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Rate law of elementary reaction

  1. Oct 18, 2014 #1

    Maylis

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    Gold Member

    Hello,

    I am working through some reaction engineering problems, and something in particular has caught my attention related more to chemistry I think than reaction engineering.

    Suppose we have a gas phase reaction
    ## A + 3B \rightarrow 2C ##

    If this is an elementary reaction, the rate law is

    ##-r_{A} = k_{A}C_{A}C_{B}^{3}##

    However, if species B is my limiting reactant, I wish to divide through to make the reaction

    ##\frac {1}{3} A + B \rightarrow \frac {2}{3} C##

    Why is it that the rate law cannot be
    ##-r_{A} = k_{A}C_{A}^{1/3}C_{B}##?

    It seems like all other parameters are allowed to (and must be changed) to calculate the concentration of a species as a function of conversion. For example, in a flow system that is isothermal and isobaric, the concentration of species ##i## is given as

    [tex] C_{i} = C_{A0} \frac {\theta_{i} + \nu_{i} X}{1 + \epsilon X} [/tex]

    where ##C_{i}## is the concentration of species ##i## at conversion ##X##, ##C_{A0}## is the initial concentration of species ##A##, ##\theta_{i} \equiv \frac {C_{i0}}{C_{A0}} = \frac {y_{i0}}{y_{A0}}##, and ##\epsilon \equiv y_{A0} \delta##, where ##y_{A0}## is the initial mole fraction of species ##A##, and ##\delta \equiv \sum_{i} \nu_{i}##, in this case ##\delta = \frac {2}{3} - \frac {1}{3} - 1 = - \frac {1}{3}##.

    So my question can be summarized as this:
    For the gas phase reaction,
    ## A + 3B \rightarrow 2C ##

    t's perfectly fine to divide through to change the stoichiometric coefficients of the reaction to ##\frac {1}{3} A + B \rightarrow \frac {2}{3} C##, however the rate law must remain as ##-r_{A} = k_{A}C_{A}C_{B}^{3}##, and cannot be changed to ##-r_{A} = k_{A}C_{A}^{1/3}C_{B}##?

    However, the ##\delta## term has to be based on the divided coefficients,
    ##\delta = \frac {2}{3} - \frac {1}{3} - 1 = - \frac {1}{3}##., and cannot be ##\delta = 2 - 3 - 1 = -2##.

    This discrepancy is confusing to me because now I can't tell why I can just arbitrarily choose some coefficients to do some calculations, and other times I have to use different ones.
     
    Last edited: Oct 18, 2014
  2. jcsd
  3. Oct 18, 2014 #2

    Borek

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    Staff: Mentor

    First of all: in general (without getting into details of your specific case) - we decided long ago that by convention correctly balanced reaction has a set of the lowest integer coefficients. That's not the only convention possible. However, changing it will change values of all tabulated constants (think equilibrium, reaction enthalpy), it will also require rederiving almost every equation using stoichiometric coefficients. In many cases only ratio of these coefficients matter, so there will be no difference, in other cases exact values matter, so we will get an alternative, but absolutely equivalent set of equations.

    This doesn't apply to kinetics:

    Because that's not what we observe experimentally, and the rate law must follow reality.
     
  4. Oct 18, 2014 #3

    Borek

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    Staff: Mentor

    First of all: in general (without getting into details of your specific case) - we decided long ago that by convention correctly balanced reaction has a set of the lowest integer coefficients. That's not the only convention possible. However, changing it will change values of all tabulated constants (think equilibrium, reaction enthalpy), it will also require rederiving almost every equation using stoichiometric coefficients. In many cases only ratio of these coefficients matter, so there will be no difference, in other cases exact values matter, so we will get an alternative, but absolutely equivalent set of equations.

    This doesn't apply to kinetics:

    Because that's not what we observe experimentally, and the rate law must follow reality.
     
  5. Oct 18, 2014 #4

    epenguin

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    That reaction cannot itself be an elementary reaction. All elementary reactions are monomolecular or bimolecular. (A collision of four molecules is too rare to be a reaction mechanism). The reaction mechanism must be made of a number of elementary reactions.

    If you are still thinking of the reaction of hydrogen and nitrogen,:) it is not a reaction between gas molecules as gas, but molecules adsorbed on a catalyst.
     
  6. Oct 18, 2014 #5

    Maylis

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    Thanks for the info regarding elementary reactions only be monomolecular or bimolecular, I was not aware of that. However, the confusion of why the rate law can only be in terms of the original reaction, however the sum of the stoichiometric coefficients have to be changed such that the basis species has a coefficient of 1, remains unresolved.

    Just to clarify this, it seems misleading to me. I know you can have elementary rate laws of the form ##-r_{A} = kC_{A}^{1/2}C_{B}## or some other power that is less than one.
     
  7. Oct 19, 2014 #6

    Borek

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    I think you are again mistaking convention of the balancing reactions with the observed rates. They don't have to follow reaction equation as we write it.

    If memory serves me well observed fractional powers in kinetics mean the underlying mechanism is not just a simple, mono- or bimolecular reaction, but involves some intermediate steps. Which is actually another reason to treat kinetics separately from the stoichiometry (and equilibrium).
     
  8. Oct 19, 2014 #7

    epenguin

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    I am confused about what your confusion is, so it is probably netter to forget it, except to say stoichiometric numbers must've always appear in equilibrium equations. E.g. for our usual reaction we must have

    [NH3]2/[N2][H2]3 = Keq .

    But for kinetic equations we can have a wide variety. The only limitation is that the ratio of the expression for the forward reaction to that of the back reaction must correspond to the equilibrium expression - see any textbook.

    It is easy to imagine a mechanism giving your kinetic expression quoted above: if A dissociated to a small extent to two different species their equilibrium concentration would be proportional to CA1/2. If this dissociation product reacts with B at a rate proportional to its concentration the quoted expression would be the rate law.
     
  9. Oct 19, 2014 #8

    Borek

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