# Vapour Pressure of A Solution

I study in class 12, and have a doubt in vapour pressure that my school teacher and my friends are not able to clarify.

My doubt is the reason why the partial vapour pressure of each individual component of a solution is not equal to their respective standard vapour pressures($$P°$$)?

My reasoning is that consider you have a pure liquid A in a closed container, the vapour pressure of it shall be equal to $$P°_A$$. Now when you add another volatile liquid B, a solution should be formed; however, the equilibrium of the reaction

$$\ce{A_{liq} <=> A_{vap}}$$ should not be affected since the concentration of reactants and products remains the same since B is different from A. Eventually, the reaction of B must also reach equilibrium, independently, and thus the partial pressure of B must be equal to $$P°_B$$ .

However, this does not seem to be the case. Could someone please help in clarifying my doubt through the viewpoint of equilibrium(Mathematically if possible) and not the standard explanation that the surface of the solvent get's blocked by the particles of the solute, hence the vapour pressure reduces.

Regards, Arun

• I assume you are trying to apply Dalton's law of partial pressures on liquids? Also have you seen Raoult's law? In ideal gases, the molecules are independent of each other. They do not "know" the presence of each other. The situation for liquid is very different, things are so close to each other that you cannot neglect intermolecular interaction. – M. Farooq Apr 9 at 16:09
• @M.Farooq Thanks for your response sir...however, I am afraid that I still have not entirely understood your response. Are you trying to say that the adhesive forces between molecules of A and B in the solution will be different compared to the cohesive forces between pure A molecules and pure B molecules, which would have existed if they had been present separately, which is in turn affecting the individual partial pressures of each of the components of the solution? I would greatly appreciate a response to this... – Arun Apr 10 at 14:56
• They keyword you should be searching is the molecular origin of the Raoult's law. See, by definition of the Raoult's law (from Wiki) "It states that the partial 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." What you suggest is exactly $against$ this law. In your proposition, you are neglecting the mole fraction of component A and B in the liquid state. Even in the gases, the total pressure is their partial pressure times their mole fraction, so why change for liquids. – M. Farooq Apr 10 at 15:08
• @M.Farooq Yes sir, I know the definition of Raoult's law, however, I am unable to understand the reasoning behind it for we have not been taught the detailed derivation, just the fact that the partial pressure is proportional to the mole fraction . I would appreciate if you could read through my reasoning in my original question posted(not the comments) and highlight where I am right and where I am wrong(The main part being whether the equilibrium of reaction involving ' A' would be affected). Thank You – Arun Apr 10 at 15:14
• @M.Farooq Also to add on, I do agree that the partial pressure of a gas depends on it's mole fraction in vapour phase. However, I am unable to wrap my head around how it depends on the mole fraction of the component in the solution. – Arun Apr 10 at 15:22

## 1 Answer

This is where your logic fails:

since the concentration of reactants and products remains the same since B is different from A

As soon as you mix two components you change the concentration of both, you effectively dilute all components.

Otherwise your intuition is correct. The gases can often be assumed to act independently. This is true if the solution is ideal (that is, assuming interactions between all molecules in solution are roughly the same). Then each component equilibrates independently with its vapor. However, the vapor pressure of each component is in general not the same as it would be if it were pure. Instead, for an ideal solution the partial pressure of each component follows Raoult's law $$p_i=p^*_i\chi_i$$, where $$p^*_i$$ is the partial pressure of the pure component and $$\chi_i$$ is the mole fraction in solution.

The total pressure above the solution is then the sum of the partial pressures of the components (following Dalton's law, where $$\chi_i^g$$ is the mole fraction in the gas phase):

\begin{align} p &= \sum_i p_i = \sum_i \chi_i^g p \\ &= \sum_i \chi_i p^*_i \end{align}