Three erlenmeyers contain each an identical mixture of nitrogen dioxide ($NO_2$, a brown gas ), and dinitrogen tetroxide $(N_2O_4$, a colourless gas).
They have the same temperature and, other things being equal, they have the same color intensity. The three Erlenmeyer are then put at different temperatures, and you see the result here:
One has to admit - that at $50^oC$, as the brown colour (of $NO_2)$ has turned to be more intense, the following reaction has preferably proceeded: $N_2O_4\longrightarrow 2NO_2 (1)$ - that at $0^oC$, as the brown colour (of $NO_2$) has turned to be less intense, the following reaction has preferably proceeded: $N_2O_4\longleftarrow 2NO_2 (2)$ - that at $20^oC$, the two reactions proceed simultaneously : $N_2O_4$ $ \rightleftarrows $ $ 2NO_2$
An equilibrium will soon be established: $NO_2$ disappears as rapidly by reaction (2) as it appears by reaction (1) and for $N_2O_4$ the opposite is true.