[mookxi]

Equilibrium equation:

2NO₂ <=> N₂O₄
(brown)   (colourless)

by the way, this reaction is exothermic.

At a particular temperature, the gas mixture in the container is compressed by adjusting the piston. In order to maintain equilibrium, the ratio [NO₂]/[N₂O₄]  ...

the answer is that it becomes smaller. Why is that? I thought that if the gas mixture is "compressed" (increase pressure I assume, but I don't know what a 'piston' is) that that would result in producing more moles so that the K values becomes larger.

Could someone please explain? thanks :-)

[Borek]

K doesn't change (at least as long as you don't change temperature) and pressures (or concentrations) must change to reflect that.

[Will]

This is just Le Chatelier's Principle in action. If you increase the pressure (or compress the container) then the equilibrium will shift to reduce the pressure- which means it will shift to the side with less moles of gas- in this case thats the right hand side- as there is one mole of gas instead of two on the left. So as the conc. of N₂O₄ increases and the conc. of NO₂ decreases the ratio [NO₂]/[N₂O₄] will fall.
Hope that helps.

[mookxi]

This is just Le Chatelier's Principle in action. If you increase the pressure (or compress the container) then the equilibrium will shift to reduce the pressure- which means it will shift to the side with less moles of gas- in this case thats the right hand side- as there is one mole of gas instead of two on the left. So as the conc. of N₂O₄ increases and the conc. of NO₂ decreases the ratio [NO₂]/[N₂O₄] will fall.
Hope that helps.

Ahh that makes sense, thanks very much ;-) I seem to be getting stuck in many of these equilibrium problems tonight. I probably doesn't help that it's 12:20am too ;-x