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Most liquids follow Trouton's rule, which states that the molar entropy of vaporization lies in the range of \(88 \pm 5 \mathrm{~J} / \mathrm{mol}-\mathrm{K}\). The normal boiling points and enthalpies of vaporization of several organic liquids are as follows: $$ \begin{array}{lrl} \hline \text { Substance } & \begin{array}{l} \text { Normal Boiling } \\ \text { Point }\left({ }^{\circ} \mathrm{C}\right) \end{array} & \begin{array}{l} \Delta H_{\text {vap }} \\ \text { (kJ/mol) } \end{array} \\ \hline \text { Acetone, }\left(\mathrm{CH}_{3}\right)_{2} \mathrm{CO} & 56.1 & 29.1 \\ \text { Dimethyl ether, }\left(\mathrm{CH}_{3}\right)_{2} \mathrm{O} & -24.8 & 21.5 \\ \text { Ethanol } \mathrm{C}_{2} \mathrm{H}_{5} \mathrm{OH} & 78.4 & 38.6 \\ \text { Octane, } \mathrm{C}_{8} \mathrm{H}_{18} & 125.6 & 34.4 \\ \text { Pyridine, } \mathrm{C}_{5} \mathrm{H}_{5} \mathrm{~N} & 115.3 & 35.1 \\\ \hline \end{array} $$ (a) Calculate \(\Delta \mathrm{S}_{\mathrm{vap}}\) for each of the liquids. Do all of 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

Expert verified
The calculated molar entropy of vaporization for Acetone, Dimethyl ether, Ethanol, Octane, and Pyridine are \(88.35\), \(86.55\), \(109.72\), \(86.31\), and \(90.36 \mathrm{J/mol~K}\) respectively. Only ethanol does not obey Trouton's rule due to its hydrogen bonding capability. Water also does not obey Trouton's rule, as it has molar entropy of vaporization \(109.08 \mathrm{J/mol~K}\). The estimated enthalpy of vaporization for chlorobenzene is \(35.64\) kJ/mol using Trouton's rule.

Step by step solution

01

Calculate the molar entropy of vaporization for each liquid

Recall that the molar entropy of vaporization can be calculated using the formula: \[\Delta S_{vap}=\frac{\Delta H_{vap}}{T_{b}} \] Where \(T_b\) is the normal boiling point in Kelvin and \(\Delta H_{vap}\) is the enthalpy of vaporization. First, convert the boiling point from Celsius to Kelvin by adding 273.15. Then, calculate the molar entropy of vaporization by dividing the enthalpy of vaporization by the boiling point in Kelvin. Acetone: \(T_{b} = 56.1 + 273.15 = 329.25 K\) \(\Delta S_{vap}=\frac{29.1 \times 10^3}{329.25} = 88.35 \mathrm{J/mol~K}\) Similarly, perform calculations for other liquids: Dimethyl ether: \(T_{b} = -24.8 + 273.15 = 248.35 K\) \(\Delta S_{vap}=\frac{21.5 \times 10^3}{248.35} = 86.55 \mathrm{J/mol~K}\) Ethanol: \(T_{b} = 78.4 + 273.15 = 351.55 K\) \(\Delta S_{vap}=\frac{38.6 \times 10^3}{351.55} = 109.72 \mathrm{J/mol~K}\) Octane: \(T_{b} = 125.6 + 273.15 = 398.75 K\) \(\Delta S_{vap}=\frac{34.4 \times 10^3}{398.75} = 86.31 \mathrm{J/mol~K}\) Pyridine: \(T_{b} = 115.3 + 273.15 = 388.45 K\) \(\Delta S_{vap}=\frac{35.1 \times 10^3}{388.45} = 90.36 \mathrm{J/mol~K}\)
02

Check if liquids obey Trouton's rule

Now we need to check if the calculated molar entropy of vaporization lies within the range of \(88\pm5\) J/mol-K. Acetone: \(88.35 \rightarrow\) Yes Dimethyl ether: \(86.55 \rightarrow\) Yes Ethanol: \(109.72 \rightarrow\) No Octane: \(86.31 \rightarrow\) Yes Pyridine: \(90.36 \rightarrow\) Yes All the liquids, except ethanol, obey Trouton's rule.
03

Explain exceptions

Ethanol is exceptional because of its capability to form hydrogen bonds, making it harder to break the intermolecular forces during vaporization. This additional energy requirement is responsible for the higher molar entropy of vaporization compared to the range dictated by Trouton's rule.
04

Predict if water obeys Trouton's rule

Since water can also form hydrogen bonds, we can predict that it will not obey Trouton's rule. To check the accuracy of our conclusion, we will use the given data in Appendix B.
05

Verify the conclusion using given data

From Appendix B, we can find \(\Delta H_{vap}\) for water as \(40.7\) kJ/mol and the normal boiling point as \(100^\circ C\) or \(373.15 K\). Calculate the molar entropy of vaporization for water: \[\Delta S_{vap}=\frac{40.7 \times 10^3}{373.15} = 109.08 \mathrm{J/mol~K}\] As expected, since the value exceeds the range of \(88 \pm 5\) J/mol-K, water does not obey Trouton's rule.
06

Estimate enthalpy of vaporization for chlorobenzene using Trouton's rule

Chlorobenzene boils at \(131.8^\circ C\), convert to Kelvin: \[T_{b} = 131.8 + 273.15 = 404.95 K\] Using Trouton's rule, we assume that \(\Delta S_{vap} = 88\ \mathrm{J/mol~K}\). Now, we can estimate the enthalpy of vaporization using the formula: \[\Delta H_{vap} = \Delta S_{vap} \times T_{b}\] \[\Delta H_{vap} = 88 \times 404.95 = 35636.60 \mathrm{J/mol}\] Converting to kJ/mol: \[\Delta H_{vap} = \frac{35636.60}{1000} = 35.64 \mathrm{kJ/mol}\] So, the estimated enthalpy of vaporization for chlorobenzene is \(35.64\) kJ/mol.

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Key Concepts

These are the key concepts you need to understand to accurately answer the question.

Trouton's Rule
Trouton's Rule is a fascinating observation in thermodynamics that applies to many liquids. It states that the molar entropy of vaporization, which measures how much disorder is introduced when a liquid turns into gas, is typically around 88 J/mol-K. This holds true as long as the conditions are not affected strongly by intermolecular forces, such as hydrogen bonding.

This rule is quite useful for estimating the enthalpy of vaporization for various substances, especially when it's difficult to measure directly. Its approximate nature means it works best for non-polar and non-hydrogen-bonded liquids. Trouton's Rule is also a great indicator for identifying outliers in molecular behavior. For example, substances like ethanol and water deviate from this rule because of their strong hydrogen bonding.
  • The standard range is 88 ± 5 J/mol-K.
  • Works well for non-polar substances without hydrogen bonding.
  • Used to estimate other thermodynamic properties.
Entropy of Vaporization
Entropy of Vaporization (\(\Delta S_{vap}\) ) is a crucial thermodynamic concept that helps us understand how energy is distributed when a liquid turns into vapor. It essentially measures the increase in disorder or randomness as molecules transition from the condensed state to a more disorganized gaseous state.

This can be calculated by dividing the enthalpy of vaporization by the absolute temperature (in Kelvin) at which the phase change occurs:\[\Delta S_{vap} = \frac{\Delta H_{vap}}{T_b}\]The unit for entropy is typically J/mol-K.

Every substance has a characteristic entropy of vaporization. Most substances follow Trouton's rule, but there are exceptions. For molecules with strong intermolecular forces, such as hydrogen bonds, the additional energy required disrupts Trouton's typical range for \(\Delta S_{vap}\)
  • Calculates disorder from liquid to vapor phase.
  • Often aligns with Trouton's range unless influenced by strong forces.
  • Expressed in J/mol-K.
Hydrogen Bonding
Hydrogen Bonding is a powerful type of dipole-dipole attraction that plays a crucial role in the physical properties of substances. This type of bonding occurs when there is a strong attraction between a hydrogen atom, which is covalently bonded to an electronegative atom (such as oxygen or nitrogen), and another electronegative atom nearby.

This bond is much stronger than a typical dipole interaction but weaker than covalent or ionic bonds. It significantly increases the enthalpy of vaporization and entropy of vaporization for substances like water and ethanol because breaking these intermolecular forces requires more energy. This additional energy also means these liquids often do not follow Trouton's Rule.

Hydrogen bonds lead to distinctive properties, such as the high boiling point of water, ice's lower density than liquid water, and providing secondary and tertiary structures to proteins and nucleic acids.
  • Occurs between molecules with hydrogen attached to strong electronegative atoms.
  • Increases both \(\Delta H_{vap}\) and \(\Delta S_{vap}\) .
  • Leads to deviations from Trouton's Rule.
Enthalpy of Vaporization
The Enthalpy of Vaporization (\(\Delta H_{vap}\) ) denotes the amount of energy needed to convert a liquid into vapor at a constant temperature and pressure. This encompasses more than just the energy needed to overcome cohesive forces; it's an essential factor in understanding phase changes in thermodynamics.

To determine \(\Delta H_{vap}\) for a given substance, we reference data from experiments or use quantitative rules like Trouton's Rule. Knowing the enthalpy of vaporization helps in the design of chemical processes and industrial applications, as it tells us how much energy is consumed during phase changes.

For substances without strong hydrogen bonding or other strong intermolecular forces, Trouton's Rule can be used to estimate \(\Delta H_{vap}\) from the molar entropy of vaporization. Substances with strong intermolecular forces, such as hydrogen bonds, however, require more detailed consideration due to additional energy needed to break these bonds.
  • Measures energy needed for liquid to vapor phase change.
  • Critical for process design and analysis.
  • Affected by strength of intermolecular forces.

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Most popular questions from this chapter

Predict the sign of \(\Delta S_{\text {sys }}\) for each of the following processes: (a) Gaseous Ar is liquefied at \(80 \mathrm{~K}\). (b) Gaseous \(\mathrm{N}_{2} \mathrm{O}_{4}\) dissociates to form gaseous \(\mathrm{NO}_{2}\). (c) Solid potassium reacts with gaseous \(\mathrm{O}_{2}\) to form solid potassium superoxide, \(\mathrm{KO}_{2}\). (d) Lead bromide precipitates upon mixing \(\mathrm{Pb}\left(\mathrm{NO}_{3}\right)_{2}(a q)\) and \(\mathrm{KBr}(a q)\)

Using data from Appendix \(C\), calculate the change in Gibbs free energy for each of the following reactions. In each case indicate whether the reaction is spontaneous under standard conditions. (a) \(\mathrm{H}_{2}(\mathrm{~g})+\mathrm{Cl}_{2}(g) \longrightarrow 2 \mathrm{HCl}(g)\) (b) \(\mathrm{MgCl}_{2}(s)+\mathrm{H}_{2} \mathrm{O}(l) \longrightarrow \mathrm{MgO}(s)+2 \mathrm{HCl}(g)\) (c) \(2 \mathrm{NH}_{3}(g) \longrightarrow \mathrm{N}_{2} \mathrm{H}_{4}(g)+\mathrm{H}_{2}(g)\) (d) \(2 \mathrm{NOCl}(g) \longrightarrow 2 \mathrm{NO}(g)+\mathrm{Cl}_{2}(g)\)

Consider the following equilibrium: $$ \mathrm{N}_{2} \mathrm{O}_{4}(g) \rightleftharpoons 2 \mathrm{NO}_{2}(g) $$ Thermodynamic data on these gases are given in Appendix C. You may assume that \(\Delta H^{\circ}\) and \(\Delta S^{\circ}\) do not vary with temperature. (a) At what temperature will an equilibrium mixture contain equal amounts of the two gases? (b) At what temperature will an equilibrium mixture of 1 atm total pressure contain twice as much \(\mathrm{NO}_{2}\) as \(\mathrm{N}_{2} \mathrm{O}_{4} ?\) (c) At what temperature will an equilibrium mixture of \(10 \mathrm{~atm}\) total pressure contain twice as much \(\mathrm{NO}_{2}\) as \(\mathrm{N}_{2} \mathrm{O}_{4} ?\) (d) Rationalize the results from parts (b) and (c) by using Le Châtelier's principle. \(\infty\) (Section 15.7)

Write the equilibrium-constant expression and calculate the value of the equilibrium constant for each of the following reactions at \(298 \mathrm{~K}\), using data from Appendix \(\mathrm{C}\) : (a) \(\mathrm{NaHCO}_{3}(s) \rightleftharpoons \mathrm{NaOH}(s)+\mathrm{CO}_{2}(g)\) (b) \(2 \mathrm{HBr}(g)+\mathrm{Cl}_{2}(g) \rightleftharpoons 2 \mathrm{HCl}(g)+\mathrm{Br}_{2}(g)\) (c) \(2 \mathrm{SO}_{2}(g)+\mathrm{O}_{2}(g) \rightleftharpoons 2 \mathrm{SO}_{3}(g)\)

About \(86 \%\) of the world's electrical energy is produced by using steam turbines, a form of heat engine. In his analysis of an ideal heat engine, Sadi Carnot concluded that the maximum possible efficiency is defined by the total work that could be done by the engine, divided by the quantity of heat available to do the work (for example from hot steam produced by combustion of a fuel such as coal or methane). This efficiency is given by the ratio \(\left(T_{\text {high }}-T_{\text {low }}\right) / T_{\text {high }}\), where \(T_{\text {high }}\) is the temperature of the heat going into the engine and \(T_{\text {low }}\) is that of the heat leaving the engine. (a) What is the maximum possible efficiency of a heat engine operating between an input temperature of \(700 \mathrm{~K}\) and an exit temperature of \(288 \mathrm{~K} ?\) (b) Why is it important that electrical power plants be located near bodies of relatively cool water? (c) Under what conditions could a heat engine operate at or near \(100 \%\) efficiency? (d) It is often said that if the energy of combustion of a fuel such as methane were captured in an electrical fuel cell instead of by burning the fuel in a heat engine, a greater fraction of the energy could be put to useful work. Make a qualitative drawing like that in Figure \(5.10\) that illustrates the fact that in principle the fuel cell route will produce more useful work than the heat engine route from combustion of methane.

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