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ocr a f325 revision thread

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Original post by zef1995
Thank you very much :smile:

That's spot on, well done!


No worries!


Describe the factors that affect Kc and how Kc can be increased (3)
got my mock for this paper soon!
really enjoying mod 2 and 3, mod 1 with pH buffers etc is tedious sometimes
Reply 702
Original post by MedMed12
got my mock for this paper soon!
really enjoying mod 2 and 3, mod 1 with pH buffers etc is tedious sometimes


Same here, I have mine on Wednesday! Are you doing the Jan 13 paper as your mock?

Good luck, I hope it goes well :smile:
Original post by zef1995
Same here, I have mine on Wednesday! Are you doing the Jan 13 paper as your mock?

Good luck, I hope it goes well :smile:



yep I think so :smile:
thank you!! you too :h:
Reply 704
Original post by otrivine
No worries!


Describe the factors that affect Kc and how Kc can be increased (3)


Kc is not changed by concentration, pressure, or the use of catalysts - it is only affected by temperature. If for example, we have a reaction in which the forward reaction is exothermic:
-if the temperature is increased, the endothermic (reverse) reaction will increase in rate, increasing the yield of reactants. As there are more reactants and less products, Kc will decrease.
-if temperature is decreased, the rate of the exothermic reaction increases, increasing the yield of the products and decreasing that or the reactants. Kc will increase.

What can increase the entropy of a system?
Reply 705
Original post by MedMed12
yep I think so :smile:
thank you!! you too :h:


Same here, don't really know what to expect (although I'm pretty sure hat's the idea) :P
Thank you :biggrin:
Original post by zef1995
Kc is not changed by concentration, pressure, or the use of catalysts - it is only affected by temperature. If for example, we have a reaction in which the forward reaction is exothermic:
-if the temperature is increased, the endothermic (reverse) reaction will increase in rate, increasing the yield of reactants. As there are more reactants and less products, Kc will decrease.
-if temperature is decreased, the rate of the exothermic reaction increases, increasing the yield of the products and decreasing that or the reactants. Kc will increase.

What can increase the entropy of a system?


excellent, nothing to add again :tongue:


Entropy could be increased by increasing more gaseous molecules on RHS of equation than LHS,

when you dissolve a solid in solution/liquid

increasing temperature
Original post by zef1995
Same here, don't really know what to expect (although I'm pretty sure hat's the idea) :P
Thank you :biggrin:


my teacher gave us a 'hint' and said its mostly mod 2 and 3
Original post by MedMed12
my teacher gave us a 'hint' and said its mostly mod 2 and 3


you mean the exam in summer ? :smile:
Reply 709
Original post by otrivine
excellent, nothing to add again :tongue:


Entropy could be increased by increasing more gaseous molecules on RHS of equation than LHS,

when you dissolve a solid in solution/liquid

increasing temperature


Wow, thanks again :smile:

That's perfect, well done! :biggrin:
Reply 710
Original post by MedMed12
my teacher gave us a 'hint' and said its mostly mod 2 and 3


Ah okay, thank you for that! I've been told that this paper is hard, have you been told the same thing?
Original post by zef1995
Wow, thanks again :smile:

That's perfect, well done! :biggrin:




Give the difference between Ka and Kc (2)
Reply 712
Original post by otrivine
Give the difference between Ka and Kc (2)


Kc is the equilibrium constant, so will give an idea of where the position of equilibrium of a system lies (but not how fast the reaction is going). High value of Kc = equilibrium to the right.
Ka is the acid dissociation constant, and will indicate the extent of dissociation for an acid. High Ka value = strong acid.

I hope that's on the right lines :P
Original post by zef1995
Kc is the equilibrium constant, so will give an idea of where the position of equilibrium of a system lies (but not how fast the reaction is going). High value of Kc = equilibrium to the right.
Ka is the acid dissociation constant, and will indicate the extent of dissociation for an acid. High Ka value = strong acid.

I hope that's on the right lines :P


wow! you are on a roll excellent :biggrin:
Reply 714
Original post by otrivine
wow! you are on a roll excellent :biggrin:


That was correct? Wow, thank you :biggrin:

Please describe some limitations using hydrogen in a fuel cell :smile:
I miss doing these questions but I have to reframe myself this weekend as I should be doing some Maths instead even though its incredibly tedious :-/


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Original post by zef1995
That was correct? Wow, thank you :biggrin:

Please describe some limitations using hydrogen in a fuel cell :smile:


1) fuel cell uses toxic chemicals in their productions
2) fuel cells have limited life time and requires regular replacement and disposal following high production costs
3) Current adsorbers and absorbers have a limited life time
4) the feasbility of storing a pressurised liquid
5) the storage and trasnport of hydrogen is expensive
Reply 717
Original post by otrivine
1) fuel cell uses toxic chemicals in their productions
2) fuel cells have limited life time and requires regular replacement and disposal following high production costs
3) Current adsorbers and absorbers have a limited life time
4) the feasbility of storing a pressurised liquid
5) the storage and trasnport of hydrogen is expensive


I'm sorry for the late reply, sleep called!

Yeah that sounds good to me :smile:
Original post by otrivine
you mean the exam in summer ? :smile:


Original post by zef1995
Ah okay, thank you for that! I've been told that this paper is hard, have you been told the same thing?


I mean the January paper -its our mock haha I think if our teacher knew what was on the real exam she wouldn't tell-the amount of trouble she'd be in!

Ahh yes I've also heard that its the worst one yet, so hopefully June 2013 will be nicer ;D
Original post by zef1995
I'm sorry for the late reply, sleep called!

Yeah that sounds good to me :smile:


Not a problem :smile:


Define:Lattice Enthalpy

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