Lowering of Vapor Pressure

The phenomenon of lowering of vapor pressure is an important colligative property observed in solutions. It is based on the fact that when a non-volatile solute is added to a solvent, the vapor pressure of the solvent above the solution decreases.

Raoult's Law

To understand the lowering of vapor pressure, we must first understand Raoult's Law. Raoult's Law states that the partial vapor pressure of each component in an ideal solution is directly proportional to its mole fraction. Mathematically, it is expressed as:

$$ P_A = X_A \cdot P_A^0 $$

Where:

• $P_A$ is the partial vapor pressure of component A in the solution.
• $X_A$ is the mole fraction of component A in the solution.
• $P_A^0$ is the vapor pressure of pure component A.

For a solution containing a non-volatile solute, the vapor pressure of the solvent is given by:

$$ P_{solvent} = X_{solvent} \cdot P_{solvent}^0 $$

Since the solute is non-volatile, it does not contribute to the vapor pressure, and thus the total vapor pressure of the solution is solely due to the solvent.

Lowering of Vapor Pressure

The lowering of vapor pressure is defined as the difference between the vapor pressure of the pure solvent ($P_{solvent}^0$) and the vapor pressure of the solvent in the solution ($P_{solvent}$):

$$ \Delta P = P_{solvent}^0 - P_{solvent} $$

Substituting the expression for $P_{solvent}$ from Raoult's Law, we get:

$$ \Delta P = P_{solvent}^0 - (X_{solvent} \cdot P_{solvent}^0) $$

Since the mole fraction of the solvent ($X_{solvent}$) plus the mole fraction of the solute ($X_{solute}$) equals 1, we can write:

$$ X_{solvent} = 1 - X_{solute} $$

Substituting this into the equation for $\Delta P$, we get:

$$ \Delta P = P_{solvent}^0 - ((1 - X_{solute}) \cdot P_{solvent}^0) $$ $$ \Delta P = P_{solvent}^0 \cdot X_{solute} $$

This shows that the lowering of vapor pressure is directly proportional to the mole fraction of the solute.

Table of Differences and Important Points

Property	Pure Solvent	Solution with Non-Volatile Solute
Vapor Pressure	$P_{solvent}^0$	$P_{solvent} = X_{solvent} \cdot P_{solvent}^0$
Mole Fraction	$X_{solvent} = 1$	$X_{solvent} < 1$
Lowering of Vapor Pressure	$\Delta P = 0$	$\Delta P = P_{solvent}^0 \cdot X_{solute}$

Examples

Example 1: Calculating Lowering of Vapor Pressure

Suppose we have a solution of a non-volatile solute in water at 25°C. The vapor pressure of pure water at this temperature is 23.8 mmHg. If the mole fraction of water in the solution is 0.9, what is the lowering of vapor pressure?

Using the formula for lowering of vapor pressure:

$$ \Delta P = P_{solvent}^0 \cdot X_{solute} $$

Since $X_{solvent} = 0.9$, then $X_{solute} = 1 - X_{solvent} = 0.1$. Therefore:

$$ \Delta P = 23.8 \text{ mmHg} \cdot 0.1 $$ $$ \Delta P = 2.38 \text{ mmHg} $$

The vapor pressure of the solution is thus 23.8 mmHg - 2.38 mmHg = 21.42 mmHg.

Example 2: Effect of Adding More Solute

If more non-volatile solute is added to the solution, the mole fraction of the solvent decreases, and thus the vapor pressure of the solution decreases further. This is because the presence of solute particles hinders the escape of solvent molecules into the vapor phase, effectively lowering the vapor pressure.

Conclusion

Lowering of vapor pressure is a colligative property that depends on the quantity of solute particles rather than their identity. It is a direct consequence of the presence of non-volatile solute molecules in a solvent, which leads to a decrease in the solvent's vapor pressure. Understanding this concept is crucial for explaining various phenomena such as boiling point elevation and freezing point depression, which are also colligative properties related to vapor pressure.