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In attachment I have tried to derive Raoult's law by using the concept of equilibrium constant. There is nothing mentioned in my text book about how this equation comes about, so I decided to do it myself. I may like to know whether there is some mistake in what I have done.

  • $\begingroup$ No, it looks fine. $\endgroup$
    – Buck Thorn
    May 25, 2019 at 8:04
  • $\begingroup$ If you want the derivation to be more rigorous, it can be done by using the concept of Chemical Potential. You can look into any standard text book of Undergrad Chemistry to find the derivation or you can search in the web too. $\endgroup$
    – Soumik Das
    May 25, 2019 at 9:41
  • $\begingroup$ Hey, we were having a discussion about this in class, and I went hunting on this website for questions and found this. Could you explain why concentration cannot be ignored in the second case? $\endgroup$ Dec 17, 2021 at 3:42

2 Answers 2


The partial pressure of a component in a mixture is always less than that of the pure component at the same temperature. (Calculating the chemical potential shows that if it were not the case the mixture would be unstable compared to the pure liquid). The question then becomes what is the partial pressure of the component A above a mixture $p_A$ related to that above the pure liquid $p_A^*$. Raoult's law states that the ratio is equal to the mole fraction $x_A$ of A, $p_A/p_A^*=x_A$.

Physically, this is justified by considering that the chance of finding a molecule of A in the vapour, relative to the chance for the pure liquid, is equal to its mole fraction. This is reasonable if dealing with ideal mixtures and assuming that the vapour behaves as an ideal gas. Thus there is no derivation as such, we proceed directly from the physical argument to the law.


As in the case of Henry's law, it is possible to use Raoult's law to derive a more formal relation between an equilibrium constant and the standard Gibbs free energy for evaporation of solvent in a solution containing solvent at a mole fraction $\chi$.

At equilibrium between the vapor and solution, the chemical potential of the solvent in the two phases is identical:

$$\begin{align} \mu(l)&=\mu(g)\end{align}$$

For the gas phase (assuming ideal behavior) we can write that

$$\begin{align} \mu(g)&=\mu^\circ(g)+RT\log\left(\frac{p_i}{p^\circ}\right)\end{align}$$

Raoult's law states that the partial pressure and mole fraction $\chi$ of solvent in the solution are related as

$$p_i=p_i^\bullet \chi$$

where $p_i^\bullet$ is the vapour pressure of the pure solvent at the same temperature (compare this to Henry's law). As before - and as you show - this equation suggests the definition of an equilibrium constant expression. Introducing this last expression into that for the chemical potential results in

$$\begin{align} \mu(l)&=\mu^\circ(g)+RT\log\left(\frac{p_i^\bullet \chi}{p^\circ }\right)\end{align}$$


$$\begin{align} \mu(l)^\circ&=\mu^\circ(g)+RT\log\left(\frac{p_i^\bullet}{p^\circ }\right)\end{align}$$

we can see that

$$\Delta G^\circ_m = \mu^\circ(g)-\mu^\circ(l)=-RT\log\left(\frac{p_i^\bullet}{p^\circ }\right)=-RT\log\left(K_{eq}\right)$$

As in the derivation involving Henry's law, this expression relates a standard Gibbs free energy change and an associated equilibrium constant (for the equilibrium $\ce{solvent <=> vapor}$) provided we define the equilibrium constant as

$$K_{eq} = \frac{p_i^\bullet}{p^\circ} = \frac{(p_i/p^\circ)}{\chi}$$


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