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Ch.19 - Chemical Thermodynamics
All textbooksBrown 14th EditionCh.19 - Chemical ThermodynamicsProblem 103
Chapter 19, Problem 103

Most liquids follow Trouton’s rule (see Exercise 19.93), which states that the molar entropy of vaporization is approximately 88 J/mol⋅K. The normal boiling points and enthalpies of vaporization of several organic liquids are as follows: (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 B, check the accuracy of your conclusion.
Verified step by step guidance
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Step 1: Understand Trouton's Rule, which states that the molar entropy of vaporization (ΔS_vap) for most liquids is approximately 88 J/mol⋅K. This rule is based on the observation that the entropy change when a liquid vaporizes is relatively constant for many substances.
Step 2: Consider the role of intermolecular forces in determining the enthalpy of vaporization (ΔH_vap). Stronger intermolecular forces require more energy to overcome, leading to higher ΔH_vap values. This can affect whether a substance follows Trouton's Rule.
Step 3: Identify exceptions to Trouton's Rule by considering substances with strong hydrogen bonding or other significant intermolecular forces. These substances may have higher ΔS_vap values due to the additional energy required to overcome these forces.
Step 4: Evaluate whether water is an exception to Trouton's Rule. Water has strong hydrogen bonds, which could lead to deviations from the typical ΔS_vap value of 88 J/mol⋅K. Use Appendix B to find the ΔH_vap and boiling point of water.
Step 5: Calculate the molar entropy of vaporization for water using the formula ΔS_vap = ΔH_vap / T_b, where T_b is the boiling point in Kelvin. Compare this calculated value to 88 J/mol⋅K to determine if water follows Trouton's Rule.

Key Concepts

Here are the essential concepts you must grasp in order to answer the question correctly.

Trouton's Rule

Trouton's Rule states that the molar entropy of vaporization for most liquids is approximately 88 J/mol·K. This rule is based on the observation that the entropy change associated with the phase transition from liquid to vapor is relatively consistent across different substances, reflecting the degree of disorder in the vapor phase compared to the liquid phase.
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Intermolecular Forces

Intermolecular forces are the attractive forces between molecules that influence their physical properties, including boiling points and enthalpies of vaporization. These forces include hydrogen bonding, dipole-dipole interactions, and London dispersion forces. Variations in these forces can lead to exceptions to Trouton's Rule, as stronger intermolecular attractions typically result in higher enthalpies of vaporization and lower entropies.
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Water's Unique Properties

Water exhibits unique properties due to its strong hydrogen bonding, which significantly affects its boiling point and enthalpy of vaporization. Unlike most liquids, water has a higher molar entropy of vaporization than predicted by Trouton's Rule, making it an exception. This is due to the extensive hydrogen bonding in liquid water, which requires more energy to break during vaporization, leading to a lower entropy change.
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Related Practice
Textbook Question

The conversion of natural gas, which is mostly methane, into products that contain two or more carbon atoms, such as ethane (C2H6), is a very important industrial chemical process. In principle, methane can be converted into ethane and hydrogen: 2 CH4(g) → C2H6(g) + H2(g) In practice, this reaction is carried out in the presence of oxygen: 2 CH4(g) + 12 O2(g) → C2H6(g) + H2O(g) (b) Is the difference in ΔG° for the two reactions due primarily to the enthalpy term (ΔH) or the entropy term (-TΔS)?

Textbook Question

The conversion of natural gas, which is mostly methane, into products that contain two or more carbon atoms, such as ethane (C2H6), is a very important industrial chemical process. In principle, methane can be converted into ethane and hydrogen: 2 CH4(g) → C2H6(g) + H2(g) In practice, this reaction is carried out in the presence of oxygen: 2 CH4(g) + 1/2 O2(g) → C2H6(g) + H2O(g) (c) Explain how the preceding reactions are an example of driving a nonspontaneous reaction, as discussed in the 'Chemistry and Life' box in Section 19.7.

Textbook Question

The potassium-ion concentration in blood plasma is about 5.0⨉10-3 M, whereas the concentration in muscle-cell fluid is much greater (0.15 M ). The plasma and intracellular fluid are separated by the cell membrane, which we assume is permeable only to K+. (a) What is ΔG for the transfer of 1 mol of K+ from blood plasma to the cellular fluid at body temperature 37 °C? (b) What is the minimum amount of work that must be used to transfer this K+?

Textbook Question
In chemical kinetics, the entropy of activation is the entropy change for the process in which the reactants reach the activated complex. Predict whether the entropy of activation for a bimolecular process is usually positive or negative.
Textbook Question

At what temperatures is the following reaction, the reduction of magnetite by graphite to elemental iron, spontaneous? Fe3O4(s) + 2 C(s, graphite) → 2 CO2(g) + 3 Fe(s)