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Equilibrium constant vs rate constant - where kinetics meets thermodynamics?

  1. Sep 17, 2012 #1
    Equilibrium constant vs rate constant - where kinetics meets thermodynamics??

    The Equilibrium Law for aA + bB ⇌ cC + dD gives Kc = [C]c [D]d / [A]a x b at a given temperature. Kc* is also equal to the ratio of forward to reverse rate constants kfwd/krev. The rate expression for the forward and reverse reactions often does not follow the stoichiometric equation, and is of the form: rate = kfwd [A]mn rather than rate = kfwd[A]ab ie where m, n often does not equal a, b. At equilibrium, kfwd[A]ab = krev[C]c[D]d, so how is it that Kc can equal kfwd/krev … and how can the units of Kc equal those of kfwd / krev??
     
  2. jcsd
  3. Sep 17, 2012 #2
    Re: Equilibrium constant vs rate constant - where kinetics meets thermodynamics??

    maybe this will help you: the equilibrium constant is not actually supposed to be written with concentrations. It is supposed to be written with a unitless parameter called the activity, which is numerically equal to the concentration in dilute mixtures, but can change with anything that affects the chemical potential, which means not just concentration changes, but temperature, pressure, etc. it is just written as concentration because it is numerically equal at STP for most dilute solutions.

    So the equilibrium constant should be unitless and that solves your problem.
     
  4. Sep 18, 2012 #3
    Re: Equilibrium constant vs rate constant - where kinetics meets thermodynamics??

    … sorry I meant: At equilibrium kfwd[A)mn = krev[C]p[D]q ….
     
  5. Sep 18, 2012 #4
    Re: Equilibrium constant vs rate constant - where kinetics meets thermodynamics??

    The rate law for a reaction does not always follow its stoichiometry for reasons quite different from the activity/concentration relationship.

    For example, the reaction H2(g) + I2(g) ⇔ 2 HI(g)
    when carried out with low pressure reactants diluted into 1 atm nitrogen gas, follows the initial rate law

    Rate (HI) = const * Intensity of light * p(H2) * √p(I2)

    and an even more complicated rate law as HI starts to form. The important thing is that for any reversible reaction system, a complicated rate law for a forward reaction must be matched by an equally complicated rate law for the reverse reaction, such that at the activities represented in the usual equilibrium constant expression, the forward and reverse rates will always be exactly the same as one another.
     
  6. Sep 18, 2012 #5
    Re: Equilibrium constant vs rate constant - where kinetics meets thermodynamics??

    Thanks JohnRC - so are you saying that if we have A+B ⇌ C+D and eg rate(fwd) = kfwd[A], ie first order wrt A and zero order wrt B, then rate(reverse) must equal krev[C] [D] ^-1 , to give kfwd/krev = [C] [D] / [A] = Kc?
     
  7. Sep 18, 2012 #6
    Re: Equilibrium constant vs rate constant - where kinetics meets thermodynamics??

    In effect that is the case, although this particular rate law would only need to apply close to the equilibrium point.

    Your example has actually given me a bit of a problem, because the rate law you have written cannot possibly be an initial rate law -- that is:
    if we were to take a mixture of products free of reactants and start the reaction, the reaction would initially have an infinite rate, and that is an impossibility.

    I will think some more and get back later.
     
  8. Sep 18, 2012 #7
    Re: Equilibrium constant vs rate constant - where kinetics meets thermodynamics??

    In the case of the HI example that I presented as an example in my earlier post, the initial rate of the reverse reaction is zero; although HI(g) is significantly decomposed to H2(g) and I2(g) in an equilibrium mix, HI is perfectly stable in the absence of a trace of I2 (or some other source of free radicals that can get the reaction started).

    I think it might be the case that any reaction that is strictly zero order in a reactant cannot be reversible.

    The only reason that a reaction can have a rate law for the forward reaction that does not match the numerator of the equilibrium constant expression is that it is really a complex reaction system made up of a number of "elementary reaction steps".

    I will use the example I have drawn on already. For the HI reaction we have:

    I2(g) + light(λ = 500-650 nm) → 2 I
    Rate = ki.I(light).p(I2)

    I + H2(g) ⇔ H + HI(g)
    Forward rate = kf1.p(I).p(H2)
    Reverse rate = kr1.p(H).p(HI)

    H + I2(g) ⇔ I + HI(g)
    Forward rate = kf2.p(H).p(I2)
    Reverse rate = kr2.p(I).p(HI)

    2 I + N2(g) → I2(g) + N2(g)
    Rate = kt.p(I)2.p(N2)

    We then have to do a lot of algebra to arrive at the rate law that I gave, which is actually a simplified rate law that will only really apply after the intermediate free radicals have built up to steady state concentrations.
     
  9. Sep 19, 2012 #8
    Re: Equilibrium constant vs rate constant - where kinetics meets thermodynamics??

    Hi - it seems your theory works in your example. For the forward reaction, your second step is the slow step preceded by a fast equilibrium, which gives the rate expression. Then for the reverse reaction the first step is rate determining with a rate equation of Rate = k√p(I2])-1. Perhaps we have to assume a small but finite amount of reactant is present (by dissociation or reaction) when starting with "pure products".
     
  10. Sep 20, 2012 #9
    Re: Equilibrium constant vs rate constant - where kinetics meets thermodynamics??

    If I'm not mistaken I think the problem lies in the original question and a small misconception (or perhaps confusion) in notation/nomenclature.

    When you write kForwarda[A]b = kReversec[C]d[D] you're describing the condition for equilibrium in terms of kinetics, that is, when the forward rate and the reverse rate are equivalent. The equilibrium constant for the reaction, however, is not expressed as a ratio of the rates of reaction (concentration dependent) but as the ratio of the forward and reverse rate constants (concentration independent): K = kForward/kReverse.

    Let's take the a hypothetical ionisation reaction as an example:

    RX ⇌ R+ + X-

    with the forward rate constant being k1, the reverse rate constant as k-1.

    At equilibrium k1[RX] = k-1[R+][X-] and one can write an equilibrium constant (K) for the process as K = ([R+][X-])/[RX] and in terms of rate constants of the forward and reverse reaction you write it as K = k1/k-1.

    This can be extended to a more complicated system with multiple equilibria (i happened to come across the following example earlier today). For example, if we take the example above and add a second equilibrium (as in an exo-endo isomerisation reaction):

    RXexo ⇌ RXendo

    with equilibrium constant K, forward rate constant kexo and reverse rate constant kendo and involves ionisation as an intermediate step:

    RXexo ⇌ R+ + X- ⇌ RXendo

    In this case the equilibrium constant K = RXendo/RXexo and one will have to take into account all the individual rate constants in order to express the equilibrium constant in terms of rate constant. Suppose the forward and backward rates with respect to RXexo are k1 and k-1, respectively and, similarly, k-2 and k2 with respect to RXendo; then one would describe the equilibrium constant of the isomerisation reaction as:

    K = (k1k-2)/(k2k-1)

    I hope that helps. c:
     
    Last edited: Sep 20, 2012
  11. Sep 20, 2012 #10
    Re: Equilibrium constant vs rate constant - where kinetics meets thermodynamics??



    You are quite right, GhostlyPaw, and your exposition is sound. The problem, though, that was originally raised can be expressed in the following way:

    ------
    K = k(forward)/k(reverse) = (product{multiplication} of product{substance} activities) / (product{multiplication} of reactant activities)

    But for many reactions the equilibrium constant still takes this activity product form when the rate law for the forward reaction is different to
    Rate = k(forward) * (product of reactant activities)

    How come? and how do we reconcile units?
    ------

    Your illustration works well for a simple situation with consecutive or competing reactions. The most usual situation, though, is a chain reaction (nearly all gas reactions and polymerization reactions) where a much more complicated analysis is needed, but will still yield an equilibrium constant expression of the same form in terms of substance activities.
     
    Last edited: Sep 20, 2012
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