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Aqa chem 4/ chem 5 june 2016 thread

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Original post by Kay Fearn
It wanted enthalpy change of the reaction not standard enthalpy of formation?


Thats what I remember they asked for, I'm getting so stressed atm 😂😂


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Original post by Suits101
Enthalpy change can be calculated from standard enthalpy of formation, which is what you were given and used.



Yes but people are saying you have to divide the so3 and so2 values by 2

it asked for the enthalpy change of the reaction so i don't think you had to divide by 2?
Original post by Osalim1998
thing is though, the given values for the question if you didnt get an answer was -250. not dividing by two gave you a closer answer, thats why im thinking you dont have to divide

those two values are normally irelevant.I just could not get over STANDARD
Original post by Kay Fearn
It wanted enthalpy change of the reaction not standard enthalpy of formation?


I genuinely don't recall them asking for standard either, dunno if Azzer is right or not.

confirm anyone who remembers what the deltaH question asked for?
Original post by tbh12
No it wasn't. -250 was given value for delta H.


and for delta s aswell
Original post by girl :D
omd no i didn't hear anything. where did u hear this? i thought that it might be on ATP
gud luk


some of my classmates were saying it, I am not sure but I think that it may be possible.
Original post by Kay Fearn
Yes but people are saying you have to divide the so3 and so2 values by 2

it asked for the enthalpy change of the reaction so i don't think you had to divide by 2?


It asked for standard enthalpy change

Original post by hi-zen-berg
I genuinely don't recall them asking for standard either, dunno if Azzer is right or not.

confirm anyone who remembers what the deltaH question asked for?


Calculate the standard enthalpy change for this reaction.
Original post by Kay Fearn
Yes but people are saying you have to divide the so3 and so2 values by 2

it asked for the enthalpy change of the reaction so i don't think you had to divide by 2?


I agree - the end result (delta G) is still the same. I didn't divide enthalpy or entropy by 2 because it said 'for THIS reaction' in the question.
(edited 7 years ago)
I don't think you divide by 2 because you are finding the enthalpy change for 2SO3 not SO3 ,
it is not working out the enthalpy of formation,that would be making one mole
Did anyone else get delta G as -134 or something close
yes
Original post by koolgurl14
Did anyone else get delta G as -134 or something close
Reply 2691
Avatar for Anam
Anam
11
Did anyone get I think -135 for the delta G question
For the gold jewellery one did anyone say:

E (Au+/Au) > E (O2/H2O) so Au+ ions oxidise water to oxygen and Au+ ions are reduced to Au
Original post by Aerosmith
What are you on about? You don't reach the same answers from the equation they gave you. I was trying to tell people what the correct reasoning was as a number of people appeared to be confused


I thought you were talking about a different question, never mind.

The Gibbs question asked for the free energy change for the given reaction, not the standard Gibbs energy change... Awkward.
We need someone to clear up this enthalpy/entropy dilemma ahaha. If pictures of the paper are released, we'll be able to see.
Original post by Suits101
For the gold jewellery one did anyone say:

E (Au+/Au) > E (O2/H2O) so Au+ ions oxidise water to oxygen and Au+ ions are reduced to Au


I said, in presence of H+, oxygen CANNOT oxidise Au to Au+ because E<0 and, hence, as delta G = -nFE, reaction is not feasible as delta G>0.

I don't think you need to know the Nernst equation and all that, but it may get a mark, who knows.
This paper was almost identical to Jan 13. Expecting similar grade boundaries.

The dividing by 2 bit for the enthalpy change question seems to have caught most of us out. Anyone know whether ecf will be allowed for consequent questions, a it could cost us 10 MARKS for that small error
Waiting on the unofficial mark scheme :biggrin:
Original post by Suits101
For the gold jewellery one did anyone say:

E (Au+/Au) > E (O2/H2O) so Au+ ions oxidise water to oxygen and Au+ ions are reduced to Au


It had nothing to do with oxidizing water, the question asked why solid gold would not react with moisture in the air and since E (Au+/Au) > E (O2/H2O) no reaction would take place under standard conditions.
Original post by Suits101
The question asked you to calculate the STANDARD enthalpy and entropy change, which was asked on Jan 13 I believe and you had to divide by two.


What does STANDARD have to do with it? If it asked for the enthalpy change of the REACTION (which I'm pretty sure it did) rather than formation, you wouldn't need to divide by 2? A Hess cycle would just include 2x the enthalpy of formation for SO3 (or whatever it was at the top, can't really remember).

Edit: All standard means is that the reaction is taking place under standard conditions.
(edited 7 years ago)

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