OCR A2 CHEMISTRY F324 and F325- 14th and 22nd June 2016- OFFICIAL THREAD

Hi guys , you know for stability constant , why when a high value of stability constant equalilibrium lies to the right? I don't get why that is please explain

Hi guys , you know for stability constant , why when a high value of stability constant equalilibrium lies to the right? I don't get why that is please explain

if you look at the Kc equation, its product divided by reactants...but for a second ignore that and imagine 1 divided by 1 =1 ....now if your kc value is less than 1..you can see that there is more reactant than product...so it turns to 1 divided by 2 where 1 is product and 2 is reactants = 0.5 kc value...but if KC is very large than it would be e.g 5 divided by 1 = 5 so kc of 5....once you get this all you need to know if KC is very large the equilibrium must lie to right as the top number in division needs to be bigger than the bottom in kc=product/reactant

Like all equilibrium constants...

Simply, the equation is K = [Products]/[Reactants] where the concentration is taken when the reaction is at equilibrium.

It becomes a maths question after this. How do you get a large K value? Well, by either having a larger numerator or a small denominator. It's essentially a ratio so either one will work. I'll use a large numerator to explain though.
What does a large numerator represent? It represents a large concentration of products at equilibrium. You have to think logically that if there is a large concentration of products at equilibrium then the reaction must favour the forward reaction and so the equilibrium lies to the right.

Another sanity check would be, if there were equal concentrations of reactants and products where would it lie? Obviously it is in the middle. And as [reactants]=[products], then [products]/[reactants] = 1.
So you know a stability constant of 1 lies in the middle, a large constant (being greater than 1) must lie to the right of it. It's called a stability constant because the reactions are reversible, therefore if you have a large number of products in equilibrium then the product must be stable! Something that is stable is unlikely to change.

Specifically for Kstab, the reaction is a ligand substitution and so really what it's telling us is which complex is more stable. You should know that in chemistry things like to be in it's lowest energy state where it's most stable. If the value of Kstab was less than 1 then you know that the reactant is more stable and therefore unlikely to undergo ligand substitution into a less stable complex.

I've rambled on. I'm trying to show you a few ways of looking at it. If it still goes over your head just rote learn "high Kstab value, equilibrium lies to the right" and you might get lucky.

if you look at the Kc equation, its product divided by reactants...but for a second ignore that and imagine 1 divided by 1 =1 ....now if your kc value is less than 1..you can see that there is more reactant than product...so it turns to 1 divided by 2 where 1 is product and 2 is reactants = 0.5 kc value...but if KC is very large than it would be e.g 5 divided by 1 = 5 so kc of 5....once you get this all you need to know if KC is very large the equilibrium must lie to right as the top number in division needs to be bigger than the bottom in kc=product/reactant

Does anyone know if they will ask for any of the AS mechanisms in the A2 exams?

Does anyone know if they will ask for any of the AS mechanisms in the A2 exams?

It's not impossible.

Maybe because the syllabus is synoptic

Will have to get my AS knowledge up to scratch then

I get that nitration of benzene has to be carried out at temperatures below 50 degrees to prevent multi-substitution but could anyone explain WHY multi-substitution occurs at temperatures above 50 degrees??

The value of 50 degrees is fairly random tbh. To rationalise it though just think about how rate increases with T. More collisions attain greater than activation energy.