I am learning about COSMO model used in chemical engineering for calculating chemical potential and similiar quantites (actitvity coefficients) for solutes in solutions/mixtures.

In the model, it is assumed that solvent is a continuum with some dielectric constant. Solute is placed inside cavities of such medium. Solute molecules are modeled as having certain shape and size with particular surface area and volume characteristic for molecules of different compounds.

On wikipedia page: COSMO Solvation Model, it is said that if solvent were perfect conductor than electric potential on the surface of the cavity ,where solute is placed, is zero. I don't understand why this is. What law of nature supports this claim?

I suppose it has something to do with electrostatic condition which says that if no current flows in a conductor, electric field must be zero at its every point or to be more precise electrochemical potential of charge carriers is the same in every point of the conductor since differences in chemical potential can drive current as well.

I am not sure if electrostatic condition has something to do with this question.


1 Answer 1


First of all, the

method used in chemical engineering for calculating chemical potential and similar quantities (actitvity coefficients) for solutes in solutions/mixtures

is called COSMO-RS and not COSMO. Almost the same name, but very different meaning :-)

COSMO is a continuum solvation model often used in quantum chemistry. To learn more about it either try to get a copy of the book that was already mentioned in the comment, or google also for CPCM which is just another name for the same model. COSMO-RS uses the output of COSMO calculations to do thermodynamics from which chemical potentials in liquids are derived. OK, but that was not your question...

COSMO generates a cavity around the molecule (based on some kind of solvent-accessible surface) and assumes it to have a dielectric constant of infinity. That's a property metals have (and what "perfect conductor" refers to: an ideal metal).

So why is the electrostatic potential on a metal zero? The correct answer is that it is not zero but constant. As there is no absolute value for potentials (as only differences are physically meaningful), COSMO simply sets this constant to zero. Another wording for that is: It defines the reference state. So all values of the electrostatic potential at each point in space in a COSMO calculation are relative to the value on the cavity surface (which is set to zero for practical reasons).

The reason why an electrostatic potential is constant on a metal indeed stems from standard electrostatics (check your physics book):

The work (W) caused by an electric force (= electric field E) on a test charge q moved in space from $r_1$ to $r_2$ ($dr=r_2-r_1$): $$ dW = - F dr $$ with the force F: $$ F = q E$$ So you need to 'spend' a work $dW$ to move a charge in an electric field.

The work done by an electrostatic force is the negative of the change in potential energy, $dU$ (so $dW = dU$ ):
$$dU = -q E dr$$

The potential $V$ (or better: the difference in the potential $V$, so something like $V(r_2) - V(r_1)$ ) is defined as the change in potential energy divided by the test charge q: $V_{12} = - E (r_2-r_1)$ for a constant electric field E, otherwise you need the integral of E with respect to r...

$$dV = dU / q$$

On a metal the electrons can move freely, so there is no difference in potential energy $dU$ - because if there WAS a force between two points in the same metal, the electrons would simply move until the force vanishes. We are looking at a conductor in equilibrium...

So $dU$ is zero. And if the difference in potential energy is zero, also the difference in the electrostatic potential is zero, $dV=0$, and hence the potential $V$ is constant at any place on the metal cavity (that is called an "equipotential surface").

  • $\begingroup$ Just what I thought! Thank you for the detailed explanation. I thought it was in fact constant or that electric field is zero at this point, but was not sure if this is what I thought. $\endgroup$ Apr 17, 2022 at 12:55

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