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    "Nitrogen dioxide reacts with carbon monoxide emitted from car exhausts in the following reaction.

    NO2 + CO ----> NO + CO2

    rate equation for this reaction is rate = k[NO2]2

    This is a multi-step reaction. The first step is the rate determining step.

    Suggest a two step reaction mechanism for this reaction mechanism for this reaction that is consistent with the overall reaction"

    So, i know in the first step there will be "2NO2" cause it's slow step but not sure what to do from there. Had a look at the mark scheme and i see where they are coming vaguely but if someone else can explain this please.

    Thanks in advance.
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    2NO2 ---> N2O4
    maybe? Don't know where it would go from there
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    (Original post by Norby)
    2NO2 ---> N2O4
    maybe? Don't know where it would go from there
    I think what they want you to realize is that, the product formed is NO and CO2.

    First step is the rate determining step, so.....

    2NO2 -----> is something
    something -----> NO + CO2

    That's what they went because they want a two step reaction????
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    (Original post by boromir9111)
    "Nitrogen dioxide reacts with carbon monoxide emitted from car exhausts in the following reaction.

    NO2 + CO ----> NO + CO2

    rate equation for this reaction is rate = k[NO2]2

    This is a multi-step reaction. The first step is the rate determining step.

    Suggest a two step reaction mechanism for this reaction mechanism for this reaction that is consistent with the overall reaction"

    So, i know in the first step there will be "2NO2" cause it's slow step but not sure what to do from there. Had a look at the mark scheme and i see where they are coming vaguely but if someone else can explain this please.

    Thanks in advance.
    2NO2 ----> N2O4; they form a dimer

    N2O4 + 2CO ----> 2NO + 2CO2

    add eqn 1 and 2; you'd get your final stated overall eqn. N2O4 dimer is formed, single sigma N-N bond is easily cleaved at high temperature and overall process is favourable because of forming strong double bonded C=O (relative to breaking N-N)

    The N2O4 cancels out, and you'd get twice the original eqn given but that is just simple division.
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    (Original post by boromir9111)
    I think what they want you to realize is that, the product formed is NO and CO2.

    First step is the rate determining step, so.....

    2NO2 -----> is something
    something -----> NO + CO2

    That's what they went because they want a two step reaction????
    Yeah but I think they must want the first intermediary.

    Thinking about the products

    2NO2---> NO+NO3

    EDIT - Above post looks more likely
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    Okay, thank you for your input people..... you've been very helpful!!!!
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    (Original post by shengoc)
    2NO2 ----> N2O4; they form a dimer

    N2O4 + 2CO ----> 2NO + 2CO2

    add eqn 1 and 2; you'd get your final stated overall eqn. N2O4 dimer is formed, single sigma N-N bond is easily cleaved at high temperature and overall process is favourable because of forming strong double bonded C=O (relative to breaking N-N)

    The N2O4 cancels out, and you'd get twice the original eqn given but that is just simple division.
    This second step is not possible (or at least highly improbable) as you have a three particle collision.
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    Hmm, i accept without much complaints. what would be more likely then, charco?
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    (Original post by shengoc)
    Hmm, i accept without much complaints. what would be more likely then, charco?
    Given the information that the mechanism is multistep and the order wrt NO2 = 2 then the only things you can state with certainty is that there are two NO2 particles in the (first) RDS. We know also that the sum of the steps must equal the stoichiometric equation:

    NO2 + CO ----> NO + CO2

    So as a previous poster said we have:

    step 1: 2NO2 --> products
    ---- > intermediate steps --->
    final step: --> NO + CO2

    It really doesn't matter what you invent for the intermediate steps providing that you follow the basic rules and rememeber that no three particle collisions are possible.

    For example:

    STEP1: 2NO2 --> N2O4
    STEP 2: N2O4 + CO --> N2O3 + CO2
    STEP 3: N2O3 --> NO2 + NO

    if you add all of these steps up you get:

    2NO2 --> N2O4
    N2O4 + CO --> N2O3 + CO2
    N2O3 --> NO2 + NO
    --------------------
    NO2 + CO --> CO2 + NO

    which fits the stoichiometry and fulfills all of the basic requirements.

    Notice though that this does not make it correct. It remains a hypothetical mechanism until more evidence is obtained.
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    (Original post by charco)
    Given the information that the mechanism is multistep and the order wrt NO2 = 2 then the only things you can state with certainty is that there are two NO2 particles in the (first) RDS. We know also that the sum of the steps must equal the stoichiometric equation:

    NO2 + CO ----> NO + CO2

    So as a previous poster said we have:

    step 1: 2NO2 --> products
    ---- > intermediate steps --->
    final step: --> NO + CO2

    It really doesn't matter what you invent for the intermediate steps providing that you follow the basic rules and rememeber that no three particle collisions are possible.

    For example:

    STEP1: 2NO2 --> N2O4
    STEP 2: N2O4 + CO --> N2O3 + CO2
    STEP 3: N2O3 --> NO2 + NO

    if you add all of these steps up you get:

    2NO2 --> N2O4
    N2O4 + CO --> N2O3 + CO2
    N2O3 --> NO2 + NO
    --------------------
    NO2 + CO --> CO2 + NO

    which fits the stoichiometry and fulfills all of the basic requirements.

    Notice though that this does not make it correct. It remains a hypothetical mechanism until more evidence is obtained.
    i'm reminded of first year kinetics now, thanks for that charco. but the question asks to suggest two steps instead of three.
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    (Original post by charco)
    Given the information that the mechanism is multistep and the order wrt NO2 = 2 then the only things you can state with certainty is that there are two NO2 particles in the (first) RDS. We know also that the sum of the steps must equal the stoichiometric equation:

    NO2 + CO ----> NO + CO2

    So as a previous poster said we have:

    step 1: 2NO2 --> products
    ---- > intermediate steps --->
    final step: --> NO + CO2

    It really doesn't matter what you invent for the intermediate steps providing that you follow the basic rules and rememeber that no three particle collisions are possible.

    For example:

    STEP1: 2NO2 --> N2O4
    STEP 2: N2O4 + CO --> N2O3 + CO2
    STEP 3: N2O3 --> NO2 + NO

    if you add all of these steps up you get:

    2NO2 --> N2O4
    N2O4 + CO --> N2O3 + CO2
    N2O3 --> NO2 + NO
    --------------------
    NO2 + CO --> CO2 + NO

    which fits the stoichiometry and fulfills all of the basic requirements.

    Notice though that this does not make it correct. It remains a hypothetical mechanism until more evidence is obtained.
    Dude, sometimes you complicate things lol.
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    (Original post by boromir9111)
    Dude, sometimes you complicate things lol.
    That's strange!

    I thought I was simplifying things :confused:
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    (Original post by charco)
    That's strange!

    I thought I was simplifying things :confused:
    True, you are but it asks for a two step reaction and you've given 3, i think you need to "dumb" down your answer to A level chemistry lol. I get the part that 2NO2 = N2O4 but now you're saying that N2O4 + CO does not equal 2NO + 2CO? this is what essentially got me because you're meant to get CO2 but ended up with 2CO from an earlier post by "shengoc"?
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    (Original post by shengoc)
    i'm reminded of first year kinetics now, thanks for that charco. but the question asks to suggest two steps instead of three.
    You are quite correct - the question says:

    This is a multi-step reaction. The first step is the rate determining step.

    Suggest a two step reaction mechanism for this reaction mechanism for this reaction that is consistent with the overall reaction"
    Which is rather contradictory, or at least I would not call a two step process 'multi-step'

    But if you want a two step process:

    2NO2 --> NO + NO3
    NO3 + CO --> NO2 + CO2

    which also fulfills the requirements... :yep:
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    (Original post by charco)
    You are quite correct - the question says:
    Which is rather contradictory, or at least I would not call a two step process 'multi-step'

    But if you want a two step process:

    2NO2 --> NO + NO3
    NO3 + CO --> NO2 + CO2

    which also fulfills the requirements... :yep:
    Wait we need NO and CO2?

    edit - lol, it cancels out, i get it.
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    (Original post by boromir9111)
    Wait we need NO and CO2?
    the equation overall is:

    NO2 + CO ----> NO + CO2

    If you add up the steps I gave:


    2NO2 --> NO + NO3
    NO3 + CO --> NO2 + CO2
    -------------------------
    2NO2 + NO3 + CO --> NO + NO3 + NO2 + CO2

    and cancel out common factors

    NO2 + CO --> NO + CO2

    So what's the problem?
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    (Original post by charco)
    the equation overall is:

    NO2 + CO ----> NO + CO2

    If you add up the steps I gave:


    2NO2 --> NO + NO3
    NO3 + CO --> NO2 + CO2
    -------------------------
    2NO2 + NO3 + CO --> NO + NO3 + NO2 + CO2

    and cancel out common factors

    NO2 + CO --> NO + CO2

    So what's the problem?
    Yeah, just before you sent that i edited saying that it cancels out but thanks for helping me out on this, much appreciated!!!
 
 
 
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