 There's two LaChavage Principles problems up already, but this is the third one. Okay, and this one's good because it really isolates the effect of pressure, what pressure does to a system at equilibrium. Okay, so will an increase in pressure decrease, or sorry, will an increase in pressure decrease or have no effect on the hydrogen concentration in the reaction? So, huh? Yeah, I think this is a homework problem. Huh? It's 89, I think. 89. But anyways, so if you look here, on this side of the reaction, we have one mole of gas, right? So in phase two O's and gaseous forms, we have one mole. On this side we have two moles of gas combined. So when this system gets at equilibrium, it's at whatever concentrations that these are, okay? But then, if you increase the pressure, so say you have some sort of a piston, and this is in, this reaction is in this system here, and you push the piston down to increase the pressure like that, what's going to happen, okay? Well, you've got two moles of gas here and one mole of gas here. So what's going to happen is that the reaction is going to try to get to less moles of gas because it's compressed, okay? So after equilibrium will have happened is that if you increase the pressure, the reaction would go to the left towards the reactant side, okay? And of course, if we're looking at what happens to the concentration of H2, it's actually going to go down because of that increase in pressure, okay? So the concentration of H2 decreases due to the increase in pressure.