# What is the molecular interpretation of Raoult's law?

I was reading about Raoult's law which states:

It states that the partial vapour pressure of each component of an ideal mixture of liquids is equal to the vapour pressure of the pure component multiplied by its mole fraction in the mixture.

Regarding the molecular origin of this law, Peter Atkins in his book writes:

The molecular origin of Raoult's law is the effect of the solute on the entropy of the solution. In the pure solvent, the molecules have a certain disorder & a corresponding entropy; the vapour pressure then represents the tendency of the system & its surroundings to reach a higher entropy.

[...] Because the entropy of the solution is higher than the pure solvent, the solution has a lower tendency to acquire an even high entropy by the solvent vaporising. In other words, the vapour pressure of the solvent in the solution is lower than that of the pure solvent.

I've not understood his explanation especially the bold lines above.

$$\bullet$$ Firstly, how could vapour pressure represent the tendency of the system & its surroundings to reach a higher entropy?

$$\bullet$$ Why does the solution has a lower tendency to acquire an even high entropy by the solvent vaporising ? After-all every system has greater tendency to acquire high entropy, isn't it? Is not this statement contradictory to the Second law by saying lower tendency to acquire high entropy?

Could anyone please explain that to me?

## 3 Answers

I believe what the quote is saying is that once equilibrium is reached between the vapor-liquid mixture, then there will be no tendency to increase entropy. This is because entropy is not the only factor to consider. In any system, one must always consider the interplay between both entropy and energy. For instance, at equilibrium, $$\Delta G=\Delta H-T\Delta S=0$$ which means that at equilibrium, $$\Delta H=T\Delta S$$So, while you raise a good point that in most cases, a system will tend to maximize entropy, in a closed system, one must consider the interplay between entropy and energy which in this case are exactly equal (technically they differ by a scaling factor which is the absolute temperature).

Additionally, what is meant by "vapour pressure represents the tendency of the system and its surroundings to reach a higher entropy" is that one can make an inference about the energetic and entropic properties of a solution based on the vapor pressure of that solution. For instance, if a system has a very low vapor pressure (consider a solid as an extreme example) then the solution is very energetically stable and thus it has a very low tendency to reach higher entropy. But, if you think about acetone, which has a vapor pressure of $30.6 ~\text{kPa}$ at $25^{\circ}~\text{C}$ (ten times that of water at the same temperature), one could say that this then has a strong tendency to increase its own entropy and the entropy of the surroundings.

Peter Atkins in his book writes:

The molecular origin of Raoult's law is the effect of the solute on the entropy of the solution. In the pure solvent, the molecules have a certain disorder & a corresponding entropy; the vapour pressure then represents the tendency of the system & its surroundings to reach a higher entropy.

What is meant by "vapour pressure represents the tendency of the system and its surroundings to reach a higher entropy" is that one can make an inference about the energetic and entropic properties of a solution based on the vapor pressure of that solution

Raoult's law states that the partial vapor pressure of each component of an ideal mixture of liquids is equal to the vapor pressure of the pure component multiplied by its mole fraction in the mixture

p=($\ce{pA^⋆}$) x ( $\ce{xA}$)

In this equation, $\ce{pA^⋆}$ is the vapor pressure of the pure solvent and $\ce{xA}$ is the mole fraction of the solvent. This law allows us to calculate the vapor pressure of a given solution, and shows the influence of all of the molecules in the solution.

Source: Boundless. “Vapor Pressure of Nonelectrolyte Solutions.” Boundless Chemistry. Boundless, 08 Aug. 2016. Retrieved 23 Oct. 2016 from https://www.boundless.com/chemistry/textbooks/boundless-chemistry-textbook/solutions-12/colligative-properties-of-nonelectrolyte-solutions-95/vapor-pressure-of-nonelectrolyte-solutions-406-7509/

I don't think the Atkins book's explanation is correct, but if I explain it locally, it's as follows.

The molecular origin of Raoult's law is the effect of the solute on the entropy of the solution. In the pure solvent, the molecules have a certain disorder & a corresponding entropy; the vapour pressure then represents the tendency of the system & its surroundings to reach a higher entropy.

Vapor pressure means the number of molecules in the vapor phase, and since gas phase generally has much higher entropy than liquid phase, large vapor pressure means that the entropy of the system is that high.

[...] Because the entropy of the solution is higher than the pure solvent, the solution has a lower tendency to acquire an even high entropy by the solvent vaporising. In other words, the vapour pressure of the solvent in the solution is lower than that of the pure solvent.

Assuming that the entropy $$S$$ is the same whether there is a solute or not, the existence of solute creates mixing entropy, which additionally contributes to $$S$$. In order for the $$S$$ value to be the same, the vapor pressure must be reduced, reducing what this contributes to $$S$$.

However, these arguments have one major problem and one minor problem. The state is different between the two systems with only pure solvent and when solute is added. Therefore, the second argument that the entropy $$S$$ should be constant for both cases is not valid.

In addition, according to Raoult's law (1), the more solute you add, the less mole fraction of the solvent and the less vapor pressure you should have.

$$p_A = x_A p_A^* \tag{1}$$ This is qualitatively the same as the conclusion of the above argument, but quantitatively, it is not known why the changed vapor pressure should be multiplied by the mole fraction in the vapor pressure of the pure solvent.

The simple molecular interpretation of Raoult's law is as follows. Raoult's law is a law established in an ideal solution that is mixed with similar chemical species. In other words, both in liquid phase and in vapor phase, solvent A and solute B cannot distinguish each other.

Now suppose that system1 contains 100 solvent A and 10 vapor pressure (A) of these occurs. Let's also suppose that system 2 contains 50 solvent A and 50 solute B. At this time, A and B are not distinguished, so the total vapor pressure (A+B) of system2 will have to be generated by 10.

But how much of this should be the vapor pressure corresponding to "solvent A"? I hope the answer to this question will help you understand why (1) holds.