Question

Question: Ethanol $\left({\text{CH}}_3{\text{CH}}_2\text{OH}\right)$has a normal boiling point of $\mathbf{78}{\mathbf{.4}}^{\mathbf{°}}\mathbf{\text{C}}$and a molar enthalpy of vaporization of $\mathbf{38}\mathbf{.74}\mathbf{}{\mathbf{\text{kJmol}}}^{\mathbf{-}\mathbf{1}}$. Calculate the molar entropy of vaporization of ethanol and compare it with the prediction of Trouton's rule.

Short answer

The molar entropy of vaporization of ethanol $\Delta S_v$ is $110.24{\text{Jmol}}^{-1}$. The value of molar entropy does not obey the Trouton's rule.

Step-by-step solution

Given data

The boiling point of ethanol $T_b={78.4}^{°}C=351.4\text{K}$.

Molar enthalpy of vaporization of ethanol $\Delta H_v=38.74{\text{kJmol}}^{-1}$.

Concept of vapor pressure

The entropy of vaporization is the increase inentropyupon thevaporizationof a liquid. The entropy of vaporization is then equal to the heat of vaporization divided by the boiling point. According to Trouton's rule, the entropy of vaporization (at standard pressure) of most liquids has similar values.

Calculation of entropy

The entropy has been calculated as follows:

$\Delta S_v=\frac{\Delta H_v}{T_b}$ ........ (1)

Where, $\Delta H_v$ is the heat or enthalpy of vaporization and $T_b$refers to the boiling point of ethanol (measured in kelvins (K)).

$T_b={78.4}^{°}C=351.4\text{K}$.

$\Delta H_v=38.74{\text{kJmol}}^{-1}$.

Substitute these values in equation (1):

$\begin{array}{c}\Delta S_v=\frac{38.74\times {10}^3}{351.4}\\ =110.24{\text{Jmol}}^{-1}\end{array}$

Comparison with the Trouton's rule

According to this rule, most liquids have similar values of the molar entropy of vaporization. This value is given by the interval 88 give or take $5\text{J}/\text{mol}$.

The value of molar entropy does not obey Trouton's rule.

This can be the fault of the strong hydrogen bonds which is responsible for the level of randomness.