AP Chemistry: Vapor Pressure and Raoult's Law

Why does adding salt to water make it boil at a higher temperature? The answer lies in a fundamental colligative property: vapor pressure lowering. Understanding how and why a solvent's vapor pressure decreases when a solute is dissolved is crucial not only for your AP Chemistry exam but also for applications ranging from designing car antifreeze to formulating medical IV solutions. This principle, quantitatively described by Raoult's Law, bridges microscopic particle interactions with macroscopic phenomena like boiling point elevation.

Vapor Pressure: The Foundation

Vapor pressure is the pressure exerted by a vapor in equilibrium with its liquid (or solid) phase in a closed system. Imagine a sealed container partially filled with a pure liquid, like water. Some molecules at the surface have enough kinetic energy to escape into the gas phase. As the number of gas molecules increases, some will condense back into the liquid. At equilibrium, the rate of evaporation equals the rate of condensation, and the pressure of the vapor at this point is the equilibrium vapor pressure. A key point: vapor pressure depends only on the substance's identity and the temperature. A volatile substance (one that evaporates easily) has a high vapor pressure.

When you dissolve a non-volatile solute (like table salt or sugar) into a solvent, the vapor pressure of the solution decreases. This occurs because solute particles occupy space at the surface of the liquid. Fewer solvent molecules are on the surface, so fewer have the opportunity to escape into the gas phase. The rate of evaporation slows, while the rate of condensation remains initially unchanged. A new, lower equilibrium vapor pressure is established. This vapor pressure lowering is a colligative property, meaning it depends only on the number of solute particles dissolved, not their chemical identity.

Introducing Raoult's Law for Ideal Solutions

For an ideal solution, the quantitative relationship is given by Raoult's Law. It states that the partial vapor pressure of a solvent component ($P_A$) in an ideal solution is equal to the vapor pressure of the pure solvent ($P_A^{\circ }$) multiplied by its mole fraction (${\chi }_A$) in the solution.

$$P_A={\chi }_A{P}_A^{\circ }$$

For a solution with a single non-volatile solute, the vapor pressure lowering ($\Delta P$) is:

$$\Delta P=P_A^{\circ }-P_A=P_A^{\circ }-{\chi }_A{P}_A^{\circ }=P_A^{\circ }(1-{\chi }_A)$$

Since $(1-{\chi }_A)$ is the mole fraction of the solute (${\chi }_B$), this simplifies to:

$$\Delta P={\chi }_B{P}_A^{\circ }$$

This equation confirms the colligative nature: the vapor pressure lowering is directly proportional to the mole fraction of the solute particles. The more solute particles you add, the greater the effect.

Applying Raoult's Law: A Worked Example

Let's calculate the vapor pressure lowering when 45.0 g of glucose ($C_6{H}_{12}{O}_6$, molar mass = 180.16 g/mol) is dissolved in 250.0 g of water at 25°C. The vapor pressure of pure water at 25°C is 23.8 mmHg.

1. Find moles of solute and solvent.

• Moles of glucose: $45.0\text{ g}÷180.16\text{ g/mol}=0.250\text{ mol}$
• Moles of water: $250.0\text{ g}÷18.02\text{ g/mol}=13.9\text{ mol}$

1. Calculate the mole fraction of water (${\chi }_{water}$).

• Total moles = $0.250+13.9=14.15\text{ mol}$
• ${\chi }_{water}=13.9\text{ mol}/14.15\text{ mol}=0.982$

1. Apply Raoult's Law.

• $P_{solution}={\chi }_{water}{P}_{water}^{\circ }=(0.982)(23.8\text{ mmHg})$
• $P_{solution}=23.4\text{ mmHg}$

1. Find the vapor pressure lowering.

• $\Delta P=23.8\text{ mmHg}-23.4\text{ mmHg}=0.4\text{ mmHg}$

This solution now has a vapor pressure of 23.4 mmHg, slightly lower than pure water.

Deviations from Raoult's Law: Non-Ideal Solutions

Raoult's Law holds true for ideal solutions, where solute-solvent interactions are very similar in strength to the pure solvent-solvent and pure solute-solute interactions. In reality, many solutions show deviations.

• Negative Deviations: The observed vapor pressure of the solution is lower than predicted by Raoult's Law. This happens when the solute-solvent interactions (e.g., hydrogen bonding, ion-dipole forces) are stronger than the pure interactions. An example is acetone and chloroform, which can form hydrogen bonds. This stronger attraction makes it harder for solvent molecules to escape, further lowering the vapor pressure.
• Positive Deviations: The observed vapor pressure is higher than predicted. This occurs when solute-solvent interactions are weaker. The components "prefer their own company," so it's easier for molecules to escape into the vapor phase. An example is ethanol and hexane; the polar ethanol has much stronger self-interactions (hydrogen bonding) than its interactions with nonpolar hexane.

Connecting to Boiling Point Elevation

Boiling occurs when a liquid's vapor pressure equals the external atmospheric pressure. Because a solution has a lower vapor pressure than the pure solvent, you must heat it to a higher temperature to get its vapor pressure to match atmospheric pressure. This is boiling point elevation.

The quantitative link between vapor pressure lowering and boiling point elevation is provided by the Clausius-Clapeyron equation, which describes how vapor pressure changes with temperature. From this relationship, we derive the boiling point elevation formula:

$$\Delta T_b=i{K}_{b}m$$

Where $\Delta T_b$ is the boiling point elevation, $i$ is the van't Hoff factor (number of particles per formula unit), $K_b$ is the molal boiling point elevation constant (unique to each solvent), and $m$ is the molality of the solution. This equation is a direct, practical consequence of Raoult's Law and vapor pressure lowering. For example, adding ethylene glycol (antifreeze) to your car's radiator lowers the water's vapor pressure and significantly elevates its boiling point, preventing overheating.

Common Pitfalls

1. Using the Wrong Concentration Unit: Raoult's Law uses mole fraction, not molarity or molality. Confusing these will lead to incorrect calculations. Always convert masses to moles first.
2. Ignoring the Solute's Volatility: The standard Raoult's Law equation $P_A={\chi }_A{P}_A^{\circ }$ applies to the solvent in a solution with a non-volatile solute. If the solute is also volatile (e.g., in a mixture of two liquids), you must calculate the total vapor pressure as the sum of the partial pressures of both components: $P_{total}={\chi }_A{P}_A^{\circ }+{\chi }_B{P}_B^{\circ }$.
3. Forgetting the van't Hoff Factor ($i$) for Electrolytes: For ionic compounds like NaCl, which dissociate into multiple particles (Na⁺ and Cl⁻, so $i\approx 2$), the effective mole fraction of solute particles is larger. You must account for this in both Raoult's Law (${\chi }_B$ becomes $i\times {\chi }_B$) and the boiling point elevation equation. Using the formula unit moles instead of particle moles is a frequent error.
4. Assuming All Solutions are Ideal: Automatically applying Raoult's Law without considering the nature of the intermolecular forces can lead to incorrect predictions. Always consider whether strong specific interactions (like hydrogen bonding between different molecules) might cause negative or positive deviations.

Summary

• Vapor pressure lowering is a colligative property: the vapor pressure of a solvent decreases when a non-volatile solute is added because solute particles reduce the surface area available for solvent evaporation.
• Raoult's Law quantifies this for ideal solutions: $P_{solution}={\chi }_{solvent}{P}_{solvent}^{\circ }$. The vapor pressure lowering is $\Delta P={\chi }_{solute}{P}_{solvent}^{\circ }$.
• Non-ideal solutions exhibit deviations from Raoult's Law: negative deviations arise from strong solute-solvent attractions, while positive deviations arise from weak attractions.
• This lowering of vapor pressure directly causes boiling point elevation ($\Delta T_b=i{K}_{b}m$). To boil, the solution must be heated to a higher temperature so its vapor pressure can reach atmospheric pressure.
• For electrolytes, remember to use the van't Hoff factor ($i$) to account for dissociation into multiple particles, which amplifies all colligative effects.