# Is the mole fraction used in Raoult's law calculated initially or at equilibrium?

Raoult’s law states that:

The partial pressure of any volatile constituent of a solution at a constant temperature is equal to the vapour pressure of pure constituent multiplied by the mole fraction of that constituent in the solution

My question:

Is the mole fraction of the constituent of the solution to be considered at the time when equilibrium between vapours dissolving into the solution and rate of evaporation has been achieved or at the beginning when the solution is just formed and evaporation has just begun change the mole fraction of the constituent which we have just calculated?

Also, experimentally, is there much difference in the mole fractions at equilibrium and at the beginning when there is apparently no vapour pressure (for zero vapour pressure in the beginning, I think we can just close the lid of the container)?

• Jun 16 '20 at 14:52

Application of Raoult's law assumes an equilibrium has been established between gas and liquid phases (otherwise the standard equations of thermodynamics, of which it forms a part, would not apply). For a mixture of components one can write a set of equations $$\gamma_i p = \chi_i p_i^\circ$$ that can be solved for a self-consistent composition. For instance, for two components one can write two equations in two unknowns:
\begin{align} \gamma_1 p &= \chi_1 p_1^\circ \\(1-\gamma_1) p &= (1-\chi_1) p_2^\circ \end{align}
\begin{align} \chi_1 &= \frac{p-p_2^\circ}{p_1^\circ-p_2^\circ} \\ \gamma_1 &= \left(\frac{p-p_2^\circ}{p_1^\circ-p_2^\circ}\right)\left(\frac{p_1^\circ}{p}\right) \end{align}