Chapter 15, Problem 15.58P

Interpretation Introduction

Interpretation:

The $\Delta H_{\text{vap}}$ for chloromethane, water, and hydrogen sulfide through Trouton’s rule has to be determined. The molecular explanation for discrepancy in $\Delta H_{\text{vap}}$ values should be determined.

Concept Introduction:

Molecules of liquid and solid are held together by a force of attraction. And the value of enthalpy of vaporization and sublimation reflects the magnitude of force between the molecules. Greater the value of $\Delta H_{\text{vap}}$ or $\Delta H_{\text{sub}}$, greater will be attractive force between the molecules.

If a molecule has hydrogen bonds, it increases it $\Delta H_{\text{vap}}$ because hydrogen bonds yield $\Delta H_{\text{vap}}$ of $20–40\text{ kJ}\cdot {\text{mol}}^{\text{–1}}$, approximately.

Answer to Problem 15.58P

The $\Delta H_{\text{vap}}$ for chloromethane, water and hydrogen sulfide are $21170.1\text{J}\cdot {\text{mol}}^{-\text{1}}$, $31717.75\text{J}\cdot {\text{mol}}^{-\text{1}}$, and $18156\text{J}\cdot {\text{mol}}^{-\text{1}}$, respectively.

Explanation of Solution

As per Trouton’s rule, compounds that don’t have strong molecular interactions like hydrogen bonding have molar enthalpy of vaporization as follows:

  $\Delta H_{\text{vap}}=\left(85\text{J}\cdot {\text{K}}^{-\text{1}}\cdot {\text{mol}}^{-\text{1}}\right)T_{\text{b}}$        (1)

Here,

$T_{\text{b}}$ is normal boiling point of liquid in kelvin.

$\Delta H_{\text{vap}}$ is enthalpy of vaporization.

Substitute $249.06\text{K}$ for $T_{\text{b}}$ in equation (1) to determine $\Delta H_{\text{vap}}$ for chloromethane.

  $\begin{array}{c}\Delta H_{\text{vap}}=\left(85\text{J}\cdot {\text{K}}^{-\text{1}}\cdot {\text{mol}}^{-\text{1}}\right)\left(249.06\text{K}\right)\\ =21170.1\text{J}\cdot {\text{mol}}^{-\text{1}}\end{array}$

Substitute $373.15\text{K}$ for $T_{\text{b}}$ in equation (1) to determine $\Delta H_{\text{vap}}$ for water.

  $\begin{array}{c}\Delta H_{\text{vap}}=\left(85\text{J}\cdot {\text{K}}^{-\text{1}}\cdot {\text{mol}}^{-\text{1}}\right)\left(373.15\text{K}\right)\\ =31717.75\text{J}\cdot {\text{mol}}^{-\text{1}}\end{array}$

Substitute $213.60\text{K}$ for $T_{\text{b}}$ in equation (1) to determine $\Delta H_{\text{vap}}$ for hydrogen sulfide.

  $\begin{array}{c}\Delta H_{\text{vap}}=\left(85\text{J}\cdot {\text{K}}^{-\text{1}}\cdot {\text{mol}}^{-\text{1}}\right)\left(213.60\text{K}\right)\\ =18156\text{J}\cdot {\text{mol}}^{-\text{1}}\end{array}$

The formula to calculate the percentage error in $\Delta H_{\text{vap}}$ is as follows:

  $\text{percentage}\text{error}=\frac{\left[\begin{array}{l}{\left(\Delta H_{\text{vap}}\right)}_{\text{theoretical}\text{value}}-\\ {\left(\Delta H_{\text{vap}}\right)}_{\text{experimental}\text{value}}\end{array}\right]}{{\left(\Delta H_{\text{vap}}\right)}_{\text{theoretical}\text{value}}}\times 100$        (2)

Substitute $21170.1\text{J}\cdot {\text{mol}}^{-\text{1}}$ for ${\left(\Delta H_{\text{vap}}\right)}_{\text{experimental}\text{value}}$ and $21.40\text{kJ}\cdot {\text{mol}}^{-\text{1}}$ for ${\left(\Delta H_{\text{vap}}\right)}_{\text{theoretical}\text{value}}$ in equation (2) to determine percentage error in $\Delta H_{\text{vap}}$ of water.

  $\begin{array}{c}\text{percentage}\text{error}=\frac{\left(21.40\text{kJ}\cdot {\text{mol}}^{-\text{1}}\right)\left(\frac{1000\text{J}}{1\text{kJ}}\right)-21170.1\text{J}\cdot {\text{mol}}^{-\text{1}}}{\left(21.40\text{kJ}\cdot {\text{mol}}^{-\text{1}}\right)\left(\frac{1000\text{J}}{1\text{kJ}}\right)}\times 100\\ =0.01074\times 100\\ =1.074\%\end{array}$

Substitute $31717.75\text{J}\cdot {\text{mol}}^{-\text{1}}$ for ${\left(\Delta H_{\text{vap}}\right)}_{\text{experimental}\text{value}}$ and $40.65\text{kJ}\cdot {\text{mol}}^{-\text{1}}$ for ${\left(\Delta H_{\text{vap}}\right)}_{\text{theoretical}\text{value}}$ in equation (2) to determine percentage error in $\Delta H_{\text{vap}}$ of water

  $\begin{array}{c}\text{percentage}\text{error}=\frac{\left(40.65\text{kJ}\cdot {\text{mol}}^{-\text{1}}\right)\left(\frac{1000\text{J}}{1\text{kJ}}\right)-31717.75\text{J}\cdot {\text{mol}}^{-\text{1}}}{\left(40.65\text{kJ}\cdot {\text{mol}}^{-\text{1}}\right)\left(\frac{1000\text{J}}{1\text{kJ}}\right)}\times 100\\ =0.21973\times 100\\ =21.973\%\end{array}$

Substitute $18156\text{J}\cdot {\text{mol}}^{-\text{1}}$ for ${\left(\Delta H_{\text{vap}}\right)}_{\text{experimental}\text{value}}$ and $18.67\text{kJ}\cdot {\text{mol}}^{-\text{1}}$ for ${\left(\Delta H_{\text{vap}}\right)}_{\text{theoretical}\text{value}}$ in equation (2) to determine percentage error in $\Delta H_{\text{vap}}$ of hydrogen sulfide

  $\begin{array}{c}\text{percentage}\text{error}=\frac{\left(18.67\text{kJ}\cdot {\text{mol}}^{-\text{1}}\right)\left(\frac{1000\text{J}}{1\text{kJ}}\right)-18156\text{J}\cdot {\text{mol}}^{-\text{1}}}{\left(18.67\text{kJ}\cdot {\text{mol}}^{-\text{1}}\right)\left(\frac{1000\text{J}}{1\text{kJ}}\right)}\times 100\\ =0.02753\times 100\\ =2.753\%\end{array}$

Water molecule has hydrogen bonding so the error is maximum as Trouton’s rule can not be applied to water molecule.