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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.
    4s is filled before 3d, and emptied before 3d as its the outermost shell, so I think the Co2+ config i put is correct

    1s2 2s2 2p6 3s2 3p6 3d7
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    (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.
    I wrote
    Na + H2O ---> NaOH + 1/2H2
    As it didnt say it reacted with hot steam but just water so i assumed Na2O wouldnt be produced
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    (Original post by Ginpls)
    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
    Was there a question about advantage and disadvantage?? I don't remember that at aII?
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    (Original post by MarimaKauser)
    It's 3d7 you lose 4s electrons first


    Posted from TSR Mobile
    Yeah, just realised that I made a tiny mistake, forgot to remove 4s first due to lower energy compared to the 3d... :/
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    (Original post by Ginpls)
    I wrote
    Na + H2O ---> NaOH +H2
    As it didnt say it reacted with hot steam but just water so i assumed Na2O wouldnt be produced
    Boo... Ok thanks
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    how many marks do u think ill lose if i forgot to put the hydrogens on the 12 ethandiamie structure that we had to draw
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    (Original post by hi-zen-berg)
    4s is filled before 3d, and emptied before 3d as its the outermost shell, so I think the Co2+ config i put is correct

    1s2 2s2 2p6 3s2 3p6 3d7
    Brilliant.

    (Original post by Ginpls)
    I wrote
    Na + H2O ---> NaOH +H2
    As it didnt say it reacted with hot steam but just water so i assumed Na2O wouldnt be produced
    That's wrong, it needs to be balanced.
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    (Original post by Suits101)

    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.
    Chemguide had this helpful definition:

    Standard enthalpy change of reaction, ΔH°r
    'The standard enthalpy change of a reaction is the enthalpy change which occurs when equation quantities of materials react under standard conditions, and with everything in its standard state.'

    Alevelchemistry notes say this:
    'The standard enthalpy change for a reaction is the heat energy change measured under standard conditions: 100 kPa and a stated temperature (usually 298K).'

    and then goes on to further clarify:

    'Given a reaction: A + 3B -> 2C + 4D

    The standard enthalpy change for this reaction is taken to be the enthalpy change under standard conditions when one mole of A reacts with three moles of B to give two moles of C and four moles of D.'

    From this I'm now certain that dividing by 2 was not required for the marks in this question.

    (It is also mentioned in these notes that it is convention to leave the answer in kJmol^-1 which means those who did divide by two to get these units have a strong case for also gaining full marks.)
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    (Original post by 26december)
    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
    "Standard enthalpy change of reaction, ΔH°r - The standard enthalpy change of a reaction is the enthalpy change which occurs when equation quantities of materials react under standard conditions, and with everything in its standard state."

    ^ The question asked for the standard enthapy change of the reaction given and you were given the correct ratio for the reaction stated, which was 2 SO2 + O2 -> 2SO3 and the standard enthalpy of formation values for these substances so you were expected to use D.H = D.H(Products) - D.H(Reactants) to work out the enthalpy change for the reaction given. The D.H calculated would give you the standard enthalpy change of such a reaction under standard conditions, which is what the question asked for.

    Then you worked out the entropy change for the reaction given and put the figures into the gibbs free energy equation to work out the free energy change
    for the reaction given.

    tl;dr: there was no reason to divide by two, if you didn't you are correct.
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    (Original post by 26december)
    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
    I agree I put both answers down and specifically said for one mole the answer is :
    And divided by 2
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    (Original post by JCleggy)
    Chemguide had this helpful definition:

    Standard enthalpy change of reaction, ΔH°r
    'The standard enthalpy change of a reaction is the enthalpy change which occurs when equation quantities of materials react under standard conditions, and with everything in its standard state.'

    Alevelchemistry notes say this:
    'Thestandard enthalpy change for a reaction is the heat energy change measuredunder standard conditions: 100 kPa and a stated temperature (usually 298K).'

    and then goes on to further clarify:

    'Given a reaction: A + 3B à 2C + 4D

    The standard enthalpy change for thisreaction is taken to be the enthalpy change under standard conditions when onemole of A reacts with three moles of B to give two moles of C and four moles ofD.'

    From this I'm now certain that dividing by 2 was not required for the marks in this question.

    (It is also mentioned in these notes that it is convention to leave the answer in kJmol^-1 which means those who did divide by two to get these units have a strong case for also gaining full marks.)
    I divided by 2 and put units as kJ/mol.

    Definition when I googled standard enthalpy change:

    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.

    I don't know anymore! Haha
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    (Original post by Suits101)
    Brilliant.



    That's wrong, it needs to be balanced.
    Well yeah, that eqn but balanced
    I was just explaining how Na2O wouldnt have been made lmao
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    You could draw a skeletal formula for the [Co(H2NCh2Ch2Nh2)3]2+, right?
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    For when i drew the compIex i didn't draw arrows to show co-ordinate bonds? I aIso didn't put brackets over the whoIe compIex and just put in brackets (+2 on overaII compIex) on top of the drawing. WouId i Iose marks for this??
    • Welcome Squad
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    Welcome Squad
    Eh what i thought electronic configuration will be 3d5 4s2 for the Co 2+ due it being more stable ?
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    (Original post by Ginpls)
    Well yeah, that eqn but balanced
    I was just explaining how Na2O wouldnt have been made lmao
    Brilliant

    Exactly right.
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    The equation to calculate Gibbs free energy, ΔG, looks like this:
    ΔG = ΔH - TΔS
    It is the *changes* in enthalpy and entropy which give us Gibbs free energy, ΔG, and from which we can determine if a reaction is spontaneous or nonspontaneous at a particular temperature.

    We need to compute the enthalpy change, ΔH. ΔHrx = ∑ΔHf(prod) - ∑ΔHf(react)
    ΔHrx = 2mol(-46.1 kJ/mol) = -92.2 kJ .... the sum of the heats of formation of the reactants are zero because they are in the elemental state.

    ΔSrx = ∑ΔSf(prod) - ∑ΔSf(react) = 2 mol(0.1923 kJ/Kmol) - (3 mol(0.1306 kJ/Kmol) + 1mol(0.1915 kJ/Kmol)) = -0.1987 kJ/K

    ΔG = ΔH - TΔS
    ΔG = -92.2 kJ - 573K(-0.1987 kJ/k)
    ΔG = +21.7 kJ
    Off Yahoo answers is this wrong or right I think its right?
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    (Original post by palacefloor)
    For when i drew the compIex i didn't draw arrows to show co-ordinate bonds? I aIso didn't put brackets over the whoIe compIex and just put in brackets (+2 on overaII compIex) on top of the drawing. WouId i Iose marks for this??
    Maybe 1 for co-ordinate bonds

    (Original post by koolgurl14)
    Eh what i thought electronic configuration will be 3d5 4s2 for the Co 2+ due it being more stable ?
    Google it:

    Co: [Ar] (or electron configuration) 4s2 3d7
    Co2+: [Ar] 3d7 (4s electrons are lost first)
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    (Original post by RME11)
    "Standard enthalpy change of reaction, ΔH°r - The standard enthalpy change of a reaction is the enthalpy change which occurs when equation quantities of materials react under standard conditions, and with everything in its standard state."

    ^ The question asked for the standard enthapy change of the reaction given and you were given the correct ratio for the reaction stated, which was 2 SO2 + O2 -> 2SO3 and the standard enthalpy of formation values for these substances so you were expected to use D.H = D.H(Products) - D.H(Reactants) to work out the enthalpy change for the reaction given. The D.H calculated would give you the standard enthalpy change of such a reaction under standard conditions, which is what the question asked for.

    Then you worked out the entropy change for the reaction given and put the figures into the gibbs free energy equation to work out the free energy change
    for the reaction given.

    tl;dr: there was no reason to divide by two, if you didn't you are correct.
    Exactly this. It 100% said for the reaction given.
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    Hey guys, will they accept [Ar] 3d7 4s2 and [Ar] 3d7 for the electron configurations or did you have to do the full ones??
 
 
 
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