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Law of Mass Action vs The Reaction Rate Law

  1. Feb 1, 2015 #1
    Why in the Law of Mass Action we raise the concentrations of the reactants to the power of their coefficients according to the stoichiometry of the chemical reaction,
    while the order of the reaction must be determined experimentally ?
    I may not understand the difference between them.
     
  2. jcsd
  3. Feb 1, 2015 #2

    Borek

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    It can be derived from thermodynamics and the way equilibrium is related to the changes in the Gibbs free energy.

    That's because these things are completely unrelated. For some simple cases equilibrium can be explained with the kinetic approach, but in general these are separate things.

    Order of the reaction depends on its mechanism, equilibrium doesn't.
     
  4. Feb 1, 2015 #3

    epenguin

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    I was thinking over how to explain this, but the problem is my explanation would not be better than the one you have in your textbook written by someone who thought for longer about how to explain it. We don't know what you know and therefore what problem you have with it.

    Maybe you haven't done thermodynamics yet. Maybe you have and have some problem with it. The thermodynamic proof of the law consists in imagining a reaction at equilibrium, then a virtual displacement from there where the amounts of the components vary. But the variation of the different substances is necessarily constrained by the stoichiometry. Equilibrium means The free energy is at a minimum, I.e. It's varation by the virtual displacement is zero. Consequence: the equilibrium law you call Mass Action is constrained by the stoichiometry. A very general (thermodynamic) argument which is independent of anything about the rates at which equilibrium is arrived at.

    It does not constrain the way rates starting from non-equilibrium depend on concentrations. But for any experimental or mechanistic law or equation for the rate in one direction, the Mass Action law does then constrain that for the reaction in the opposite direction.

    Perhaps you can find some examples of such equations in one direction and work out what they have to be in the other.
     
    Last edited: Feb 1, 2015
  5. Feb 2, 2015 #4
    Thanks a lot for your replies.
    Yes, I haven't yet done thermochemistry and my textbook, A Molecular Approach, doesn't explain how Law of Mass Action is derived.
    All what I want to know for now : Is the rate of the reaction at equilibrium is directly proportional to the concentration of a reactant raised to a power according to the stoichiometry of the chemical equation, while the rate of the reaction at any other point of the reaction is directly proportional to the concentration of the reactants raised to a power that is determined experimentally?
     
  6. Feb 2, 2015 #5

    Borek

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    There is no difference between order of the reaction at equilibrium and at any other moment.

    Rate does change with the concentrations, but the order (and underlying mechanism) does not.
     
  7. Feb 2, 2015 #6

    epenguin

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    OK it is also possible to understand this without the thermodynamics which can be learned later for a deeper understanding.

    I trust you do know what you mean by "rate of the reaction at equilibrium"? :oldsmile:
    In one sense the rate of reaction at equilibrium, that is rate of change of concentration of anything, is zero!

    But that does not mean no chemical reaction is going on, just that the rate of the forward reaction is equal to the rate of the back reaction.

    This is not theoretical explaining away, it can be observed. In a reaction say A + B ⇔ C + D at equilibrium you can add a tiny amount of radioactive A as tracer to the equilibrium mixture and observe it in time being incorporated into C, without overall change of concentration of anything. You can then observe how changing [A] of etc. changes the rate. (It also changes the equilibrium. If you don't want that you have to change also [C] or [D] correspondingly.) It is then found that the forward rate may not be proportional to [A].

    I think to take it any farther you need actual examples of mechanisms and equations. There is an almost trivial example - a catalyst. The reaction rate may be zero in its absence, so say the forward rate can be dependent on [X] catalyst concentration. This may be proportionality to [X] or it may be some more complicated function. But the backwards rate will have exactly the same dependence on [X] with the result [X ] appears in the mechanism-dependent rate laws but does not appear in the "Mass action" or equilibrium expression.

    It all depends in the end on 'microscopic reversibility' which is that the steps and structures molecules go through in the forward reaction are exactly the same as those they go through in the reverse order in the back reaction - anything else is physically unimaginable
     
  8. Feb 2, 2015 #7
    I think I do understand this. :)


    Yeah, It might be proportional to [A]2 or [A]3, however, it's determined experimentally.
    But why do we just use [A] when expressing the equilibrium constant.

    I'm sorry, if I'm asking too much.
     
  9. Feb 2, 2015 #8

    Borek

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    You were already told several times - why it would be just [A] can be explained in terms of thermodynamics.
     
  10. Feb 2, 2015 #9
    Oh, ok, thanks alot.
    It seems that I must wait till I do some thermodynamics.:)
     
  11. Apr 26, 2015 #10
    yes but can you prove it?
     
  12. Apr 26, 2015 #11

    Borek

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    You can derive it by calculating free energy changes during the reaction. There is a derivation even in wikipedia.
     
  13. Apr 27, 2015 #12

    DrDu

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    Two points: 1. reaction rates can also be derived in principle ab initio, so they aren't mere experimental input.
    2. in the equilirium constant, the two reaction rates go into the numerator and denominator, respectively, and terms simply cancel. In principle, this happens already when writing down the reaction A->B, you could also write 2A->B+A.
     
  14. Apr 27, 2015 #13

    epenguin

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    If the equilibrium is for instance A ⇔ B then there will be an equilibrium constant Keq = [ Beq]/[Aeq]. The forward rate equation might still be rf = kf[A]3, to take your example, but then the backwards rate law would have to be rb = kb[A]2[ B], giving still the above equilibrium law.

    It is not true that rate equations have always integral powers. That is only true generally of the elementary steps, not the overall resulting rate equations.

    For instance the simplest solid- or enzyme-catalysed reaction rates have an equation like

    rf = k1[E][A]/{KA{1 + [A]/KA + [ B]/KB)}

    where [E] is total enzyme concentration or catalyst site concentration.

    For backwards reaction

    rb = k-1[E][ B]/{KB{1 + [A]/KA + [ B]/KB)}

    But the result at equilibrium when rf = rb is

    [ Beq]/[Aeq] = k1KB/k-1KA = Keq, the equilibrium constant, independent of reaction mechanism.

    Another way to say all this is if you write out the forward and backwards rate equations then always the expression for
    -d[A]/dt = d[ B]/dt = (rf - rb) will have (Keq[A] - [ B]) as a factor.

    At this point I think we have told you everything we usefully can and the most useful thing you can do is what I recommended earlier: think of some reaction mechanisms, work out the rate equations they predict, learning how to do it from your book if necessary - it is not very difficult -and try and see if you can find a mechanism for which the expected equilibrium law is untrue (remembering microscopic reversibility mentioned above). If you think you have found an example come back with it and we will surely find what you have missed; if you don't find one, but find the expected equilibrium law always confirmed, come back and confirm this finding to us.
     
    Last edited: Apr 28, 2015
  15. May 10, 2015 #14

    James Pelezo

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    Last edited: May 10, 2015
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