Question

Question 13. The normal boiling point of acetone is $\mathbf{56}\mathbf{.}\mathbf{2}\mathbf{°}\mathbf{}\mathbf{C}$. Use Trouton's rule to estimate its molar enthalpy of vaporization.

Short answer

The molar enthalpy of vaporization of acetone is$29{\text{kmol}}^{-1}$ .

Step-by-step solution

Given information

The normal boiling point of acetone${\mathrm{T}}_{\mathrm{b}}=562°\mathrm{C}$ .

Concept of Trouton’s rule

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

Convert unit of temperature

The unit of temperature is converted from degree Celsius to kelvin as:

The normal boiling point of acetone${\mathrm{T}}_{\mathrm{b}}=562°\mathrm{C}$ .
$$\begin{gathered} T_b=56.2+273.15 \\ {T}_b=329.35K \end{gathered}$$

Calculate molar enthalpy of vaporisation

Thus, the molar entropy of given acetone is$88{\text{JK}}^{-1}{\text{mol}}^{-1}$.

The molar entropy of change of vaporization of acetone ${\mathrm{S}}_{\mathrm{vap}}={\mathrm{H}}_{\mathrm{vap}}/{\mathrm{T}}_{\mathrm{b}}$.

Rearrange, molar enthalpy changes of vaporization,$H_{vap}=S_{vap}\times T_b$.
$$\begin{gathered} H_{vap}={88}^{*}329.35\backslash \mathrm{hfill} \\ {H}_{vap}=29kmo{l}^{-1}\backslash \mathrm{hfill} \end{gathered}$$