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# Re: [Phys-L] Electrolysis under Pressure

For an equilibrium to exist, there must exist some of both the
reactants and products at equilibrium.

For the reaction
2 H2(g) + O2(g) --> 2 H2O(l)
deltaG°(298 K) = -474.2 kJ
K = 1·10^83 bar^3

This 'equilibrium constant' is so high that not a single reactant
molecule will remain once the reaction proceeds.

Similarly, the reverse reaction has
2 H2O(l) --> 2 H2(g) + O2(g)
has K = 1·10^-83 bar^-3
Not a single molecule of water will dissociate with this K.

So, while it is theoretically possible to calculate K, it is not
actually possible to achieve equilibrium for this system.

Now, during the electrolysis, applying exactly 1.229 V at 298 K will
take the system to deltaG° = 0 kJ
This corresponds to an equilibrium system with K = 1: pure water and
1 bar of each gas. Applying double the voltage will drive the reaction
to beyond equilibrium state again.

So, I concede that there is a voltage range where an equilibrium will
exist with that applied voltage. But once that potential is removed,
the final state of the system will not be an 'equilibrium'.

Dr. Roy Jensen
(==========)-----------------------------------------¤
Lecturer, Chemistry
E5-33F, University of Alberta
780.248.1808

On Sun, 15 Jun 2014 12:54:27 -0700, you wrote:

In the context of:

A friend has reverted to childhood by wishing to make balloon bombs using
electrolysis of waster. I suggested that compression of the gasses would
"slow" the electrolysis due to the Le Chatelier principle.

On 06/15/2014 11:34 AM, rjensen@ualberta.ca wrote:

This reaction is not in equilibrium, so le Chatelier's principle
doesn't apply. The gas pressure will not effect the electrolysis rate.

I wouldn't have said that. The reaction /could/ be carried
out under close-to-equilibrium conditions, in which case
you /could/ see the effect of pressure.

There is energy in the pressurized products, and this enters
directly into the energy budget of the reaction. In the
case of electrolysis, I would expect to see it in the cell
voltage. In other situations, I would expect it to shift
the equilibrium point.

As a familiar example, consider the equilibrium between
gaseous CO2 and an aqueous solution of dissolved CO2 plus
carbonic acid. The equilibrium point is a strong function
of pressure.
-- Hint: champagne
-- Hint: seltzer water
-- Hint: mentos + diet coke

===========================

As a separate matter, Le Châtelier's so-called principle is
not the smart way to think about the issue. In his lifetime,
Le Châtelier gave two inconsistent statements, one of which
is trivially tautological, and the other of which is just
wrong. For details, see
http://www.av8n.com/physics/thermo/spontaneous.html#sec-le-chatelier

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