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.

  • $\begingroup$ Because you made mistakes in calculations? $\endgroup$ – Mithoron Jan 1 '18 at 13:24
  • $\begingroup$ How did you calculate these enthalpies? $\endgroup$ – Gert Jan 1 '18 at 14:21
  • $\begingroup$ @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. $\endgroup$ – Gert Jan 1 '18 at 14:52
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    $\begingroup$ 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. $\endgroup$ – Gert Jan 1 '18 at 15:09
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    $\begingroup$ 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. $\endgroup$ – Gert Jan 1 '18 at 16:17

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