Problem 94 - Chemistry An Atoms First Approach

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?

Step-by-Step Solution

Answer

The enthalpy of vaporization for diethyl ether is approximately \(\Delta H_{vap} = 26.33\:\mathrm{kJ\:mol^{-1}}\).

Step 1: Write the Clausius-Clapeyron Equation

The Clausius-Clapeyron Equation relates vapor pressure, temperature, and enthalpy of vaporization as follows: \[ \ln \left(\frac{P_2}{P_1}\right) = -\frac{\Delta H_{vap}}{R} \left(\frac{1}{T_2} - \frac{1}{T_1}\right) \] Where \(P_1\) and \(P_2\) are the vapor pressures at temperatures \(T_1\) and \(T_2\) in Kelvin, respectively. \(\Delta H_{vap}\) is the enthalpy of vaporization, and R is the gas constant with the appropriate units (\(\mathrm{J\:mol^{-1}K^{-1}}\) in this case).

Step 2: Convert temperatures to Kelvin

We need to convert the given temperatures from Celsius to Kelvin to use the Clausius-Clapeyron equation. We can do this using the following formula: \[T(K) = T(^\circ C) + 273.15\] So, \[T_1 = 34.6 + 273.15 = 307.75\: K\] \[T_2 = 17.9 + 273.15 = 291.05\: K\]

Step 3: Substitute values into the Clausius-Clapeyron Equation and solve for \(\Delta H_{vap}\)

We have all of the necessary values, and we can now substitute into the Clausius-Clapeyron equation to solve for \(\Delta H_{vap}\): \[ \ln\left(\frac{760}{400}\right) = -\frac{\Delta H_{vap}}{8.314} \left(\frac{1}{291.05} - \frac{1}{307.75}\right) \] Now, solve for \(\Delta H_{vap}\): \[ \Delta H_{vap} = -8.314 \cdot \ln\left(\frac{760}{400}\right) \cdot \frac{1}{\left(\frac{1}{291.05} - \frac{1}{307.75}\right)} \] Calculate the values: \[ \Delta H_{vap} = 26,328.55 \: \mathrm{J\:mol^{-1}} \] or \[ \Delta H_{vap} = 26.33 \: \mathrm{kJ\:mol^{-1}} \] So, the enthalpy of vaporization for diethyl ether is approximately \(\:\Delta H_{vap} = 26.33\:\mathrm{kJ\:mol^{-1}}\).

Key Concepts

Clausius-Clapeyron Equation

The Clausius-Clapeyron Equation is a fundamental principle in physical chemistry that explains how the vapor pressure of a substance changes with temperature, given the substance's enthalpy of vaporization.

Considered a cornerstone for understanding phase transitions, this equation is particularly useful for predicting how much energy is needed to convert a liquid into a gas (vaporization) at a constant pressure.

In practical terms, the equation helps us calculate the enthalpy of vaporization, \(\Delta H_{vap}\), by measuring the vapor pressures at two different temperatures. It's a logarithmic relationship, signifying that small changes in temperature can lead to exponential changes in vapor pressure—critical for various industrial applications involving distillation or refrigeration.

Vapor Pressure

Vapor pressure is an intrinsic property of each liquid that represents the tendency of its particles to escape into the vapor phase.

It's determined by the kinetic energy of the liquid particles, which in turn depends on the temperature: as the temperature increases, more particles have enough energy to escape, and the vapor pressure rises.

Understanding vapor pressure is crucial not just in high-level chemistry, but also for common tasks, such as checking the pressure of car tires, where temperature changes can significantly affect pressure readings.

Diethyl Ether

Diethyl ether, \(\mathrm{CH}_3\mathrm{CH}_2\mathrm{OCH}_2\mathrm{CH}_3\), holds an important place in both chemical history and present-day practices. As one of the earliest anesthetics, it marked a revolution in surgical medicine.

In terms of physical properties, diethyl ether is known for its relatively high volatility, which is the ability of a substance to vaporize. This characteristic implies that diethyl ether has a significant vapor pressure even at low temperatures, making it an excellent case study for examining phase transitions.

Temperature Conversion

Temperature conversion is crucial in scientific calculations, where consistent units are necessary to ensure accuracy.

Converting Celsius to Kelvin is straightforward: simply add 273.15 to the Celsius temperature to obtain the Kelvin equivalent. This conversion is key when applying the Clausius-Clapeyron equation since it requires temperature measures in an absolute scale where zero represents the absence of thermal energy.

Practically, this means that even small miscalculations in temperature can lead to large errors in calculating properties such as enthalpy of vaporization, highlighting the importance of exact temperature conversion.