Exercise 9 | Chemical Thermodynamics

Clapeyron-Clausius equation :

ln $\left(\frac{{\mathrm{P}}_2}{{\mathrm{P}}_1}\right)$ = $\frac{∆{{\mathrm{H}}^0}_{\mathrm{rxn}}}{\mathrm{R}}$ x $\left(\frac{1}{{\mathrm{T}}_1} - \frac{1}{{\mathrm{T}}_2}\right)$

ΔH⁰_rxn = R ln $\left(\frac{{\mathrm{P}}_2}{{\mathrm{P}}_1}\right)$ x $\frac{{\mathrm{T}}_1{\mathrm{T}}_2}{{\mathrm{T}}_2 - {\mathrm{T}}_1}$

Situation 1: P₁ = 133 Torr and T₁ = 293K

Situation 2: P₂ = 48.1 Torr and T₂ = 273K

ΔH⁰_vap = 8.314 x ln $\left(\frac{48.1}{133}\right)$ x $\frac{273 \times 293}{\left(- 20\right)}$

ΔH⁰_vap = 33.8 kJ.mol^-1