How do repulsive solute interactions affect the van't Hoff factor?

Question

Context. Consider an ideal solution, one in which the intermolecular solvent-solvent, solute-solute, and solute-solvent interactions are all equal. The presence of a solute therefore produces (solely) an increase in entropy due to the increase in configurational microstates. This increase in entropy leads to the well-known colligative properties, characterized by the van't Hoff factor, the number of effective solute particles per molecule or salt of solute.

Now consider a solution in which the solute-solute intermolecular interaction is changed by $+\epsilon$, so that solutes are more attracted to each other. This decreases the van't Hoff factor due to clumping of the solutes, which we can rationalize from two different perspectives:

• probabilistic: solute particles are more likely to be found close to each other (due to the favorable attraction between them), so we effectively have less solute particles as a whole, resulting in a lower van't Hoff factor
• statistical-mechanical: microstates in which solute particles are close to each other are weighted more heavily; this perturbation from the uniform distribution of microstates decreases the entropy of the system, effectively decreasing the "strength" of the colligative effect and hence resulting in a lower van't Hoff factor

Both factors agree, so that's good.

Now consider a solution in which the solute-solute intermolecular interaction is changed by $-\epsilon$, so that solutes are less attracted to each other. Consider the two perspectives again:

• probabilistic: solute particles are less likely to be found close to each other (due to the unfavorable attraction between them), so we effectively have more solute particles as a whole, resulting in a higher van't Hoff factor
• statistical-mechanical: microstates in which solute particles are close to each other are weighted less heavily; this perturbation from the uniform distribution of microstates decreases the entropy of the system, effectively decreasing the "strength" of the colligative effect and hence resulting in a lower van't Hoff factor

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Questions.

• Why don't the probabilistic and statistical-mechanical viewpoints agree in the latter case?
• How can the incorrect viewpoint(s) be modified to make them correct?

I am inclined to believe the statistical-mechanical viewpoint is correct, since it's more fundamental. I'm also somewhat leery of the implication that solute particles being less likely to be close to each other leads to effectively more solute particles.

• If this implication is indeed wrong, why is it wrong? How about the converse, that because solute particles are more likely to be found close to each other, we effectively have less solute particles as a whole?
• If this implication is actually right, why is it right?

Answer 1

I think that your first probabalistic argument is correct, i.e that as the solute interaction increases the effective number of them is reduced. In the second case, as the interaction become repulsive, the effective number of them cannot increase over the maximum number there. They are restricted in their configurational entropy because the repulsion means that the chance of finding one next to another is greatly reduced. Thus the probabilistic aligns with your stat-mech one.

You can calculate the entropy of mixing in an ideal solution then in a 'regular' one in which the components interact with one another.

In the ideal solution a statistical argument based on number of ways of placing particles on a grid, leads to the entropy being given by

$$\Delta S_{mix}= -R[n_1\ln(x_1)+n_2\ln(x_2)]$$

for $n_1,\,n_2$ moles and $x_1,\,x_2$ mole fractions of species 1 and 2.

In a solution in which there is interaction between (equal sized) molecules the free energy is

$$\Delta G_{mix} = R[n_1\ln(x_1)+n_2\ln(x_2)] +(n_1+n_2)x_1x_2E$$

where E is is an energy of mixing. This can be calculated by considering that one solute molecule interacts with energy $\epsilon _{aa}$, the solvent with energy $\epsilon _{bs}$ and between solute and solvent with $\epsilon _{ab}$. Again a statistical argument based on molecules being on lattice sites means that nearest neighbour interactions of the type a-a and b-b occur with probability $x_a^2$, and $x_b^2$ respectively and between a and b as $2x_ax_b$ and the 2 arise because a_b and b-a interactions are counted.

The total mixed energy is then $E_{tot} \approx (\epsilon_{aa}N_a^2+ 2\epsilon_{aa}N_aN_b+\epsilon_{bb}N_b^2 )$$

but the change on mixing is this energy less that due to the two components unmixed which is $\approx (N_a\epsilon_{aa}+N_b\epsilon_{bb})$

which means that the extra mixing energy is $E = zN_0(2\epsilon_{ab}-\epsilon_{aa}-\epsilon_{bb})$

where $z$ is related to the coordination number on the lattice.

The entropy on mixing is then

$$\Delta S_{mix}= -R[n_1\ln(x_1)+n_2\ln(x_2)] -(n_1+n_2)x_1x_2\frac{\partial w}{\partial T}$$

where the second term $\displaystyle \Delta S^E =-(n_1+n_2)x_1x_2\frac{\partial w}{\partial T}$ is a positive number about $0.2R$ J /mol/K for benzene/cyclohexane and $0.01R$ for benzene /$\ce{CCl4}$

Changing $\epsilon_{bb}$ as you suggest will still increase or decrease $\Delta S^E$ acting through the derivative.