Question

Given the data, calculate \(\Delta S_{\text {vap }}\) for each of the first four liquids. \(\left(\Delta S_{\text {vap }}=\Delta H_{\text {vap }} / T,\right.\) where \(T\) is in \(\left.K\right)\) $$ \begin{array}{llcc} \text { Compound } & \text { Name } & \text { BP }\left({ }^{\circ} \mathrm{C}\right) & \Delta H_{\text {vap }}(\mathrm{kJ} / \mathrm{mol} \text { ) at BP } \\ \mathrm{C}_{4} \mathrm{H}_{10} \mathrm{O} & \text { Diethyl ether } & 34.6 & 26.5 \\ \hline \mathrm{C}_{3} \mathrm{H}_{6} \mathrm{O} & \text { Acetone } & 56.1 & 29.1 \\ \hline \mathrm{C}_{6} \mathrm{H}_{6} \mathrm{O} & \text { Benzene } & 79.8 & 30.8 \\ \hline \mathrm{CHCl}_{3} & \text { Chloroform } & 60.8 & 29.4 \\ \hline \mathrm{C}_{2} \mathrm{H}_{5} \mathrm{OH} & \text { Ethanol } & 77.8 & 38.6 \\ \hline \mathrm{H}_{2} \mathrm{O} & \text { Water } & 100 & 40.7 \\ \hline \end{array} $$ All four values should be close to each other. Predict whether the last two liquids in the table have \(\Delta S_{\text {vap }}\) in this same range. If not, predict whether it is larger or smaller and explain. Verify your prediction.

Short answer

\(\Delta S_{\text{vap}}\) values for diethyl ether, acetone, benzene, and chloroform are within a similar range due to comparatively weaker intermolecular forces than water and ethanol, which have stronger hydrogen bonds, thus resulting in higher \(\Delta S_{\text{vap}}\) values.

Step-by-step solution

Convert Boiling Points from Celsius to Kelvin

To calculate the entropy of vaporization, \(\Delta S_{\text{vap}}\), the temperature must be in Kelvin. To convert from Celsius to Kelvin, add 273.15 to each boiling point (BP) given in Celsius: \[ T(K) = BP(°C) + 273.15 \]

Calculate \(\Delta S_{\text{vap}}\) for Diethyl Ether

Using the formula \(\Delta S_{\text{vap}} = \frac{\Delta H_{\text{vap}}}{T}\), substitute the values for Diethyl Ether: \[ \Delta S_{\text{vap}} = \frac{26.5 \text{ kJ/mol}}{34.6 + 273.15} \]

Calculate \(\Delta S_{\text{vap}}\) for Acetone

Use the same formula for Acetone: \[ \Delta S_{\text{vap}} = \frac{29.1 \text{ kJ/mol}}{56.1 + 273.15} \]

Calculate \(\Delta S_{\text{vap}}\) for Benzene

Apply the formula for Benzene: \[ \Delta S_{\text{vap}} = \frac{30.8 \text{ kJ/mol}}{79.8 + 273.15} \]

Calculate \(\Delta S_{\text{vap}}\) for Chloroform

Finally, calculate it for Chloroform: \[ \Delta S_{\text{vap}} = \frac{29.4 \text{ kJ/mol}}{60.8 + 273.15} \]

Compare the \(\Delta S_{\text{vap}}\) Values

After finding \(\Delta S_{\text{vap}}\) for the first four liquids, compare these values and note that they should be close to each other. This will set a basis for predicting the \(\Delta S_{\text{vap}}\) for the last two liquids and then calculate and verify these predictions.

Predict the \(\Delta S_{\text{vap}}\) for the Last Two Liquids

Using the concept that \(\Delta S_{\text{vap}}\) values are related to the intermolecular forces present in the liquid, we can predict whether the values for ethanol and water will be larger or smaller than the ones calculated above. Given that ethanol and water have strong intermolecular hydrogen bonding, it's likely that their \(\Delta S_{\text{vap}}\) values will be higher.

Calculate \(\Delta S_{\text{vap}}\) for Ethanol

Calculate \(\Delta S_{\text{vap}}\) for Ethanol to verify the prediction: \[ \Delta S_{\text{vap}} = \frac{38.6 \text{ kJ/mol}}{77.8 + 273.15} \]

Calculate \(\Delta S_{\text{vap}}\) for Water

Calculate \(\Delta S_{\text{vap}}\) for Water to verify the prediction: \[ \Delta S_{\text{vap}} = \frac{40.7 \text{ kJ/mol}}{100 + 273.15} \]

Final Analysis

With the values of \(\Delta S_{\text{vap}}\) for all substances at hand, compare them to the predictions made earlier. Check if ethanol and water have higher \(\Delta S_{\text{vap}}\) values as anticipated due to their strong hydrogen bonding compared to the first four compounds.

Key concepts

Enthalpy of Vaporization

Enthalpy of vaporization, also known as the heat of vaporization, is the energy required for a substance to transform from a liquid state to a gaseous state at constant pressure. It is a crucial concept in thermodynamics and provides insights into the energy needed to overcome intermolecular forces within a liquid.

When a substance undergoes vaporization, heat is absorbed to break the bonds or forces holding the molecules together in the liquid phase. This absorbed heat is the enthalpy of vaporization and is typically expressed in kilojoules per mole (kJ/mol). The higher the enthalpy of vaporization, the stronger the intermolecular forces that must be overcome.

For instance, in the provided exercise, substances like diethyl ether and acetone have lower enthalpy of vaporization values, indicating weaker intermolecular forces compared to substances like water, which has a relatively higher value, reflecting its strong hydrogen bonds. Thus, the enthalpy of vaporization serves as a key indicator of the intermolecular interactions present in a liquid.

Conversion from Celsius to Kelvin

Temperature scales play an essential role in scientific calculations, especially in chemistry and physics. The Celsius and Kelvin scales are two temperature scales used widely. The Kelvin scale is the SI unit for temperature and is favored in scientific calculations because it is an absolute temperature scale, starting at absolute zero.

To convert a temperature from Celsius to Kelvin, which is a necessary step in many thermodynamic equations, one simply adds 273.15 to the Celsius temperature: \[ T(K) = T(°C) + 273.15 \]
For example, the boiling point of acetone at 56.1°C is equivalent to 329.25 K. This conversion ensures that all temperature-dependent calculations in thermodynamics are consistent and accurate, such as calculating the entropy of vaporization in the aforementioned exercise.

Intermolecular Forces

Intermolecular forces are the forces of attraction or repulsion which act between neighboring particles (molecules, atoms, ions). These forces are responsible for many physical properties of substances, such as boiling points, melting points, and vapor pressures.

There are several types of intermolecular forces, with varying strengths:

• Van der Waals Forces (including London dispersion forces) - the weakest interactions due to temporary dipoles.
• Dipole-Dipole Interactions - occur between two polar molecules.
• Hydrogen Bonds - unusually strong dipole-dipole interactions that occur when hydrogen is bonded to a highly electronegative atom like oxygen, nitrogen, or fluorine.

All these forces significantly influence a substance's enthalpy of vaporization. Stronger intermolecular forces result in a higher enthalpy of vaporization as more energy is required to separate the molecules.

Hydrogen Bonding

Hydrogen bonding is a special type of dipole-dipole interaction, and it's one of the strongest intermolecular forces. This bond forms when a hydrogen atom, which is covalently bonded to one electronegative atom (like oxygen, nitrogen, or fluorine), is attracted to another electronegative atom in a different molecule.

Water and alcohols, such as ethanol, exhibit significant hydrogen bonding due to the presence of hydrogen atoms attached to oxygen. These strong bonds lead to higher boiling points and higher enthalpies of vaporization, as observed in the solution to the original problem. This characteristic also impacts other physical properties such as increased surface tension and viscosity.

Understanding hydrogen bonding is not only crucial for interpreting the high enthalpy of vaporization of substances but also for comprehending the structure and properties of many biological molecules, including DNA and proteins.