Exercice 8 | Thermodynamique chimique.

ΔH⁰_rxn = 2 x ΔH⁰_f [HI (g)] – ΔH⁰_f [H₂ (g)] – ΔH⁰_f [I₂ (g)]

ΔH⁰_rxn = 2 x 26.5 – 62.4

ΔH⁰_rxn = - 9.40 kJ.mol^-1

van’t Hoff Equation:

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

Situation 1 : T₁ = 400°C = 673K and K₁ = 58.0

Situation 2 : T₂ = 500°C = 773K and K₂ = ?

ln $\left(\frac{{\mathrm{K}}_2}{58.0}\right)$ = - $\frac{9.10 \times {10}^3}{8.314}$ x  $\frac{100}{673 \times 773}$

ln $\left(\frac{{\mathrm{K}}_2}{58.0}\right)$ = -0.217

K₂ = 46.7