1
$\begingroup$

Carboxylic acids such as acetic acid do not dissociate much in the gas phase. In aqueous solution, however, they act as acids (proton donors).

$$\ce{CH3COOH(aq) <=> H+(aq) + CH3COO-(aq)}$$

Are reaction entropy and reaction enthalpy positive or negative, and what are there magnitudes?

In the gas phase, we have clear expectations:

$$\ce{CH3COOH(g) <=> H+(g) + CH3COO-(g)}$$

The reaction enthalpy will be positive (we need to break a bond) while the reaction entropy will be positive (more gas particles). As the reaction does not go forward in the gas phase, the enthalpic contribution must be of larger magnitude.

So another way of stating the question is to ask what the effect of solvation is on the enthalpy and entropy of reaction.

$\endgroup$
2
$\begingroup$

Electrostatic screening associated with solvation is obviously the key factor enabling dissociation in aqueous solution when compared to the gas phase. Without such dielectric screening, dissociation would be too costly ($\approx \pu{1 MJ/mol}$ or $\approx\pu{14 eV}$, roughly the energy required to ionize a hydrogen atom) to occur to a significant extent at RT. The dielectric constant of water reduces the energy of the gas phase electrostatic interaction by a factor of $\approx 80$. This is an enthalpic effect. This energy is further reduced by water oxygen bonding to $\ce{H^+}$ to form $\ce{H3O+}$ (proton to hydronium).

Differences between dissociation constants of different carboxylic acid groups originate from enthalpy or entropy changes due to anion formation (since the formation of protium is something shared in common by all acids, its solvation cannot explain differences in behaviour of different acids). The effects on the solvent are often described in terms of "structure making or breaking". As a rule there is an entropic penalty ($\Delta S^\circ<0$) with dissociation that can be attributed to solvent rearrangement around the ensuing charged species. The penalty is seen in thermodynamic data for many weak acids. Trends are otherwise not quite as clearcut. Beyond the entropic penalty, other factors (enthalpic or entropic) may play a role in encouraging or discouraging dissociation, for instance charge delocalization (inductive, mesomeric effects) or intramolecular rearrangements.

$\endgroup$
  • $\begingroup$ Thanks for the link to the thermodynamic data for a range of weak acids. The lack of entropic driving force for this dissociation is not what I had expected and this has really helped me gain a better understanding of the dissociation process. $\endgroup$ – Withnail May 16 at 12:00
  • $\begingroup$ @Withnail I'm glad you found it useful. It admittedly presents only a summary of the picture $\endgroup$ – Buck Thorn May 16 at 12:43

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

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