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Equilibrium constant changes with pressure?

  1. Feb 11, 2010 #1
    Why doesn't the equilibrium constant change with change in pressure and concentration ?Why does it depend only on tempurature ?
  2. jcsd
  3. Feb 11, 2010 #2


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    How do you know this is true? Did you find it in a textbook or your class notes or a website? If so, please quote the statement exactly as it appears in its original form.

    Then, in addition to posing the question, show us what you think about it.
  4. Feb 11, 2010 #3
    I found it in my text book.Here is the statement
    "The value of the equilibrium constant for a particular reaction is always constant depending only upon the temperature of the reaction and is independent of the concentrations of the reactants with which we start or the direction from which the equilibrium is approached."

    And i want to know WHY?
  5. Feb 11, 2010 #4
    At least for gases,

    [tex]K=e^{\frac{\Delta G^0}{RT}}[/tex]

    Then you can remember that [tex]RT=zP\bar{V}[/tex], where [tex]\bar{V}[/tex] is molar volume.

    In any case, that statement doesn't seem to take into account Le Châtelier's principle.
  6. Feb 28, 2010 #5


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    Let me give a stab at the answer because I think it is such a good question, and I hope that proper teachers can refine it.

    The OP does not define 'Equilibrium Constant' but mentions concentrations, so means things like e.g.

    Keq = [A]/[C][D] , the square brackets being molarities of reactants and products. (Can't understand why this site won't give me capital B back there :confused:) If you go back into your textbook derivations you will find that this is the condition under which the free energy change with respect to a virtual change in the composition of the mixture is minimised, taking account of constancy of mass. You will find equations like

    [tex]\delta[/tex] [tex]\Sigma[/tex]i [tex]\mu[/tex]i = 0 .

    This is an algebraic sum with the [tex]\mu[/tex] 's being affected with a sign according as they are reactants or products.

    [tex]\mu[/tex]i = ([tex]\mu[/tex]0)i + RT ln ci.

    From that you work out the equilibrium law. Now assuming we have that straight :uhh: then we trace the answer to your question to this last equation, which we then trace to the assumption that the substances whose equilibrium we are treating are 'ideal'. E.g. perfect gases. There are substances, e.g. dilute gases and even solutions that are close enough to this ideality.

    But you may object, for things that are not ideal, then it doesn't work. You are right, in that case the equilibrium constant as defined above does vary with pressures or concentrations. You can talk of a variable equilibrium constant. You can and do keep the idea of a constant when you redefine the law as

    Keq = aAaB/aCaD

    The a's are the thermodynamic activities of the substances. They are related to the concentrations ci by ai = [tex]\gamma[/tex]ici, the [tex]\gamma[/tex]i are called the activity coefficients, which are 1 in the case of ideality.

    You may object you are cheating and have merely redefined terms to make your law come true. No, because the thermodynamic activities are something physical that will be manifested in all thermodynamic properties and can be tied to something measurable.

    For example suppose these reactants and products in solution are not very ideal, then the law in terms of concentrations will not be perfectly right either. But suppose these dissolved reactants are in equilibrium with gases that are more nearly ideal (not reacting in gas phase). Even if the relation between gas concentration and amounts dissolved is also not ideal, you should have an experimentally verifiable prediction between gas concentrations without any fudges. Likewise if you have ionic reactions, quite non-ideal relations with ionic concentrations, if you can couple these to electrical potentials you should have a non-fudged prediction. So activity coefficients are not in fact a fudge.

    Unfortunately I cannot call to mind any clear experimental examples of these principles, and for this and for clearer explanation I call for other contributors, because I think this is an excellent question and without answers and examples there is a didactic gap and appearance of fudge.
    Last edited: Feb 28, 2010
  7. Mar 1, 2010 #6
    I'd like to point out that it is,

    -[tex]{\frac{\Delta G^0}{RT}}[/tex]

    The answer to the question lies in the Le Chatelier Principle itself. For example take the reaction,
    A + B [tex]\rightleftharpoons[/tex] C + D
    The equilibrium constant is,
    Keq = [tex]\frac{[C][D]}{[A]}[/tex]
    Now at a given temperature, if you increase the volume of one of the products, the value of Keq should increase by the above equation. However, by the Le Chatelier Principle, the system will oppose this change to keep Keq constant. Thus, there will be a shift in the position of the equilibrium; the rate of backward reaction will increase; more reactants will be formed; and Keq will remain constant.
    The realtions between K and temperature and pressure can be obtained from the thermodynamic considerations. K is a function of temperature. And for ideal gases it does not depend upon pressure conditions. However, for real gases it does change with change in pressure, but the change is too small to be considered. You need very high pressure to produce an observable effect.

    P.S. refer Laidler & Meiser Physical Chemistry, 2nd ed. That should be helpful
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