Question

Most liquids follow Trouton's rule (see Exercise 19.93 ), which states that the molar entropy of vaporization is approximately \(88 \pm 5 \mathrm{~J} / \mathrm{mol}-\mathrm{K}\). The normal boiling points and enthalpies of vaporization of several organic liquids are as follows: \begin{tabular}{lcc} \hline & Normal Boiling & \\ Substance & Point \(\left({ }^{\circ} \mathrm{C}\right)\) & \(\Delta H_{\text {vap }}(\mathrm{k} / / \mathrm{mol})\) \\ \hline Acetone, \(\left(\mathrm{CH}_{3}\right)_{2} \mathrm{CO}\) & 56.1 & 29.1 \\\ Dimethyl ether, \(\left(\mathrm{CH}_{3}\right)_{2} \mathrm{O}\) & -24.8 & 21.5 \\\ Ethanol, \(\mathrm{C}_{2} \mathrm{H}_{5} \mathrm{OH}\) & 78.4 & 38.6 \\ Octane, \(\mathrm{C}_{\mathrm{s}} \mathrm{H}_{18}\) & 125.6 & 34.4 \\ Pyridine, \(\mathrm{C}_{5} \mathrm{H}_{\mathrm{S}} \mathrm{N}\) & 115.3 & 35.1 \\\ \hline \end{tabular} (a) Calculate \(\Delta S_{\text {vap }}\) for each of the liquids. Do all the liquids obey Trouton's rule? (b) With reference to intermolecular forces (Section 11.2), can you explain any exceptions to the rule? (c) Would you expect water to obey Trouton's rule? By using data in Appendix \(\mathrm{B}\), check the accuracy of your conclusion. (d) Chlorobenzene \(\left(\mathrm{C}_{6} \mathrm{H}_{5} \mathrm{Cl}\right)\) boils at \(131.8^{\circ} \mathrm{C}\). Use Trouton's rule to estimate \(\Delta H_{\text {vap }}\) for this substance.

Short answer

In this exercise, the molar entropy of vaporization (\(\Delta S_{vap}\)) was calculated for various organic liquids. Comparing the calculated values to Trouton's rule, it was found that not all the liquids obey the rule, with ethanol being an exception due to stronger hydrogen bonding. Water also doesn't follow Trouton's rule because of its strong hydrogen bonding network, resulting in a higher \(\Delta S_{vap}\) value. For chlorobenzene, assuming it follows Trouton's rule, its estimated enthalpy of vaporization is approximately 35.63 kJ/mol.

Step-by-step solution

Part (a): Calculate \(\Delta S_{vap}\) for each liquid

To calculate the molar entropy of vaporization \(\Delta S_{vap}\) for each liquid, we'll use the formula: \[\Delta S_{vap} = \frac{\Delta H_{vap}}{T_b}\] where \(\Delta H_{vap}\) is the enthalpy of vaporization, and \(T_b\) is the normal boiling point in Kelvin. First, let's convert the boiling points from Celsius to Kelvin: \[T_b(K) = T_b(^{\circ}C) + 273.15\] For each liquid, calculate the entropy of vaporization and check if it follows Trouton's rule: 1. Acetone: \[T_b = 56.1 + 273.15 = 329.25 \mathrm{~K}\] \[\Delta S_{\text{vap}} = \frac{29.1 * 10^3}{329.25} = 88.34 \mathrm{~J/mol~K}\] 2. Dimethyl Ether: \[T_b = -24.8 + 273.15 = 248.35 \mathrm{~K}\] \[\Delta S_{\text{vap}} = \frac{21.5 * 10^3}{248.35} = 86.54 \mathrm{~J/mol~K}\] 3. Ethanol: \[T_b = 78.4 + 273.15 = 351.55 \mathrm{~K}\] \[\Delta S_{\text{vap}} = \frac{38.6 * 10^3}{351.55} = 109.69 \mathrm{~J/mol~K}\] 4. Octane: \[T_b = 125.6 + 273.15 = 398.75 \mathrm{~K}\] \[\Delta S_{\text{vap}} = \frac{34.4 * 10^3}{398.75} = 86.26 \mathrm{~J/mol~K}\] 5. Pyridine: \[T_b = 115.3 + 273.15 = 388.45 \mathrm{~K}\] \[\Delta S_{\text{vap}} = \frac{35.1 * 10^3}{388.45} = 90.33 \mathrm{~J/mol~K}\] Comparing the calculated \(\Delta S_{\text{vap}}\) values to Trouton's rule limits (88 ± 5 J/mol K), we can observe that not all the liquids obey the rule. Ethanol is the exception with a higher entropy of vaporization than the expected range.

Part (b): Intermolecular forces and exceptions to Trouton's rule

Ethanol doesn't obey Trouton's rule due to the presence of hydrogen bonding, which is a stronger intermolecular force than van der Waals forces experienced by other organic liquids. This hydrogen bonding increases the energy required for vaporization, leading to a higher molar entropy of vaporization.

Part (c): Water and Trouton's rule

We would not expect water to obey Trouton's rule since it has a very strong hydrogen bonding network, which would result in a higher molar entropy of vaporization than predicted by the rule. Using data from Appendix B: \[T_b = 100^{\circ} C \Rightarrow 373.15 \mathrm{~K}\] \[\Delta H_{\text{vap}} = 40.7 \mathrm{~kJ/mol}\] Calculate \(\Delta S_{\text{vap}}\) for water: \[\Delta S_{\text{vap}} = \frac{40.7 * 10^3}{373.15} = 109.07 \mathrm{~J/mol~K}\] As expected, the \(\Delta S_{\text{vap}}\) value for water (109.07 J/mol K) doesn't obey Trouton's rule due to its strong hydrogen bonding network.

Part (d): Estimate \(\Delta H_{vap}\) for Chlorobenzene using Trouton's rule

According to Trouton's rule, we can estimate the molar entropy of vaporization \(\Delta S_{\text{vap}}\) to be 88 J/mol K (assuming it follows the rule). We are given the boiling point of chlorobenzene: \[T_b = 131.8^{\circ} C \Rightarrow 404.95 \mathrm{~K}\] Using the estimated \(\Delta S_{\text{vap}}\) and the boiling point, we can calculate \(\Delta H_{\text{vap}}\): \[\Delta H_{\text{vap}} = \Delta S_{\text{vap}} * T_b\] \[\Delta H_{\text{vap}} = 88 * 404.95 \approx 35.63 \mathrm{~kJ/mol}\] So, the estimated enthalpy of vaporization for chlorobenzene is approximately 35.63 kJ/mol using Trouton's rule.

Key concepts

Molar Entropy of Vaporization

The molar entropy of vaporization, denoted as \( \Delta S_{\text{vap}} \), is a measure of the randomness or disorder added to a system when a unit amount of substance transitions from the liquid phase to the vapor phase. This concept is crucial because it helps us understand how energy, in the form of heat, affects this transition.
In simpler terms, it tells us how much disorder increases when a liquid becomes a gas.

• Entropy is a thermodynamic property that quantifies the degree of disorder in a system.
• Vaporization is the process of a liquid becoming a gas.

According to Trouton's rule, for many liquids, the molar entropy of vaporization is approximately \( 88 \pm 5 \, \text{J/mol K} \). This rule suggests that the energy needed to transform a liquid into a vapor is consistent across different substances, due to the similar increase in disorder.
However, exceptions to this rule exist, primarily due to variations in intermolecular forces.

Intermolecular Forces

Intermolecular forces are the attractions that occur between molecules, holding them together. These forces play a significant role in determining a substance's physical properties, such as boiling point and vaporization behavior. They are the reason why different substances require varying amounts of energy to vaporize.

There are several types of intermolecular forces, including:

• London dispersion forces, which are the weakest and present in all molecules.
• Dipole-dipole interactions, which occur between polar molecules.
• Hydrogen bonding, a particularly strong type of dipole-dipole interaction occurring when hydrogen is bonded to electronegative atoms like oxygen, nitrogen, or fluorine.

These forces influence how much energy is needed for a molecule to overcome them and go from a liquid state to a gaseous state. For instance, a liquid with strong hydrogen bonds requires more energy to vaporize than one with only weak London dispersion forces.

Enthalpy of Vaporization

Enthalpy of vaporization, represented as \( \Delta H_{\text{vap}} \), is the heat required to vaporize one mole of a liquid at its boiling point under constant pressure. Essentially, it is the energy needed to break the intermolecular forces holding the liquid molecules together.
In a molecular sense, when a liquid is heated to its boiling point, its molecules gain enough energy to overcome the attractions (or intermolecular forces) between them.

• A higher \( \Delta H_{\text{vap}} \) indicates stronger intermolecular forces requiring more energy to vaporize the liquid.
• Substances like water, with significant hydrogen bonding, typically have higher enthalpies of vaporization.

Calculating \( \Delta H_{\text{vap}} \) provides insights into the strength of the intermolecular forces present within a substance. This calculation is crucial for understanding a substance's phase transitions and energy requirements during processes like boiling and evaporation.

Hydrogen Bonding

Hydrogen bonding is a form of strong dipole-dipole interaction typically occurring in molecules where hydrogen is directly bonded to highly electronegative atoms such as oxygen, nitrogen, or fluorine. These bonds are significant for several reasons, particularly in relation to Trouton's rule and its exceptions.
Owing to its strength, hydrogen bonding greatly increases a liquid's enthalpy of vaporization.

• Substances like water and ethanol exhibit hydrogen bonding, which makes them exceptions to Trouton’s rule.
• Because hydrogen bonds are stronger than other types of dipole interactions, they increase the energy required for vaporization and subsequently influence the boiling point.

Understanding hydrogen bonding helps in explaining why some substances have unusual thermodynamic properties, such as elevated boiling points and enthalpies of vaporization compared to what Trouton’s rule anticipates.

Boiling Points

The boiling point of a liquid is the temperature at which its vapor pressure equals the external pressure surrounding the liquid. At this point, the liquid changes to vapor. It is a direct indicator of how strong the intermolecular forces in a liquid are, as stronger forces will result in higher boiling points.
Different liquids have varying boiling points because of the diverse types of intermolecular forces at play.

• A liquid with only weak London dispersion forces might boil at a lower temperature, while one with strong hydrogen bonds, like water, will have a higher boiling point.
• The boiling point also depends on external pressure, which is why water boils at lower temperatures at higher altitudes.

Boiling points are crucial in practical applications, like distillation, where they are used to separate substances based on their volatility. These points also relate to Trouton's rule, as a substance’s boiling point is used in the calculation of its molar entropy of vaporization.