Question

Question14. The molar enthalpy of vaporization of liquid hydrogen chloride is$\mathbf{16}\mathbf{.}\mathbf{15}{\mathbf{\text{kJmol}}}^{\mathbf{-}\mathbf{1}}$. Use Trouton's rule to estimate its normal boiling point.

Short answer

The normal boiling point of liquid hydrogen chloride is $183.52\text{K}$

Step-by-step solution

Given information

The molar enthalpy of vaporization of liquid hydrogen chloride is $16.15{\text{kJmol}}^{-1}$.

Concept of Trouton’s rule

The Trouton's rule, the molar entropy of vaporization of most liquids are same, which is equal to${\mathbf{S}}_{\mathbf{vap}}\mathbf{=}\mathbf{88}{\mathbf{JK}}^{\mathbf{-}\mathbf{1}}{\mathbf{mol}}^{\mathbf{-}\mathbf{1}}$.

Calculate normal boiling point

The molar entropy of given Liquid$\text{HCl}$is$88J{K}^{-1}mo{l}^{-1}$.

The molar entropy of change of vaporization, $S_{vap}=H_{vap}/T_b$.

Rearrange, boiling point, $T_b=H_{vap}/S_{vap}$.

$$\begin{gathered} T_b=16.{15}^{*}{10}^3{J_{mol}}^{-1}/88J{K}^{-1}mo{l}^{-1} \\ T=183.52K \end{gathered}$$

Unit conversion.
$$\begin{gathered} {\mathrm{T}}_{\mathrm{b}}=183.52-273.15 \\ {\mathrm{T}}_{\mathrm{b}}=-89.6°\mathrm{C} \end{gathered}$$