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Equilibrium constant and concentration

  1. Jul 9, 2011 #1
    I am aware that the equilibrium constant, Kc for the reaction
    [itex] {aA}~~+~~{bB}~~~\mathop{\rightleftharpoons}^{k_f}_{k_b}~~~{cC}~~+~~{dD}[/itex] is

    [itex]\frac{k_f}{k_b} = \frac{[A]^a ^b}{[C]^c [D]^d}[/itex]

    Now my question is, are the stoichiometric coefficients a,b,c and d, written in simplest whole number ratios? Because if they can be written as we like, then the exponents of the concentrations of the reactants and products in the formula for the equilibrium constant will be different and we will have different values for the same reaction. My textbook says the relationship between the equilibrium constants for the two reactions

    [itex] {aA}~~+~~{bB}~~~\mathop{\rightleftharpoons}^{k_f}_{k_b}~~~{cC}~~+~~{dD}[/itex] and

    [itex] {naA}~~+~~{nbB}~~~\mathop{\rightleftharpoons}^{k'_f}_{k'_b}~~~{ncC}~~+~~{ndD}[/itex] is

    [itex]\frac{k'_f}{k'_b} = \bigg(\frac{k_f}{k_b}\bigg)^n[/itex]

    What I don't understand is whether it is trying to tell that the stoichiometric coefficients of a reaction should be converted into a simple whole number ratio before finding out the equilibrium constant or whether they will be different values altogether.

    EDIT: I don't know why I can't get the rate constants above and below the harpoons but hope you get the point.
    Last edited: Jul 10, 2011
  2. jcsd
  3. Jul 11, 2011 #2
  4. Jul 12, 2011 #3
    I'm not sure what you mean by the simplest whole number possible.
    Each of the exponents is unique.

    Perhaps if you understood where these come from by an example?

    Consider the reaction between Hydrogen (H2) gas and Iodine (I2) gas to form hydrogen iodide

    The forward reaction (sorry can't do the arrow)

    H2 + I2 = 2HI and has coefficient kf such that the forward velocity, vf is

    vf = [H2][I2]

    this is because it requires one molecule of each gas to collide to create a reaction.

    The reverse reaction requires two molecules of the same type to collide so

    vr = kr [HI][HI] = kr[HI]2

    Does this help
  5. Jul 12, 2011 #4
    I do understand why the exponents come into picture. But that is where I am also confused. If suppose instead of the reaction [itex]2HI \rightleftharpoons H_2 + I_2[/itex] I write the equation [itex]HI \rightleftharpoons (1/2)H_2 + (1/2) I_2[/itex], do I put the exponents as 1, 1/2 and 1/2? Or do we simply convert it into the simple ratio of 2:1:1?
  6. Jul 12, 2011 #5
    How would you get half a molecule of hydrogen?

    Notice I mean molecule not mole. You can obviously divide up a mole.
  7. Jul 12, 2011 #6
    I know, but the stoichiometric coefficients are written in number of moles and not molecules. Anyways consider this reaction then.

    [itex]4HI \rightleftharpoons 2H_2 + 2I_2[/itex]

    Do we write 4, 2 and 2 as the exponents of their respective concentrations? Or do we still apply the simplest ratio of 2:1:1?
  8. Jul 12, 2011 #7

    You can obviously divide/ multiply through by any number for calculating quantities.

    For reaction kinetics the key one is the one that gives you whole molecules.

    Think about this.

    What the equation is saying is that for it to proceed (forwards) a molecule of hydrogen must collide with a molecule of iodine. Not half or three quarters or fiftythree molecules. One molecule.

    Clearly the greater the concentration of each the greater the chance this will happen so the rate is proportional to the concentration of each and overall proportional to the concentration of the product of the individual concentrations.

    Now look to the reverse if you are really saying that the decomposition cannot proceed unless four molecules of hydrogen iodide simultaneously collide then the

    rate is proportional to [HI][HI][HI][HI] = [HI]4

    (of course since this is the reverse reaction this goes on the bottom of the rate constant fraction)


    If you only need 2 molecules of hydrogen iodide to collide to effect a decomposition then

    rate is proportional to [HI][HI] = [HI]2
  9. Jul 12, 2011 #8
    All right thanks
  10. Jul 12, 2011 #9


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    The expression for the equilibrium constant does depend on how your write the chemical equation. For 2HI <--> H2 + I2, your equilibrium constant would be K = [H2][I2]/[HI]2. For 4HI <--> 2H2 + 2I2, the equilibrium constant would be K' = [H2]2[I2]2/[HI]4. Clearly, these two equilibrium constants share a simple relationship K' = K2. When considering only the concentrations present at equilibrium, the exact stoichiometry of the reaction does not matter so much. Either value of the equilibrium constant will give you the correct answer (clearly if K is constant over all concentrations then K' = K2 will be constant over all concentrations). However, by convention and for convenience, chemists generally will write reactions with the smallest integer coefficients possible.

    When considering the kinetics of a reaction, the relationship between the stoichiometric coefficients and the rate law is not so simple. For some reactions, you can write the rate of the forward reaction as Vf = kf[HI]a and the reverse reaction as Vr = kr[H2]b[I2]c. However, unless the reaction is a special kind of reaction called an elementary reaction, the a, b, and c are not guaranteed to match up with the stoichiometric coefficients.
  11. Jul 13, 2011 #10
    But the Equilibrium constant of a reaction is the ratio of forward to backward reaction isn't it, which has a fixed value. Why then should the value of Kc depend on how we write the reaction?
  12. Jul 13, 2011 #11


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    I think the answer to that question has to do with the fact that while you can derive an equilibrium constant from knowledge of the rate constants, knowing only the equilibrium constant is not information to give you rate constants.
  13. Jul 13, 2011 #12
    I think it is important to emphasize that

    For 4HI <--> 2H2 + 2I2, the equilibrium constant would be K' = [H2]2[I2]2/[HI]4.

    is a different chemical equation from

    For 2HI <--> H2 + I2, your equilibrium constant would be K = [H2][I2]/[HI]2

    This reaction does not require four molecules of hydrogen iodide to coincide before it can take place - only two are required

    The writing and interpretation of chemical equations follows rules special to chemistry. For some purposes they follow the a+b=c+d just the rules of ordinary arithmetic and for other purposes they follow these additional special chemical rules.
  14. Jul 13, 2011 #13


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    That 4HI <--> 2H2 + 2I2 and 2HI <--> H2 + I2 represent different chemical reactions is true only if you are claiming that these are elementary reactions. In most cases, we are not talking about elementary reactions; rather, the equation represents the overall endpoint of a series of elementary reactions.

    NB: According to my chemistry text (Laidler, Chemical Kinetics, 3rd Ed), the reaction mechanism for the hydrogen-iodine reaction is:

    I2 <--> 2I
    I + H2 -> HI + H
    H + I2 -> HI + I

    So, the reaction never actually involves two molecules of HI combining or a molecule of H2 combining with I2
  15. Jul 13, 2011 #14
    I am trying to keep this simple for the moment (avoiding multistep calculations) and also use a well studied reaction for which I have plenty of data to hand.

    More can always be added in later posts.
  16. Jul 16, 2011 #15
    I didn't understand. What are these special rules?
  17. Jul 17, 2011 #16
    I called them special rules but really it is an acknowledgement of the fact that a chemical equation (can) contains more information than a simple arithmetic expression.

    So in ordinary arithmetic

    If a+b=c

    Then 2a+b=a+c

    is always true, as is the reverse equation a+c=2a+b

    But in Chemistry this may or may not be true.

    So 1 Kg-mole of sodium hydroxide + 1Kg-mole of Hydrochloric acid or half a Kg-mole of each will always end up with 1Kg mole of dissolved sodium chloride pus a deal of water.

    Consider the following reactions:

    Reactions involving a change of state eg precipitation or gas evolution.

    CaCl2 (aq) + Na2CO3(aq) = 2NaCl (aq) + CaCO3(s)

    So the reverse reaction may not proceed at all.

    Divalent acid reactions.

    For instance carbonate reactions are a mixture of carbonate and bicarbonate, the relative proportions of which vary with the concentrations of the initial reactants.
  18. Jul 17, 2011 #17
    All right. I get it now.
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