: On a 100 mL d’une solution de NaOH à 0,5 M. On ajoute 400 mL d’eau. Quelle est la nouvelle concentration ?
On cherche 2C + H₂ → C₂H₂. On utilise (2)×2 : 2C + 2O₂ → 2CO₂ ΔH = –787,0 kJ On garde (3) : H₂ + ½ O₂ → H₂O ΔH = –285,8 kJ On inverse (1) : 2CO₂ + H₂O → C₂H₂ + 5/2 O₂ ΔH = +1299,6 kJ chimie generale cours et exercices corriges pdf
: On a 100 mL d’une solution de NaOH à 0,5 M. On ajoute 400 mL d’eau. Quelle est la nouvelle concentration ?
On cherche 2C + H₂ → C₂H₂. On utilise (2)×2 : 2C + 2O₂ → 2CO₂ ΔH = –787,0 kJ On garde (3) : H₂ + ½ O₂ → H₂O ΔH = –285,8 kJ On inverse (1) : 2CO₂ + H₂O → C₂H₂ + 5/2 O₂ ΔH = +1299,6 kJ
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