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

1. 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
2. (Original post by Miminfl)
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
3. (Original post by Miminfl)
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
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.
4. (Original post by 76584)
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
You beat me to it
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.
That makes so much sense, I wish the book explained it to us this way thank you I got it now
6. (Original post by 76584)
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
thankyou ) I understood it now appreciate your help
7. 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??
8. Does anyone know if they will ask for any of the AS mechanisms in the A2 exams?
9. (Original post by rory58824)
Does anyone know if they will ask for any of the AS mechanisms in the A2 exams?
It's not impossible.
10. (Original post by rory58824)
Does anyone know if they will ask for any of the AS mechanisms in the A2 exams?
Maybe because the syllabus is synoptic
11. (Original post by alow)
It's not impossible.
(Original post by Miminfl)
Maybe because the syllabus is synoptic
Will have to get my AS knowledge up to scratch then
12. (Original post by rory58824)
Will have to get my AS knowledge up to scratch then
Same, I'm only sitting F325 so I'll be ****ed if they include a lot of it in.
13. Anyone know if the edexcel and aqa papers really help in your revision? or is it best to just stick to ocr?
14. (Original post by Hunnybeebee)
Anyone know if the edexcel and aqa papers really help in your revision? or is it best to just stick to ocr?
AQA is decent for the by topic questions on the same ones covered by OCR. They go deeper in some aspects (transition metal reactions) but not so much in others. I wouldn't bother so much with doing full papers though.
They mostly ask the same questions anyway so as long as you have the standard answers all right then just attempt any difficult question that makes you stop and actually think. For the AQA papers, if I can confidently answer it, I just say it out loud and move on rather than wasting time writing the whole solution or answer.

I find the legacy papers better than doing AQA.
AQA is decent for the by topic questions on the same ones covered by OCR. They go deeper in some aspects (transition metal reactions) but not so much in others. I wouldn't bother so much with doing full papers though.
They mostly ask the same questions anyway so as long as you have the standard answers all right then just attempt any difficult question that makes you stop and actually think. For the AQA papers, if I can confidently answer it, I just say it out loud and move on rather than wasting time writing the whole solution or answer.

I find the legacy papers better than doing AQA.
I just skimmed through the AQA ones and I agree, they're quite similar. Although haven't looked at the mark schemes. Do you know where I can hold of legacy papers or are they the only ones on the ocr website? thank you
16. (Original post by Hunnybeebee)
I just skimmed through the AQA ones and I agree, they're quite similar. Although haven't looked at the mark schemes. Do you know where I can hold of legacy papers or are they the only ones on the ocr website? thank you
Pastpapers.org
17. (Original post by AqsaMx)
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.

18. Could anyone explain question 3? I'm a little confused on what I'm even supposed to do and for question 4, I got 3+ but not sure if this is right, could anyone check it for me? Thank you )
19. (Original post by TeachChemistry)
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.
So because the temperature is increased, it reaches a high enough activation energy to add on two nitro groups or more?

Okay makes sense thank you
20. (Original post by AqsaMx)

Could anyone explain question 3? I'm a little confused on what I'm even supposed to do and for question 4, I got 3+ but not sure if this is right, could anyone check it for me? Thank you )
For question 3 you may need to work out unknown info about KI but we're not sure if it's conc or vol or mols that they want to know

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