Join TSR now and get all your revision questions answeredSign up now
    Offline

    1
    ReputationRep:
    (Original post by Super199)
    Right got the majority of that. Why can't the halogen be any of the other halogens? Why specifically 2-chloropropane or 2-bromopropane?

    I'm guessing its the secondary carbocation = more stable
    Offline

    3
    ReputationRep:
    (Original post by Super199)
    Right got the majority of that. Why can't the halogen be any of the other halogens? Why specifically 2-chloropropane or 2-bromopropane?
    Even though fluorine is highly electronegative (so there will be a strong partial charge on carbon), the C-F bond is extremely strong meaning it's actually quite unreactive, so most exam boards ignore it (the reaction will be very slow and doesn't happy as easily as Cl or Br). Iodine isn't that electronegative (same as Carbon) but it is a very weak bond. Because it's a weak bond, it is easily broken and reacts quickly. I don't see why they might not ask you about Iodoalkanes (perhaps there something going on behind the scenes that is beyond a-level...) Chloro and bromo are just the standard halogens they use

    (Original post by shhh123)
    I'm guessing its the secondary carbocation = more stable
    Although this is generally correct, in this case the reason why it is "2" is simply because the product (propan-2-ol) has its hydroxyl (alcohol) group on the second carbon. Since it's a substitution reaction, the original halogen must also have been on the second carbon.

    However carbocation stability is something you should definitely always consider during mechanisms - especially electrophilic addition - but it is not a factor in every mechanism. I think the Sn1 mechanism (different to the Sn2 mechanism taught at a-level) involves ionisation of the halogen in which carbocation stability IS a factor, but in Sn2 it is not.
    Offline

    3
    ReputationRep:
    Anyone care to explain the propagation step in free radical substitution. With an example, other than the methane and chlorine one? I understand the initiation and termination step but this one is a bit hit and miss...
    Offline

    9
    ReputationRep:
    (Original post by kira1)
    So I just had a mock exam and got an E, at the end of this year I need a high A, is that possible? If so how ? I feel so stressed and keep imagining myself failing. God forbid
    Hi there!

    I am an A2 Chemistry student.

    I got an A last year at AS level.

    In the March MOCKs I got a U grade for Chemistry but that was because I lacked understanding at that time.

    To help me understand better I bought the CGP text book which helped contribute to my A grade.

    I recommend you get it if you really want that high A grade.

    CGP AS Chemistry

    SAMPLE PAGES

    I also recommend you do lots of past papers so help strengthen your exam technique.
    Offline

    3
    ReputationRep:
    (Original post by haemo)
    Using chloromethane (CClH3) as an example:

    Initiation step: Cl2 -> 2Cl•

    Propagation step 1: CClH3 + Cl• -> •CClH2 + HCl

    Propagation step 2: •CClH2 + Cl2 -> CCl2H2 + Cl•

    Termination step (1): •Cl + •Cl -> Cl2
    Termination step (2): •Cl + CClH2 -> CCl2H2
    Termination step (3): •CClH2 + •CClH2 -> CClH2CClH2 (NOT C2Cl2H4)
    Care to explain what is happening with the propagation step?
    Offline

    15
    ReputationRep:
    Hey guys.
    So i was doing some a level chemistry questions (http://www.a-levelchemistry.co.uk/AQ...2.1%20home.htm)
    exercise 2.
    We are given bond disassociation energies and are asked to calculate an enthalpy of formation of HF? I am unsure of how to go about this.. The question before asks to calculate an enthalpy of reaction where HF is a product
    All help is greatly appreciated
    Offline

    3
    ReputationRep:
    (Original post by Gladiatorsword)
    Hey guys.
    So i was doing some a level chemistry questions (http://www.a-levelchemistry.co.uk/AQ...2.1%20home.htm)
    exercise 2.
    We are given bond disassociation energies and are asked to calculate an enthalpy of formation of HF? I am unsure of how to go about this.. The question before asks to calculate an enthalpy of reaction where HF is a product
    All help is greatly appreciated
    Total up all the bond enthalpies from all the bonds of all the reactants.
    Total up all of the bond enthalpies from all the bonds of the products.

    Do reactants minus products.
    This equals delta H.

    Bob's your uncle.


    Posted from TSR Mobile
    • Welcome Squad
    Offline

    3
    ReputationRep:
    Can anyone help me pleeeease!!

    Calculate the enthalpy change for thermal decomposition of lithium carbonate (li2c03) given enthalpy of formation for li2c03- -1216 kj mol-1, li02- -598 kj mol-1 and c02- -394 kj mol-1. I got 224 kj mol-1, but I don't think it's right.

    Calculate the enthalpy change for the reaction; c(graphite)>> c(diamond)
    Enthalpy of combustion of carbon- graphite is -393.5 kj mol-1 and carbon-diamond is -395.4 kj mol-1.
    Offline

    15
    ReputationRep:
    (Original post by danconway)
    Total up all the bond enthalpies from all the bonds of all the reactants.
    Total up all of the bond enthalpies from all the bonds of the products.

    Do reactants minus products.
    This equals delta H.

    Bob's your uncle.


    Posted from TSR Mobile
    So is this delta h the same as enthalpy of formation?
    Offline

    3
    ReputationRep:
    (Original post by Gladiatorsword)
    S is this delta h the same ad enthalpy of formation?
    It'd be the enthalpy of reaction. You may need to put further work in e.g. draw a Hess cycle.
    Offline

    3
    ReputationRep:
    (Original post by Gladiatorsword)
    So is this delta h the same as enthalpy of formation?
    Ah, my bad. Serves me right for trying to assume what the question is!

    How you do (c):
    Formation of HF will be H2 + F2 -> 2HF, which you have to change to
    0.5H2 + 0.5F2 -> HF

    Now do the bond enthalpy thing with this reaction – this reaction is the formation of HF.
    Offline

    15
    ReputationRep:
    1. Given the data:
    4NH3(g) + 3O2(g) à 2N2(g) + 6H2O(l), DH = -1530kJmol-1
    H2(g) + 1/2O2(g) à H2O(l), DH = -288 kJmol-1
    Calculate the enthalpy of formation of ammonia.
    Offline

    6
    ReputationRep:
    (Original post by Super199)
    Care to explain what is happening with the propagation step?
    The •Cl is highly reactive, so it reacts with the neutral compound. It therefore forms HCl and CClH2, in this example. The •C(Cl)H2 is a free radical and so, is highly reactive. This reacts with Cl2, to form C(Cl2)H2 and a free radical, •Cl.

    Hopefully that explained it?
    Offline

    3
    ReputationRep:
    (Original post by Gladiatorsword)
    1. Given the data:
    4NH3(g) + 3O2(g) à 2N2(g) + 6H2O(l), DH = -1530kJmol-1
    H2(g) + 1/2O2(g) à H2O(l), DH = -288 kJmol-1
    Calculate the enthalpy of formation of ammonia.
    Have you tried drawing a Hess cycle?

    Posted from TSR Mobile
    Offline

    15
    ReputationRep:
    (Original post by Dylann)
    Have you tried drawing a Hess cycle?

    Posted from TSR Mobile
    oh....that never occurred to me. thanks
    Offline

    15
    ReputationRep:
    how would i draaw the hess's cycle?
    Offline

    3
    ReputationRep:
    Need some help with 1c and 2a for now . kind of cropped out the 2 but should be able to see which question it is. Is 2a the enthalpy of formation? Grr I hate this topic. Any help would be appreciated
    Attached Images
     
    Offline

    6
    ReputationRep:
    (Original post by Super199)
    Need some help with 1c and 2a for now . kind of cropped out the 2 but should be able to see which question it is. Is 2a the enthalpy of formation? Grr I hate this topic. Any help would be appreciated
    1c)
    Okay, so you're trying to find out the C-H bond.
    If you were given the value, and your question was to work out the value of the enthalpy change, you do the breaking bonds - forming bonds = enthalpy change.

    Now, we use that here -
    Lets say X = C-H, just to make it easier to write
    Therefore,
    (4X + 612 + 436) - (6X + 348) = -136
    You can rearrange this to -2X = -836
    You then divide by -2, to get X = 418
    Therefore C-H = 418kJ/mol

    I attached a picture of my workings, if that helps.


    Attached Images
     
    Offline

    3
    ReputationRep:
    (Original post by Jmedi)
    1c)
    Okay, so you're trying to find out the C-H bond.
    If you were given the value, and your question was to work out the value of the enthalpy change, you do the breaking bonds - forming bonds = enthalpy change.

    Now, we use that here -
    Lets say X = C-H, just to make it easier to write
    Therefore,
    (4X + 612 + 436) - (6X + 348) = -136
    You can rearrange this to -2X = -836
    You then divide by -2, to get X = 418
    Therefore C-H = 418kJ/mol

    I attached a picture of my workings, if that helps.


    This is brilliant!! Thank you Any help with 2a?
    Offline

    6
    ReputationRep:
    (Original post by Super199)
    This is brilliant!! Thank you Any help with 2a?
    I'm pretty sure it's meant to be H2O2 -> H2O + 0.5O2
    (I'm pretty sure as the one in the picture doesn't make sense unless I'm missing something?)

    Therefore, using sum of bonds broken - sum of bonds formed

    (2*463+146) - (2*463 + 496/2)
    496/2 because the oxygen molecule is O=O
    [ this can be simplified to
    146-(496/2) ]
    So, answer = -102 kJmol^-1
 
 
 
Poll
Which web browser do you use?

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.