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    (Original post by Suits101)
    The question asked you to calculate the STANDARD enthalpy and entropy change, which was asked on Jan 13 I believe and you had to divide by two.
    Oh, yes, your right, it did.
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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
    Q8. First equation was just reacting it with 6 water moleculesP was Cr(H2O)3(OH)3Reagent was anything with OH- ions

    NH3 can be used as a reagent here too!

    Reagent for cr3+ to cr2+ is Zn and H+

    Hydrogen fuel cell half eqn was

    O2 + 4e- + 2H2O ---> 4OH- postive electrode

    H2 + 2OH- ---> 2H2O + 2e- negative electrode

    Overall: o2 + 2H2 ---> 2H2O

    No effect if pressure increases as you are only increasing the pressure of oxygen so the same number of electron are released so same number of redox reactions??? (Not sure though)

    Graph was a straight horizontal line

    No effect on emf for increase in surface area of pt

    Environmental advantage was h2o doesnt contribute to global warming as much as co2 and/or acid rain
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    Someone make a proper unofficial markscheme?
    • Welcome Squad
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    Welcome Squad
    (Original post by EverydayHell)
    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.
    You are right however i did it 2Na + 2H2O -> 2NaOH + H2
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    Example 4Using the values in the above table of standard enthalpies of formation, calculate the ΔHreactiono for the formation of NO2(g).SOLUTIONNO2(g) is formed from the combination of NO(g) and O2(g) in the following reaction:
    2NO(g)+O2(g)⇋2NO2(g)
    To find the ΔHreactiono, use the formula for the standard enthalpy change of formation:
    ΔHoreaction=∑ΔHof(products) ˆ’∑ΔHof(Reactants)
    The relevant standard enthalpy of formation values from Table 1 are:
    • O2(g): 0 kJ/mol
    • NO(g): 90.25 kJ/mol
    • NO2(g): 33.18 kJ/mol
    Plugging these values into the formula above gives the following:
    ΔHoreaction=(2mol)(33.18kJ/mol)−[(2mol)(90.25 kJ/mol)+(1mol)(0kJ/mol)]ΔHoreaction=−114.1kJ
    To stop people losing about 7 marks in total bad luck for the clever clogs
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    (Original post by RME11)
    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.
    We said the exact same thing with E (Au+/Au) > E (O2/H2O) = 1 mark

    I said Au+ would react with water instead of gold hence gold does not react with with Au...

    It said standard enthalpy change.
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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
    I actually think 87 is A*, 81 is A and it caps at around 95

    Posted from TSR Mobile
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    (Original post by hi-zen-berg)
    So delta G is negative and rxn is feasible
    Structure is attached photo with 2+ charge around entire complex in brackets

    Attachment 554912
    Shouldnt the arrows be going towards to Co since the Nitrogens are donating the electron pairs?
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    (Original post by Suits101)
    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)



    H2SO4

    Boundaries will not be that high, people still found it hard and many stupid mistakes were made.
    I don't agree with Co2+ ion electronic configuration, its supposed to be 4s2 3d5 and because its a transition element, that does have variable oxidation state due to its incomplete 3d sub-shell.
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    (Original post by FireBLue97)
    Example 4Using the values in the above table of standard enthalpies of formation, calculate the ΔHreactiono for the formation of NO2(g).SOLUTIONNO2(g) is formed from the combination of NO(g) and O2(g) in the following reaction:
    2NO(g)+O2(g)⇋2NO2(g)
    To find the ΔHreactiono, use the formula for the standard enthalpy change of formation:
    ΔHoreaction=∑ΔHof(products)â ˆ’∑ΔHof(Reactants)
    The relevant standard enthalpy of formation values from Table 1 are:
    • O2(g): 0 kJ/mol
    • NO(g): 90.25 kJ/mol
    • NO2(g): 33.18 kJ/mol
    Plugging these values into the formula above gives the following:
    ΔHoreaction=(2mol)(33.18kJ/mol)−[(2mol)(90.25 kJ/mol)+(1mol)(0kJ/mol)]ΔHoreaction=−114.1kJ
    To stop people losing about 7 marks in total bad luck for the clever clogs
    The question said STANDARD enthalpy change, not enthalpy change as this question says.

    Regardless, one mark for each question would be dropped when calculating values, which means -2.

    Gibbs free-energy change would allow ecf meaning no marks dropped.
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    (Original post by isaisababy)
    Shouldnt the arrows be going towards to Co since the Nitrogens are donating the electron pairs?
    yh ur 100% right mate, I just googled the image and attached it.

    Anyone remembering specific marks now? MS pls someone
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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

    Attachment 554912

    Would it be alright to use [Ar] for the electron configurations ???
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    Which acid was it for the redox titration?
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    (Original post by thomaarrss)
    i put this as well
    Yea same its right because the further to the right that the equilibrium lies the more positive the voltage (emf) of the cell so because equlibrium shift left emf would decrease
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    (Original post by User2334541)
    I don't agree with Co2+ ion electronic configuration, its supposed to be 4s2 3d5 and because its a transition element, that does have variable oxidation state due to its incomplete 3d sub-shell.
    Hi,

    No that's wrong.

    Co is [Ar] 4s2 3d7
    Co2+ is [Ar] 3d7 (you lose 4s electrons first - google it)

    Sorry.
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    (Original post by koolgurl14)
    You are right however i did it 2Na + 2H2O -> 2NaOH + H2
    Thanks you might be right there the question threw me off a bit because it the said the oxide dissolved in water, but only the mark scheme will tell.
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    (Original post by Suits101)
    The question said STANDARD enthalpy change, not enthalpy change as this question says.

    Regardless, one mark for each question would be dropped when calculating values, which means -2.

    Gibbs free-energy change would allow ecf meaning no marks dropped.
    If I could rep this man x10 then I would
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    (Original post by User2334541)
    I don't agree with Co2+ ion electronic configuration, its supposed to be 4s2 3d5 and because its a transition element, that does have variable oxidation state due to its incomplete 3d sub-shell.
    It's 3d7 you lose 4s electrons first


    Posted from TSR Mobile
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    (Original post by isaisababy)
    Shouldnt the arrows be going towards to Co since the Nitrogens are donating the electron pairs?
    was looking for this comment i drew the arrows the other way round. I drew the lone pairs as well so they should know what i mean
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    (Original post by K2Cr2O7)
    Would it be alright to use [Ar] for the electron configurations ???
    It didn't say "full" electron configuration so you're probably okay.
 
 
 
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