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Edexcel A2 Chemistry Exams -6CH04 (14th June) and 6CH05 (22nd June) Discussion Thread

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Original post by Don Pedro K.
Well in the Jan 15 (IAL) paper for example I failed to realise that the NaOH used for titration reacted with the equilibrium mixture of ethanoic acid + HCl acid rather than the initial mixture.

I also got this one completely wrong (section C):

https://c838cff4741acb48ae1ed62e59927ff4c2073557.googledrive.com/host/0B1ZiqBksUHNYbVpyRTRJN1lFS00/June%202014%20QP%20-%20Unit%204%20Edexcel%20Chemistry.pdf


June 14 UK or June 14 R has a good question like this. In general, most of the past paper questions are straightforward so you'll have to look at other resources (like PMT quite possibly)
Guys, has anyone got the jan 16 (IAL)? I will really appreciate it if you could send me the link - its locked on the edexcel website
https://c838cff4741acb48ae1ed62e59927ff4c2073557.googledrive.com/host/0B1ZiqBksUHNYbVpyRTRJN1lFS00/June%202011%20QP%20-%20Unit%204%20Edexcel%20Chemistry.pdf


how do we do Q6?
https://c838cff4741acb48ae1ed62e59927ff4c2073557.googledrive.com/host/0B1ZiqBksUHNYbVpyRTRJN1lFS00/June%202014%20(IAL)%20QP%20-%20Unit%204%20Edexcel%20Chemistry.pdf

can someone please explain how to do Q16?
Original post by imnoteinstein


Okay firstly work out the number of moles of HCl to begin with. This is
(volume * concentration / 1000) = (10*1.0)/1000 = 0.01 moles of HCl.

Since HCl is a strong acid, it will dissociate completely, meaning that the concentration of HCl = H+ ion concentration.

What will the concentration of [H+] be in 990cm^3 of water?

Rearranging the formula above, you get:

Concentration = (moles * 1000) / volume.
= (0.01 * 1000) / 990 = 0.01010101.....

pH = -log([H+]) = -log(0.0101010101...)

Therefore the pH = 1.99... = 2 when you round it up. So the answer should be C!
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Avatar for n2697
n2697
11
Original post by Wunderbarr
My thoughts are:

a), if ΔS(soln) is exothermic then ΔS(surr) will be positive, so ΔS(total) will be more likely to be positive overall.

b), at the equivalence point, all of the initial base/acid has been neutralised by the added acid/base, so at the half equivalence point half of the initial base/acid has been neutralised. So [HA] = [A-] .


Thanks for replying but one quick question - how is ΔS(soln) linked to ΔS(surr)? Is there an equation?
Original post by Don Pedro K.
Okay firstly work out the number of moles of HCl to begin with. This is
(volume * concentration / 1000) = (10*1.0)/1000 = 0.01 moles of HCl.

Since HCl is a strong acid, it will dissociate completely, meaning that the concentration of HCl = H+ ion concentration.

What will the concentration of [H+] be in 990cm^3 of water?

Rearranging the formula above, you get:

Concentration = (moles * 1000) / volume.
= (0.01 * 1000) / 990 = 0.01010101.....

pH = -log([H+]) = -log(0.0101010101...)

Therefore the pH = 1.99... = 2 when you round it up. So the answer should be C!


Ahhhh i get it now, thanks so much!
Original post by n2697
Thanks for replying but one quick question - how is ΔS(soln) linked to ΔS(surr)? Is there an equation?


Well, ΔS(soln) can be thought of as the enthalpy change, which would then be used in calculating ΔS(surr) using the forumula:
-ΔH/1000 where ΔH is in joules per mole. That's how I think they're linked :smile:
https://c838cff4741acb48ae1ed62e59927ff4c2073557.googledrive.com/host/0B1ZiqBksUHNYbVpyRTRJN1lFS00/June%202013%20(R)%20QP%20-%20Unit%204%20Edexcel%20Chemistry.pdf

could someone please explain Q10?

does B come out the fastest bec its the most polar one and the eluent is polar and polar substances 'dissolve' in polar substances?
(edited 7 years ago)
Original post by n2697
x


Original post by Don Pedro K.
x


Yeah I just realised it was supposed to say ΔH(soln), not ΔS(soln). :colondollar:

But yes I do believe The Don P. K. is correct in that ΔH(soln) can be used in the formula to calculate ΔS(surr).
Does anyone know where I can find Jan 16 IAL Chem4? Thanks

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Is Gibbs Free Energy part of our specification, it's mentioned in the textbook but not in the specification? :redface:
Original post by Adorable98
Is Gibbs Free Energy part of our specification, it's mentioned in the textbook but not in the specification? :redface:


Nar fam.
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samb1234
OP
13
Original post by Adorable98
Is Gibbs Free Energy part of our specification, it's mentioned in the textbook but not in the specification? :redface:


It's just another way of doing feasibility questions, don't worry about it
Original post by samb1234
It's just another way of doing feasibility questions, don't worry about it


I see! So I guess we could be asked to use to use the formula ΔGθsys = ΔHθsys – TΔSθsys in the exam?
Hey thanks very much
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samb1234
OP
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Original post by Adorable98
I see! So I guess we could be asked to use to use the formula ΔGθsys = ΔHθsys – TΔSθsys in the exam?


Only if they told you what it was
Original post by samb1234
Only if they told you what it was


I see thank you!
Original post by samb1234
Only if they told you what it was

Do you know where I could find the jan 16 IAL paper ?
Original post by Wunderbarr
Yeah I just realised it was supposed to say ΔH(soln), not ΔS(soln). :colondollar:

But yes I do believe The Don P. K. is correct in that ΔH(soln) can be used in the formula to calculate ΔS(surr).


^_^

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