3
$\begingroup$

Which of the following has greater entropy?

a. $\ce{CH3COOH~ (aq)}$

or

b. $\ce{CH3COO^{-}~ (aq) + H+~(aq)}$

They are both in the same state, and have the same number of particles.. so how do I figure this out? My instinct says a, because the products will be more "random", but I'm not completely sure.. Any ideas?

Improve this question
$\endgroup$
1
  • $\begingroup$ Your welcome, glad to help. $\endgroup$
    – ron
    Commented Nov 25, 2014 at 18:59

1 Answer 1

Reset to default
1
$\begingroup$

In the gas phase I would suspect that the entropies would be equal as both systems have the same number of translational, rotational and vibrational modes (18).

However you specifically ask about the liquid phase system, since you show that both "a" and "b" are solvated in aqueous media. In this case, the ionized molecule (b) has charges on both ions and this will cause increased ordering of the solvent molecules surrounding the ions, more so than neutral molecule "a". This increased ordering of the system in case "b" will reduce the entropy of the system. Consequently system "a" will have greater entropy.

Improve this answer
$\endgroup$
11
  • 1
    $\begingroup$ I don't think your statement for the gas phase is correct. Even though both systems might have the same number of degrees of freedom the entropy is not the same for translation, rotation and vibration. I plan to provide an answer to your question "Can Entropy Increase During Bond Formation?" when I get the time and this answer will also deal with this topic. $\endgroup$
    – Philipp
    Commented Nov 25, 2014 at 19:35
  • $\begingroup$ @Philipp Great, I look forward to reading it! $\endgroup$
    – ron
    Commented Nov 25, 2014 at 19:38
  • $\begingroup$ entropy != disorder. I also hope you are not seriously suggesting orderly arrangement has any meaning in a normal aqueous solution. The solvent separated ion pair will have more solvent molecules involved in the interaction than the neutral molecule leading to a greater number of available microstates, leading to a higher entropy. $\endgroup$
    – Lighthart
    Commented Nov 25, 2014 at 19:43
  • $\begingroup$ @Lighthart Yes, I am suggesting that the orderly arrangement has entropic significance. Are you saying that system B is less ordered than A and consequently B has higher entropy? If so, perhaps you could elaborate your comment into an answer. $\endgroup$
    – ron
    Commented Nov 25, 2014 at 23:00
  • $\begingroup$ I am not saying anything about order. Order is not related to entropy in any way. Entropy is the number of available microstates-- the number of possible ways to arrange items in the system. When there are more parts, there are necessarily more ways to arrange the parts, and hence larger entropy. Thus, entropy favors dissociation. $\endgroup$
    – Lighthart
    Commented Nov 26, 2014 at 22:08

Not the answer you're looking for? Browse other questions tagged or ask your own question.