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 = ΔPΔT

= P2 − P1T2 −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 (

€

T2 = 54.3 K ,

€

P2 = 1.14 torr ), the slope is

€

slope = ΔPΔT

= P2 − P1T2 −T1

= 1.14 torr − 760 torr54.3 K − 54.75 K

slope = 1686 torr/K . 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

€

dPdT

= ΔH fus,mT fusΔVm

.

We can approximate the left side of the equation as

€

dPdT

≈ ΔPΔT

.

Substituting,

€

ΔPΔT

= ΔH fus,mT fusΔVm

.

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 = T fus ΔVm ΔPΔH fus,m

,

or T2 −T1 = T fus ΔVm P2 − P1( )

ΔH fus,m.

Solving for

€

T2,

€

T2 = 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,

€

Vs,m = MD⋅

1L1000 mL

⎛

⎝ ⎜

⎞

⎠ ⎟

= 18.016 g mol−1

0.92 g mL−1

⎛

⎝ ⎜ ⎜

⎞

⎠ ⎟ ⎟ ⋅

1L1000 mL

⎛

⎝ ⎜

⎞

⎠ ⎟

Vs,m = 0.01958 L/mol.

For liquid water, the molar volume is

€

Vl,m = MD⋅

1L1000 mL

⎛

⎝ ⎜

⎞

⎠ ⎟

= 18.016 g mol−1

1.00 g mL−1

⎛

⎝ ⎜ ⎜

⎞

⎠ ⎟ ⎟ ⋅

1L1000 mL

⎛

⎝ ⎜

⎞

⎠ ⎟

Vl,m = 0.01802 L/mol.

4

2. Continued Substituting,

€

T2 = T1 + T fus ΔVm P2 − P1( )

ΔH fus,m

= 273.15K + 273.15K( ) 0.01802− 0.01958 L/mol( ) 50−1bar( )

6010 J/mol( )

= 273.15K − 0.00348 K L bar J -1 100 J1L bar

⎛

⎝ ⎜

⎞

⎠ ⎟

= 273.15K − 0.35K

T2 = 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

€

dPdT

= ΔH fus,mT fusΔVm

.

Using an approximation for the left side in terms of finite changes leads to the relation

€

ΔPΔT

≈ ΔH fus,mT fusΔVm

.

Solving for the molar entropy of fusion, we have

€

ΔH fus,m = ΔPΔT⎛

⎝ ⎜

⎞

⎠ ⎟ T fus ΔVm .

Substituting

€

ΔP = 1.334 ×106 bar,

€

ΔT = −100K ,

€

T fus = 495.15 K , and

€

ΔVm = −3.22×10−3 L/mol yields

€

ΔH fus,m = ΔPΔT⎛

⎝ ⎜

⎞

⎠ ⎟ T fus ΔVm

= 1.334 ×106 bar−100 K

⎛

⎝ ⎜ ⎜

⎞

⎠ ⎟ ⎟ 495.15K( ) −3.22×10−3 L/mol( )

= 21270 L bar/mol 100 J1L bar

⎛

⎝ ⎜

⎞

⎠ ⎟

ΔH fus,m = 2.127×106 J/mol or 2127 kJ/mol.

5

4. The vapor pressure of liquid benzene obeys the relation

€

ln P = 29.411 − 5893.5T

,

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 = − ΔHvap,mRT

+ C .

Since the equation above is in the same form, we have that

€

−ΔHvap,m

R = − 5893.5K ,

or ΔHvap,m = R 5893.5K( )

= 8.314 J mol−1K−1( ) 5893.5K( )ΔHvap,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,

€

ln P = 29.411 − 5893.5T

ln P − 29.411 = − 5893.5T

T = −5893.5ln P − 29.411

.

Then, for P = 760 torr,

€

T = −5893.5ln 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

€

ln P2P1

⎛

⎝ ⎜

⎞

⎠ ⎟ =

ΔHvap,mR

1T1

− 1T2

⎛

⎝ ⎜

⎞

⎠ ⎟ .

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 (

€

ΔHvap,m = 44.03 kJ/mol ). Substituting,

€

ln P2 = ln P1 + ΔHvap,m

R1

T1 − 1

T2

⎛

⎝ ⎜

⎞

⎠ ⎟

= ln 1atm( ) + 44.03×103 J/mol( )

8.314 J/molK( )1

373.15K − 1

573.15K

⎛

⎝ ⎜

⎞

⎠ ⎟

ln P2 = 4.9524 .

Taking the exponential gives the pressure,

€

P2 = e4.9524

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

€

ln P2P1

⎛

⎝ ⎜

⎞

⎠ ⎟ = ΔHm

R1T1

− 1T2

⎛

⎝ ⎜

⎞

⎠ ⎟ .

Solving for the enthalpy change,

€

ΔHm = R ln P2 /P1( )

1T1

− 1T2

⎛

⎝ ⎜

⎞

⎠ ⎟

.

7

6. continued For the S-V equilibrium, the equation is

€

ΔHsub,m = R ln P2 /P1( )

1T1

− 1T2

⎛

⎝ ⎜

⎞

⎠ ⎟

= 8.314 J mol−1K−1( ) ln 35Pa

352 Pa

⎛

⎝ ⎜

⎞

⎠ ⎟

1161.15K

− 1146.65K

⎛

⎝ ⎜

⎞

⎠ ⎟

ΔHsub,m = 31280 J/mol or 31.28 kJ/mol. For the L-V equilibrium, the equation becomes

€

ΔHvap,m = R ln P2 /P1( )

1T1

− 1T2

⎛

⎝ ⎜

⎞

⎠ ⎟

= 8.314 J mol−1K−1( ) ln 7830 Pa

1590 Pa

⎛

⎝ ⎜

⎞

⎠ ⎟

1173.15K

− 1193.15K

⎛

⎝ ⎜

⎞

⎠ ⎟

ΔHvap,m = 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

€

ΔHsub,m = ΔH fus,m + ΔHvap,m . Solving for the enthalpy of fusion,

€

ΔH fus,m = ΔHsub,m − ΔHvap,m

= 31.28 kJ/mol − 22.16 kJ/molΔH 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

€

ln P2P1

⎛

⎝ ⎜

⎞

⎠ ⎟ =

ΔHvap,mR

1T1

− 1T2

⎛

⎝ ⎜

⎞

⎠ ⎟ .

Solving for

€

ln P2 yields

€

ln P2 = ln P1 + ΔHvap,m

R1

T1 − 1

T2

⎛

⎝ ⎜

⎞

⎠ ⎟

= ln 1.0 atm( ) + 20.25×103 J/mol( )8.314 J mol−1K−1( )

1243.95K

− 1313.15K

⎛

⎝ ⎜

⎞

⎠ ⎟

ln P2 = 2.2063. The pressure

€

P2 is then

€

P2 = e2.2063

P2 = 9.08 atm.

8. The vapor pressure of solid uranium hexafluoride, UF6, follows the equation

€

ln P = 29.411 − 5893.5T

,

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 − 3479.9T

.

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 − 5893.5T

= 22.254 − 3479.9T

7.157 = 2413.6T

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,

€

ln P = 29.411 − 5893.5T

ln P = 29.411 − 5893.5337.24 K

ln P = 11.935

P = 1.526×105 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

€

ln P2P1

⎛

⎝ ⎜

⎞

⎠ ⎟ =

ΔHvap,mR

1T1

− 1T2

⎛

⎝ ⎜

⎞

⎠ ⎟ .

Solving for the enthalpy of vaporization,

€

ΔHvap,m = R ln P2 /P1( )

1T1

− 1T2

⎛

⎝ ⎜

⎞

⎠ ⎟

.

Substituting,

€

ΔHvap,m = R ln P2 /P1( )

1T1

− 1T2

⎛

⎝ ⎜

⎞

⎠ ⎟

= 8.314 J mol−1K−1( ) ln 0.533bar

0.133bar

⎛

⎝ ⎜

⎞

⎠ ⎟

1533.15K

− 1603.15K

⎛

⎝ ⎜

⎞

⎠ ⎟

ΔHvap,m = 53020 J/mol or 53.02 kJ/mol. Therefore, assuming that

€

ΔHvap,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

€

ΔGvap,m = ΔHvap,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 = −ΔHvap,mRT

+ ΔSvap,m

R.

Solving for the entropy of vaporization,

€

ΔSvap,m = R ln P + ΔHvap,m

T.

Substituting 0.133 bar and 533.15 K for the pressure and temperature,

€

ΔSvap,m = 8.314 J mol-1K−1( ) ln 0.133bar( ) + 53020 J/mol533.15K

ΔSvap,m = 82.70 J mol-1K−1 . Finally, the Gibbs free energy at 25˚C is

€

ΔGvap,m = ΔHvap,m − T ΔSvap,m

= 53020 J/mol − 298.15K( ) 82.70 J mol−1K−1( )ΔGvap,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 298 K 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,

€

dPdT

> 0 .

From the Clapeyron equation, we have

€

dPdT

= ΔSmΔVm

.

The slope of the coexistence curve is positive,

€

ΔSmΔVm

> 0 .

We can assume that

€

ΔVm is positive because 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 . And from the Second Law, a positive entropy change corresponds to a spontaneous process in an isolated system.