Question

A) Use the Clapeyron equation to estimate (in bar/K) the slope of the solid-liquid phase boundary of water given the enthalpy of fusion is 6.008 kJ/mol and the densities of ice and water at T=0^oC are 0.91671 and 0.99984 gm/cm³ respectively.

B) In the preceding problem, what is the pressure required to lower the melting point of ice by 1C, in bars?

C) The standard molar entropy of rhombic sulfur is 31.80 J/K mol and that of monoclinic sulfur is 32.6 J/K mol. At what temperature will the transition occur at 1 bar? Assume the enthalpy and entropy do not depend on the temperature within the temperature range of interest. Give answer in Kelvins. Note you will need data on the enthalpy of the transition. To get that, you will have to use the table data that the standard Gibbs formation energy and density of rhombic and monoclinic sulfur, at T=25C, are equal to 0 kJ/mol, 2.070 g/cm3 and +0.33 kJ/mol, 1.957 g/cm3 respectively.

Answer 1

A) The equation to use will be the following:

For the solid-liquid phase boundary:

dP/dT = ∆Strans/∆Vtrans = ∆Hf/Tf /∆Vf

To get ∆Vf:

Vm(s) = MM/d(s) = 18 g/mol / 0.91671 g/cm³ = 19.64 cm³/mol or 19.64x10^-6 m³/mol

Vm(l) = 18 / 0.99984 = 18.003 cm³/mol or 1.8x10^-5 m³/mol

So, ∆Vm = 1.8x10^-5 - 1.964x10^-5 = -1.637x10^-6 m³/mol

finally:

dP/dT = 6008 / 273 x -1.637x10^-6 = -13,443,693.35 Pa/K

To convert Pa to bar: -13,443,693.55 x 1x10^-5 = -134.44 bar/K

b) This is actually easy, because if we want to lower that in 1 °C, the equilibrium must be changed. This is achieved by the definition of the slope, so the pressure needed would be 134.44 bar.

c) Assuming molar entropies are independent of temperature:

∆Gm = -Sm∆T

In the transition: Gm(rhom) = Gm(mono)

For Gm relative to the ∆Gf of the rhombic form (i.e., the monoclinic form lies 0.33 kJ/mol above the rhombic form:

0 + ∆Gm(rhom) = 0.33 + ∆Gm(mono)

-Sm(tf - 298) = 0.33 - Sm(tf-298)

-31.8(tf - 298) = 330 - 32.6(tf-298)

-31.8tf + 9476.4 = 330 - 32.6tf + 9714.8

(32.6 - 31.8)tf = 330 + 9714.8 - 9476.4

0.8tf = 568.4

tf = 710.5 K