Question

From the following data for liquid nitric acid, determine its heat of vaporization and normal boiling point.

$\begin{array}{c}\mathbf{Temperature}\mathbf{(}\mathbf{°}\mathbf{C}\mathbf{)}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{Vapor}\mathbf{pressure}\mathbf{\left(}\mathbf{mm}\mathbf{Hg}\mathbf{\right)}\\ \mathbf{0}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{14}\mathbf{.4}\\ \mathbf{10}\mathbf{.}\mathbf{26}\mathbf{.6}\\ \mathbf{20}\mathbf{.}\mathbf{47}\mathbf{.9}\\ \mathbf{30}\mathbf{.}\mathbf{81}\mathbf{.3}\\ \mathbf{40}\mathbf{.}\mathbf{133}\\ \mathbf{50}\mathbf{.}\mathbf{208}\\ \mathbf{80}\mathbf{.}\mathbf{670}\mathbf{.}\end{array}$

Short answer

Enthalpy of vaporization: ${\mathrm{\Delta H}}_{\mathrm{vap}}=39.418\mathrm{kJ}/\mathrm{mol}$

The normal boiling point of the nitric acid is $89.79°\mathrm{C}$.

Step-by-step solution

Calculation of enthalpy of vaporization

The enthalpy of vaporization can be calculated as shown below:

$\ln \left(\frac{{\mathrm{P}}_1}{{\mathrm{P}}_2}\right)=-\frac{{\mathrm{\Delta H}}_{\mathrm{vap}}}{\mathrm{R}}\left(\frac{1}{{\mathrm{T}}_2}-\frac{1}{{\mathrm{T}}_1}\right)$

$\begin{array}{c}\ln \left(\frac{26.6}{14.4}\right)=\frac{{\mathrm{\Delta H}}_{\mathrm{vap}}}{8.314\mathrm{J}.{\mathrm{K}}^{-1}.{\mathrm{mol}}^{-1}}\left(\frac{1}{283\mathrm{K}}-\frac{1}{273\mathrm{K}}\right)\\ \ln \left(\frac{26.6}{14.4}\right)=\frac{{\mathrm{\Delta H}}_{\mathrm{vap}}}{8.314\mathrm{J}.{\mathrm{K}}^{-1}.{\mathrm{mol}}^{-1}}\left(\frac{10}{283\times 273}\right)\\ {\mathrm{\Delta H}}_{\mathrm{vap}}=\frac{8.314\times 283\times 273}{10}\ln \left(\frac{26.6}{14.4}\right)\\ {\mathrm{\Delta H}}_{\mathrm{vap}}=39.418\mathrm{kJ}/\mathrm{mol}\end{array}$

Enthalpy of vaporization (∆H_vap ) =39.418kJ/mol

Calculation of normal boiling point

This can be solved using the Clausius-Clapeyron equation.

P₂= 36.6 mmHg

T₂ = 10⁰C= 283K

P₁= 14.4 mm Hg

T₁= 0⁰ C = 273K

To find the normal boiling point, use the value of ∆H_vap. In the case of this value, we take P₂as 760 mm Hg (Normal boiling point is the temperature at which a liquid boils at atm pressure = 760 mm Hg )

$\begin{array}{c}\ln \left(\frac{760}{14.4}\right)=\frac{39418.78}{8.314}\left(\frac{1}{273}-\frac{1}{{\mathrm{T}}_2}\right)\\ \ln \left(\frac{760}{14.4}\right)\times \frac{8.314}{39418.78}=\left(\frac{1}{273}-\frac{1}{{\mathrm{T}}_2}\right)\\ {\mathrm{T}}_2=\frac{1}{2.826\times {10}^{-3}}\\ {\mathrm{T}}_2=353.79\mathrm{K}\end{array}$

The normal boiling point of the nitric acid is $353.79Kor89.79°C$.