- #1

- 47

- 0

## Main Question or Discussion Point

This question was prompted by reflecting after reading the standard textbook explanation that "the greater acidity of RCOOH vs ROH is due to the greater stability of the delocalised RCOO- ion causing the position of equilibrium to be further to the right". The equilibria can be written as:

RCOOH(aq) ⇄ RCOO-(aq) + H+(aq) Eq 1

ROH(aq) ⇄ RO-(aq) + H+(aq) Eq 2

It seems an oversimplified explanation. Charge delocalisation clearly means entropy is increasing from more electron disordering and enthalpy is decreasing from lower mutual repulsion forces. However, we aren’t just comparing the theoretical stability of delocalised RCOO-(g) vs RO-(g), but the Kc for an overall process in aqueous solution. Taking simple example typical values:

pKa of ethanoic acid = 4.8; pKa ethanol = 16, both assumed at 298K.

ΔGϴ = -RTlnKc gives ΔGϴ(Eq 1) = +27 kJ mol-1 and ΔGϴ(Eq 2) = +91 kJ mol-1

Using ΔG = ΔH – TΔS, the enthalpy component of ΔG should be a function of the difference in O-H bond energies in RCOOH(aq) and ROH(aq) and their respective anion hydration energies. These values for RCOOH would be expected to be lower on both counts (due to the presence of the carbonyl group withdrawing electron density), but their sum cannot be easily predicted. H-bonding in aqueous solution will also have a different effect. Some literature values are approximately

ΔHhydration CH3COOH = -9kJ mol-1 and ΔHhydration C2H5OH = -11kJ mol-1

Even accepting some variation in these data, and that the enthalpy changes for Equations 3 & 4

CH3COOH(l) ⇄ CH3COOH(aq) Eq 3

C2H5OH(l) ⇄ C2H5OH(aq) Eq 4

are included in these values, it still means that Equation 2 must have a significantly more negative ΔS to account for its larger ΔG, one assumes from greater ordering of water by RO- than by RCOO- due to its greater charge density / lower delocalisation / higher entropy. I cannot find any data to quantify this aspect, but from the signs and magnitudes of ΔG and ΔH it appears we can say that the delocalisation stability which gives rise to the differing pKa’s is indeed driven more by entropy changes rather than enthalpy changes! Any one like to agree or disagree?!

RCOOH(aq) ⇄ RCOO-(aq) + H+(aq) Eq 1

ROH(aq) ⇄ RO-(aq) + H+(aq) Eq 2

It seems an oversimplified explanation. Charge delocalisation clearly means entropy is increasing from more electron disordering and enthalpy is decreasing from lower mutual repulsion forces. However, we aren’t just comparing the theoretical stability of delocalised RCOO-(g) vs RO-(g), but the Kc for an overall process in aqueous solution. Taking simple example typical values:

pKa of ethanoic acid = 4.8; pKa ethanol = 16, both assumed at 298K.

ΔGϴ = -RTlnKc gives ΔGϴ(Eq 1) = +27 kJ mol-1 and ΔGϴ(Eq 2) = +91 kJ mol-1

Using ΔG = ΔH – TΔS, the enthalpy component of ΔG should be a function of the difference in O-H bond energies in RCOOH(aq) and ROH(aq) and their respective anion hydration energies. These values for RCOOH would be expected to be lower on both counts (due to the presence of the carbonyl group withdrawing electron density), but their sum cannot be easily predicted. H-bonding in aqueous solution will also have a different effect. Some literature values are approximately

ΔHhydration CH3COOH = -9kJ mol-1 and ΔHhydration C2H5OH = -11kJ mol-1

Even accepting some variation in these data, and that the enthalpy changes for Equations 3 & 4

CH3COOH(l) ⇄ CH3COOH(aq) Eq 3

C2H5OH(l) ⇄ C2H5OH(aq) Eq 4

are included in these values, it still means that Equation 2 must have a significantly more negative ΔS to account for its larger ΔG, one assumes from greater ordering of water by RO- than by RCOO- due to its greater charge density / lower delocalisation / higher entropy. I cannot find any data to quantify this aspect, but from the signs and magnitudes of ΔG and ΔH it appears we can say that the delocalisation stability which gives rise to the differing pKa’s is indeed driven more by entropy changes rather than enthalpy changes! Any one like to agree or disagree?!