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Why is equilibrium constant K a product of the concentrations?

  1. Sep 23, 2013 #1
    So in a given amount of solution, the equilibrium constant is the product of the concentrations of the ionic species present, raised to the power of the stoichiometric coefficient. Wouldn't it make more sense if it was the sum of the concentrations though,since basically the net number of ions of each species needs to be constant for equilibrium?
  2. jcsd
  3. Sep 23, 2013 #2
    The theoretical explanation for the formula comes from the kinetic theory.

    Let's look at a simple reaction: A+ + B- [itex]\rightleftharpoons[/itex] AB. Assume this occurs in aqueous solution.

    In aqueous solution, you have a zillion molecules -- a term I will use to refer to ions and atoms as well -- whizzing around and bashing into each other. Let's guess that the average molecule collides with another molecule once every femtosecond (don't assume this is accurate), and suppose there are n molecules (including water and the species we're interested in) in the entire solution.

    Look at a particular A+ molecule. How likely is it to react to form AB in this femtosecond? Well, first, it has to bash into a B- molecule; otherwise, it is definitely not going to react. The probability that it hits a B- molecule is [B-], if the concentration is given in units of (moles of B)/(moles of everything). Even if it does hit, there is only a certain probability that anything happens (it might not have enough energy to react, or it might hit at the wrong angle, etc.), which I'll write P1. So the probability that a particular A+ molecule reacts in a given femtosecond is P1[B-]. Therefore, the total NUMBER of A+ molecules that react every femtosecond is nP1[A+][B-]. (Of course this is also equal to the number of B- molecules that will react every femtosecond.)

    Now look at the number of AB molecules that break up every femtosecond. The total number of AB molecules in the container is n[AB], but in a given femtosecond, each one has a probability of only P2 of actually breaking up (something has to hit it hard enough, and in the right direction, to break the bond, and this might not happen). So the number that break up each femtosecond is n[AB]P2.

    If you assume the solution is at equilibrium, then the concentrations of the various species must not be changing. So in every femtosecond, the number of AB molecules forming should be equal to the number of AB molecules breaking up. Therefore, at equilibrium, we should have nP1[A+][B-] = n[AB]P2. The "n"s cancel and, manipulating, we get the familiar expression [tex]{[\mathrm{AB}] \over [\mathrm{A}^+][\mathrm{B}^-]} = {P_1 \over P_2}[/tex] where P1/P2 is the equilibrium constant.

    So that's why you might expect the rule to be what it is.
  4. Sep 24, 2013 #3
    Thanks a lot for that very detailed explanation, eigenperson!!
  5. Sep 25, 2013 #4


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    Please note it doesn't matter how do you feel something will make sense - as Feynmann put it, in the end it is nature that is always right. So if we have experimentally tested that K works the way it works (and we did it zillions of times), we don't have to care whether it seems to be making sense or not. If anything, we can try to find why it works they way it does.
  6. Sep 25, 2013 #5
    There is also a solution thermodynamics development of the equilibrium constant that leads to this same mathematical form.
  7. Sep 26, 2013 #6


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    If memory serves me well it can be also derived using statistical mechanics, but I believe "that's the way nature does it" closes any discussion :smile:
  8. Nov 27, 2013 #7
    I've been getting a whole new perspective of K, the 'equilibrium' constatnt...

    It turns out that:

    Affinity of a reaction (A)=R*T*ln(K/((x1^v1)*(x2^v2)))

    Where v1, v2 are stoichiometric coefficients, and x1, x2 are mole fractions of each reactant...

    So if the reaction is in equilibrium, A=0 and K= ((x1^v1)*(x2^v2))

    How my question is what happens if the reaction is not in chemical equilibrium? What does the K mean in that context?
  9. Nov 27, 2013 #8
    The equilibrium constant only applies to systems in equilibrium. For systems not at equilibrium you can calculate something called the reaction quotient which can then be used to predict the composition of a system once it does reach equilibrium, assuming the equilibrium constant is known.

    The derivation using thermodynamics is the more "fundamental" one because using rate laws and such is only valid when speaking of elementary reactions. Trying to derive an equilibrium constant expression using rate laws which describe a multi-step mechanism will not necessarily give you the correct equilibrium constant expression because the order of each reactant in a rate law need not equal the stoichiometric coefficient of that species in the chemical equation. On the other hand, using the condition of equilibrium (Ʃiμiγi = 0, μ being the chemical potential and γ being the stoichiometric coefficient of species i) allows us to quickly determine an expression for the equilibrium constant.
  10. Nov 28, 2013 #9


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    Basically it is due to the chemical potential of gases and ideal solutions depending on the logarithm of pressure/ concentration. These logarithms are added to calculate Gibbs enthalpy of the reaction, which corresponds to a multiplication of the partial pressures/ concentrations.
    That's clearest for an ideal gas. The thermodynamical properties of an ideal gas only depend on its partial pressure p:
    ##dU=TdS-pdV+\mu dN##
    As ##\mu=(\partial U/\partial N)_{S,V}## and ##p=(\partial U/\partial V)_{S,N}##,
    so ##(\partial \mu/\partial V)_{S,N}=-(\partial p/\partial N)_{S,V}=-kT/V=-p/N## where in the last steps I used the ideal gas law. From it you also get ## dV=NkT/p^2 dp##,
    so, at constant S, ##d\mu=\partial \mu/\partial V)_{S,N}dV=(-p/N)(NkT/p^2) dp=-kTdp/p##.
    Upon integration over p you get a nice logarithm.
    Last edited: Nov 28, 2013
  11. Nov 28, 2013 #10


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    Note: K describes an equilibrium. Identical formula, denoted Q and named reaction quotient, describes any solution. Q doesn't have to equal K, but if it doesn't, system will be reacting and will Q change till it equals K.
  12. Dec 8, 2013 #11
    Borek, you're saying in the equation for affinity, (A)=R*T*ln(K/((x1^v1)*(x2^v2)))
    The K is basically the Q if the reaction is not in equilibrium?
  13. Dec 9, 2013 #12


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    No idea what you mean by chemical affinity in this context.

    What are x1, v1, x2 and v2?

    No. K is K. If equation uses K, it means K, period.
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