From the bonding energy of following bonds that I found in a Wikipedia article, I calculated the reaction energy that would be released when oxygen and hydrogen are reacted to produce water. The phase of each component is unknown.

$$ \begin{array}{lc} \hline \text{Bond} & D_0/\pu{kJ mol-1} \\ \hline \ce{H-H} & 436 \\ \ce{O-H} & 497 \\ \ce{O=O} & 498 \\ \hline \end{array} $$


$$\ce{2 H2 + O2 -> 2 H2O} + \pu{618 kJ}$$

However, this Wikipedia article suggests that $\pu{241.8 kJ}$ of energy is released for every mole of hydrogen, which is $\pu{483.6 kJ}$ per $\pu{2 mol}$ for the context of the equation above.

Even when I consider the vaporization of water, which is $\pu{40.65 kJ mol-1}$, the total energy will amount to $\pu{536.7 kJ}$ per $\pu{2 mol}$, which is still far away from $\pu{483.6 kJ}$ per $\pu{2 mol}$ that is suggested in the Wikipedia article.

Where could this extra energy be going?

  • 1
    $\begingroup$ Bond energies are a great deal less universal than you seem to be thinking. $\endgroup$ Jan 20, 2019 at 2:10
  • $\begingroup$ The catch may be the cleavage of OH in water. Also, you need to be careful with signs, vaporization is endothermic. $\endgroup$
    – Buck Thorn
    Jan 20, 2019 at 11:28
  • $\begingroup$ I am aware that it is endothermic. Therefore I have performed the following calculation: 2H2 + O2 = 2H2O(liquid) + 618kJ = (2H2O(gas) - 40.65kJ*2) + 618kJ = 2H2O(gas) + 536.7kJ $\endgroup$
    – martian17
    Jan 20, 2019 at 16:16
  • $\begingroup$ Please tell me. If my way is unconventional, what method is normally used to solve thermochemical equations in chemistry? $\endgroup$
    – martian17
    Jan 20, 2019 at 16:24
  • $\begingroup$ @Y.Yoshii Maybe this is useful: pubs.acs.org/doi/10.1021/jp013909s $\endgroup$
    – Buck Thorn
    Jan 20, 2019 at 20:23

1 Answer 1


Part of the discrepancy appears to be due to the need for two $\ce{O-H}$ dissociation energy terms to compute the $\ce{OH}$ bond energy for water: one for the $\ce{OH}$ bond in water (i.e. $\ce{H-OH}$), and another for the $\ce{OH}$ bond in the $\ce{OH}$ radical (i.e. $\ce{^.O-H}$). Values for these are reported by Ruscic et al. [1]. Using the average value of $\ce{D_0}$ for $\ce{H2O}$ (the $\ce{OH}$ bond energy for water), namely $\pu{459 kJ/mol}$, gives

$$\ce{2H2 + O2 -> 2H2O} \quad \Delta H = \pu{-466 kJ/mol}$$

This point is addressed in the Wikipedia page for "bond dissociation energy", the reference cited therein providing a slightly different bond energy for water of $\pu{461.5 kJ/mol}$.

Given various uncertainties common to the measurement of bond dissociation energies, and the potential effect of temperature on the bond energy, the final discrepancy seems reasonable.


  1. Ruscic, B.; Wagner, A. F.; Harding, L. B.; Asher, R. L.; Feller, D.; Dixon, D. A.; Peterson, K. A.; Song, Y.; Qian, X.; Ng, C.-Y.; et al. On the Enthalpy of Formation of Hydroxyl Radical and Gas-Phase Bond Dissociation Energies of Water and Hydroxyl. The Journal of Physical Chemistry A 2002, 106 (11), 2727–2747. https://doi.org/10.1021/jp013909s.
  • $\begingroup$ That is brilliant. Thank you for your information. $\endgroup$
    – martian17
    Jan 20, 2019 at 23:06

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