DEMONSTRATIONS          Faculty &
Research 		 |  Academic
Programs		  |  Industrial
Relations		  |  Research
Facilities		  |  Department
Administration		      

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 Kmole

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


Official Web Page of the University of California, San Diego
For questions or comments contact webmaster@chem.ucsd.edu
Last Updated: 5/23/02