Find enthalpy of solvation $\Delta_\mathrm{solv}H$ using Hess's law, and show the expression that allows to calculate the enthalpy of solvation.

$$\ce{Ca^2+(g) -> Ca2+(aq)}$$

I am not told which enthalpies of reaction to use. I don't need the exact value, but I do need to show the sum of each enthalpy. I managed to find the enthalpy of the following reactions:

$$ \begin{align} \ce{Ca(g) &-> Ca+(g) + e-} &\quad \Delta H &= \pu{590 kJ mol-1} \\ \ce{Ca+(g) &-> Ca^2+(g) + e-} &\quad \Delta H &= \pu{1145 kJ mol-1} \\ \ce{Ca(s) + 2 H+(aq) &-> Ca^2+(aq) + H2(g)} &\quad \Delta H &= \pu{1925.9 kJ mol-1} \\ \ce{Ca(s) &-> Ca(g)} &\quad \Delta H &= \pu{178 kJ mol-1} \end{align} $$

I nearly have the reactions I need, but I am still needing

$$\ce{H2 -> 2 H+ + 2 e^-}$$

I haven't been able to find the answer.

P.S. I am aware of the enthalpy of formation of $\ce{H2}$, but don't know how/if I can use it here.

  • 1
    $\begingroup$ By definition, the reaction $\ce{H2 -> 2 H+}$ has an enthalpy of reaction equal to zero kJ/mol $\endgroup$
    – Maurice
    Commented Mar 21, 2021 at 20:07

1 Answer 1


One approach to compute the enthalpy change for the reaction

$$\ce{H2(g)->2H+(aq) + 2e-}$$

is to combine the bond dissociation enthalpy and hydrogen ionization energy (both refer to gas phase species) and the hydration enthalpy of the hydrogen ion:

$$\ce{H2(g)->2H(g)}~~~~~~~~{\Delta H = \pu{439 kJ/mol} }\\ \ce{2H(g)->2H+(g) +2e-(g)}~~~~~~~~{\Delta H = \pu{262 kJ/mol} }\\ \ce{2H+(g)->2H+(aq)}~~~~~~~~{\Delta H = \pu{-2260 kJ/mol} }$$

If you add up the enthalpies you obtain as the solvation enthalpy of $\ce{Ca^{2+}}$ the value $\Delta_{\text{solv}} H = -1546 \pu{kJ/mol}$ which is quite reasonable (see eg a table here)

The result is clearly very different from my earlier answer which I am still pondering:

The catch is that $\Delta H^\circ=0$ for this reaction. The reason is because $\Delta G^\circ=0$ for the reaction (by definition, see standard reduction potential) at all temperatures, and then, since

$$\Delta H = -T^2\left( \frac{\partial (\Delta G/T)}{\partial T}\right)_p$$ it follows that $\Delta H^\circ=0$. I suspect the problem with this approach is due to a missing term involving enthalpy of solvation of the electrons (or a similar term involving the electrons) but I haven't put my finger on it.

  • $\begingroup$ Hm, but is not the standard Delta G = 0 the trivial consequence of considering the same but the opposite half reactions ? The each half reaction has non zero but the opposite standard Delta G and Delta H. The above reaction would not have zero reaction enthalpy just because chemists decided SHE to have the reference potential. $\endgroup$
    – Poutnik
    Commented Mar 21, 2021 at 20:28
  • $\begingroup$ I would also like to ask, does that also apply to the change needed? I didn't add symbols, but the reaction I need for the hydrogen isn't in a single phase. To make it more accurate to what I need, it'd be this: $$\ce{H2_{(g)}->2H^+_{(aq)} + 2e^-}$$ $\endgroup$ Commented Mar 21, 2021 at 20:32
  • $\begingroup$ That's what I'm saying in my answer, you should just use $\Delta H = 0$ for that reaction. @Poutnik what else would you suggest? $\endgroup$
    – Buck Thorn
    Commented Mar 21, 2021 at 20:55
  • $\begingroup$ @Buckthorn I say "Why should be delta G and Delta H of that reaction accidentally zero ?“ It cannot be zero by convention. If we created convention 235U fission has zero enthalpy, it would not have any effect on real enthalpy of fission. That Delta G=0 is for reaction 2H+(aq) + 2e- + H2(g) -> H2(g) + 2 H+(aq) + 2e-. Similarly H2 formation enthalpy is zero, because it relates to another trivial reaction H2 -> H2. $\endgroup$
    – Poutnik
    Commented Mar 22, 2021 at 6:03
  • 1
    $\begingroup$ @BuckThorn By other words, evaluation of Delta G of redox reaction assumes having one of half-reactions from SHE. Evaluation H2(g)->2 H+(aq) + 2e- is evaluating 2 SHE electrodes together, for which Delta G = Delta G_forward + Delta G_backward = obviously 0, whatever value the single terms have. $\endgroup$
    – Poutnik
    Commented Mar 22, 2021 at 12:51

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