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    did anyone get 94.4% on the 7 marker at the end??
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    50.7% purity 7 mark defo rite ???
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    (Original post by JCleggy)
    Regarding the Enthalpy/Entropy calculation -

    I was conscious of similarly-worded questions in the past that required you to divide to get enthalpy and entropy values for 1 mole.
    However, the question asked for the 'standard enthalpy change for the reaction', with that reaction being the one given in the above equation, not the 'standard enthalpy of formation of SO3'.
    The use of the word standard here applied to the conditions of the reaction being 100kPa and 298K, and not the creation of 1 mole of product.

    This, combined with the fact that both questions were worth 2 marks rather than 3 (previously released questions requiring you to do that extra division step were worth 3 marks) leads me to believe the values were the initial undivided ones and that it's just a poorly worded question on aqa's behalf.

    This said I think if you divided by 2 and gave the units kJmol^-1 you will get both marks, and if you did not divide by 2 but gave the units kJ you will equally get both marks.
    Bravo! I think you're spot on here
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    (Original post by FireBLue97)
    50.7% purity 7 mark defo rite ???
    i got 94.4% mate? not sure
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    Had the enthalpy question asked for the 'enthalpy of formation of SO3', the dividing by 2 would be necessary. however, since we were asked to find Delta H FOR THE REACTION, dividing by 2 would not be required.
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    If they wanted you to calculate the change for one mle of s03 - why would they give you the equation?
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    (Original post by Cadherin)
    I said, in presence of H+, oxygen CANNOT oxidise Au to Au+ because E<0 and, hence, as delta G = -nFE, reaction is not feasible as delta G>0.

    I don't think you need to know the Nernst equation and all that, but it may get a mark, who knows.
    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?

    (Original post by OloMed)
    This paper was almost identical to Jan 13. Expecting similar grade boundaries.

    The dividing by 2 bit for the enthalpy change question seems to have caught most of us out. Anyone know whether ecf will be allowed for consequent questions, a it could cost us 10 MARKS for that small error
    Regardless of what's correct, only one mark would be deducted from each incorrect answer so -2 max.

    (Original post by RME11)
    It had nothing to do with oxidizing water, the question asked why solid gold would not react with moisture in the air and since E (Au+/Au) > E (O2/H2O) no reaction would take place under standard conditions.
    I said exactly that lol?

    I just went into more detail.

    (Original post by Parallex)
    What does STANDARD have to do with it? If it asked for the enthalpy change of the REACTION (which I'm pretty sure it did) rather than formation, you wouldn't need to divide by 2? A Hess cycle would just include 2x the enthalpy of formation for SO3 (or whatever it was at the top, can't really remember).

    Edit: All standard means is that the reaction is taking place under standard conditions.
    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.
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    (Original post by hi-zen-berg)
    I genuinely don't recall them asking for standard either, dunno if Azzer is right or not.

    confirm anyone who remembers what the deltaH question asked for?
    STANDARD enthalpy change
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    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

    Q7. 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-

    Zn/HCl, final thing was a blue solution.

    Q8. Credit for this Q to hi-zen-berg
    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

    Not sure if this was needed:
    DeltaH roughly 0 as same type and number of bond broken

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



    Q9.
    Calculation was 94.4%
    Colorimetry talk about MnO4- ions being coloured use the intensity of the colour to measure the concentration etc.
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    (Original post by Azzer11)
    We need someone to clear up this enthalpy/entropy dilemma ahaha. If pictures of the paper are released, we'll be able to see.
    There was no dividing required, the question asked for:

    a) Enthalpy change of the reaction given
    b) Entropy change of the reaction given
    c) The Gibbs-free energy change using the calculated values
    d) Due to the magnitude of the free energy change calculated what was reaction feasible at 50 degrees.
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    no you dont divide by 2 for the enthalpy change reaction, because it never said work out the enthalpy for formation. it gave you enthalpy of formation values which means you just take away sum of enthalpy change of products from the reactants,

    its come up before and you dont need to divide values by 2
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    good work mate!
    (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
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    (Original post by jaackq1221)
    yes
    What units are you in? I got -134000 roughly in J, which works out the same in kJ doesn't it?
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    how did ppl get 94.4% for the last question
    i didnt get anything of the sort lol explain pls
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    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
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    (Original post by jjsnyder)
    Q7.
    Enthalpy change -196
    Entropy ~-160
    G was around -185
    Reaction is feasible as G is less than or equal to zero

    Posted from TSR Mobile
    Entropy change is -189 JK-1mol-1
    And delta G is -135 Kj mol-1
    • Welcome Squad
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    Welcome Squad
    (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
    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?
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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
    Classic jjsnyder
    Reagent would be Zn and H2SO4
    wouldnt have to be an excess of OH- as youve got a precipitate there already so even adding small amounts of Oh would cause the reaction
    Could have NH3 as the reagent aswell for the first one
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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
    Zinc (in HCl) to reduce chromium down to 2+. Note: Fe2+ etc, would not be strong enough to reduce beyond Cr3+ so this would not be acceptable.
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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?
    It asked for element X which would have just been P
 
 
 
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