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Equilibrium Between Nitrogen Dioxide and Dinitrogen Tetroxide

Procedure

Show class gas tubes. Immerse one tube in hot water, lay the other on a piece of dry ice. Cold tube clears up with green liquid forming. Hot tube gets dark brown. Melting point of nitrogen (IV) oxide is -11.2 C and boiling point is 21.2 C. The solid state is exclusively N2O4, while the liquid is a mixture.

2 NO₂ (g) <---> N₂O₄ (g)     dH = -57.20 kJ/mole

Nitrogen (IV) oxide gas is in equilibrium as a mixture of a monomer (NO₂) and a dimer (N₂O₄). Nitrogen dioxide is reddish brown, dinitrogen tetraoxide is colorless. Impurities can impart a blue-green color to it.

Dimerization is exothermic, however dimerization leads to a decrease in entropy, and because of the entropy decrease, the free energy increases as the temperature increases, shifting the equilibrium toward NO₂, as indicated by dG = dH - TdS.

Dark tube = LA smog

Room temp tube = San Diego smog

Cold tube = Smog during Rose Bowl

2NO₂(g) = N₂O₄(g)     enthalpy of dimerization = -57.2 kJ/mole

Dimerization causes a decrease in entropy = -175.83 J/degrees K_mole

dG = dH -TdS increases as temps increases, indicates that equilibrium shifts toward NO₂ as temp increases.