# Using Raoult's law... and knowing that A and B form an ideal binary mixture but R and S have negative deviation...?

## a) Show that ${P}_{A B} = {P}_{B}^{\circ} + {\chi}_{A} \left({P}_{A}^{\circ} - {P}_{B}^{\circ}\right)$. b) If ${P}_{A}^{\circ} > {P}_{B}^{\circ}$, ${P}_{A}^{\circ} = {P}_{R}^{\circ}$ and ${P}_{B}^{\circ} = {P}_{S}^{\circ}$, sketch a diagram against the mol fraction of $B$ for solution $A B$ and of $S$ for solution $R S$, showing how ${P}_{A}$, ${P}_{B}$, ${P}_{R}$, and ${P}_{S}$ vary, and how ${P}_{A B}$ and ${P}_{R S}$ look.

May 15, 2016

Okay, so the idea with this is to compare an ideal binary mixture with a nonideal binary mixture wherein negative deviation occurs upon creation of the mixture.

"SATURATED" VAPOR PRESSURE

If you remember Raoult's law, it dealt with the following relationship:

$\setminus m a t h b f \left({P}_{j} = {\chi}_{j} {P}_{j}^{\circ}\right)$

where:

• ${P}_{j}$ is the vapor pressure of the solution containing solute (i.e. non-pure).
• ${\chi}_{j} = \frac{{n}_{j}}{{n}_{i} + {n}_{j}}$ is the mol fraction of solvent $j$ in solution. When ${\chi}_{j} \downarrow$, ${\chi}_{j} < 1$, and ${\chi}_{i}$, the mol fraction of solute $i$, is increasing.
• ${P}_{j}^{\circ}$ is the vapor pressure of the solution containing NO solute at all (i.e. purely solvent). This is known as the vapor pressure of the pure solvent, or the "saturated" vapor pressure.

Let's just do the derivation part of the problem right here. Starting from Dalton's law of partial pressures, we just have, for the ideal binary mixture $A B$:

${P}_{\text{tot}} = \textcolor{b l u e}{{P}_{A B}^{\circ}} = {P}_{A} + {P}_{B}$

$= {\chi}_{A} {P}_{A}^{\circ} + {\chi}_{B} {P}_{B}^{\circ}$

$= {\chi}_{A} {P}_{A}^{\circ} + \left(1 - {\chi}_{A}\right) {P}_{B}^{\circ}$

$= {\chi}_{A} {P}_{A}^{\circ} + {P}_{B}^{\circ} - {\chi}_{A} {P}_{B}^{\circ}$

$= \textcolor{b l u e}{{P}_{B}^{\circ} + {\chi}_{A} \left({P}_{A}^{\circ} - {P}_{B}^{\circ}\right)}$

The physical interpretation of this result is that the total pressure can be computed for the mixture of any two substances, assuming ideality, by using their pure vapor pressures and by knowing how much $A$ you add to $B$.

If you look back at how I did this derivation, you should notice that ${\chi}_{B} + {\chi}_{A} = 1$. That makes sense though, since the mol fractions of the only two components of the ideal binary mixture should add up to 100% of the components in the mixture.

Thus, you should remember that ${\chi}_{A} = 1 - {\chi}_{B}$.

ENERGIES OF INTERACTION IN AN IDEAL BINARY MIXTURE

For the ideal binary mixture $A B$, we say that the intermolecular force of attraction between liquid $A$ and liquid $B$ are equal to those between $A$ with itself and $B$ with itself. This can be expressed as:

$2 {\epsilon}_{A B} = {\epsilon}_{A A} + {\epsilon}_{B B}$

This says that the energy of the system after mixing $A$ and $B$ with each other is equal to the energy of the system before mixing $A$ and $B$ with each other.

POSITIVE AND NEGATIVE DEVIATIONS FROM IDEAL MOLAR VOLUMES

Now, we can have two variations on this.

$\setminus m a t h b f \left(2 {\epsilon}_{A B} < {\epsilon}_{A A} + {\epsilon}_{B B}\right)$ (1) In (1), the energy $\epsilon$ for the interaction between $A$ and $B$ with themselves is greater than that for the interaction between $A$ and $B$ with each other.

That means the particles, after mixing, have a lower overall energy, and thus, mixing is favorable. Thus, the average distance between $A$ and $B$ is smaller, and we have what's called negative deviation---the volume of the solution is smaller than for the ideal solution.

That leads to a dip relative to Raoult's law behavior when the mol fraction of the solvent is less than $1$ but that of the solute is not yet $1$.

$\setminus m a t h b f \left(2 {\epsilon}_{A B} > {\epsilon}_{A A} + {\epsilon}_{B B}\right)$ (2) Similarly, in (2), since the energy for the $A B$ interaction is greater than for the $A A$ and $B B$ interactions combined, mixing is unfavorable, and the average distance between particles is larger; thus, the volume increases and we have positive deviation.

That leads to a "hump" relative to Raoult's law behavior when the mol fraction of the solvent is less than $1$ but that of the solute is not yet $1$.

THE AB VS THE RS SOLUTION

Based on your problem setup, what you're dealing with is ideality for the AB solution and negative deviation for the RS solution.

For the ideal mixture, the total vapor pressure, ${P}_{A B}^{\circ}$, of the solution should add up to be horizontal with the addition of solute $B$ (and thus decrease ${P}_{A}$ relative to ${P}_{A}^{\circ}$) into a solution that contained only $\setminus m a t h b f \left(A\right)$ at first. This is because there is no favorability in either the positive or negative direction when it comes to mixing $A$ and $B$; the resultant energy of their combination, in any combination ($A A$, $B B$, or $A B$), doesn't make a difference.

Note that since we defined $\setminus m a t h b f \left({P}_{A}^{\circ} > {P}_{B}^{\circ}\right)$, it makes sense that the vapor pressure of $A$, ${P}_{A}$, decreases from ${P}_{A}^{\circ}$ as more $B$ is added to the solution.

For the $R S$ mixture, we noted that it had negative deviation. This manifests itself as a dip downwards in the graph of total vapor pressure, ${P}_{R S}$, as you add $S$ (and thus an increase in ${P}_{S}$ relative to ${P}_{S}^{\circ}$), and an increase again to ${P}_{R S}$ as usual as you add more $S$. When there is 50% $R$ and 50% $S$, or whatever particular quantity of $R$, the solution is most favorably mixed and the volume contracts relative to ideal (Raoult's law) behavior.

When there is 100% $R$, note that we defined ${P}_{A}^{\circ} > {P}_{B}^{\circ}$, ${P}_{R}^{\circ} = {P}_{A}^{\circ}$, and ${P}_{S}^{\circ} = {P}_{B}^{\circ}$.

Thus, $\setminus m a t h b f \left({P}_{R}^{\circ} > {P}_{S}^{\circ}\right)$, and it makes sense that ultimately, with 100% $R$ (0% S), ${P}_{R}^{\circ} > {P}_{S}^{\circ}$.