OCR Chemistry A - Equilibria, Energetics and Elements (F325). Watch

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TwirlGirl
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#501
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#501
(Original post by lekky)
This is the final A2 unit right? I dont' keep track of their names/numbers. Like the mahoosive 1:45 hour one?

IN THAT CASE I shall do some bugging of people





Basically I still don't get Kc and pressure. I think I've got my head around concentration by explaining it like this in my notes.

Kc = [A]^a [B]^b / [C]^c [D]^d

(by bottom number/top number I mean like bottom concentration etc)

If the top number increases/bottom number decreases
-- Kc will increase
-- bottom number will increase
-- top number will decrease
-- equilibrium position will shift right to left
-- Kc value will be restored

If the top number decreases/bottom number increases:
-- Kc will increase
-- bottom number will decrease
-- top number will increase
-- equilibrium position will shift left to right
-- Kc value will be restored


I HAD done some notes like that for changes in pressure but when I was going through them again I realised that I can't because BOTH the bottom/top number increase/decrease just by different factors? So I can't get my head around how to write it down.. I think once I've written it down in a form like I have for concentration I might get it more. IDK if I get it now... but it feels like a superficial understanding so I don't feel confident. Help? In return you will get rep and appreciation and gratitude .. I'd offer to help you in the future but meh that's most likely worthless.

I think I understand it...

-- both the top and the bottom will increase if pressure increases. the level raised to the highest power will increase the most
-- if the top increases the most (see if the top increases/bottom decreases for concentration)
-- if the bottom increases the most (see if the top decreases/bottom increases for concentration)
-- if pressure decreases then just reverse arguements for the above
-- if top and bottom are to the same power then no changes will occur

Ummm /confused and tired/ sorry ....

Ahh i love you for this! I have to have things generalised! I need a pattern to follow :p: This has cleared things up a lot
Edit - if the bottom increases most (pressure bit) doesn't Kc decrease?
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discodan1
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(Original post by cyborg)
This is not in our spec. The spec states that calculations will have to be done for monobasic strong and weak acids only.
stretch and challenge was mentioned specifically....
it is easy though
you just add them together
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discodan1
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(Original post by win2kpro)
Hints pretty please
i will PM them so some people today
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Presidential
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ive completely lost track of this thread. why were people posting pictures with random signatures inside??
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student92
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forward reaction endothermic, increase in temp, increase in kc, product yield increases, reactant yield decreases... shifts position of equil from LHS to RHS. forward reaction exothermic, increase in temp, decrease in kc, product yield decreases, reactant yield increases... shifts position of equil from RHS to LHS. Kc is more than 1 - product favoured, Kc is less than 1 - reactant favoured. is that ryt?
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cyborg
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(Original post by student92)
forward reaction endothermic, increase in temp, increase in kc, product yield increases, reactant yield decreases... shifts position of equil from LHS to RHS. forward reaction exothermic, increase in temp, decrease in kc, product yield decreases, reactant yield increases... shifts position of equil from RHS to LHS. Kc is more than 1 - product favoured, Kc is less than 1 - reactant favoured. is that ryt?
:yep: Looks ok to me.
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TwirlGirl
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What happens if the pressure is increased, and there is a larger amount of gas in the reactants? What does Kc do and how is eqm restored?

Is it even possible!? :p:
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student92
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(Original post by TwirlGirl)
What happens if the pressure is increased, and there is a larger amount of gas in the reactants? What does Kc do and how is eqm restored?

Is it even possible!? :p:
well if there is an increase in pressure the position of the equilibria would shift to the side with least moles - the RHS in this case. And Kc im guessing would increase - not entirely sure. EDIT: by larger amount do u mean larger amount of moles?
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andy892
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(Original post by student92)
well if there is an increase in pressure the position of the equilibria would shift to the side with least moles - the RHS in this case. And Kc im guessing would increase - not entirely sure.
Kc values do not change. They must always remain the same for changes in concentration and pressure.
So, for the equilibrium:
2X (g) <---> Y(g)
Let's say to begin with X has a concentration of 1mol and Y has a conc of 1mol too.
If the concentration of Y is increased to 2 mol. The value for Kc would now not be the same - i.e. it would be 0.5, instead of 1 to begin with. The equilibrium will move to the LHS to restore this change, increasing the concentration of X to ensure that the Kc value is always at 1.

A change in pressure is effectively the same.
Equilibrium will always move to the side with the fewest gas molecules to best minimise the change.
Thus Kc value will not change again.
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andy892
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(Original post by student92)
well if there is an increase in pressure the position of the equilibria would shift to the side with least moles - the RHS in this case. And Kc im guessing would increase - not entirely sure. EDIT: by larger amount do u mean larger amount of moles?
or larger number of gas molecules even.
ONLY gas molecules count for that.
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andy892
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(Original post by Presidential)
ive completely lost track of this thread. why were people posting pictures with random signatures inside??
Haha, well the OCR A2 Chemistry book's authors - Dave Gent & Rob Ritchie - carried out a revision session yesterday at some school / college. And a couple of people on here got their books signed and a photo taken with them.
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student92
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(Original post by andy892)
or larger number of gas molecules even.
ONLY gas molecules count for that.
true! taaaaa . I don't think this is ryt but still: if the position of the equil shifts to the RHS - kc always increases, and when positon of the equil shifts to LHS kc always decreases... because I always follow that rule and I get the ryt answer, well from the questions ive done
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Presidential
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(Original post by andy892)
Haha, well the OCR A2 Chemistry book's authors - Dave Gent & Rob Ritchie - carried out a revision session yesterday at some school / college. And a couple of people on here got their books signed and a photo taken with them.
oh i see, was the session useful?
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andy892
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(Original post by student92)
true! taaaaa . I don't think this is ryt but still: if the position of the equil shifts to the RHS - kc always increases, and when positon of the equil shifts to LHS kc always decreases... because I always follow that rule and I get the ryt answer, well from the questions ive done
Kc values do not change for changes in conc / pressure
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andy892
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(Original post by Presidential)
oh i see, was the session useful?
Well I didn't go unfortunately, nor did I have the chance to do so!
However, it does sound like it was pretty useful tbh.
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TwirlGirl
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(Original post by andy892)
All depends upon which side now has the highest value.
If bigger value on bottom of Kc expression (hence more reactants) then Kc will decrease.
If bigger value on top of expression (hence more products) then Kc will increase.
So I guess in your example, Kc would increase.
Okay, so if for example the equation was something like:

Kc = [X]/[y]^2

(and the pressure was increased...)

meaning there were more moles of gas on the bottom...Kc would decrease?
1) [bottom] would decrease
2) [top] would increase
3) and equilibrium would shift from left to right so the Kc value would be restored?

(or would 1) and 2) be the other way around? )
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student92
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(Original post by andy892)
Yeah that has to be correct.
If it moves to the RHS, there will be more products, so will be a larger number on top of expression than before, thus Kc increases.
If it moves to the LHS, there will be more reactants, so larger number on bottom of expression than before, thus Kc decreases.
thanks!
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andy892
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(Original post by TwirlGirl)
Okay, so if for example the equation was something like:

Kc = [X]/[y]^2

(and the pressure was increased...)

meaning there were more moles of gas on the bottom...Kc would decrease?
1) [bottom] would decrease
2) [top] would increase
3) and equilibrium would shift from left to right so the Kc value would be restored?

(or would 1) and 2) be the other way around? )

Kc values are not affected by changes in concentration or pressure.
They always remain the same.
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andy892
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Effectively it is a simple recall of Le Chatelier's principle for this.
If concentration increases, the equilibrium will move to the side that best minimises the change.
If pressure is increased, the equilibrium will move to the side with the fewer gas molecules.

So:

Kc values do not change. They must always remain the same for changes in concentration and pressure.
So, for the equilibrium:
2X (g) <---> Y(g)
Let's say to begin with X has a concentration of 1mol and Y has a conc of 1mol too.
If the concentration of Y is increased to 2 mol. The value for Kc would now not be the same - i.e. it would be 0.5, instead of 1 to begin with. The equilibrium will move to the LHS to restore this change, increasing the concentration of X to ensure that the Kc value is always at 1.

A change in pressure is effectively the same.
Equilibrium will always move to the side with the fewest gas molecules to best minimise the change.
Thus Kc value will not change again.
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andy892
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(Original post by student92)
thanks!
See above post. Need to check this is correct. It wasn't entirely correct what I put before.
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