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    (Original post by j5994)
    did anyone get +2404 or something for born haber?

    I got something like 2328 KJ Mol-1 at first, then I halved the atomisation of oxygen to get 2404, but I think the original answer of 2328 was correct, because atomisation is for one mole, and you got 1 mole of O
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    (Original post by Cadherin)
    You're wrong.

    The question asked 'for this reaction' - this is the standard reaction that occurs when OXYGEN is in its standard state (O2) is the one that was given.

    Halving the values would be meaningless and would not be for the standard reaction when reagents and products are in their standard states.
    I now think both routes are valid
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    what did you have to write for the 12 diaminoethane thermodynamics question?
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    (Original post by Suits101)
    You'll be fine.



    Hey just wanted to confirm something:

    Does SA effect emf of cell?

    Would saying increase in pressure increase emf value, equilibrium shifts to RHS be ok for 2 marks do you think?

    Thanks.
    SA has no effect as it referred to BOTH platinum electrodes.

    Yes, emf did increase at greater pressure and your reasoning is correct, although I think I worded it slightly differently with reference to the p.d. produced in a given time with an increased reaction rate.
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    (Original post by Spectral)
    The reaction in the contact process is not delta Hf, SO2 and O2 are not 'elements'
    It is delta Hr, there is no need to divide by 2, in fact technically it is wrong to divide by 2

    As a simple analogy:
    For this reaction A + B -> C I'm sure you'd be fine with this being standard
    About this one: A + B -> C + 2D Well C is 1 mol, about D?

    That's where the argument of 1 mol breaks down. In fact, standard enthalpy of reaction specifically does not specify 1 mol for that very reason - because it's incorrect
    I get your point but in the question it stated delta H f because I was working out it without halving till I saw that it had the symbol there was no definition on the paper and you had to apply your knowledge of that why would they state delta Hf if they didn't want one mol
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    (Original post by Suits101)
    It came up!!!
    Thank you so much man, you really did me a solid there
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    Aerosmith

    Explain to me how this is any different to the question. You're provided with enthalpy of formation values and asked to work out the standard enthalpy change of the following reaction exactly as we did.

    Just telling you now, there's absolutely no diving in the mark scheme.
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    (Original post by Abc321zxc)
    what did you have to write for the 12 diaminoethane thermodynamics question?
    4 moles or reactants forming 7 moles produces therefore entropy increasing.
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    (Original post by RME11)
    Nice back-peddle, I suggest you read up on your definitions because that is wrong:

    http://chemwiki.ucdavis.edu/Core/Phy...ed_Definitions
    http://www.chemguide.co.uk/physical/...finitions.html

    If it wanted the DH for one mole, it would have asked for it.
    I got that quote from the internet... The top thing that comes up when when I googled standard enthalpy change? Lol.

    Well I put kJ/mol so it should be right if what you say is right regardless.
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    (Original post by Cadherin)
    SA has no effect as it referred to BOTH platinum electrodes.

    Yes, emf did increase at greater pressure and your reasoning is correct, although I think I worded it slightly differently with reference to the p.d. produced in a given time with an increased reaction rate.
    Excellent - my thoughts exactly!

    Ok I might deduct a mark for my explanation then, thanks
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    Could you use sulfur for question 1 instead of phosphorous?
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    (Original post by RME11)
    Nice back-peddle, I suggest you read up on your definitions because that is wrong:

    http://chemwiki.ucdavis.edu/Core/Phy...ed_Definitions
    http://www.chemguide.co.uk/physical/...finitions.html

    If it wanted the DH for one mole, it would have asked for it.
    I agree. They always ask for 'one mole of product', whereas this question specified 'for this reaction'.
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    (Original post by sean.17)
    For the reagents of Cr3+ to 2+, could you put HCl instead of H+?
    Yeah by h+ i meant any strong acid!
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    (Original post by Ohnis)
    Thank you so much man, you really did me a solid there
    No problem, it was about time it came up again
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    (Original post by Suits101)
    Excellent - my thoughts exactly!

    Ok I might deduct a mark for my explanation then, thanks
    No, don't, they should be lenient on that, you'll probably get 2/2.
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    Is there an unofficial markscheme??
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    (Original post by Parallex)


    Aerosmith

    Explain to me how this is any different to the question. You're provided with enthalpy of formation values and asked to work out the standard enthalpy change of the following reaction exactly as we did.

    Just telling you now, there's absolutely no diving in the mark scheme.
    Na isn't in its standard state so formation of 1 mol would not be required. Do you now see the logic?
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    I put increasing SA would increase emf

    ALL SETTLED- You did not need to divide by 2.

    THE END of that
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    (Original post by Darthsion100)
    I got something like 2328 KJ Mol-1 at first, then I halved the atomisation of oxygen to get 2404, but I think the original answer of 2328 was correct, because atomisation is for one mole, and you got 1 mole of O
    oh **** yeah ur right, ah well should only be one mark right?
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    (Original post by Suits101)
    I got that quote from the internet... The top thing that comes up when when I googled standard enthalpy change? Lol.

    Well I put kJ/mol so it should be right if what you say is right regardless.
    No it wont be right, because you have divided your enthalpy by 2 and I presume your entropy by 2 - you can expect to drop 2/3 marks.

    Next time don't take to TSR trying to tell the 95% of people who didn't get your answer that they are wrong, causing unnecessary stress for people.
 
 
 
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