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AQA CHEM5 A2 Chemistry - 19th June 2013

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Original post by Yetix
For the Q=MxCxdeltaT I got 284K

What was the answer as people have been suggesting 311K but no one has confirmed it.

thanks


I got the same answer as you. I was about to write down 311k but realised that enthalpy of solution is endothermic for silver chloride so it must gain heat from the surrounding water hence water temperature goes down from 298k.
(edited 10 years ago)
Reply 3021
There was a question on the lattice dissociation of AgCl being same, greater or equal to
Reply 3022
Original post by teen1234
There was a question on the lattice dissociation of AgCl being same, greater or equal to


was the answer greater?
Original post by Hamburglar
Overall in my opinion, tougher than any past paper on this specification. Most of it should have been well answered but if you compare this paper to the other papers, the others were all the same and this one tried to put a different spin on things. I can see a lot of candidates dropping marks on the Q=mcT question, the question with Co2+ as a catalyst, and the very last question. These questions had quite a bit of weight to them as well, 2 of them were 5 markers.

Not only that, I can imagine that some people would lose marks on the E=hv questions by not understanding that a red solution actually absorbs blue light.

However, I can't see the boundaries changing that much to be honest because it's AQA :/

Lowest it has been so far is 79/100 for an A* on past papers. I can possibly imagine 76-77 being an A* on this paper? I could be wrong though :s



I think we all agree this is the toughest paper to date. What do you mean that the grade boundaries won't be changing because it is AQA? :confused:
Reply 3024
Original post by Dedicated1
I got the same answer as you. I was about to write down 311k but realised that enthalpy of solution is endothermic for silver chloride so it must gain heat from the surrounding water hence water temperature goes down from 298k.


That was my thinking behind it too. Lets hope it's right and that's 5 marks in the bag!
Original post by Hippokrates
Sorry :smile:
Where is this markscheme, I've only seen bits of it floating around?



I am waiting for it too. Is someone doing it?
Original post by brittanna
On 8)a), would just saying that cobalt ions have variable oxidation states and so provide an alternate reaction route with a lower activation energy get you the first mark or two?

thats what i put
Original post by Anshul91
i think it asked about temp above 500k


It was below, i read it initally and wrote above then read below.
Original post by mulac1
2) calculate the enthalpy of solution for KBr (2)
why is the hydration enthalpy of the Cl- ion larger than that of the Br- ion? (3)
the enthalpy change for this reaction is +17.3kJmol-1 , explain why the dissolution of KBr is feasible above x degrees (2)
calculate the temperature after the KBr dissolves (5)

3) write an equation linking deltaG, deltaH and deltaS (1)
why is entropy=0 at 0K (2)
why does the line slope at the start of the graph (2)
Mark on the boiling point (1)
Why is the change L2 greater than change L1 (2)
some questions missing here...

why can this graph be represented as y=mx+c (2)
what is the significance of the deltaG value below 500K (1)
and again some questions missing here..


For KBr now being a suitable salt bridge did anyone write the Pottassium would react with the water in the solution?
for the delta s question, did you have to double the answer to find 1 mole of the product? i think it was -44.5 doubled?
Original post by MrMeep2580
For KBr now being a suitable salt bridge did anyone write the Pottassium would react with the water in the solution?


Didn't it ask about a salt bridge with Cl- in?
Reply 3031
Original post by fifaking2
for the delta s question, did you have to double the answer to find 1 mole of the product? i think it was -44.5 doubled?


values given per mol already wasnt it?
What a joke! came on here looking for the mark scheme to get a bit of confidence due to a sea a crying faces at my school caused by this exam to find that, so far i think most of my answers are wrong... pretty **** way to start the summer. Please can someone who isn't predicted a D (me) and thinks they only dropped a few marks tell the rest of us where we went wrong?? would be a massive help and would defiantly fix the sinking feeling I have about kissing university good bye.
Original post by Kutie Karen
I think we all agree this is the toughest paper to date. What do you mean that the grade boundaries won't be changing because it is AQA? :confused:


It's just from experience that a lot of people find a paper really difficult, and boundaries don't tend to shift as much as people would think. Just don't want anyone to get their hopes up of the boundaries being super low or anything, I'm only expecting 2-3 marks of a drop :frown: Of course, I'd be happy to be wrong
Reply 3034
Original post by teen1234
a. Highest state of order, no vibration
b. As T increases, more kinetic energy, more vibration, greater disorder, higher entropy
c. Below L2
d. From liquid to gas at L2, larger change in entropy
e. easy
f. ( 1 mark) becomes feasable


I thought f was it isn't feasible because on the graph DG was bigger than zero between 300 and 500? :-S xx


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Reply 3035
what did people put about the y=mx+c question ? i put temp was gradient and it symbolised a negative correlation , thus negative gradient which it refers to in the equation??
Original post by tomj2013
what did people put about the y=mx+c question ? i put temp was gradient and it symbolised a negative correlation , thus negative gradient which it refers to in the equation??


Wasn't sure about this one. I just talked about proportions because a straight line graph shows proportionate increases between changes of two variables. Not sure though.
Original post by tomj2013
what did people put about the y=mx+c question ? i put temp was gradient and it symbolised a negative correlation , thus negative gradient which it refers to in the equation??


Personally, I just said G = -TS + H. G was plotted on the y axis, and T on the x-axis. The gradient was negative, and the y-intercept represented H. Not sure if it would score 2 marks though

(Can't do the delta symbol, sorry).
Reply 3038
I said gradient is= to negative entropy :s-smilie:
Original post by tomj2013
what did people put about the y=mx+c question ? i put temp was gradient and it symbolised a negative correlation , thus negative gradient which it refers to in the equation??


I wrote about the negative gradient too (not about the correlation). But I put the gradient as deltaH/deltaS. Dumb :frown:

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