m = ΔH sub,m − ΔH vap,m
= 31.28 kJ/mol − 22.16 kJ/mol
ΔH fus,m = 9.12 kJ/mol.
8
7.
Freon-12 (CF2Cl2) was commonly used in spray cans prior to the discovery that it was harmful to the
ozone layer. Its enthalpy of vaporization is 20.25 kJ/mol and its normal boiling point is –29.2˚C.
Determine the vapor pressure of Freon-12 at 40˚C.
The Clausius-Clapeyron equation for the liquid-vapor equilibrium is
"P %
ΔH vap,m " 1
1 %
ln $ 2 ' =
−
$
'.
R
T2 &
# P1 &
# T1
Solving for ln P2 yields
€
ΔH vap,m $ 1
1 '
ln P2 = ln P1 +
−
&
)
R
T2 (
% T1
€
(20.25× 10 J/mol) $& 1 −
(8.314 J mol K ) % 243.95K
3
= ln (1.0 atm) +
−1 −1
'
1
)
313.15K (
ln P2 = 2.2063.
The pressure P2 is then
€
P2 = e 2.2063
P2 = 9.08 atm.
€
€
8.
The vapor pressure of solid uranium hexafluoride, UF6, follows the equation
ln P = 29.411 −
5893.5
,
T
where the pressure is in Pa and temperature is in Kelvin. The vapor pressure of liquid uranium
hexafluoride follows the equation
€
ln P = 22.254 −
3479.9
.
T
Determine the temperature and pressure of the triple point.
€
The triple point occurs at the intersection
of the S-V and L-V coexistence curves. Equating these from above,
29.411 −
5893.5
3479.9
= 22.254 −
T
T
2413.6
7.157 =
T
T = 337.24 K .
Thus, 337.24 K is the triple point temperature. Substituting this temperature back into either of the two
coexistence equations and solving for pressure give the triple point pressure.
€
9
8.
continued
Using the S-V equation,
5893.5
T
5893.5
ln P = 29.411 −
337.24 K
ln P = 11.935
ln P = 29.411 −
P = 1.526 × 10 5 Pa.
€
9.
The vapor pressure of liquid mercury is 0.133 bar at 260˚C and 0.533 bar at 330˚C. Assume that the
mercury vapor can be treated as an ideal gas and that the enthalpy of vaporization is independent of
temperature. Calculate the molar enthalpy and the molar Gibbs free energy of vaporization at 25˚C.
The Clausius-Clapeyron equation for the liquid-vapor equilibrium is
"P %
ΔH vap,m
ln $ 2 ' =
R
# P1 &
"1
1 %
−
$
'.
T2 &
# T1
Solving for the enthalpy of vaporization,
€
ΔH vap,m =
R ln ( P2 / P1)
.
$1
1 '
−
&
)
T2 (
% T1
Substituting,
€
R ln ( P2 / P1)
ΔH vap,m =
$1
1 '
−
&
)
T2 (
% T1
(8.314 J mol
−1 −1
=
ΔH vap,m
K
$
'
)
) ln &% 0.533bar
0.133bar (
$
'
1
1
−
&
)
603.15K (
% 533.15K
= 53020 J/mol or 53.02 kJ/mol.
Therefore, assuming that ΔH vap,m is independent of temperature, the molar enthalpy of vaporization at 25˚C is
53.02 kJ/mol.
€
To determine the Gibbs free energy, we can use the equation
€
ΔG vap,m = ΔH vap,m − T ΔSvap,m .
€
10
9.
continued
The entropy of vaporization can be determined from the alternate form of the Clausius-Clapeyron equation,
ln P = −
ΔH vap,m
RT
+
ΔSvap,m
R
.
Solving for the entropy of vaporization,
€
ΔSvap,m = R ln P +
ΔH vap,m
T
.
Substituting 0.133 bar and 533.15 K for the pressure and temperature,
€
53020 J/mol
ΔSvap,m = 8.314 J mol-1K−1 ln ( 0.133bar ) +
533.15K
(
)
ΔSvap,m = 82.70 J mol-1K−1 .
Finally, the Gibbs free energy at 25˚C is
€
ΔG vap,m = ΔH vap,m − T ΔSvap,m
(
= 53020 J/mol − ( 298.15 K) 82.70 J mol−1K−1
ΔG vap,m = 28360 J/mol or 28.36 kJ/mol.
€
)
11
10. Consider the phase diagram of sulfur given below (rhombic and monoclinic are two different solid forms
of sulfur). Starting at 298K and 1 atm pressure and considering an increase in the temperature (at
constant pressure), comment on the entropy change as the sulfur goes from the rhombic solid phase to
the monoclinic sold phase. Is the entropy change expected to be positive or negative? On the basis of the
Second Law of Thermodynamics, is the phase transition expected to be spontaneous in an isolated
system?
For the Solid-Rhombic to Solid-Monoclinic phase transition, the phase diagram shows that the slope of the
coexistence curve dP/dT is positive,
dP
> 0.
dT
From the Clapeyron equation, we have
€
dP
ΔSm
=
.
dT
ΔVm
Since the slope of the coexistence curve is positive,
€
ΔSm
> 0.
ΔVm
We can assume that ΔVm is positive since the molar volume of the monoclinic phase would be predicted to be
larger than the molar volume of the rhombic phase at 25°C. This is because at a given temperature, the phase
€
with the smaller molar volume (and therefore higher density) will be stable at higher pressure. This means that
ΔSm is also positive,
€
ΔSm > 0 .
€
From the Second Law, a positive entropy change corresponds to a spontaneous process in an isolated system.
€