MadSci Network: Physics

Re: Energy required to disassociate the water molecule: photon and/or temp

Date: Wed Mar 26 21:01:05 2003
Posted By: Todd Whitcombe, Associate Professor, Chemistry
Area of science: Physics
ID: 1039068382.Ph

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 

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 

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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