Solvent Free Energy
The solvents of all real solutions approach obedience to Raoult’s law in the dilute solution limit, as can be seen in the generalised example where it follows that that the solvent of any solution, ideal or nor, follows ideal behavior in the dilute solution limit. Thus the partial molal free energy given by the eq. G^{l}_{A} = G°_{A} + RT in x_{A}, describes the partial molal free energy of the solvent of any dilute solution – as well as that of the components of ideal solutions.
Activities and activity coefficients of solvents: it is convenient to have an expression of the form of the eq. even when the component of interest is not conforming to ideal behavior. This can be done by introducing the activity a, defined as a_{A} = P_{A}/P°_{A}, so that the partial molar free energy is given, even for nonideal solutions, by:
G_{A} = G°_{A} = RT in a_{A}
Since we expect a_{A} to approach x_{A} for the solvent of a dilute solution, it is convenient to introduce the activity coefficient γ as
= a_{A}/x_{A} or a_{A} = x_{A}
By its variation from unity the activity coefficient shows the nonideality of the solvent. Since the solvents of all solutions that are sufficiently dilute obey Raoult’s law, the value of γ approaches unity for the solvent of any solution as the solute concentration approaches zero.
The vapour pressure data of the fig. provide the necessary information for the calculation of activity and activity coefficients according to the above equations. The activities and activity coefficients for acetone chloroform solutions are shown in fig. The factors that operate to make the activity coefficients less than unity are, of course, the same as those mentioned in connection with the energy of mixing in the formation of solutions.
Raoult’s law is obeyed by the solvent of a solution as the infinite dilution, or pure solvent is approached. It followed that the activity coefficient of the solvent becomes equal to unity as the solution composition approaches that of the pure component.
Example: at 30˚C the vapor pressure of a 1M aqueous solution of sucrose is 31.207 mmHg and the density of the solution is 1.1256 g mL^{-1}. The vapor pressure of water at 30˚C is 31.824 mmHg. Calculate (a) the activity of the water;(b) the activity coefficient of water in the sugar solution; (c) the free energy that water in the solution would have, compared with pure water, if the solution were ideal and (d) the free energy of water in the sugar solution compared with that pure water.
Solution: one L of the 1 M sucrose solution contains 1 mol of sucrose, for which the molar mass is 342.3 g. the mass of water is 1125.6 g – 342.3 g = 783.3 g. the number of moles of water is 943.47 mol)/ (44.47 mol) = 0.9775. (a) The activity of water is the solution is given by the vapor-pressure ratio P_{A}/P˚_{A} as:
a = 31.207 mmHg/31.824 mmHg = 0.981
(b) The activity coefficient enters as = a_{A}/x_{A} = (P_{A}/P˚_{A}). Here, with a_{A} = 0.981 and x_{A} = 0.977, we obtain:
= 1.004
(c) If ideal solution behavior is assumed, the free energy of the solution component is related to the mole fraction of that component by:
G_{A }= G˚_{A} + RT In x_{A}
Now we can calculate;
(G_{A} - G˚_{A})_{ideal} = (8.3143 JK^{-1 }mol^{-1}) (303.15 K) In 0.9775 = 57.4 J mol^{-1}
(d) The vapor pressure of sucrose at this temperature is neglible and the solution vapor pressure is due to water vapor. The molar free energy difference between pure water and the water in the solution can be obtained from the pressures of equilibrium vapors. Assuming ideal vapor behavior and using the subscript A to identify water, the solvent in this solution, we have:
G_{A} = G_{A} + RT In P_{A}/P˚_{A}
And, G_{A} - G˚_{A} = (8.3143 J K^{-1} mol^{-1}) (303.15 K) In [(31.207 mmHg)/ (31.824 mmHg)]
= - 49.3 J mol^{-1}
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