In a previous section, we learned that a lower vapor pressure is due to stronger intermolecular forces of attraction and results in a higher boiling point. A higher vapor pressure means the intermolecular forces of attraction are weaker, and the boiling point is lower.
Below is a plot of vapor pressure, in mmHg, vs. Temperature in degrees Celsius for diethyl ether, ethanol, and water.
It can be seen from the plot that diethyl ether has the highest vapor pressure while water has the lowest vapor pressure. Ethanol is in between diethyl ether and water. Looking at the structures of diethyl ether, ethanol, and water, it is obvious that water has the stronger intermolecular forces mainly due to hydrogen bonding. As for diethyl ether, there is no hydrogen bonding, only London dispersion forces and weak dipole-dipole forces of attraction.
Included in the plot, are the normal boiling points of the three substances. The normal boiling point is defined as the boiling point at 1 atm (760 mmHg) of pressure. For diethyl ether, ethanol, and water the normal boiling points are 34.5°C, 78.5°C, and 100.0°C, respectively. Water has the highest boiling point while diethyl ether has the lowest boiling point. Recall, boiling points are dependent on the strengths of the interparticle attractions. The weaker the attractions, the lower the boiling point. The stronger the attractive forces, the higher the boiling point.
If ln(P) is plotted against 1/T, Kelvin temperature, a linear relationship is observed between ln(P) and 1/T as shown in the plot below.
The equation of a line is y = mx + b where m is equal to the slope.
The equation is:
\(\displaystyle ln(P)=\frac{ΔH_{vap}}{R}\times 1/T + C\) Clausius-Clapeyron Equation
This is called the Clausius-Clapeyron equation. The temperature must be in Kelvin, and C is a constant equal to the y intercept. The slope in this case is equal to \(\displaystyle \frac{ΔH_{vap}}{R}\) where R = 8.314 J/(mol⋅K).
Below I derive the 2-point form of the Clausius-Clapeyron equation.
We have P1 and P2 at two different temperatures.
\(\displaystyle ln(P_1)=-\frac{ΔH_{vap}}{R}\times 1/T_1\;+\;C\;\text{and}\;ln(P_2)=-\frac{ΔH_{vap}}{R}\times 1/T_2\; +\;C\)
The constant, C, is the same at any two pressures and temperatures. Solve the above equations for C.
\(\displaystyle C=ln(P_1) +\frac{ΔH_{vap}}{R}\times 1/T_1 \text{and}\;C=ln(P_2)+\frac{ΔH_{vap}}{R}\times 1/T_2\)
Next, we set both equations equal to one another and we collect the pressure terms on one side of the equation.
\(\displaystyle ln(P_1) +\frac{ΔH_{vap}}{R}\times 1/T_1 =ln(P_2)+\frac{ΔH_{vap}}{R}\times 1/T_2\)
\(\displaystyle ln(P_1)-ln(P_2)=\frac{ΔH_{vap}}{R}\biggl(\frac{1}{T_2}-\frac{1}{T_1}\biggr)\)
If you know the pressure values at two different temperatures, you can calculate ΔHvap.
Watch the following video:
Worksheet: The Clausius Clapeyron Equation
Exercises
Exercise 1. What is the enthalpy of vaporization of a compound that has a vapor pressure of 154 torr at -20.0°C K and 0.350 torr at -82.2°C?
Check Answer/Solution to Exercise 1
Exercise 2. A liquid has an enthalpy of vaporization of 45.6 kJ/mol. Its vapor pressure at 78.0 °C is 224 mmHg. What is the normal boiling point of the liquid?
Check Answer/Solution to Exercise 2
Exercise 3. A liquid has a vapor pressure of 14.5 mmHg at 25.0°C and 150.2 mmHg at 150.0°C. What is the enthalpy of vaporization for the liquid?
Check Answer/Solution to Exercise 3
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