Question

Diethyl ether \(\left(\mathrm{CH}_{3} \mathrm{CH}_{2} \mathrm{OCH}_{2} \mathrm{CH}_{3}\right)\) was one of the first chemicals used as an anesthetic. At \(34.6^{\circ} \mathrm{C}\) , diethyl ether has a vapor pressure of 760 . torr, and at \(17.9^{\circ} \mathrm{C},\) it has a vapor pressure of 400 . torr. What is the \(\Delta H\) of vaporization for diethyl ether?

Short answer

The enthalpy of vaporization for diethyl ether is approximately \(26,089\,\mathrm{J/mol}\).

Step-by-step solution

Write down the Clausius-Clapeyron equation

The Clausius-Clapeyron equation is given by: \[\ln{\frac{P_2}{P_1}} = \frac{\Delta H_\text{vap}}{R} \left(\frac{1}{T_1} - \frac{1}{T_2}\right)\] where \(P_1\) and \(P_2\) are the vapor pressures at temperatures \(T_1\) and \(T_2\), respectively, and \(R\) is the gas constant.

Convert the given temperature values to kelvin

To use the Clausius-Clapeyron equation, we need to convert the given temperature values from Celsius to Kelvin. \[T_1 = 17.9^{\circ}C + 273.15 = 291.05 K\] \[T_2 = 34.6^{\circ}C + 273.15 = 307.75 K\]

Plug in the given values and solve for \(\Delta H_\text{vap}\)

Using the given vapor pressures and converted temperature values, we can plug them into the Clausius-Clapeyron equation: \[\ln{\frac{760\,\mathrm{torr}}{400\,\mathrm{torr}}} = \frac{\Delta H_\text{vap}}{8.3145\, \mathrm{J/mol\cdot K}} \left(\frac{1}{291.05\,\mathrm{K}} - \frac{1}{307.75\,\mathrm{K}}\right)\] Now, solve for \(\Delta H_\text{vap}\): \[\Delta H_\text{vap} = 8.3145\, \mathrm{J/mol\cdot K} \cdot \left(\frac{1}{291.05\,\mathrm{K}} - \frac{1}{307.75\,\mathrm{K}}\right)^{-1} \cdot \ln{\frac{760\,\mathrm{torr}}{400\,\mathrm{torr}}}\] \[\Delta H_\text{vap} \approx 26,089\,\mathrm{J/mol}\] So, the enthalpy of vaporization for diethyl ether is approximately \(26,089\,\mathrm{J/mol}\).

Key concepts

Clausius-Clapeyron Equation

The Clausius-Clapeyron equation is a powerful tool used to relate the vapor pressure of a substance to its temperature, specifically by providing insight into the enthalpy of vaporization (\(\Delta H_\text{vap}\)). This equation is essential for understanding how a liquid transits to a gas (vaporization) and quantifies the energy required for this phase change. The equation is written as:

• \(\ln{\frac{P_2}{P_1}} = \frac{\Delta H_\text{vap}}{R} \left(\frac{1}{T_1} - \frac{1}{T_2}\right)\)

In the equation:

• \(P_1\) and \(P_2\) are the vapor pressures at temperatures \(T_1\) and \(T_2\) respectively.
• \(R\) is the universal gas constant, typically \(8.3145\, \mathrm{J/mol\cdot K}\).
• \(T_1\) and \(T_2\) must be in Kelvin for proper computations.

The Clausius-Clapeyron equation is integral in calculating how vapor pressure changes with temperature, therefore allowing the prediction of a liquid's behavior under different conditions.

Vapor Pressure

Vapor pressure is a fundamental concept in the study of liquids and their transition to gaseous states. It refers to the pressure exerted by a vapor in equilibrium with its solid or liquid phase at a given temperature.
For any liquid, the molecules escape into the vapor phase until equilibrium is achieved, meaning the rate of evaporation equals the rate of condensation.

• Substances with higher vapor pressures at a given temperature are more volatile, meaning they evaporate more easily.
• Vapor pressure depends on the nature of the liquid and the temperature but not on the amount of the liquid.

In the context of the Clausius-Clapeyron equation, knowing the vapor pressures at two different temperatures allows us to determine the enthalpy of vaporization, which describes the energy needed to vaporize a unit amount of the substance.

Temperature Conversion

Temperature conversion is a crucial step when using scientific equations, as different equations demand temperatures in specific units. The most common conversion in thermodynamics is from Celsius to Kelvin.
This can be achieved using the following formula:

• \(T(\mathrm{K}) = T(\degree\mathrm{C}) + 273.15\)

Kelvin is the standard temperature unit utilized in scientific equations because it begins at absolute zero, offering a more natural scale for scientific calculations.

Converting temperatures ensures consistency in calculations and prevents errors like using Celsius in equations designed for Kelvin. For the problem involving diethyl ether, it was necessary to convert the given Celsius temperatures to Kelvin before applying the Clausius-Clapeyron equation.

Diethyl Ether

Diethyl ether (\(\mathrm{CH}_{3}\mathrm{CH}_{2}\mathrm{OCH}_{2}\mathrm{CH}_{3}\)) is an important compound historically used as an anesthetic. Known for its high volatility and flammability, it quickly evaporates, which is why it's often used when early boiling is desired.
Understanding its vapor pressure at various temperatures sheds light on its energetic properties and behaviors.

• Diethyl ether has a relatively low boiling point due to its lower enthalpy of vaporization compared to water.
• It has been used in various chemical processes, not only for anesthesia but also as a solvent in laboratories.

In our exercise, the concept of calculating its enthalpy of vaporization involves understanding how much energy is needed to transform it from the liquid to gaseous phase - crucial for its applications in both historical and modern contexts.