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How do you calculate how much energy is needed to break a water molecule into its constituent atoms?

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This is not homework. I just want to know what is required to get hydrogen out of water.

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  • $\begingroup$ Please see the homework policy. Indicate what work you've done so far in trying to ascertain the answer. To get you started, you might find it useful to read up on bond energies, enthalpy of formation, and Hess's law. $\endgroup$
    – Greg E.
    Aug 11, 2013 at 9:06
  • $\begingroup$ @GregE: it's not homework. If you don't want to answer, that's ok, maybe someone else will. I thought the whole idea of stackexchange is that you can ask questions you want to know about, and people help you out. Note that I'm not asking the solution, I want to learn how you do it. $\endgroup$
    – Johan.A
    Aug 11, 2013 at 11:48
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    $\begingroup$ @JohanA, the tag is intended for a broad type of question, not just ones that are actually assigned by a teacher/professor in class. I don't mean any offense, I simply think your question fits the template of a "homework" question. I won't edit the tags again, but, in my view, it would be very good if you did some research prior to asking. This is not at all an obscure topic. $\endgroup$
    – Greg E.
    Aug 11, 2013 at 11:55
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    $\begingroup$ @JohanA, the other advantage of attempting to answer the question yourself first and showing the work is that somebody interested in answering your question can better gauge your prior knowledge. If you have no general chemistry knowledge, then what you're asking for is either (a) an indication of the relevant topics to read up on, which I gave you in my first comment, or (b) a concise summary of a hodgepodge of introductory gen. chem knowledge, which, AFAIK, isn't appropriate for this forum. $\endgroup$
    – Greg E.
    Aug 11, 2013 at 12:02

1 Answer 1

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The answer is given by the standard enthalpy of formation, $\Delta H_\mathrm f^\circ$ for water:

$$\begin{alignat}{2} \ce{H2 + 1/2O2 &-> H2O(l)}\quad&\Delta H_\mathrm f^\circ&=-285.8\ \mathrm{kJ/mol}\\[6pt] \ce{H2 + 1/2O2 &-> H2O(g)}\quad&\Delta H_\mathrm f^\circ&=-241.8\ \mathrm{kJ/mol} \end{alignat}$$

The energy required for the following reaction, your point of interest: $\ce{H2O -> H2 + 1/2O2}$ is simply a positive value for either one of the above $\Delta H_\mathrm f^\circ$ values depending on the state of the reactant. It means that per mole water splitting requires at least $241.8\ \mathrm{kJ}$ of energy.

Any method(thermal,electrolysis etc) followed, this much of energy will be always required. When using electrolysis, it may be convenient to express it in $\mathrm{eV}$.

Regarding calculation, I have to mention that $\Delta H_\mathrm f^\circ$s are experimental values. To know how they are calculated, here is a nice example of $\ce{NaCl}$.

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  • $\begingroup$ Just out of curiosity: One could regard the question in terms of the Gibbs free enthalpy as well. Then the situation would change as a favourable entropy term would have to be taken into account as well. What energy would we calculate then? $\endgroup$ Oct 13, 2016 at 13:57

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