The pressure-temperature phase diagrams of pure substances describe the conditions at which various — Physical Chemistry — Thermodynamics Chemistry Question
The dense and hot ice
The pressure-temperature phase diagrams of pure substances describe the conditions at which various equilibrium phases exist. The phase diagram of water is shown below (pressure is given in the logarithmic scale).
The phase diagram of water in the semi-log scale
[VISUAL]
Using this diagram and the appropriate thermodynamic equations describing phase transitions, answer the following questions.
Necessary data:
Densities of ordinary ice and water: 0.917 and 1.000 g cm-3, respectively;
Enthalpy of fusion of ordinary ice: +6010 J mol-1;
Triple point «water – ice VI – ice VII»: pressure 2200 MPa, temperature 355 K.
Hint. Assume that the densities of condensed phases and the enthalpies of phase transitions do not vary with pressure and temperature.
How do the boiling point of water and the melting points of ordinary ice (ice I) and ice V vary with pressure? Explain this qualitatively applying the Le Chatelier principle.
Model Answer
The boiling point of water and the melting point of ice V increase, and the melting point of ordinary ice decreases with the increasing pressure. This can be easily explained using the Le Chatelier principle. In the phase transitions:
H2O(l) ⇆ H2O(g)
and
H2O(ice,V) ⇆ H2O(l)
the volume increases and heat is absorbed (ΔV > 0, ΔH > 0). Hence, with the increasing pressure both equilibria are shifted to the left; consequently, temperature should be increased to keep the equilibria.
In the phase transition:
H2O(ice,I) ⇆ H2O(l)
the volume decreases and heat is absorbed (ΔV < 0, ΔH > 0). Hence, with the increasing pressure the phase equilibrium is shifted to the right, and temperature should be decreased to keep the equilibrium.
What would happen with water vapor if the pressure is gradually increased from 10 Pa to 10 GPa at a temperature: a) 250 К, b) 400 К, c) 700 К ?
Model Answer
a) 250 К: vapor → ice I → ice III → ice V → ice VI
b) 400 К: vapor → liquid → ice VII
c) 700 К: only vapor (at high pressure it may be called “supercritical fluid”), no phase transitions occur.
The lowest possible temperature at which equilibrium liquid water still exists is achieved in the triple point between water, ice I, and ice III. The pressure in this point is 210 MPa, estimate the temperature.
Model Answer
Phase transitions between condensed phases are described by the Clapeyron equation:
dp/dT = ΔH / (T * ΔV)
or, in approximate form:
Δp/ΔT = ΔH / (T * ΔV).
We calculate the right side of this equation for the ice I ⇆ water transition. The volume change is determined from the densities:
ΔV = V(water) – V(ice) = M/ρ(water) - M/ρ(ice) = 18 / 1.000 - 18 / 0.917 = -1.63 cm3 mol-1
Δp/ΔT = ΔH / (T * ΔV) = 6010 J mol-1 / (273 K * (-1.63 * 10^-6 m3 mol-1)) = -1.35 * 10^7 Pa K-1 = -13.5 MPa K-1.
If this slope does not depend on pressure and temperature then at the pressure of 210 MPa the temperature of liquid water in equilibrium with ice I and Ice III is approximately:
T = 273 + ΔT = 273 + (210 - 0.1) / (-13.5) = 257.5 K = -15.5 °C.
This is an estimate; the real value is –22 °C. The difference between the estimated and real values is due to the fact that the enthalpy of fusion and densities vary with pressure. For example, at 210 MPa the enthalpy of fusion of ice I is 4230 J/mol (instead of 6010 at normal pressure), and the volume change is ΔV = –2.43 cm3/mol (instead of –1.63 cm3 mol-1 at normal pressure).
Several forms of ice can exist in equilibrium with liquid water. Assuming that the heat of fusion is approximately the same for all forms, determine, which of the ices has the largest density. What is the melting point of this ice at a pressure of 10 GPa?
Model Answer
From the Clapeyron equation it follows that the slope of the p(T) dependencies for the melting points of ice III to ice VII is determined by ΔH, T, and ΔV. The first quantity is assumed to be the same for all transitions, the temperature is comparable in all cases, hence the main contribution to the slope comes from ΔV.
For ice VII, the slope is the smallest, hence, the ΔV = V(water) – V(ice) is the largest, whereas V(ice) is the smallest. It means that ice VII is the densest form of ice (among those forms that are shown on the phase diagram).
From the phase diagram we see that the melting point of ice VII at a pressure of 10 GPa is about 630 K. This is, indeed, a very “hot” ice.
The densest ice has the cubic crystal structure with two water molecules per one unit cell. The edge of the unit cell is 0.335 nm. Calculate the density of ice.
Model Answer
Determine the molar volume of ice VII. One mole contains NA/2 cubic unit cells:
Vm = NA/2 * d^3 = 3.01 * 10^23 * (0.335 * 10^-7)^3 = 11.3 cm3 mol-1.
The density of ice VII is:
ρ = M / Vm = 18 / 11.3 = 1.59 g cm-3.
Estimate the enthalpy of fusion of the densest ice.
Model Answer
Knowing the density of ice VII, we use the Clapeyron equation to estimate its enthalpy of fusion. Comparing the triple point “water – ice VI – ice VII” and the melting point of ice VII at pressure 10 GPa we estimate the slope: Δp / ΔT = (10^4 – 2200) / (630 – 355) = 28 MPa / K. The volume change during melting is: ΔV = (18/1.00) – 11.3 = 6.7 cm3 mol-1. Substituting these data into the Clapeyron equation, we get:
ΔH = T * ΔV * (Δp / ΔT) = 355 K * (6.7 * 10^-6 m3 mol-1) * (28 * 10^6 Pa K-1) = 66 000 J mol-1.
This value is by an order of magnitude larger than the exact value 6400 J mol-1. The reason is probably due to a low resolution of the phase diagram at high pressures, which leads to a rough estimate of the slope. This result also shows that the approximations used are not valid at high pressures and temperatures.