Understanding the Logarithm of K in DeltaG

[aniketp]

Well I read in a book that:
\DeltaG= RTln (K) (Where K is the eqm const)
H "K" is not necessarily dimensionless so how can we take a log of "K"?

 

[MathematicalPhysicist]

Do you mean Gibbs free energy?

then it's given by:
\tau log(Z_G)
after taking the derivative you can get the difference or differential form of this energy.
Z_G is the Grand canonical partition function, and it's indeed dimensionless.

 

[Andy Resnick]

The full Nernst-Planck equation is:

\Delta\mu=RTln(\frac{[x_{i}]}{[x_{o}]})+ZF(\psi_{i}-\psi_{o}), where

\Delta\mu is the change in chemical potential for a particular species

[x_{i}] is the concentration of species 'x' on one side of a dividing surface (and [x_{o}] the concentration on the other side)

\psi_{i} the electrical potential on one side of a dividing surface (and \psi_{o} the potential on the other side)

And R, T, Z, F the usual gas constant, temperature, charge per molecule and Faraday constant.

It's worth understanding this equation- it governs diffusive processes of charged solutes in solution and leads to a remarkable (IMO) result: the membrane potential. There's various simplifications, it looks like you have uncharged solutes (Z = 0), and instead of \Delta\mu you are using \Delta G, which also changes the \frac{[x_{i}]}{[x_{o}]} term to the equilibrium constant. But, since it's still dimensionless, there's no problem.

Does that help? This is a really fundamental concept- make sure you understand it.

 

[Mapes]

It's often ignored, but the K inside the logarithm is divided by K₀, which has magnitude of 1 and the same units. The quotient is therefore dimensionless and equal to the magnitude of K.

EDIT: Whoops, Andy got there first with a more complete answer.