| MadSci Network: Physics |
That is a difficult question. Simply put, the energy required to break up a water molecule is the bond dissociation energy of the H2O molecule. And this has been recently re-measured: "Several new experiments presented in Ruscic et al (J.Phys.Chem. A 106 (2002) 2727 - 2747) such as mass-selected photoionization measurements, pulse-field-ionization photoelectron spectroscopy measurements, and photoelectron-photoion coincidence measurements, which utilize the power fo the positive ion cycle to derive the O-H bond energy, produce a consensus value of the bond dissociation energy of water D(H-OH) = 41128 +/- 24 cm-1 = 117.59 +/- 0.07 kcal/mol, with a value of D(O-H) - 35593 +/- 24 cm-1 = 101.76 +/- 0.07 kcal/mol." In other words, the breaking up of water into its constituent atoms is a two step process. The first step involves pulling off the first proton to leave a "hydroxyl radical" (represented as OH). This requires a wavelength of 243 nm. Translating that into temperature is a little harder as it requires that we define temperature and that means that we need to consider the vibrational modes of the molecule. That is more difficult than space would allow, so simply put, it is 243 nm or 117.59 kcal/mol. The second stage of turning H2O into 2H + O is the breaking up of the hydroxyl radical which occurs at a slightly different energy and wavelength (281 nm; 101.76 kcal/mol). Having said this, this is the bare minimum energy required, according to the bond dissociation energy, to break up the molecule to the atoms. Another text (Bernath, Spectra of Atoms and Molecules, Oxford University Press, 1995) gives the actual photodissociation value as 186 nm. The extra energy is presumably a consequence of the real shape of the potential energy surface - that is, taking into account that the bond dissociation energy would just be enough to break the bonds. Both of these answers assume that we are talking about water in isolation - gaseous water molecules at a lower enough pressure to react without interference or solvation. Solvation would and does change everything. In aqueous solution - that is, in water - the water molecule breaks up into the hydrogen ion (H+) and the hydroxide ion (OH-). This is a heterolytic cleavage of the H-OH bond. That is, both electrons end up on the hydroxide ion and the proton is "stripped bare". This process occurs in pure water such that the concentration of H+ equals that of OH- at 1 x 10-7 M (0.0000001 mol/litre) and gives us a pH of 7. It can't be avoided. Water breaks up. And generating Hydrogen and Oxygen gas is a simple matter involving very little energy - 1.23 volts or the aid of an ordinary battery. But this is not breaking the molecule up into its constituent atoms. And the profile here is different than for the single molecule in the gaseous state. The answer to the second part of your question deals with the idea that there is a threshold value for dissociation - that is, there is a wavelength below which dissociation will not occur but increasing the energy of the light will result in further or more extensive dissociation. But it is intensity - the total flux of photons - that, to a first approximation, determines the extent of dissociation. Intensity is not the same thing as wavelength. More energy does not necessarily translate to more water molecules breaking up. A higher wavelength from a more intense source shining up the water vapour for a longer period of time may result in more extensive dissociation than a shorter wavelength for a shorter period of time. But having said that, different wavelengths at the same intensity result in different amounts of dissociation occuring with the above wavelengths as the threshold values. Hopefully, this has provided you with an answer to your question.
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