Solid CaCO_3 exists in two crystalline forms: calcite with a density of 2.710 g/

Question

Solid CaCO_3 exists in two crystalline forms: calcite with a density of 2.710 g/cm^3 and argonite with a density of 2.930 g/cm^3. For this conversion between the two form: delta_rG degree = 1.04 kJ/mol at 25 degree C. At what pressure will the two forms be in equilibrium at 25 degree C? Using the data in problem 24-1 in the text for a solution of 1-propanol and 2-propanol, determine a pressure/composition curve for the solution using Dalton's law. Note the vapor pressure of the pure components are 20.9 Torr and 45.2 Torr respectively at 25 degree C.

Explanation / Answer

First we will calculate Vm

Vm = (100.1g/mol /2.930 g/cm3) - (100.1g/mol /2.710 g/cm3) = -2.773 cm3/mol

Now, T is constant at 298 K.

dG = VdP

dG = 1.04 kJ/mol

V = -2.773 cm3/mol

dP = (1.04 kJ/mol) / (-2.773 cm3/mol)

      = 3775 bar

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