The vapor pressure of pure water at its melting point, 0.0 °C, is 611 Pa (6.11 × 10-3 bar). — Colligative Properties and Thermodynamics Chemistry Question
Problem Context
The vapor pressure of pure water at its melting point, 0.0 °C, is 611 Pa (6.11 × 10-3 bar).
0.880 mol MgCl2 is added to 1.00 kg liquid water at 0.0 °C. Calculate the vapor pressure of this solution.
Model Answer
Pvap,soln = (xH2O)(Pvap H2O, pure)
nH2O = 1000 g ´ 1 mol/18.02 g = 55.49 mol
Because each mol MgCl2 forms 3 moles of ions, the total moles are 55.49 mol H2O + 3(0.880) mol ions = 58.13 mol. So xH2O = 55.49/58.13 = 0.955.
Pvap,soln = (0.955)(611 Pa) = 584 Pa
The enthalpy of sublimation of ice is 51.1 kJ mol-1 at 0.0 °C and the enthalpy of vaporization of liquid water is 45.1 kJ mol-1 at 0.0 °C. Assuming these enthalpies are independent of temperature, calculate the temperature at which pure ice has the same vapor pressure as the magnesium chloride solution at 0.0 °C calculated in part a.
Model Answer
ln(P2/P1) = (–∆H°subl/R)(1/T2 – 1/T1)
ln(584 Pa/611 Pa) = (–51100 J mol-1/8.314 J mol-1 K-1)(1/T2 – 1/273.15 K)
7.353 ´ 10-6 K-1 = 1/T2 – 1/273.15 K
T2 = 272.6 K = –0.5 °C
At the freezing point temperature of the aqueous magnesium chloride solution, the vapor pressure of the solution is equal to the vapor pressure of pure ice. Explain why this statement is true.
Model Answer
When the solution freezes, the solid formed is pure ice. Unlike the flexible structure of water in solution, the ice crystal has a rigid structure that cannot accommodate impurities such as magnesium or chloride ions. If pure ice is in equilibrium with the aqueous solution, then the vapor in equilibrium with ice must be in equilibrium with the vapor in equilibrium with the solution. This is the same as saying that the vapor pressures are equal.
Using the principle enunciated in part c, calculate the freezing point temperature of the solution of 0.880 mol MgCl2 dissolved in 1.00 kg water.
Model Answer
At the freezing point temperature, the vapor pressure of pure ice is equal to the vapor pressure of the solution, which is 0.955 times the vapor pressure of pure water at this temperature (as in part a). If we call this vapor pressure Pf, then:
From the temperature-dependence of the vapor pressure of ice:
ln(Pf/611 Pa) = (–∆H°subl/R)(1/Tf – 1/273.15 K)
ln(Pf) = (–∆H°subl/R)(1/Tf – 1/273.15 K) + ln(611 Pa)
The vapor pressure of pure water at this temperature is Pf/0.955, so:
ln(1.047•Pf/611 Pa) = (–∆H°vap/R)(1/Tf – 1/273.15 K)
ln(Pf) = (–∆H°vap/R)(1/Tf – 1/273.15 K) + ln(611 Pa) – ln(1.047)
Equating the two expressions for ln(Pf) gives:
ln(1.047) = ([∆H°subl–∆H°vap]/R)•(1/Tf – 1/273.15 K)
6.36 ´ 10-5 K-1 = (1/Tf – 1/273.15 K)
Tf = 268.5 K = –4.7 °C
Note that this number is slightly different from the value obtained from the molal freezing point depression equation (–4.9 °C), as the derivation of that equation assumes that ln(1 – x) ≈ –x and that 1/T2 – 1/T1 ≈ (ΔT)/T1^2. Those assumptions are fine at the limit of dilute concentrations but they introduce small errors as concentration is increased.
The temperature calculated in part d is the temperature of the system when the first solid appears at equilibrium as the system is cooled. As more solid is formed, do you expect the temperature to increase, decrease, or remain constant? Briefly justify your choice.
Model Answer
As more pure ice crystallizes from the solution, the concentration of ions in solution increases, further depressing the freezing point. Thus the temperature will decrease as more solid forms.