Chemistry Vapour Pressure of Liquid Solutions and Roult's Law
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Topics Covered :

● Vapour Pressure of Liquid Solutions
● Vapour Pressure of Liquid- Liquid Solutions
● Roult's Law
● Raoult’s Law as a special case of Henry’s Law
● Vapour Pressure of Solutions of Solids in Liquids

Vapour Pressure of Liquid Solutions :

=> Liquid solutions are formed when solvent is a liquid.

=> The solute can be a gas, a liquid or a solid.

=> Such solutions may contain one or more volatile components.

=> Generally, the liquid solvent is volatile.

=> The solute may or may not be volatile.

=> Only binary solutions shall be discussed, that is, the solutions containing two components i.e., the solutions of :

(i) liquids in liquids and (ii) solids in liquids.

color{purple}♣ color{Violet} " Just for Curious"
The vapour phase is richer in more volatile component than the less volatile called as Kanawaloff's rule

Vapour Pressure of Liquid- Liquid Solutions :

-> Let us consider a binary solution of two volatile liquids and denote the two components ascolor{red}( 1 and 2)
.

-> When both liquids are taken in a closed vessel, both the components evaporate and an equilibrium is established between vapour phase and the liquid phase.

-> Let the total vapour pressure at equilibrium be p_text(total) and p_1 and p_2 be the partial vapour pressures of the two components 1 and 2 respectively.

-> And partial pressures are related to the mole fractions x_1 and x_2 of the two components 1 and 2 respectively.

Roult's Law :

=> The French chemist, Francois Marte Raoult (1886) gave the quantitative relationship between mole fraction and partial pressure of a component in a solution.

=> This relationship is known as the Raoult’s law.

color{green}("Statement") : For a solution of volatile liquids, the partial vapour pressure of each component in the solution is directly proportional to its mole fraction.

Now, for component 1

color{red}(p_1 ∝ x_1)

and color{red}(p_1 = p_1^0 x_1) .................(1)

where p_1^0 is the vapour pressure of pure component 1 at the same temperature.

Similarly, for component 2

color{red}(p_2 = p_2^0 x_2) ................(2)

where p_2^0 represents the vapour pressure of the pure component 2.

According to Dalton’s law of partial pressures, the total pressure (p_text(total)) over the solution phase in the container will be the sum of the partial pressures of the components of the solution and is given as :

color{red}(p_text(total) = p_1 + p_2) ................(3)

Substituting the values of p_1 and p_2, we get

color{red}(p_text(total) = x_1 p_1^0 + x_2 p_2^0)

color{red}(= (1 – x_2) p_1^0 + x_2 p_2^0) ......................(4)

color{red}(= p_1^0 + (p_2^0 – p_1^0) x_2) .....................(5)

Following conclusions are drawn from equation (5) :

(i) Total vapour pressure over the solution can be related to the mole fraction of any one component.

(ii) Total vapour pressure over the solution varies linearly with the mole fraction of component 2.

(iii) Depending on the vapour pressures of the pure components 1 and 2, total vapour pressure over the solution decreases or increases with the increase of the mole fraction of component 1.

=> A plot of p_1 or p_2 versus the mole fractions x_1 and x_2 for a solution gives a linear plot as shown in Fig.

=> Lines (I and II) pass through the points and respectively when x_1 and x_2 equal unity.

=> Similarly the plot (line III) of p_text(total) versus x_2 is also linear.

=> The minimum value of p_text(total) is p_1^0 and the maximum value is p_2^0, assuming that component 1 is less volatile than component 2, i.e., p_1^0 < p_2^0.

=> The composition of vapour phase in equilibrium with the solution is determined by the partial pressures of the components.

=> If y_1 and y_2 are the mole fractions of the components 1 and 2 respectively in the vapour phase then, using Dalton’s law of partial pressures :

color{red}(p_1 = y_1 p_text(total)) ..................(6)

color{red}(p_2 = y_2 p_text(total)) .....................(7)

In general

color{red}(p_i = y_i p_text(total))
Q 2916701670

Vapour pressure of chloroform (CHCl_3) and dichloromethane (CH_2Cl_2) at 298 K are 200 mm Hg and 415 mm Hg respectively. (i) Calculate the vapour pressure of the solution prepared by mixing 25.5 g of CHCl_3 and 40 g of CH_2Cl_2 at 298 K and, (ii) mole fractions of each component in vapour phase.
NCERT
Solution:

(i) Molar mass of CH_2Cl_2 = 12 xx 1 + 1 xx 2 + 35.5 xx 2 = 85 g mol^–1

Molar mass of CHCl_3 = 12 xx 1 + 1 xx 1 + 35.5 xx 3 = 119.5 g mol^-1

Moles of CH_2Cl_2 = (40 g)/(85 g mol^-1) = 0.47 mol

Moles of CHCl_3 = (25.5 g)/ (119.5 g mol^-1) = 0.213 mol

Total number of moles = 0.47 + 0.213 = 0.683 mol

x_(CH_2CL_2) = (0.47 mol)/( 0.683 mol) = 0.688

x_(CHCL_3) = 1.00 – 0.688 = 0.312

Using equation (2.16),

P_text(total) = P_1 ^0 + (p_2^0 – p_1^0) x_2 = 200 + (415 – 200) xx 0.688

= 200 + 147.9 = 347.9 mm Hg

(ii) Using the relation (2.17), y _i = p_i/(p_text(total)), we can calculate the mole fraction of the components in gas phase (yi_).

P_(CH_2Cl_2) = 0.688 xx 415 mm Hg = 285.5 mm Hg

P_(CHCl_3) = 0.312 xx 200 mm Hg = 62.4 mm Hg

y_(CH_2Cl_2) = 285.5 mm Hg//347.9 mm Hg = 0.82

y_(CHCl_3) = 62.4 mm Hg//347.9 mm Hg = 0.18

Raoult’s Law as a special case of Henry’s Law :

=> According to Raoult’s law, the vapour pressure of a volatile component in a given solution is given by color{red}(p_i = x_i p_i^0).

=> For a solution of a gas in a liquid, one of the components is so volatile that it exists as a gas and its solubility is given by Henry’s law which states that

color{red}(p = K_H x)

=> On comparing the equations for Raoult’s law and Henry’s law, it is seen that the partial pressure of the volatile component or gas is directly proportional to its mole fraction in solution.

=> Only the proportionality constant K_H differs from p_1^0.

Thus, Raoult’s law becomes a special case of Henry’s law in which K_H becomes equal to p_1^0.

Vapour Pressure of Solutions of Solids in Liquids :

=> Another class of solutions consists of solids dissolved in liquid. For example, sodium chloride, glucose, urea and cane sugar in water and iodine and sulphur dissolved in carbon disulphide.

=> Some physical properties of these solutions are quite different from those of pure solvents. For example, vapour pressure.

=> We know that under equilibrium conditions the pressure exerted by the vapours of the liquid over the liquid phase is called vapour pressure .

=> In a pure liquid the entire surface is occupied by the molecules of the liquid.

=> If a non-volatile solute is added to a solvent to give a solution, the vapour pressure of the solution is decreased as the vapour pressure is solely from the solvent alone.

=> In the solution, at surface both solute and solvent molecules are there; thereby the fraction of the surface covered by the solvent molecules gets reduced and the number of solvent molecules escaping from the surface get reduced. Hence, the vapour pressure is also reduced.

=> The decrease in the vapour pressure of solvent depends on the quantity of non-volatile solute present in the solution, irrespective of its nature.

=> For example, decrease in the vapour pressure of water by adding 1.0 mol of sucrose to one kg of water is nearly similar to that produced by adding 1.0 mol of urea to the same quantity of water at the same temperature.

=> Raoult’s law in its general form is stated as, for any solution the partial vapour pressure of each volatile component in the
solution is directly proportional to its mole fraction.

color{green}("Explanation") : In a binary solution, let us denote the solvent by 1 and solute by 2. When the solute is non-volatile, only the solvent molecules are present in vapour phase and contribute to vapour pressure. Let p_1 be the vapour pressure of the solvent, x_1 be its mole fraction, p_1^0 be its vapour pressure in the pure state. Then according to Raoult’s law

color{red}(p_1 ∝ x_1)

and color{red}(p_1 = x_1 p_1^0) .....................(8)

The proportionality constant is equal to the vapour pressure of pure solvent, p_1^0.

A plot between the vapour pressure and the mole fraction of the solvent is linear.