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Mechanism from Order of Reaction Help!!! Question iss weird. Watch

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    Please could you look at question Q3)b)iii).... the mechanism one. I do not understand the mark scheme. Why is it that only one molecule is needed when in the equation in the question, it states something else?
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    (Original post by Future_Dr)
    Please could you look at question Q3)b)iii).... the mechanism one. I do not understand the mark scheme. Why is it that only one molecule is needed when in the equation in the question, it states something else?
    The rate equation gives you the molecules that are in the rate determining step - you can see the concentration of N_{2} O_{5} is raised to the first power meaning there is only one molecule of this present in the rate determining step - it can't be mechanism B because there are 2 molecules of N_{2} O_{5}
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    It's because the rate equation is first order with respect to N2O5, mechanism B (two molecules) is second order.
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    (Original post by Felix Felicis)
    The rate equation gives you the molecules that are in the rate determining step - you can see the concentration of N_{2} O_{5} is raised to the first power meaning there is only one molecule of this present in the rate determining step - it can't be mechanism B because there are 2 molecules of N_{2} O_{5}

    (Original post by EierVonSatan)
    It's because the rate equation is first order with respect to N2O5, mechanism B (two molecules) is second order.
    But I thought that since the rate equation has the N2O5 raised to first order.. it just means that the concentration is directly proportional to the rate. Also I thought that you can determine the reaction order only experimentally. Isn't it the power next to the square bracktets which is used to show the coeeficient only used in equilirbria constants?
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    (Original post by Future_Dr)
    But I thought that since the rate equation has the N2O5 raised to first order.. it just means that the concentration is directly proportional to the rate. Also I thought that you can determine the reaction order only experimentally. Isn't it the power next to the square bracktets which is used to show the coeeficient only used in equilirbria constants?
    No, the power to which the concentration is raised also gives the number of molecules of the reactant in the rate determining step
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    (Original post by Felix Felicis)
    No, the power to which the concentration is raised also gives the number of molecules of the reactant in the rate determining step
    So given this, you can make a rate equation just by looking at the coefficients of the reactants?
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    (Original post by Future_Dr)
    Also I thought that you can determine the reaction order only experimentally.
    This is true, but the rate equation gives you information about the mechanism :yes: There are several steps in many chemical reactions and you don't know which is rate limiting until you've done some experimentation

    So given this, you can make a rate equation just by looking at the coefficients of the reactants?
    No :p:
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    (Original post by Future_Dr)
    So given this, you can make a rate equation just by looking at the coefficients of the reactants?
    You could make a rate equation by looking at the coefficient of reactants in the reaction mechanism, but not from the overall stoichiometric equation
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    (Original post by EierVonSatan)
    This is true, but the rate equation gives you information about the mechanism :yes: There are several steps in many chemical reactions and you don't know which is rate limiting until you've done some experimentation



    No :p:
    Rigghhtt.. so if I could just clear this from my head. If you were given an equation of a reaction like 2N2O5 ----> 4NO2 + O2 you will do experiments and see how the concentration of the N2O5 will effect the rate. And you find out that Rate = k[N2O5]^1. Then you wish to know the mechanism. So you would know that the Overall rate = the rate of the rate determininbg step (rds). So the rds involves only the N2O5.

    If I rate equation had [2N2O5] then would this mean that the mechanism also involves 2N2O5?

    Sorry to be a pest.
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    (Original post by Future_Dr)
    Rigghhtt.. so if I could just clear this from my head. If you were given an equation of a reaction like 2N2O5 ----> 4NO2 + O2 you will do experiments and see how the concentration of the N2O5 will effect the rate. And you find out that Rate = k[N2O5]^1. Then you wish to know the mechanism. So you would know that the Overall rate = the rate of the rate determininbg step (rds). So the rds involves only the N2O5.
    This is correct

    If I rate equation had [2N2O5] then would this mean that the mechanism also involves 2N2O5?
    It would be rate = k[N2O5]2 for mechanism B.

    Sorry to be a pest.
    No problem
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    (Original post by EierVonSatan)
    This is correct



    It would be rate = k[N2O5]2 for mechanism B.



    No problem
    Ahh.. I get it... is this just something we have to remember or there is a reason for this?
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    (Original post by Future_Dr)
    Ahh.. I get it... is this just something we have to remember or there is a reason for this?
    There is always a reason :yes:

    If you have a reaction that goes A + B ---> C in a single step then the rate of the reaction is written: rate = k[A][B] which I'm sure you're familiar with

    Compare this with a single step reaction A + A --> D then you would write this rate as: rate = k[A][A] or rate = k[A]2 since there are two molecules of A involved.

    Hoes that help?
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    (Original post by EierVonSatan)
    There is always a reason :yes:

    If you have a reaction that goes A + B ---> C in a single step then the rate of the reaction is written: rate = k[A][B] which I'm sure you're familiar with

    Compare this with a single step reaction A + A --> D then you would write this rate as: rate = k[A][A] or rate = k[A]2 since there are two molecules of A involved.

    Hoes that help?
    Yeah.. I get it. Thanks

    I just stumbled upon another question which is pretty easy but yet again strange.

    I understand the order of reaction wrt O3 BUT I don't get the ethene order.
    I used the last two experiments with ethene so that the 2/1 = 2 and when I did same with the rate.. I found 16/4 = 4. So clearly the rate of the reaction is square of the concentration of ethene. Therefore the order is 2. So why does it say 1 in the MS?
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    (Original post by Future_Dr)
    Yeah.. I get it. Thanks

    I just stumbled upon another question which is pretty easy but yet again strange.

    I understand the order of reaction wrt O3 BUT I don't get the ethene order.
    I used the last two experiments with ethene so that the 2/1 = 2 and when I did same with the rate.. I found 16/4 = 4. So clearly the rate of the reaction is square of the concentration of ethene. Therefore the order is 2. So why does it say 1 in the MS?
    Imagine that experiment 3 had an ethene concentration of 1.0 what would you expect the rate to be, knowing the order with respect to ozone?

    Compare this value with experiment 3, what is the order for ethene?
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    (Original post by EierVonSatan)
    Imagine that experiment 3 had an ethene concentration of 1.0 M what would you expect the rate to be, knowing the order with respect to ozone?

    Compare this value with experiment 3, what is the order for ethene?
    Ahh.. So the rate is 4 times because it is 2 times for the O3 and 2 times for the ethene. Rigghtt. I feel so stupid. This is a type of question.
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    (Original post by Future_Dr)
    Ahh.. So the rate is 4 times because it is 2 times for the O3 and 2 times for the ethene. Rigghtt. I feel so stupid. This is a type of question.
    Just takes practice
 
 
 
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