# 24.5: Most Solutions are Not Ideal

If we plot the partial pressure of one component say P1 above a mixture with a mole fraction $$x_1$$ we should get a straight line with a slope of $$P^*_1$$ (Raoult's law). Above non-ideal solutions the graph will no longer be a straight line but a curve. However towards $$x_1=1$$ the curve typically approaches the Raoult line. On the other extreme there often is a more or less linear region as well, but with a different slope. This means that we have two limiting laws.

Figure $$\PageIndex{1}$$: Vapor pressure above an ideal and a non-ideal solution

• For $$x \rightarrow 0$$: Henry's law is applicable

$P_1 = K_H x_1$

• For $$x \rightarrow 1$$: Raoult's law is applicable

$P_1 = P^*_1 x_1$

Note that this implies that the straight line that indicates the Henry expression will intersect the y-axis at x=1 (pure compound) at a different point than $$P^*$$. For $$x \rightarrow 0$$ (low concentrations) we can speak of component 1 being the solute (the minority component). At the other end $$x \rightarrow 1$$ it plays the role of the solvent (majority component).

Another thing to note is that $$P^*$$ is a property of one pure component, the value of $$K_H$$ by contrast is a property of the combination of two components, so it needs to be measures for each solute-solvent combination.

As you can see we have a description for both the high and the low end, but not in the middle. In general, the more modest the deviations from ideality the larger the range of validity of the two limiting laws. The way to determine $$K_H$$ would be to actually determine vapor pressures. How about the other component? Do we need to measure them too? Fortunately we can use thermodynamics to answer this question with no. There is a handy expression that saves us the trouble.

#### Gibbs-Duhem Relationship

Consider a Gibbs free energy that only includes $$μ_n$$ conjugate variables as we obtained it from our scaling experiment at $$T$$ and $$P$$ constant:

$G = μ_1n_1 + μ_2n_2$

Consider a change in $$G$$:

$dG = d(μ_1n_1) + d(μ_2n_2)$

$dG = n_1dμ_1+μ_1dn_1 + n_2dμ_2+μ_2dn_2$

However, if we simply write out a change in $$G$$ due to the number of moles we have:

$dG = μ_1dn_1 +μ_2dn_2$

Consequently the other terms must add up to zero:

$0 = n_1dμ_1+ n_2dμ_2$

$dμ_1= - \dfrac{n_2}{n_1}dμ_2$

$dμ_1= - \dfrac{x_2}{x_1}dμ_2$

In the last step we have simply divided both denominator and numerator by the total number of moles. This expression is known as the Gibbs-Duhem equation, which relates the change in one thermodynamic potential ($$dμ_1$$) to the other ($$dμ_2$$).

The Gibbs-Duhem equation relates the change in one thermodynamic potential ($$dμ_1$$) to the other ($$dμ_2$$).

### Gibbs-Duhem in the Ideal Case

In the ideal case we have:

$μ_2^{sln}=μ^*_2 + RT \ln x_2$

Gibbs-Duhem gives:

$dμ_1= - \dfrac{x_2}{x_1} dμ_2$

As

$dμ_2 = 0 + \dfrac{RT}{x_2}$

with $$x_2$$ being the only active variable at constant temperature, we get:

$dμ_1= - \dfrac{x_2}{x_1} \dfrac{RT}{x_2} = \dfrac{RT}{x_1}$

If we now wish to find $$μ_1$$ we need to integrate $$dμ_1$$, e.g. form pure 1 to $$x_1$$.

This produces:

$μ_1^{sln}=μ^*_1 + RT \ln x_1$

This demonstrates that Raoult's law can only hold over the whole range for one component if it also holds for the other over the whole range.