Hey there! Sign in to join this conversationNew here? Join for free

Edexcel A2 Chemistry Exams -6CH04 (14th June) and 6CH05 (22nd June) Discussion Thread Watch

    Offline

    3
    ReputationRep:
    (Original post by genevievelaw)
    can anyone send me the january 2016 a level paper today? it would be really useful
    Attached below.

    (Original post by samb1234)
    Essentially yeah - we know that an increased temperature is going to increase the rate as more successful collisions, so as you increase the conc of A more and more so if we wrote the rate eq wrt A and use Arrhenius for k then rate= (Ae^-Ea/RT)[A] so we are not just varying the conc of A hence why it is slightly above the straight line predicted
    How come this doesn't happen for most other reactions we consider?
    Attached Images
  1. File Type: pdf WCH04_01_que_20160111_2.pdf (516.0 KB, 80 views)
  2. File Type: pdf WCH04_01_msc_20160302.pdf (336.5 KB, 67 views)
    • Thread Starter
    Offline

    10
    ReputationRep:
    (Original post by Ayman!)
    Attached below.



    How come this doesn't happen for most other reactions we consider?
    It will. However most of the time we deal with initial rates where the temp hasn't had time to rise so hence the rate constant is actually constant.
    Offline

    2
    ReputationRep:
    samb1234 Ayman! Do you guys think it is worth doing the June 2015 IAL if we've already done June 2015 UK?
    Offline

    3
    ReputationRep:
    I'm officially done with past papers for Unit 4. Will possibly look through again from Jan 10 for the trickier stuff.

    (Original post by samb1234)
    It will. However most of the time we deal with initial rates where the temp hasn't had time to rise so hence the rate constant is actually constant.
    Brilliant! That makes so much sense now, thank you.

    (Original post by Don Pedro K.)
    samb1234 Ayman! Do you guys think it is worth doing the June 2015 IAL if we've already done June 2015 UK?
    I think it's worth a go. Like Samb said, the questions look similar but there's a different approach to them. Don't do it as a mock, though - just look through it.
    Offline

    2
    ReputationRep:
    (Original post by Ayman!)
    I'm officially done with past papers for Unit 4. Will possibly look through again from Jan 10 for the trickier stuff.



    Brilliant! That makes so much sense now, thank you.



    I think it's worth a go. Like Samb said, the questions look similar but there's a different approach to them. Don't do it as a mock, though - just look through it.
    Ah okay cool I might just skim that one and then do the specimen paper then
    Offline

    2
    ReputationRep:
    (Original post by Don Pedro K.)
    Yeah since it's second order with respect to nitrogen dioxide, the rate equation will be: rate = k[NO2]2. Since the rate equation shows the species and the number of molecules of that species involved in the rate determining step, the rate determining step will start with 2NO2
    alright got it, thanks!
    im sorry for asking so many questions buut could you please explain Q22b)iv)? how do we know this random conc would be formed at the equivalence point?
    https://c838cff4741acb48ae1ed62e5992...0Chemistry.pdf
    Offline

    2
    ReputationRep:
    (Original post by imnoteinstein)
    alright got it, thanks!
    im sorry for asking so many questions buut could you please explain Q22b)iv)? how do we know this random conc would be formed at the equivalence point?
    https://c838cff4741acb48ae1ed62e5992...0Chemistry.pdf
    Well, the way I would look at it is that the point where you get just sodium lactate on its own is at the equivalence point - when all the acid has been neutralised and no further alkali has been added yet. Therefore, the pH would be anywhere in the vertical section in the graph; I would play it safe and say pH 8 as you know that this is definitely within the vertical section

    "The concentration of sodium lactate is0.075 mol dm-3 when equal amounts of acid andbase have been mixed" is what the mark scheme says which is basically the same thing haha
    Offline

    3
    ReputationRep:
    What causes an IR peak in spectrometric analysis to have different characteristics? Like, what makes them to be broad, sharp, wide, etc?
    Offline

    3
    ReputationRep:
    (Original post by Ayman!)
    Attached below.



    How come this doesn't happen for most other reactions we consider?
    thank you so much
    Offline

    3
    ReputationRep:
    (Original post by Ayman!)
    What causes an IR peak in spectrometric analysis to have different characteristics? Like, what makes them to be broad, sharp, wide, etc?
    I found this (it suggests why OH produces a broad peak), look on page 4
    Attached Images
  3. File Type: pdf notes_14C_IR.pdf (124.0 KB, 281 views)
    Offline

    12
    ReputationRep:
    (Original post by Ayman!)
    What causes an IR peak in spectrometric analysis to have different characteristics? Like, what makes them to be broad, sharp, wide, etc?
    From what I know, it's something to do with on how the bonds react when they are exposed to infrared. When a bond is exposed to infrared, what happens is that they undergo stretching or bending. This is to do with the frequency absorbed and every particular bond absorbs particular frequencies which is why they produce distinct peaks.
    • Thread Starter
    Offline

    10
    ReputationRep:
    (Original post by Ayman!)
    What causes an IR peak in spectrometric analysis to have different characteristics? Like, what makes them to be broad, sharp, wide, etc?
    Within your large sample of alcohol you'll have hydrogen bonding to different extents in different molecules, which affects the amount of energy needed to get the bond to stretch so because of this difference for a large sample you are going to get a wide number of frequencies at which some of the bonds stretch
    Offline

    2
    ReputationRep:
    any predictions for section c on Tuesday's paper ?
    Offline

    0
    ReputationRep:
    https://c838cff4741acb48ae1ed62e5992...0Chemistry.pdf

    someone please help on 24aiii. its a calculation question. thank you
    Offline

    3
    ReputationRep:
    Did this question and I still don't get it, some help would be appreciated

    "Stage 2 of this sequence (CH3CHBrCOOH --> CH3CH(OH)COOH) was carried out in two steps. Identify the reagent required for each step in this stage,"
    The textbook just says heat with NaOH (aq)

    The mark scheme:
    First stepNaOH(aq)/KOH(aq) or names (1)
    Second mark dependent on first beingcorrect.
    Second stepHCl(aq)/hydrochloricacid/H2SO4(aq)/sulfuric acid

    Can anyone explain why you need HCl for the second step?
    Offline

    14
    ReputationRep:
    (Original post by genevievelaw)
    Did this question and I still don't get it, some help would be appreciated

    "Stage 2 of this sequence (CH3CHBrCOOH --> CH3CH(OH)COOH) was carried out in two steps. Identify the reagent required for each step in this stage,"
    The textbook just says heat with NaOH (aq)

    The mark scheme:
    First stepNaOH(aq)/KOH(aq) or names (1)
    Second mark dependent on first beingcorrect.
    Second stepHCl(aq)/hydrochloricacid/H2SO4(aq)/sulfuric acid

    Can anyone explain why you need HCl for the second step?
    You need to add NaOH/KOH to remove an HBr and form an alkene (deprotonate it) in the first step and then, you have to add HCl(aq) to protonate it and form the hydroxyl group
    Offline

    8
    ReputationRep:
    (Original post by genevievelaw)
    Did this question and I still don't get it, some help would be appreciated

    "Stage 2 of this sequence (CH3CHBrCOOH --> CH3CH(OH)COOH) was carried out in two steps. Identify the reagent required for each step in this stage,"
    The textbook just says heat with NaOH (aq)

    The mark scheme:
    First stepNaOH(aq)/KOH(aq) or names (1)
    Second mark dependent on first beingcorrect.
    Second stepHCl(aq)/hydrochloricacid/H2SO4(aq)/sulfuric acid

    Can anyone explain why you need HCl for the second step?
    1st step: Nucleophilic substitution so u replace Br- with OH- group. However, don't forget the COOH group is acidic so it reacts with NaOH to form COO-

    2nd step: Addition of acid converts COO- back to COOH
    Offline

    14
    ReputationRep:
    I thought we didn't include solid molecules in the equilibrium constant because their concentrations were too low compared to other reactants that any changes would be insignificant and that's why they're constant, however, it seems like i got something wrong, so can someone help? Why the answer is A and not C ?
    Attached Images
     
    Offline

    3
    ReputationRep:
    (Original post by setarcos)
    1st step: Nucleophilic substitution so u replace Br- with OH- group. However, don't forget the COOH group is acidic so it reacts with NaOH to form COO-

    2nd step: Addition of acid converts COO- back to COOH
    Aaah thank you! I get it now, thank you so much I forgot that the COOH would react with the NaOH - kind of silly really
    Offline

    3
    ReputationRep:
    (Original post by PlayerBB)
    You need to add NaOH/KOH to remove an HBr and form an alkene (deprotonate it) in the first step and then, you have to add HCl(aq) to protonate it and form the hydroxyl group
    thank you for answering
 
 
 
  • See more of what you like on The Student Room

    You can personalise what you see on TSR. Tell us a little about yourself to get started.

  • Poll
    Would you like to hibernate through the winter months?
  • See more of what you like on The Student Room

    You can personalise what you see on TSR. Tell us a little about yourself to get started.

  • The Student Room, Get Revising and Marked by Teachers are trading names of The Student Room Group Ltd.

    Register Number: 04666380 (England and Wales), VAT No. 806 8067 22 Registered Office: International House, Queens Road, Brighton, BN1 3XE

    Quick reply
    Reputation gems: You get these gems as you gain rep from other members for making good contributions and giving helpful advice.