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    (Original post by Aerosmith)
    I'm afraid to say guys you did have to divide by 2. It said standard enthalpy change, which is defined as forming 1 mol. So all of you please calm your tits
    No you didn't...
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    (Original post by justinamarina)
    I picked sulfur . Phosphorus or sulfur was fine
    No because it said pick something from Na - P and sulfur is after P
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    (Original post by koolgurl14)
    G is -134 i think.
    The first question you had to say it was P4O10 not just P because it said X is an oxide . there was the two questions increase in pressure how it affect emf . and how surface area of platinum will affect emf if anyone can plz confirm what the answers for these two questions?
    Incorrect. It said Element X forms an oxide.....

    Identify X

    Answer : X= P
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    (Original post by jjsnyder)
    Just a bit of an answer dump, missing loads if someone gives me question topics I'll fill if in.
    Q1.
    Covalent
    P
    P4O10 + 6H2O -> H3PO4
    Ionic
    Na
    Na2O + H2O -> 2NaOH
    Al2O3
    Reacts with an acid and a base
    3H2O + Al2O3 + 2NaOH -> 2NaAl(OH)4
    Al2O3 + 6HCl -> 2AlCl3 + 3H2O

    Q2. Standard Born Haber stuff
    Lattice dissociation = +2328? Something around this value

    Q7.
    Enthalpy change -196
    Entropy ~-160
    G was around -185
    Reaction is feasible as G is less than or equal to zero

    Q8. First equation was just reacting it with 6 water molecules
    P was Cr(H2O)3(OH)3
    Reagent was anything with OH- ions

    Q was CO2
    Reagent was anything with CO3 2- ions

    Reaction 4 product was [Cr(OH)6]3-
    Excess reagent containing OH-

    Not sure what the Reagent was for the last part, final thing was a blue solution.
    Q9.
    Calculation was 94.4%
    Colorimetry talk about MnO4- ions being coloured use the intensity of the colour to measure the concentration etc.


    Posted from TSR Mobile
    Thanks a lot;
    For last part of Chromium Q, Zn + Hcl were reagents

    I'll just add on some stuff Beginning of section B:

    Co Atom is 1s^2 2s^2 2p^6 3s^2 3p^6 4s^2 3d^7
    Co2+ 1s2 2s2p6 3s2p6d7
    Characteristic features: complex formation, coloured compounds, var. ox state (5 marks)


    [Co(h2o)6]2+ + 3NH2CH2CH2NH2 --> [Co(NH2CH2CH2NH2)3]2+ + 6H2O

    4 mol reactant to 7 mol product therefore deltaS positive

    DeltaH roughly 0 as same type and number of bond broken

    So delta G is negative and rxn is feasible
    Structure is attached photo with 2+ charge around entire complex in brackets

    Name:  7224-73-9.png
Views: 258
Size:  4.3 KB
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    (Original post by jjsnyder)
    Just a bit of an answer dump, missing loads if someone gives me question topics I'll fill if in.
    Q1.
    Covalent
    P
    P4O10 + 6H2O -> H3PO4
    Ionic
    Na
    Na2O + H2O -> 2NaOH
    Al2O3
    Reacts with an acid and a base
    3H2O + Al2O3 + 2NaOH -> 2NaAl(OH)4
    Al2O3 + 6HCl -> 2AlCl3 + 3H2O

    Q2. Standard Born Haber stuff
    Lattice dissociation = +2328? Something around this value

    Q7.
    Enthalpy change -196
    Entropy ~-160
    G was around -185
    Reaction is feasible as G is less than or equal to zero

    Q8. First equation was just reacting it with 6 water molecules
    P was Cr(H2O)3(OH)3
    Reagent was anything with OH- ions

    Q was CO2
    Reagent was anything with CO3 2- ions

    Reaction 4 product was [Cr(OH)6]3-
    Excess reagent containing OH-

    Not sure what the Reagent was for the last part, final thing was a blue solution.
    Q9.
    Calculation was 94.4%
    Colorimetry talk about MnO4- ions being coloured use the intensity of the colour to measure the concentration etc.

    Unassigned:
    Hydrogen-Oxygen fuel cell Q:
    Overall equation 2H2 + O2 -> 2H2O or multiples

    Posted from TSR Mobile
    Enthalpy/ entropy question is incorrect - should be -98, -94.5 and -65.1
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    what did everyone get for the tick box acid that should be used? i put H2SO4 and for the reducing agent for chromium??

    i reckon 88/89 for an A* (maybe even 90)
    82/83 for an A
    70 something for a B
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    (Original post by Parallex)
    No you didn't...
    Can you not read?
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    I think AQA have messed up a bit there if they asked "standard enthalpy change for this reaction"
    Standard enthalpy change would mean dividing by 2
    But the "this reaction" would suggest not
    However the reaction can be halved so I think the mark scheme will say that you have to divide by 2


    Posted from TSR Mobile
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    (Original post by chzm)
    what did everyone get for the tick box acid that should be used? i put H2SO4 and for the reducing agent for chromium??

    i reckon 88/89 for an A* (maybe even 90)
    82/83 for an A
    70 something for a B
    Lol has it ever been that high? The paper was easy but that's just how we found it.
    A lot of people will much lower which always brings the grade boundaries down. I reckon 86 for an A* max
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    (Original post by chzm)
    what did everyone get for the tick box acid that should be used? i put H2SO4 and for the reducing agent for chromium??

    i reckon 88/89 for an A* (maybe even 90)
    82/83 for an A
    70 something for a B
    doubt it, the highest its ever been for an A is 76%, it was good but like not overly good haha, 93% for full ums
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    (Original post by Aerosmith)
    Enthalpy/ entropy question is incorrect - should be -98, -94.5 and -65.1
    Agree with that

    The colorimeter question should be about using known concentrations and absorption rates surely?


    Posted from TSR Mobile
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    For the Cr2+ do you get the mark for saying H2SO4 as the acid? I bloody put HCl originally then changed it for some reason...
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    Was formation as there was this symbol Name:  image.jpeg
Views: 37
Size:  18.6 KB
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    Oh I've seen something in the 1st question about Na the equation asked for the element reacting with water not the oxide of the element I think? What did everyone else write? I did 2Na + H2O ---> Na2O + H2.
    The other equation asked for the equation with oxide which is why I did that.
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    (Original post by Suits101)
    That's what I said except I explained it the other way around e.g Au+ reduces water to oxygen hence oxygen doesn't react with gold?



    Regardless of what's correct, only one mark would be deducted from each incorrect answer so -2 max.



    I said exactly that lol?

    I just went into more detail.



    The question said to calculate the standard enthalpy change for this reaction:

    "The standard enthalpy of reaction (denoted ΔHr) is the enthalpy change that occurs in a system when one mole of matter is transformed by a chemical reaction under standard conditions."

    Maybe you're right, but my gut feeling was to divide by 2.
    No you didn't? The question didn't ask about Au+ ions the question asked about the reaction between gold solid and water because they don't react at all under standard conditions- that was the nature of the question. If it asked about what would happen if Au+ ions were placed in water then you'd be right, but it didn't so you're wrong... lol.

    The question didn't say anything about standard enthalpy of formation of SO3, it asked for the enthapy change of the reaction given which was the formation of 2 moles of SO3 from 2 moles of SO2 and one mole of O2.
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    (Original post by 26december)
    Agree with that

    The colorimeter question should be about using known concentrations and absorption rates surely?


    Posted from TSR Mobile
    Exactly right
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    It said enthalpy change for reaction not formation
    They values they gave if you didnt get enthalpy and entropy to use for the free energy change are similar to the values u get without dividing by 2 also
    You didnt need to divide lol
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    Would HNO3 be alright as a reagent with Zinc?
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    I really don't think you had to divide by 2 for the calculations. The calculations were only worth 2 marks each, there'd be another mark if there was another step involved, like there has been on previous papers.
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    (Original post by hi-zen-berg)
    Thanks a lot;
    For last part of Chromium Q, Zn + Hcl were reagents

    I'll just add on some stuff Beginning of section B:

    Co Atom is 1s^2 2s^2 2p^6 3s^2 3p^6 4s^2 3d^7
    Co2+ 1s2 2s2p6 3s2p6d7
    Characteristic features: complex formation, coloured compounds, var. ox state (5 marks)


    [Co(h2o)6]2+ + 3NH2CH2CH2NH2 --> [Co(NH2CH2CH2NH2)3]2+ + 6H2O

    4 mol reactant to 7 mol product therefore deltaS positive

    DeltaH roughly 0 as same type and number of bond broken

    So delta G is negative and rxn is feasible
    Structure is attached photo with 2+ charge around entire complex in brackets

    Name:  7224-73-9.png
Views: 258
Size:  4.3 KB
    Excellent I agree

    For delta G being negative you can have <= 0 and < 0, either one is fine.

    The arrows should be from N to Co (co-ordinate bond)

    (Original post by chzm)
    what did everyone get for the tick box acid that should be used? i put H2SO4 and for the reducing agent for chromium??

    i reckon 88/89 for an A* (maybe even 90)
    82/83 for an A
    70 something for a B
    H2SO4

    Boundaries will not be that high, people still found it hard and many stupid mistakes were made.
 
 
 
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