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Equilibrium Between Nitrogen Dioxide and Dinitrogen Tetroxide


3 Tubes of N2O4 gas

3 800 mL or 1 liter beakers

Hot Water

Dry Ice

Thermal Gloves

Gloves and Goggles



Show class gas tubes. Line up the 3 beakers in a row.  Put hot water in one, nothing in the middle one and a few small chunks of dry ice in the 3rd.  Put one tube in each beaker. The cold tube will  clear up with green liquid forming on the bottom of the tube. The hot tube turns 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 Kmole

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