Question about K/Equilibrium

[Tashi]

Could someone please explain to me why "the value of K for a particular reaction depends only on temperature. It is not affected by actions such as addition of reactants or product, changes in pressure, or the use of catalysts"? Is a shift in equilibrium not the same as changing the value of K?

[tony3272]

When you make a change in a reaction, it will no longer be in equilibrium and so the reaction will occur as to partially oppose this change. a.k.a LeChat

So you can just look at these different changes individually:

When you make a change in concentration/volume then the change will be to offset this and bring the reaction quotient back to K.

However, when a change in temperature is made, the concentrations of everything are still the same (the reaction quotient does not change). But in order to oppose this change the reaction must occur as to oppose this change in temperature. So a net forward or backwards reaction will occur and a new equilibrium will be made. In this case the value of the reaction quotient at equilibrium will not be the same as the original value of K.

[BoredSatan]

A position shift in equilibrium DOES NOT mean a change in K

K is a measure of the ratio of the concentration of products to reactants whilst the position of equilibrium is purely a measure of how much products and reactants there are.

When you increase volume/pressure, this will affect the position of equilibrium as more products/reactants are produced as a result of the change BUT K remains constant because of Le Chatelier's principle in which the system will partially oppose the initial change. Therefore in these systems, for example, the products/reactants will increase initially due to the initial change, but then one will decrease in order to restore equilibrium and this change will cause the value of K to remain constant.

However in a change in temperature, Le Chateliers works by forcing the reaction forward or backward (depending on if its endo/exothermic) and this will change the value of K as the change to oppose isnt caused by a decrease or increase in reactants or products thus changing the ratio..

I hope this is correct cause this is how i understand it

[vea]

K is a measure of the ratio of the concentration of products to reactants whilst the position of equilibrium is purely a measure of how much products and reactants there are.

The measure of the ratio of [products] to [reactants] is actually the concentration fraction. K is the concentration fraction AT equilibrium. While the concentration fraction may change when there is a change on the system (besides temperature), the system will always restore equilibrium and so the value of K is the same.

[BoredSatan]

my bad vea.. thanks 

[Tashi]

Man this is confusing me.

[Tashi]

If you increase pressure (increasing concentration of all substances so you still have the same ratio?), and the system shifts to the side with less particles in order to oppose this... doesn't the concentration of either the reactants or products decrease and the concentration of the other side increase, causing the ratio to be different?

[tony3272]

If you increase pressure (increasing concentration of all substances so you still have the same ratio?), and the system shifts to the side with less particles in order to oppose this... doesn't the concentration of either the reactants or products decrease and the concentration of the other side increase, causing the ratio to be different?

Not quite. if the concentration of everything increases then you will no longer have the same ratio (unless the sum of the stoichiometric coefficients is the same on both sides).

For example if you had C <---> A+B, then when you double the concentration of everything the reaction quotient will be . The ratio will no longer be the same and so a net backwards reaction will occur to restore the reaction quotient to K.

[Tashi]

Ohhh ok I think I get it now. Thanks guys

[vea]

If you increase pressure (increasing concentration of all substances so you still have the same ratio?), and the system shifts to the side with less particles in order to oppose this... doesn't the concentration of either the reactants or products decrease and the concentration of the other side increase, causing the ratio to be different?

If you increase pressure and hence concentration, it does not necessarily mean that the ratio is the same.
eg. A<-->2B
Start: [A]=2M and [.B]=1M at equilibrium  (K=0.5M)

Let's say the pressure is doubled so the concentrations are doubled.
[A]=4M but [.B]=4M   (CF=4M)
This is because we have the co-efficient 2 in front of the B (we can actually write the equilibrium reaction as A<-->B+B).

To oppose the increase in CF , the concentration of products must decrease and the concentration of reactants must increase again in order to get the same ratio of 0.5M that we had at the start.

Hope that made sense. :S

EDIT: Wrote [.B] for concentration of B otherwise everything would become bolded.

[Andiio]

Edit: NVM didn't see her comment!