Chapter 14, Problem 14.47QP

(a) Trouton’s rule states that the ratio of the molar heat of vaporization of a liquid (ΔH_vap) to its boiling point in kelvins is approximately 90 J/K · mol. Use the following data to show that this is the case and explain why Trouton’s rule holds true.

(b) Use the values in Table 12.5 to calculate the same ratio for ethanol and water. Explain why Trouton’s rule does not apply to these two substances as well as it does to other liquids.

Interpretation Introduction

Interpretation:

To verify Trouton's rule for the given set of substances.

Concept Introduction:

According to Trouton's rule for, various liquids attheir boiling point the change in entropy value will be the same.  The ratio of enthalpy of vaporization and boiling temperature of various liquids has the same value.

${\text{ΔS}}_{\text{vap}}\text{=}\frac{{\text{ΔΗ}}_{\text{vap}}}{{\text{T}}_{\text{bp}}}\approx {\text{90JK}}^{\text{-1}}{\text{mol}}^{\text{-1}}$

Where,

${\text{ΔS}}_{\text{vap}}$ is the entropy of vaporization

${\text{ΔΗ}}_{\text{vap}}$ is the enthalpy of vaporization

${\text{T}}_{\text{bp}}$ boiling temperature

Answer to Problem 14.47QP

Trouton's rule is verified for benzene, hexane, mercury and toluene.  All the four substances have ${\text{ΔS}}_{\text{vap}}\approx {\text{90JK}}^{\text{-1}}{\text{mol}}^{\text{-1}}$

Explanation of Solution

To record the given data

$\begin{array}{l}{\text{ΔΗ}}_{\text{vap}}\text{ of benzene}{\text{=31000Jmol}}^{\text{-1}}\\ \\ {\text{T}}_{\text{bp}}\text{ of benzene =}\text{80}\text{.1}°\text{C}\\ \text{}\text{}\text{}\text{}\text{}\text{}\text{}\text{}\text{}\text{}\text{}\text{}\text{}\text{}\text{}\text{}\text{}\text{= 80}\text{.1+273}\\ \text{= 353}\text{.1}\text{K}\end{array}$

$\begin{array}{l}{\text{ΔΗ}}_{\text{vap}}{\text{ of hexane =30800Jmol}}^{\text{-1}}\\ \\ {\text{T}}_{\text{bp}}\text{ of hexane =}\text{68}\text{.7}°\text{C}\\ \text{= 68}\text{.7+273}\\ \text{= 341}\text{.7}\text{K}\end{array}$

$\begin{array}{l}{\text{ΔΗ}}_{\text{vap}}{\text{ of mercury =59000 Jmol}}^{\text{-1}}\\ \\ {\text{T}}_{\text{bp}}\text{ of mercury =}\text{357}°\text{C}\\ =\text{357+273}\\ \text{= 630}\text{K}\\ \end{array}$

$\begin{array}{l}{\text{ΔΗ}}_{\text{vap}}{\text{ of toluene =35200 Jmol}}^{\text{-1}}\\ \\ {\text{T}}_{\text{bp}}\text{ of toluene =}\text{110}\text{.6}°\text{C}\\ \text{= 110}\text{.6 + 273}\\ \text{= 383}\text{.6}\text{K}\end{array}$

To verify Trouton's rule for benzene

Trouton's rule can be verified by plugging in the values of ${\text{ΔΗ}}_{\text{vap}}$ and ${\text{T}}_{\text{bp}}$ in the following equation.

${\text{ΔS}}_{\text{vap}}\text{=}\frac{{\text{ΔΗ}}_{\text{vap}}}{{\text{T}}_{\text{bp}}}$

$\begin{array}{l}\text{=}\frac{{\text{31000Jmol}}^{\text{-1}}}{\text{353}\text{.1K}}\\ \text{=87}\text{.8}{\text{JK}}^{\text{-1}}{\text{mol}}^{\text{-1}}\end{array}$

Since the value is approximately equal to ${\text{90JK}}^{\text{-1}}{\text{mol}}^{\text{-1}}$, Trouton's rule is followed by benzene.

To verify Trouton's rule for hexane

Trouton's rule can be verified by plugging in the values of ${\text{ΔΗ}}_{\text{vap}}$ and ${\text{T}}_{\text{bp}}$ in the following equation.

${\text{ΔS}}_{\text{vap}}\text{=}\frac{{\text{ΔΗ}}_{\text{vap}}}{{\text{T}}_{\text{bp}}}$

$\begin{array}{l}\text{=}\frac{{\text{30800Jmol}}^{\text{-1}}}{\text{341}\text{.7K}}\\ \text{=90}{\text{.1JK}}^{\text{-1}}{\text{mol}}^{\text{-1}}\end{array}$

Since the value is approximately equal to ${\text{90JK}}^{\text{-1}}{\text{mol}}^{\text{-1}}$, Trouton's rule is followed by hexane.

To verify Trouton's rule for mercury

Trouton's rule can be verified by plugging in the values of ${\text{ΔΗ}}_{\text{vap}}$ and ${\text{T}}_{\text{bp}}$ in the following equation.

${\text{ΔS}}_{\text{vap}}\text{=}\frac{{\text{ΔΗ}}_{\text{vap}}}{{\text{T}}_{\text{bp}}}$

$\begin{array}{l}\text{=}\frac{{\text{59000Jmol}}^{\text{-1}}}{\text{630K}}\\ \text{=93}{\text{.7JK}}^{\text{-1}}{\text{mol}}^{\text{-1}}\end{array}$

Since the value is approximately equal to ${\text{90JK}}^{\text{-1}}{\text{mol}}^{\text{-1}}$, Trouton's rule is followed by mercury.

To verify Trouton's rule for Toluene

Trouton's rule can be verified by plugging in the values of ${\text{ΔΗ}}_{\text{vap}}$ and ${\text{T}}_{\text{bp}}$ in the following equation.

${\text{ΔS}}_{\text{vap}}\text{=}\frac{{\text{ΔΗ}}_{\text{vap}}}{{\text{T}}_{\text{bp}}}$

$\begin{array}{l}\text{=}\frac{\text{35200}{\text{Jmol}}^{\text{-1}}}{\text{383}\text{.6K}}\\ \text{=}\text{91}\text{.8}{\text{JK}}^{\text{-1}}{\text{mol}}^{\text{-1}}\end{array}$

Since the value is approximately equal to ${\text{90JK}}^{\text{-1}}{\text{mol}}^{\text{-1}}$, Trouton's rule is followed by toluene.

Interpretation Introduction

Interpretation:

To verify Trouton's rule for the given set of substances.

Concept Introduction:

According to Trouton's rule for, various liquids attheir boiling point the change in entropy value will be the same.  The ratio of enthalpy of vaporization and boiling temperature of various liquids has the same value.

${\text{ΔS}}_{\text{vap}}\text{=}\frac{{\text{ΔΗ}}_{\text{vap}}}{{\text{T}}_{\text{bp}}}\approx {\text{90JK}}^{\text{-1}}{\text{mol}}^{\text{-1}}$

Where,

${\text{ΔS}}_{\text{vap}}$ is the entropy of vaporization

${\text{ΔΗ}}_{\text{vap}}$ is the enthalpy of vaporization

${\text{T}}_{\text{bp}}$ boiling temperature

Answer to Problem 14.47QP

Trouton's rule is not obeyed by water and ethanol due to the high entropy of vaporization.

Explanation of Solution

To record the given data

${\text{ΔΗ}}_{\text{vap}}$ of ethanol ${\text{=39300 Jmol}}^{\text{-1}}$

$\begin{array}{l}{\text{T}}_{\text{bp}}\text{of ethanol = 78}\text{.3 °C}\\ \text{= (78}\text{.3+273)}\\ \text{= 351}\text{.3K}\end{array}$

${\text{ΔΗ}}_{\text{vap}}$ of water ${\text{=40790 Jmol}}^{\text{-1}}$

$\begin{array}{l}{\text{T}}_{\text{bp}}\text{of water = 100 °C}\\ \text{= (100+273)}\\ \text{= 373K}\end{array}$

To verify Trouton's rule for ethanol

Trouton's rule can be verified by plugging in the values of ${\text{ΔΗ}}_{\text{vap}}$ and ${\text{T}}_{\text{bp}}$ in the following equation.

${\text{ΔS}}_{\text{vap}}\text{=}\frac{{\text{ΔΗ}}_{\text{vap}}}{{\text{T}}_{\text{bp}}}$

$\begin{array}{l}\text{=}\frac{\text{39300}{\text{Jmol}}^{\text{-1}}}{\text{351}\text{.3K}}\\ \text{=}\text{112}{\text{JK}}^{\text{-1}}{\text{mol}}^{\text{-1}}\end{array}$

Since, the value is greater than ${\text{90JK}}^{\text{-1}}{\text{mol}}^{\text{-1}}$, Trouton's rule is not followed by ethanol.

To verify Trouton's rule for water

Trouton's rule can be verified by plugging in  the values of ${\text{ΔΗ}}_{\text{vap}}$ and ${\text{T}}_{\text{bp}}$ in the following equation.

${\text{ΔS}}_{\text{vap}}\text{=}\frac{{\text{ΔΗ}}_{\text{vap}}}{{\text{T}}_{\text{bp}}}$

$\begin{array}{l}\text{=}\frac{\text{40790}{\text{Jmol}}^{\text{-1}}}{\text{373K}}\\ \text{=}\text{109}{\text{JK}}^{\text{-1}}{\text{mol}}^{\text{-1}}\end{array}$

Since the value is greater than ${\text{90JK}}^{\text{-1}}{\text{mol}}^{\text{-1}}$, Trouton's rule is not followed by water.

To explain the reason for the deviation from hydrogen bond in case of water and ethanol.

For water and ethanol there exist strong hydrogen bonding attraction in the liquid state.  Hydrogen bonding interaction decreases the entropy in the system. In gaseous state the hydrogen bonding interaction becomes weaker and the molecules will have more randomness in the system.  The entropy of vaporization will be high.  This is the reason why water and ethanol is showing larger deviation from Trouton's rule.