[raymanh]

I've found the value for K using Gibbs free-energy values (in KJ/mol) where Gibbs free-energy = -5.707*log(K).

Would that mean the units of K are KJ/mol as well?

[KungKemi]

I've found the value for K using Gibbs free-energy values (in KJ/mol) where Gibbs free-energy = -5.707*log(K).

Would that mean the units of K are KJ/mol as well?

Not quite, if one were to use the ΔG of a chemical process in order to determine its equilibrium expression, then one would use:

K_eq = e^-ΔG/RT, where R is the Ideal Gas constant.

This would still leave a dimensionless K-value.

n.b: This could only be applied to a reaction in which ideal gasses are involved.

KungKemi

[Borek]

I've found the value for K using Gibbs free-energy values (in KJ/mol) where Gibbs free-energy = -5.707*log(K).

Would that mean the units of K are KJ/mol as well?

Are you asking whether in a formula

[tex]A = \log B[/tex]

units of A and B are identical? No, they are not.

What is the exact formula you have used? Or, in other words, where does the -5.707 constant comes from? What are its units?

Can you take log of something that has units? What is the result of such operation?

Write several (random) reaction equations and try to find the units of K. Do you see J (or kJ) there?

[mjc123]

n.b: This could only be applied to a reaction in which ideal gasses are involved.

Why?

[KungKemi]

n.b: This could only be applied to a reaction in which ideal gasses are involved.

Why?

Hmm, the following could probably also be applied to experimentally determined data with minimal error. So...I see the contradiction in my statement.

Thanks for the pickup mjc,
KungKemi