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What is translational entropy?

  1. Jan 25, 2014 #1
    Hello, I am trying to understand a short literature article (doi: 10.1021/ja01635a030). I am not sure how much liberty I have to reproduce its contents here, and I can't explain it here because I don't understand it -- which is why I have this question.

    I believe it is proposing that a certain effect (Chelate Effect) in coordination chemistry explained by entropy changes can have the entropy component removed by a modification to the equation describing it. This allows comparisons of free energy between molecules without the contribution of entropy in this particular form overshadowing other factors. This "particular form" is referred to as translational entropy in the article.

    I know this is extremely vague, but I don't know any other detail to go into. The idea is that a metal can bond to multiple molecules. If the molecule it is bonding to has multiple bonding sites, it is called a chelate. Since this association reduces the number of molecules in the system, the entropy is reduced in the association. Because a chelate has multiple bonding sites and can bond multiple times, the lost entropy is less, so a chelate association is more stable than a single-bonding molecule. This extra stability is referred to as translational entropy and can be accounted for by a modification to the equation:

    (1) ΔS = S° - ΔnR[ln(55.5)]

    Where n is the difference in moles product to moles reactant, and R is the universal gas constant. The second term, presumably the modification to the equation, stems from:

    (2) ΔF = F° + ΔnRT[ln(55.5)]
    (3) F = -RTln(Kc)

    where Kc = K / 55.5

    The 55.5 is the concentration of water solvent in mol/L, derived from its density. From 3, separating Kc results in equation 2, and matching the units leads to equation 1.

    I don't know if this is the right forum for this, but I appreciate your time reading this and any advice you can offer! Thank you!
  2. jcsd
  3. Jan 25, 2014 #2


    Staff: Mentor

  4. Jan 25, 2014 #3
    Hi jedishrfu. Thanks for your reply! I actually already feel I have a good grasp of what a Chelate is. I simply wrote about it to provide some background info, because this is a physics forum.

    Thanks for the link to biological thermodynamics, though. It mentions translational entropy only once, regarding the assembly of a virus molecule. It also mentions rotational entropy, which was not mentioned in the literature article I was trying to understand. I always understood entropy as being proportional to the number of microstates a molecule can assume. How can a translational/rotational mode correspond to a microstate? Are microstates related to the degrees of freedom of a molecule?

    Thanks for any insight!
  5. Jan 25, 2014 #4


    Staff: Mentor

  6. Jan 25, 2014 #5
    I liked how that link explained things, but I am having difficulty assimilating such a broad concept with sufficient mastery to adapt it to my context. Thanks for bearing with me here.

    So if a microstate corresponds to an arrangement of its degrees of freedom, or its momentum and location, what exactly is translational entropy, or the removal of it?

    Is "translational entropy" microstates that have no rotational or vibrational components of its energy?

    Or does its removal mean I take all the microstates and remove their translational components? So I essentially remove all linear momentum from the particles in a system, leaving only vibrational and rotational motion?

    Thanks for your continued time
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