Hey guys, been having a little problem with relating P and T to Keq/Ksp, etc. Chad said that T is the only factor that will change K. I also saw this in Destroyer (e.g. solution to GChem #120). But this concept didn't make sense to me because I knew that pressure changes can shift equilibrium based on the #mols of gas on each side of the equation. So I looked it up and found this:
http://en.wikipedia.org/wiki/Equilibrium_constant
Which says that "All equilibrium constants depend on temperature and pressure (or volume)."
Can anyone clear this up for me?
Thanks in advance!
I think this is a tricky concept as well, and this is what I have made of it after combining my study materials. Take the equilibrium constant (Keq) for example, it is equal to products over reactants raised to their coefficients (not including solids or liquids). What Keq does is tell us what is true when we are
at equilibrium. for example:
if Keq = 10000 (essentially a 10000:1 ratio), we know that
at equilibrium there are way more products than reactants since Keq=prod/react. the opposite would also be true.
now, if we add a gas that is a reactant we are no longer at equilibrium (Keq), we are at a new point which we denote as Q. Keq is still the same, but since we have changed the
ratio (no longer 10000(prods):1(reacts)) we are not at Keq, we are now at Q which has a ratio smaller than 10000:1.
We also know that when a stress like this is placed on a system that was at equilibrium, the system responds to counteract the stress. So, at this point Keq>Q and we now need reactants to turn into products to get back to equilibrium, Keq.
Hopefully with this example you can see that adding stresses like products, reactants, etc... that the equilibrium never changes, just the ratio ([prod]/[react]) putting us at pt Q, which will in turn start to counter act the stress to return to the original equilibrium (by turning more reactants into products).
It is harder to see this with pressure though. when dealing with gasses, Keq=Kp=Partial pressure of products/Partial pressure of reactants. Its just another way to express Keq. (Notice, if you use Kp with the partial pressures for the above example, you will have the same result as adding moles of a gas that is a reactant will increase the partial pressure on the bottom of the fraction again lowering the ratio.)
So now, lets assume there are more moles of gas on the reactant side than there are on the product side of the overall reaction, and increase the pressure. Again, the Kp(equilibrium) does not change, just the
ratio of the partial pressure of products divided by the partial pressure of the reactants changes and we are now again at point Q, which is just a point not at equilibrium. Since we stated earlier that in our overall reaction there are more moles of reactant gases than moles of product gases, our ratio is again decreased and Kp>Q. And again, since Kp>Q we need to our reactants to make more products to return to our original Kp, which is why the reaction is pushed to the right when the pressure is added.
Now, think about when there are either equal moles of gas on the reactants side and products side, or no gases on either side for that matter. What happens when you increase pressure when you are already at equilibrium in this example? Nothing, because the pressure is added on the same number of moles of gas on each side of the reaction (or no moles of gas if there isnt gas present) and the ratio of Kp does not change.
And for temperature... it changes the actual
value of Keq. Im not sure exactly why, but i think u just gotta know it does... haha
hope this helps and anyone please correct me if my logic is wrong here
