Does a change in pressure change the equilibrium constant?

[Duck]

I know, and have been told countless times, that the equilibrium constant can only be changed by a change in temperature but according to LCP a change in pressure forces either the forward or backward reaction. Therefore wouldn't K change?
eg.
2S02 + 02 → 2SO3

An increase in pressure, according to LCP, would be opposed by the reaction going forwards. Thus, concentration of SO3 increases and concentrations of SO2 and O2 decrease. Wouldn't this increase the equilibrium constant?

[taiga]

nope, it shifts the concentration fraction, but not the concentration fraction at equlibrium

[matt123]

I know, and have been told countless times, that the equilibrium constant can only be changed by a change in temperature but according to LCP a change in pressure forces either the forward or backward reaction. Therefore wouldn't K change?
eg.
2S02 + 02 → 2SO3

An increase in pressure, according to LCP, would be opposed by the reaction going forwards. Thus, concentration of SO3 increases and concentrations of SO2 and O2 decrease. Wouldn't this increase the equilibrium constant?

it does seem like that.
however , don't forget , the system only PARTIALLY opposes the change.

for e.g
lets say we decrease the volume..=  the pressure/concentration is increased . then the concentration fraction increases as the system might move to the right ( net forward )
but dont forget , the overall systems pressure has increased , this also includes the pressure of the products.

so ... what im trying to say is.
if you increase pressure by decreasing volume .. you might be increasing the forward reaction , BUT dont forget , by increasing the pressure of the reactants , we are also increasing the pressure of the products , which in turn .. it kind of ..levels/cancels  out , and so the K doesn't increase , but because the system is no longer at EQM , the CF has increased.

you see , temperature , actually moves the reaction to the left or right. but it does not "level" out. because we are not increasing the overall pressure/concentration.

I hope that kind of makes sense???
some 1 please correct me if im wrong.

EDIT :
also think about the fact that .. when a change is made to the system .. it is no longer at equilibrium .. hence it dosnt even have a equilibrium constant . only a cf.
when the EQM comes back though , the kc will stay the same as the concentrations will be fixed again.

[toshibaj]

Remember that an increase in pressure initially increases the concentration of all reactants and products. At this point the system is not at equilibrium and the equilibrium fraction is different.

The shift in equilibrium position is done to restore equilibrium, thus in the end the equilibrium constant remains the same.  

For changes in temperature there is no initial change in concentration that is responded to. Therefore it changes the equilibrium constant.
That's how I think of it anyway.

[supernatural]

the concentrations change but the ratio (equilibrium constant) stays the same. in real life however there is a small effect but we ignore it in vce. like you said though, the only thing that changes the equilibrium constant is temperature

[98.40_for_sure]

I think... his question can be answered by simply:
Change in pressure changes the concentration fraction, NOT equilibrium constant. Dunno what you guys are all on about

[Duck]

Equilibrium constant is [products]/[reactants], what's concentration fraction? Doesn't an increase in yield mean more products and therefore higher K?

[jasoN-]

No, everyone should know ONLY temperature changes the K value.
Think about it this way.
H2O + H2O <-> OH- + H3O+

To increase the yield: Water is added.
INITIALLY the K value will be different (more H2O = smaller K value)
Le Chatelier's acts and it reestablishes equilibrium. H2O reacts to form more OH- and H3O+
Even though there is more yield, there is also more reactants, thus the K value is balanced

[fady_22]

Equilibrium constant is [products]/[reactants], what's concentration fraction? Doesn't an increase in yield mean more products and therefore higher K?

The equilibrium constant is a concentration fraction, except that it is the concentration fraction when the system is at equilibrium. You can calculate the concentration fraction at any time, and it indicates whether or not the system is at equilibrium and, if relevant, the direction of the shift.

If you were to increase the pressure, although the system may shift to the right, the equilibrium constant remains... constant. Mathematically, it does work. Just accept it.

[Potter]

Concentration fraction is just another name for the reaction quotient(Q). I think the authors of Chem 2 used "concentration fraction" to simplify things down? Not too sure.
K is a constant value at a certain temp, it doesn't change.

Another thing to be careful of is that inert gases don't affect equilibrium. So, even if you do increase the pressure of a system by adding neon or krypton for example it won't make any difference. The reason why is because when we say "pressure" we actually refer to the partial pressure of the gas, not the overall pressure of the system.

http://en.wikipedia.org/wiki/Partial_pressure
If you're curious about partial pressures.

[Duck]

Sorry, I'm not following.
2S02 + 02 → 2SO3
This system is in a container at equilibrium. The container is shrunk, increasing pressure, but temperature stays the same. LCP says this forces the reaction forward, increasing [S03], doesn't it?

[matt123]

Sorry, I'm not following.
2S02 + 02 → 2SO3
This system is in a container at equilibrium. The container is shrunk, increasing pressure, but temperature stays the same. LCP says this forces the reaction forward, increasing [S03], doesn't it?

yes that is right. the forward reaction occurs.
BUT DONT FORGET .... when there is more (so3) being produced .. THE SYSTEM IS NOT AT EQUILIBRIUM

thats why all industrial processes are NEVER at equilibrium .. because they are always moving the reaction forwards so that they get a higher yeild.

the kc DOSNT change .. because kc implies its at equilibrium.
when we decrease the volume and the pressure increases .. the system is not YET at eqm.

hope that helps

[m@tty]

Sorry, I'm not following.
2S02 + 02 → 2SO3
This system is in a container at equilibrium. The container is shrunk, increasing pressure, but temperature stays the same. LCP says this forces the reaction forward, increasing [S03], doesn't it?

Immediately, the concentrations of all the species increases, because the volume decreases.

Equilibrium shifts to the right to alleviate some pressure.

When the system returns to equilibrium, the value of K will be the same as it was prior to the change.

EDIT:
Before equilibrium is reached, however, the value of the concentration fraction differs from the K value.

[fady_22]

Sorry, I'm not following.
2S02 + 02 → 2SO3
This system is in a container at equilibrium. The container is shrunk, increasing pressure, but temperature stays the same. LCP says this forces the reaction forward, increasing [S03], doesn't it?

Initially the concentration of all products and reactants increases (as the volume decreases).



The unit is M-1. Therefore, the change in concentration on the bottom is essentially cubed (although this is not technically correct) and while the change on the top is squared. Hence, K is smaller than it should be. So, to increase K, there is a net forward reaction, increasing the concentration of SO3.

Overall, the fraction remains the same at equilibrium.

Also, think about the case when there are the same number of particles on either side of the equation. There is no change because the change in concentration affects the bottom and top of the equilibrium constant expression in the same way.

[Duck]

Aha. Thank you all, I believe I understand now. So the rate of yield may increase but only temporarily while the system is not at equilibrium. A round of karma for all