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Ch.18 - Thermodynamics: Entropy, Free Energy & Equilibrium
Chapter 18, Problem 126b

Trouton's rule says that the ratio of the molar heat of vaporization of a liquid to its normal boiling point (in kelvin) is approximately the same for all liquids: ∆Hvap/Tbp ≈ 88 J/(K*mol) (b) Explain why liquids tend to have the same value of ∆Hvap/Tbp.

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Trouton's rule is an empirical observation that the entropy of vaporization (\( \Delta S_{vap} \)) for many liquids is approximately the same when they are at their normal boiling point.
The entropy of vaporization is given by the formula \( \Delta S_{vap} = \frac{\Delta H_{vap}}{T_{bp}} \), where \( \Delta H_{vap} \) is the molar heat of vaporization and \( T_{bp} \) is the boiling point in Kelvin.
According to Trouton's rule, \( \Delta S_{vap} \) is approximately 88 J/(K*mol) for many liquids, indicating that the change in entropy when a liquid vaporizes is similar across different substances.
This similarity arises because the process of vaporization involves overcoming intermolecular forces to transition from a liquid to a gas, which requires a similar amount of energy per mole for many substances.
The rule holds best for non-polar and non-hydrogen-bonded liquids, as these substances have similar intermolecular forces, leading to similar \( \Delta H_{vap} \) values when normalized by their boiling points.

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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 ratio of the molar heat of vaporization (∆H<sub>vap</sub>) of a liquid to its boiling point (T<sub>bp</sub>) in Kelvin is approximately constant for many liquids, typically around 88 J/(K·mol). This empirical observation suggests that despite differences in molecular structure, many liquids exhibit similar intermolecular forces, leading to comparable energy requirements for vaporization relative to their boiling points.
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Molar Heat of Vaporization

The molar heat of vaporization (∆H<sub>vap</sub>) is the amount of energy required to convert one mole of a liquid into vapor at its boiling point. This energy is necessary to overcome the intermolecular forces holding the liquid molecules together. A higher ∆H<sub>vap</sub> indicates stronger intermolecular forces, while a lower value suggests weaker forces, influencing the boiling point and vaporization behavior of the liquid.
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Boiling Point

The boiling point (T<sub>bp</sub>) of a liquid is the temperature at which its vapor pressure equals the external pressure, allowing the liquid to transition into the gas phase. It is influenced by the strength of intermolecular forces; stronger forces typically result in higher boiling points. The relationship between boiling point and heat of vaporization is crucial for understanding why many liquids exhibit similar ratios of ∆H<sub>vap</sub>/T<sub>bp</sub> as described by Trouton's Rule.
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Related Practice
Textbook Question

Consider the Haber synthesis of gaseous NH3 (∆H°f = -46.1 kJ/mol; ∆G°f = -16.5 kJ/mol: (d) What are the equilibrium constants Kp and Kc for the reaction at 350 K? Assume that ∆H° and ∆S° are independent of temperature.

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Textbook Question
Is it possible for a reaction to be nonspontaneous yet exo-thermic? Explain.
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Textbook Question

Trouton's rule says that the ratio of the molar heat of vaporization of a liquid to its normal boiling point (in kelvin) is approximately the same for all liquids: ∆Hvap/Tbp ≈ 88 J/(K*mol) (a) Check the reliability of Trouton's rule for the liquids listed in the following table.

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Textbook Question
The normal boiling point of bromine is 58.8 °C, and the standard entropies of the liquid and vapor are S°[Br2(l) = 152.2 J/(K*mol); S°[Br2(g) = 245.4 J/(K*mol). At what temperature does bromine have a vapor pressure of 227 mmHg?
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Textbook Question
Tell whether reactions with the following values of ΔH and ΔS are spontaneous or nonspontaneous and whether they are exothermic or endothermic. (a) ΔH = - 48 kJ; ΔS = + 135 J>K at 400 K (b) ΔH = - 48 kJ; ΔS = - 135 J>K at 400 K (c) ΔH = + 48 kJ; ΔS = + 135 J>K at 400 K (d) ΔH = + 48 kJ; ΔS = - 135 J>K at 400 K
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Textbook Question
The following reaction, sometimes used in the laboratory to generate small quantities of oxygen gas, has ∆G° = -224.4 kJ/mol at 25°C:

Use the following additional data at 25 °C to calculate the standard molar entropy S° of O2 at 25°C: ∆H°f(KClO3) = -397.7 kJ/mol, ∆H°f(KCl) = -436.5 kJ/mol, S°(KClO3) = 143.1 J/(K*mol), and S°(KCl) = 82.6 J/(K*mol).
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