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Equilibrium Constants

  1. Oct 19, 2014 #1
    Consider the following generic equilibrium:

    aM + bN cO + dP

    An equilibrium constant, K, can be defined as:

    $$K = \frac{[O]^c [P]^d}{[M]^a [N]^b}$$

    But couldn't we also define another equilibrium constant similarly with coefficients that are in the same ratio as our original equation? For instance, α = 2a, β = 2b, and so on. We can then do the following:

    $$K' = \frac{[O]^γ [P]^δ}{[M]^α [N]^β}$$

    Clearly, those are two conflicting results.

    How exactly do we define a unique equilibrium constant and simultaneously grant ourselves the liberty of multiplying through the stoichiometric equation by some constant (which, in this case, was 2)?
    Where exactly am I going wrong?

    A similar conundrum arises in kinetics. Consider a similar reaction to the one I wrote above, except that now it is a one-way reaction rather than a reversible one. I will use the letter ##v## to denote rate.

    $$v = -\frac{1}{a} \frac{d[M]}{dt}$$

    To make the problem less abstract, let's consider a real reaction. The bromination of ethene (ethylene).

    $$C_2 H_4 + Br_2 → C_2 H_4 Br_2$$

    If we consider the equation in that form, then a = 1 in the above equation for rate. If we multiply both sides of the equation by 2 (and I don't see why such an action would be erroneous), then a = 2. Shouldn't the value of a in our definition of rate (not unlike our definition of equilibrium constant) be unique?

    Could someone please clarify this for me?
     
    Last edited: Oct 19, 2014
  2. jcsd
  3. Oct 19, 2014 #2

    Borek

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    Staff: Mentor

    I already told you it is a matter of convention. Using other than the lowest integers set of coefficients yields a different value of the equilibrium constant - but as it is defined differently, that's to be expected. What is important is the fact, that these systems are equivalent, and when you use them consistently for calculations, final result should be the same.

    [tex]K = \frac {A^aB^b}{C^cD^d}[/tex]

    [tex]\log K = a \log A + b \log B - c \log C - d \log D[/tex]

    Multiply by any constant you want, equation still holds.
     
  4. Oct 19, 2014 #3

    Ygggdrasil

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    Science Advisor

    With regard to the equilibrium constant, remember that the equilibrium constant of a reaction is related to the change in free energy associated with the reaciton: K = exp(-ΔG/RT)

    Clearly, the free energy change associated with the reaction of two moles of a substance will be greater than the free energy change associated with the reaction of one mole of a substance, so the equilibrium constant must be different.
     
  5. Oct 19, 2014 #4
    Thanks!
    I think you've mistaken me for someone else, though. I don't recall ever discussing this topic on PF. o_O
     
  6. Oct 20, 2014 #5

    Borek

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    Staff: Mentor

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