<p>If you increase volume, the reaction shifts to side with more moles of gas particles. But you can't say their concentration increases right? Because the volume is increasing as well. You just get more moles of gas, but not necessarily more concentrated amounts of gas?</p>
<p>After it shifts, the concentration will always be the same, since it's shifting to equilibrium, and the equilibrium concentration is constant.</p>
<p>So yes, you are correct.</p>
<p>In this case it should be constant due to it being dependent on initial concentrations which haven't changed right. If you an aqueous or gaseous component, the equilibrium would shift with a different position.</p>
<p>Actually, that's not correct. The volume changes, and pressure decreases, and the moles produced of each gas will shift to maintain equilibrium. This does not mean concentrations do not change. Concentrations do change.</p>
<p>You can see for yourself... pick any reaction, and find its Kp (partial pressures of products over partial pressures of reactants, all raised to their stoichiometric coefficients). Convert to Kc (equilibrium in terms of concentrations) using Kp=Kc(RT)^(delta n). Changing the volume changes the pressure, which also changes the concentration of each gas. That's what equilibrium does.</p>
<p>Ah, yeah, I don't know what I was thinking. It's not the initial amount, it's the initial concentration that determines K at a given temp. So yeah, you're right, changing volume changes initial concentrations.</p>