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

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Original post by AqsaMx
2SO2 + O2 > 2SO3

Explain what would happen to the pressure as the system was allowed to reach equilibrium

I wrote the pressure would decrease as wquilibrium would shift to the left, but the MS says pressure decreases and fewer moles? Could anyone explain what Is meant by this?



The pressure would decrease because at equilibrium there are only 2 moles of gas on the right hand side. If you can imagine having 2SO2 + O2 before any reaction there are 3 moles, however some of that 3 moles is converted into 2SO3 (Le Chatelier's principle) so overall there is a decrease in pressure.
anyone understand 7b from june 2012??????
Original post by KB_97
Lmao I don't even have an offer. I'm taking a gap year but still, I need the grades to maximise my chances. And it just feels like after all this hard work, I'm still not gonna get the grades I want. If only I was born 4 years earlier...the 2012 papers for all my subjects were so easy, I would've had 3A*. But Life's a b****. Good luck to you and everyone else for tomorrow anyway.


Haha ah fair enough, yeh I completely agree it feels like all the past papers/revision you do were for nothing when an exam goes like that arghhh. Oh well we go again tomoz, cheers and good luck to you too.
I feel strangely unprepared for this exam... anyone else?
Original post by Sab_007
anyone understand 7b from june 2012??????


CO2 reacts with water and forms carbonic acid and H+ ions (I only know this because of biology)
So basically CO2 is acidic in solution and the H+ would react with the OH- ions in the equation so equilibrium shifts to the right to restore the OH- ions
Could someone help me understand how to do this question please? image.jpg
Original post by tcameron
is the cathode the positive electrode?


CPR :wink:
Cathode = Positive = Reduction
Original post by rory58824
I feel strangely unprepared for this exam... anyone else?


Original post by Rust Cohle
Jan 13 Q8(c)(ii) explained and worked through



I still don't understand how you got the 5:3 ratio
Could someone pls explain biii? https://gyazo.com/9b6f88ef2b4b5858ad75ea9f3f3bfd36

Also for enthalpy change of neut do you DOUBLE volumes for m (in Q=mcdeltaT) and only do one volume for moles? ty vm
for ci https://gyazo.com/2d3b610ec0ec6ae199ede503f4da0794

how do you get first equation here:?

https://gyazo.com/053605e63dbfd229d81453c968d2033e

I thought I- would react with Fe3+ with accordance to electrode potentials?
https://gyazo.com/49069f767bf5ea95d7e870c6a979a5e2 How do we know that 2 mol X reacts with 2 mol Cr? 1.456/1.021 = 1.43, do we just round 1.43 up to 1.5? seems a bit off :/
Original post by itsConnor_
Could someone pls explain biii? https://gyazo.com/9b6f88ef2b4b5858ad75ea9f3f3bfd36

Also for enthalpy change of neut do you DOUBLE volumes for m (in Q=mcdeltaT) and only do one volume for moles? ty vm


Use your Q from previous question in this rearranged Q = mCdT

For iii:
dT = Q/(mc). Then x1000 if you used kJ (as opposed to J) for Q.
Same number of moles as in (ii) except occupying a larger volume.

For ii:
m = sum of reactants = 35+35 = 70.
For the n, yes, you use only the moles of the limiting reagent (whichever has lower moles). Here they're both equal moles (which is common for acid+base) so irrelevant.

-Q is always around -54.6 kJ for future reference

-Q/n = -4.827 kJ (exothermic)
Original post by VMD100
CPR :wink:
Cathode = Positive = Reduction


uhh cathode = negative?
Reply 2893
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taysc
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Original post by RayMasterio
guys can anyone help me with June 2015 question 2d part i, q 4c part ii, and part d iii


Im not sure if anyone has helped you yet?

2di) in step 2 you have formed something that is not in your overall equation. Hence it is an intermediate. Then if you look at step 1/step 2 reactants and compare it to the overall equation you will see that it has only used 1 mol of H2. But there is 2 mols of H2 in overall equation. So this must mean you need to react something e.g your intermediate with H2

step 3 : H2 + N2O -> N2 + H2O

(You can then check if your products in all your steps equal the overall equation)

4c) If you dilute an acid the ph increases (becomes less acidic). So your concentration of your acid has to change. The first part is calculating the change in concentration and then subbing in values.

So you need to initially workout in the no moles of acid= c x v
no moles = (25/1000)*0.480 = 0.012.

Then you take this value and work out the new concentration of acid in the 100cm3 c=n/v

c= 0.012/(100/1000) = 0.12

You can then use this a concentration of acid and sub into Ka equation

I cannot explain 4e(ii) because i dropped a mark in that question. the basis of it is that when you add metal to an acid it react

acid + metal -> salt + hydrogen

Therefore there will be a decrease in [acid] and an increase in the [conjugate base]

This is pretty poorly explained, I'm not amazing at writing my thought process down. I hope I have helped a little. If you need me to explain more I will try too.

PLEASE CORRECT ME IF I AM WRONG
(edited 7 years ago)
Original post by postexamtalk
uhh cathode = negative?


Nope
http://hyperphysics.phy-astr.gsu.edu/hbase/chemical/electrode.html

This diagram is clearer
diagram.jpg
The reactions we are doing generate electrical energy, not require it.
(edited 7 years ago)
Original post by Rust Cohle
Use your Q from previous question in this rearranged Q = mCdT

For iii:
dT = Q/(mc). Then x1000 if you used kJ (as opposed to J) for Q.
Same number of moles as in (ii) except occupying a larger volume.

For ii:
m = sum of reactants = 35+35 = 70.
For the n, yes, you use only the moles of the limiting reagent (whichever has lower moles). Here they're both equal moles (which is common for acid+base) so irrelevant.

-Q is always around -54.6 kJ for future reference

-Q/n = -4.827 kJ (exothermic)


thanks so much. your worked solution for the charge of Vn+ is amazing too :smile: good luck tomorrow :smile:
F325 will be impossible. I'm calling it. F324 was freakishly easy, and they can't make F325 easy again or too many people will get an A*. Prepare your anus lads and lasses.
Original post by itsConnor_
for ci https://gyazo.com/2d3b610ec0ec6ae199ede503f4da0794

how do you get first equation here:?

https://gyazo.com/053605e63dbfd229d81453c968d2033e

I thought I- would react with Fe3+ with accordance to electrode potentials?


In accordance to electrode potentials, the one with the highest electrode potential is forward. The one with the more negative electrode potential has the reverse reaction, balance and add together and remove electrons.
Is there a set of rules for determining half equations when given the full equation? I get confused as to which species are involved in each half equation.
Original post by Lucy985
Is there a set of rules for determining half equations when given the full equation? I get confused as to which species are involved in each half equation.


The best rule to go by is that both half equations have to be balanced, and they have to combine to give the full equation. Normally they give you one half and a full - you have to work out the other half.

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