Why does the vapor pressure of a solution decrease with the addition of a non volatile solute

For solvent at temperature $T$, surface area $A$ and having the molar fraction $x$, the rate of molecules leaving liquid is:

$$\frac {\mathrm{d}n}{\mathrm{d}t}=-C_1 \cdot A \cdot x \cdot f(T)$$

Lower molar fraction means lower surface density and lower evaporation rate per surface area.

Factor $C_1$ includes pure solvent molecular surface density and $f(T)$ includes probability of a molecule leaving liquid.

For vapour of the partial pressure $p$, the rate of molecules coming to liquid from vapour is:

$$\frac {\mathrm{d}n}{\mathrm{d}t}=C_2 \cdot A \cdot p \cdot g(T) $$

The $g(T)$ involves temperature dependent relation of pressure and collision rate with liquid surface. Higher temperature means higher kinetic energy and less collisions needed for the same pressure.

For equilibrium, the net rate must be zero, therefore:

$$p= \frac{C_1 \cdot x \cdot f(T)}{C_2 \cdot g(T)}$$

Therefore the equilibrium pressure is proportional to the molar fraction.

It is quite analogical to the steady state of water level(=pressure), when it is simultaneously being filled up(=evaporation) and drained(=condensation). When you dilute a non-volatile solute, you decrease the filling up (= surface concentration of molecules is lower), so draining (=condensation), brings the water level(=pressure) to lower value to match draining rate with the filling.

Text Solution

Solution : Evaporation always occure from the surface. The particles of the non-volatile solute occupy some surface area of the solvent. Therefore, extent of eveporation decreses and the vapour presure of the solution decreases on adding a non-volatile solute to the solvent.