@BD

Assuming gasses behave ideally, yes.

Thanks. But the equilibrium yield is what we predict by Le Chatelier's Principle, and this is affected by pressure? (i.e. let's say we've got N

_{2} + 3H

_{2} 2NH

_{3} and I increase the pressure; do I finally end up with more NH

_{3} at equilibrium, or not?)

In fact, let's take this from the perspective of K

_{c} so I can understand properly. If we define the final amount of NH

_{3} (partial pressure or concentration) as the yield, then this will move in the opposite direction to the reaction quotient Q - i.e. if we make Q larger (e.g. adding to the products), then the final yield of products will decrease and yield of reactants will decrease, whereas if we make Q smaller (e.g. adding to the reactants), the final yield of products will increase and yield of reactants will decrease. But then how is the equilibrium constant being maintained?

e.g. N

_{2} + 3H

_{2} 2NH

_{3}Let's say I add more moles of N

_{2} to the mixture. We can assume the gas is ideal so K

_{p} and K

_{c} are independent of pressure and stay constant. Now the value of Q has become smaller (as we have added reactant), and to bring this back up to K

_{c} and K

_{p} the reaction is driven forward. But what does this mean for the equilibrium amounts of products and reactant in the solution? The equilibrium ratio will always stay the same (that is, equal to the equilibrium constants) but what about the amount (i.e. number of moles)? And analogously, what if I added moles of NH

_{3} rather than N

_{2}?