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    Can anyone explain what the link is between orders of reaction and mechanism? For example, if O2 is first order and NO is second order, which of the following proposed mechanism is correct and why:

    Mechanism 1:
    2NO + O2 -> 2NO2

    Mechanism 2:
    NO + O2 -> NO3
    NO3 + O -> 2NO2

    Mechanism 3:
    O2 -> 2O
    2NO + 2O -> 2NO2


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    The order of a substance is related to the number of molecules of the substance in the rate determining step.
    Therefore there's one molecule of O2 and 2 molecules of NO in the rate determining step
    So its mechanism one.
    Though you could have seen that straight away; its the only one where they are both in the same step (rate determining step), and they both have to be because they have an order other then zero.

    Hope that makes sense lol, my explanations aren't great!
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    make an equation:

    rate = k [O2] [NO]^2

    In the rate determining step, both species from the equation have to be in the mechanisms that are given to you in the exam.

    Now we are left with mechanism 1 and 2.

    We now know that because NO is raised to the power of two, the coefficient must be two <--- (not entirely sure why this last part works, but was told by my teacher to do it this way)
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    Thanks perfectly explained
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    (Original post by Clumsy_Chemist)
    Thanks perfectly explained
    But not entirely correct.

    There are a couple of important things you need to know.

    In any step of a mechanism you can have a maximum of 2 particles colliding. Three particle collisions are statistically very unlikely and will not contribute to any mechanism.

    This immediately rules out mechanisms 1 and 3

    The rate equation shows the dependency of the reaction on the components of the reaction, but reflects only the rate determining step (RDS), PLUS any steps that lead into it that make intermediates necessary for the RDS

    The orders of the components in the rate equation show their molecularity in the RDS (plus steps leading into it)

    If you are given:

    Rate = k[O2][NO]2

    Then you can infer that as the overall order = 3, there must be at least two steps in the mechanism and that the RDS cannot be the first step.

    Your mechanism 2 (which was actually copied out incorrectly) is:

    Mechanism 2:
    ------------------
    STEP 1: NO + O2 -> NO3
    STEP 2: NO3 + NO -> 2NO2
    --------------------------------------------------------------

    Here you can see that there is an intermediate (NO3) which must be made in step 1 before it can go on to react in step 2.

    This requires collision of NO and O2.
    In step 2 (RDS) there is another NO needed

    So the order with respect to O2 is 1 and for NO it is 2
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    (Original post by charco)
    But not entirely correct.

    There are a couple of important things you need to know.

    In any step of a mechanism you can have a maximum of 2 particles colliding. Three particle collisions are statistically very unlikely and will not contribute to any mechanism.

    This immediately rules out mechanisms 1 and 3

    The rate equation shows the dependency of the reaction on the components of the reaction, but reflects only the rate determining step (RDS), PLUS any steps that lead into it that make intermediates necessary for the RDS

    The orders of the components in the rate equation show their molecularity in the RDS (plus steps leading into it)

    If you are given:

    Rate = k[O2][NO]2

    Then you can infer that as the overall order = 3, there must be at least two steps in the mechanism and that the RDS cannot be the first step.

    Your mechanism 2 (which was actually copied out incorrectly) is:

    Mechanism 2:
    ------------------
    STEP 1: NO + O2 -> NO3
    STEP 2: NO3 + NO -> 2NO2
    --------------------------------------------------------------

    Here you can see that there is an intermediate (NO3) which must be made in step 1 before it can go on to react in step 2.

    This requires collision of NO and O2.
    In step 2 (RDS) there is another NO needed

    So the order with respect to O2 is 1 and for NO it is 2
    Why cant the RDS be in the first reaction/why at least 2 steps?

    I hate this topic...

    Sorry to thread hijack OP!!

    Thanks
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    As Charco says, a reaction relying on the collision of 3 different particles for the same step is statistically improbable. Thus each step will involve a maximum of two reagents.

    As the rate equation specifies that the order with respect to NO is 2 AND with respect to O is 1, this means there are two stages to the RDS.

    The stage that produces your product is:
    NO3 + NO -> 2NO2
    however that NO3 is generated by another reaction before being used up:
    NO + O2 -> NO3
    The rate of the second stage will depend on the concentration of the product of the first, and thus the concentration of the reactants of the first.....

    This means overall the reaction can be summarised as:
    O2 + 2NO -> 2NO2
 
 
 
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