The equilibrium illustrated in this demonstration is between nitrogen dioxide (NO2) and dinitrogen tetroxide (N2O4 ). The chemical equation is shown below.
2 NO2 (g) 
N2O4 (g)
Physical Constants of Compounds of Interest [2]
| Compound | Physical state | melting point (oC) | boiling point (oC) |
|---|
| NO2 | brown gas in equil. with N2O4 | undefined | see N2O4 |
| N2O4 | colorless liquid in equil. with NO2 | -9.3 | 21.15 |
| N2O3 | blue liquid or solid | -101.1 | decomposes ~3 |
| HNO3 | colorless liquid | -41.6 | 83 |
Nitrogen dioxide is a dark reddish brown gas and dinitrogen tetroxide is a colorless gas. When the equilibrium is shifted to the left, as written, the amount of nitrogen dioxide increases, the amount of dinitrogen tetroxide decreases and the color of the tube darkens. When the equilibrium is shifted to the right, as written, the amount of nitrogen dioxide decreases, the amount of dinitrogen tetroxide increases and the color of the tube lightens.
There are two fairly easy ways to shift the equilibrium, the first makes use of the exothermicity of the reaction. As written, the reaction releases releases 58 kJ of energy ( DHo = -58 kJ). Since energy is released, adding more energy via heating shift the equilibrium to the left and removing energy via cooling will shift the equilibrium to the right. The second method utilizes changes in pressure to shift the equilibrium. This will be covered in another demonstration.
Click on any of the above thumbnails to open a window showing a larger image. Close the window and repeat to examine any of the other images. Close the window to continue with the demonstration information.
If the tubes contained pure NO2 , all that would be seen in this demonstration would be varying shades of brown, but if the last image is examined closely, a blue/geen liquid can be see in the bottom of the cold tube. This comes about because there is at least a trace of water in the tubes and this leads to the following reaction.
2 N2O4 (l) + H2O (l) 2 N2O3 (l) + 2 HNO3 (l)
From the table we see that N2O3 is a blue liquid under these conditions. Pure HNO3 is colorless, but if it is contaminated with NO2 it becomes yellow. The yellow and blue components of this reaction produce the blue/green liquid that is seen.
It has not been shown in this demonstration, but if the tube is cooled below the melting point of N2O4 , The liquid will lose its green component and become totally blue. The reason for this is as follows. In the solid phase, the mixture of NO2 and N2O4 becomes exclusively N2O4 . This means there is no longer any NO2 to color the HNO3 yellow. Without the yellow component the liquid becomes completely blue. Placing the tube in some powdered dry ice is a convenient way to see this phenomenon
;)
;)
;)
;)
;)