Molecular dissociation of water by electric field

[Capt_Maloche]

In fact not quite, it is a voltage doubler.
There are ionizers that plug into the 220V and use this principle to raise the voltage to several thousand volts thanks to a cascade of diodes / condo.

See for example http://members.aol.com/lyonelb/greinacher.html

Thank you ! I did not know this principle, it's so simple that it's great, I was looking for a way to produce 40Kv from 12V, it's great

[gegyx]

For 30KV, you can take a CRT screen power supply.

This is what I took to fuel a lifter

[nlc]

Thank you ! I did not know this principle, it's so simple that it's great, I was looking for a way to produce 40Kv from 12V, it's great

uh no, that won't work

because must not forget the diode thresholds!

To make 40kV it is better to use a transformer adapted to a high frequency.

[nlc]

But so that the 40kV does not arch, you need VERY good insulation!

[Capt_Maloche]

Hi Maloche,

I am skeptical ... In water, no free electron circulates. The current is due to the displacement of the few solvated ions, and when the water is pure, there is hardly any (H3O + 10 ^ -7 mol / L just like HO-). The electrons are captured at the anode and at the cathode by molecules, as part of an electrolysis or the discharge of a battery.

Breaking a valence bond without ionizing the atoms, it exists ... But it's called homolytic rupture.
see:
http://www.uel.education.fr/consultatio ... l_c4_3.htm
Homolytic ruptures are generated by very harsh thermal and / or radiation conditions: as in combustions, or photo-initiated chemical reactions.
see chapter 2 of this document:
https://www.econologie.info/share/partag ... ryIhE5.pdf

Certain products such as peroxides also promote radical reactions (with homolytic rupture).

With electric fields, you greatly promote ionic rupture because the electric field will help the most electronegative atom (namely O in H2O) to pull the electronic cloud towards it, H + is released and immediately captured by H2O to give H3O +.

But in my opinion, on average, it does not give any gas (HO-, H3O +)solvates and it consumes current.

Here are some ideas that may guide your research.

Thank you thank you friend!

very useful and frankly complex all that

what I note

4.3.a-2 Heterolytic rupture of a covalent bond.

The two electrons of the covalent bond remain on the more electronegative atom giving rise to two ions.

AB to A- + B +

This type of rupture is rare in the gas phase, it requires significant energy.
However, it can occur under the effect of electric discharges or ionizing radiations (X-rays, particles a or b). It is much more common in solution (water or other ionizing solvents SO2, HN3)

we come back to ionizing radiation, and we also speak of electric discharges

basically, an electric arc is an ionizing discharge,
we combine electric fields, electric discharge and radiation

UV seems to break the molecule too, it is used for sterilization

well, here are some TP to come

[Remundo]

OK Captain, I'm doing this because ...

It's to help you actually ...

[Capt_Maloche]

It's to help you actually ...

Yeah, I mostly drink ....... water

besides, I suspect that nature is well made and that the energy necessary for covalent rupture should be exactly that restored by recombination ...

unless the void has something to do with it: https://www.econologie.com/forums/energie-du ... t6213.html

[Capt_Maloche]

from https://www.econologie.com/forums/palamares- ... 83-90.html

Yes i do in bold

Here, if you have 2 minutes you could not find me the volumetric production ratio between the mass of water and H2 + O2 for an electrolysis? I don't know where I put it

so you will remind me how we use the mendeleiev table

Water has a density of 1000 kg / m3
1 mole of water has a mass of 18g / mol
H2 and O2 are formed in the gaseous state with 2 moles of H2 for one mole of O2

Each of these gases follows the PV = n RT law with:
R = 8.314 J / K / mol, T in kelvin, n number of moles in a volume V

The rate of formation of H2, O2 products is linked to the electrolysis current (Faraday's law + water electrolysis).

Go Maloche, an effort ... And this water condenser?

Correction following the Response of Com p
PFFF Well then, let's take 1L (1kg at 20 ° C) of water

1mole H2O gives 1mole of H2 + 1 / 2mole of O2

molar mass of H2O 18g / mol
molar mass of H 1.008g / mol and H2 2.016 g / mol
molar mass of O 16g / mol and O2 32g / mol

1000g of water gives 1000/18 = 55,555 mol of H2O
be broken down into
55.555 x 1 = 55.555 mol of H2 i.e. 55.555x2g = 111.111g of H2
et
55.555 x 1/2 = 27.777 Mol of O2 or 27.777x32g = 888.888g of O2

Good, and after?
PV = n RT where V = nRT / P

p is the pressure (in pascal);
V is the volume occupied by the gas (in cubic meters);
n is the quantity of material, in mole
N is the number of particles
R is the universal constant of perfect gases
R = 8,314 472 J · K-1 · mol-1
we actually have R = NA · kB where NA is the Avogadro number (6,022 × 1023) and kB is the Boltzmann constant (1,38 × 10-23);
T is the absolute temperature (in Kelvin).

OK then
V of H2 = 55.555x 8,314 472 x (20 + 273) / 101 325 = 1.3356 m3 (1336L what)

and to check

V of O2 = 27.777 x 8,314 472 x (20 + 273) / 101 325 = 0.668m3 (668 L what)


I also know that: Rho of H2 = 0,08988 kg.m-3 and Rho of O2 = 1,43.10-3 g.cm-3 to 20 ° C and http://lycees.ac-rouen.fr/galilee/oxygene.htm

For hydrogen:
the PCS is: 12,745 106 J / m3 is 286 kJ / mole
And the PCI is worth: 10,8 106 J / m3 is 242 kJ / mole

Either about 2h production with 2200W efficient or 0.5L / h
Interesting to measure the efficiency of electrolysis

source: http://uuu.enseirb.fr/~dondon/devdurabl ... stible.htm
3.3 Rounded energy balance of the decomposition of a water molecule:

H2O -> → H2 + ½ O2

The H2O water molecule consists of 2 OH bonds and each bond has a molar energy of 460 kJ (cf table § 2.2.4) which represents 2 x 460 = 920 kJ for one mole of water.

The breaking of the OH bonds of the water molecules for one mole of water requires the contribution of 920 kJ / mole (left side of the equation). However, the recombination of the hydrogen atoms H into H2 (gaseous hydrogen) produces an energy that comes, in the balance sheet, as a deduction from the previous one:

HH → H2 This recomposition brings 432 kJ / mole.

Likewise for the recomposition of the oxygen atoms:

½ OO → ½ O2 This reaction releases ½ x 494 kJ ie 247 kJ / mol.

The energy consumed for the dissociation operation of the water is finally:

920 - 432 - 247 = 241 kJ / mole

Thus the manufacture of 2 g of hydrogen by cracking of one mole of water (without taking into account the losses) requires the contribution of 241 kJ, ie 120.500 kJ to manufacture 1 kg of hydrogen.

[Remundo]

Hello Captain '

PFFF Well then, let's take 1L (1kg at 20 ° C) of water
1mole H2O gives 2 moles of H2 + 1mole of O2 (or 1mole of O or 1/2 mole of O2?)

NO;
1 mole of H2O gives 1mole of H2 and 1/2 mole of O2

1000g of water gives 1000/18 = 55,555 mol of H2O

YES !

be broken down into
55.555 x 2/18 = 6.173 mol of H2 or 6.173x2 = 12.35g of O2
et
55.555 x 16/18 = 49.383 Mol of O (or O2?) Or 49.383x16 = 790.13g of O2

NOT
55.55 moles of H2O (1kg of water) gives 55,55 mol of H2 and 27,77 moles of O2

HOLD, THE ACCOUNT IS NOT THERE, GO! (790 + 12 not equal to 1000g initial)

You surprise me !

For the rest, 1 mole of gas makes approximately a volume of 24L under ambient conditions,
Your liter of water will therefore give:
- 55.55x24 = 1333,3 Liter, i.e. 1,3 m3 of H2
et
- 666.66 Liter of O2, i.e. 0.666 m3

It's weird, I would have said the opposite ...

For energy thermal extractable:
H2 + 1/2 O2 -> H2O with 286 kJ / mol (we will not quibble PCI / PCS)

So, since 1 Liter of electrolyzed water gives 55 moles of H2:
energy = 55.55 * 286 = 15,9 MJ

4,41 kWh, i.e. 1000 W for 4h 24 min and 36 seconds ...

Maloche pupil: can do better
Must catch up on the condo to get your average back

[Remundo]

Yeah, I mostly drink ....... water

This explains things