Dinitrogen tetroxide (N₂O₄) decomposes to nitrogen dioxide (NO₂). If$\Delta$H° = 58.02 kJ/mol and$\Delta$S° = 176.1 J/mol · K, at what temperature are reactants and products in their standard states at equilibrium?

Question

Quizplus · Accepted Answer

$\Delta$G° = $\Delta$H° - T$\Delta$S° = 0 at equilibrium. Solving for T gives:T = $\Delta$H°/ $\Delta$S° = (58.02 kJ/mol)/(176.1 J/mol · K) = 329.5 K = +56.5°C

Dinitrogen Tetroxide (N2O4) Decomposes to Nitrogen Dioxide (NO2) Δ\DeltaΔ H° = 5802 KJ/mol And Decomposes

Dinitrogen Tetroxide (N₂O₄) Decomposes to Nitrogen Dioxide (NO₂) $\Delta$ H° = 5802 KJ/mol And Decomposes