# What is the origin of the huge difference in the autoprotolysis constant of heavy water versus regular water?

Basically, the gist of this question is, why is the autoprotolysis constant of heavy water different from that of ordinary water?

The autoprotolysis constant of heavy water at $\pu{25 °C}$ is $\pu{1.35 * 10^{-15}}$, while at $\pu{30 °C}$ it is about $\pu{1.47 * 10^{-14}}$. From this, I calculated that the enthalpy change for the autoprotolysis of heavy water is about $\pu{360 kJ/mol}$, which is already a huge difference from regular water's autoprotolysis enthalpy of approximately $\pu{+56 kJ/mol}$. The difference in entropy is also huge as well: by using the formula $-RT \ln(K) = \Delta G$ and $\Delta G = \Delta H - T \Delta S$ I calculated that the autoprotolysis of heavy water has an entropy change of about $\pu{+920 J K-1 mol-1}$, which is a world of difference versus water's approximately $\pu{-81 J K-1 mol-1}$. Can someone tell me why this huge difference in values is occurring?

Also could someone point me to a database that would contain this kind of thermodynamic data / equilibrium data because my textbooks aren't going to cut some of the questions I have in terms of raw data crunching.

• Because you made mistakes in calculations? – Mithoron Jan 1 '18 at 13:24
• How did you calculate these enthalpies? – Gert Jan 1 '18 at 14:21
• @Mithoron: Using Van t'Hoff I get $\Delta H^0=+358\ \mathrm{kJ/mol}$ for heavy water. And this resource: www1.lsbu.ac.uk/water/water_dissociation.html lists the value for light water as $+55.8\ \mathrm{kJ/mol}$. So it appears OP's values are correct, at least for the enthalpies. – Gert Jan 1 '18 at 14:52
• However for the $\Delta S$ values I get different results: $-283.5\ \mathrm{J/mol}$ and $-267.8\ \mathrm{J/mol}$ for heavy and light water, respectively. – Gert Jan 1 '18 at 15:09
• As regards the reason behind the different $K$ values, this is quite briefly explained here: en.wikipedia.org/wiki/Heavy_water#Effect_on_biological_systems. Quantum effects make the $\ce{OD}$ bond stronger than the $\ce{OH}$ bond. – Gert Jan 1 '18 at 16:17