I know that addition of a non-volatile solute in a pure solvent effects the rate of evaporation as the number of molecules leaving the liquid bulk per unit time decreases. However when the rate of evaporation decreases immediately , rate of condensation should automatically increase , this is because, if we consider a liquid 'X' in equilibrium as:

$$\ce{X(l) <=> X(g)}$$

Now when some amount of solute is added in the pure $\ce{X}$ , this dynamic equilibrium is obviously disturbed. $\mathrm{(Rate)_{evap}} \downarrow$ so to attain equilibrium $\mathrm{(Rate)_{cond}} \uparrow$ as reaction shifts backward. also when equilibrium is attained the number of molecules entering a liquid has decreased , so at equilibrium rate of condensation has decreased. So, overall it has been affected.

But when it browsed various sites , all agreed that it is unchanged by solute. So, where's the problem? Isn't my logic correct?


1 Answer 1


When a solvent is at isobaric conditions in equilibrium with its gaseous phase, the rate of evaporation and condensation are equal.

If we add a solute to the solvent, the activity, chemical potential, saturated vapor pressure and rate of evaporation decreases. As the rate of condensation remains the same, the net condensation increases from its equilibrium zero value.

If the system is isolated, then while there is ongoing condensation, temperature raises, saturated vapor pressure increases (due both increasing temperature and diluting by the condensate), while pressure of vapor decreases, until there is reached the new equilibrium with higher temperature but lower gas(vapor) pressure. This presure is now equal to saturated vapor pressure for the final solution and temperature.

If the system is also isothermic, then the heat from ongoing condensation is dissipated. The saturated vapor pressure and the gas (vapor) pressure at the new equilibrium are lower, compared to pure solvent, according to the Raoult law and eventual its deviations.

  • $\begingroup$ you wrote ".....saturated vapor pressure increases while pressure decreases.." also "......The saturated vapor pressure and the gas pressure...". What is this extra pressure and what here is "gas pressure"? $\endgroup$
    Dec 4, 2022 at 14:18
  • $\begingroup$ Vapor pressure = total pressure > saturated ( equilibrium) vapor pressure at given T and solute concentration. therefore net condensation. // small correction - I have not included the dilution of the solution by condensating solvent - addition inserted.. $\endgroup$
    – Poutnik
    Dec 4, 2022 at 14:20
  • $\begingroup$ that means you are saying that besides the vapor which are in equilibrium with the liquid at a given T , there exists some more vapor above the liquid surface which are never in the equilibrium with the liquid surface? $\endgroup$
    Dec 4, 2022 at 14:40
  • $\begingroup$ There are not 2 kinds of vapor. The vapor either is in equilibrium with liquid, or it is not. It was with pure liquid,but when solute was involved, the same vapor, all of it, suddenly is not. $\endgroup$
    – Poutnik
    Dec 4, 2022 at 17:04
  • $\begingroup$ [Isothermal] addition of a solute lowers the chemical potential [mole fraction etc.] of the solvent. The chemical potential of the vapor[the vapor pressure] is now higher than the solution, so vapor will condense, slowly lowering the condensation rate until equilibrium is reached. The released heat of condensation will balance everything out. Since a solution of a non volatile solute universally has a lower vapor pressure that seems to sort it all out. I am not so sure about the higher T lower VP. If adiabatic T and VP should rise to the original VP. Tough experiment to run! $\endgroup$
    – jimchmst
    Dec 5, 2022 at 22:50

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