Entropy is the driving force, but a chemical system also wants as little energy as possible. How does this relate?

Answer

Dear Sarah,

This is a very interesting question. The answer to this can be found in the laws of thermodynamics, and is in the relationship(s) found between entropy and energy of a system.

Simply put, the second law of thermodynamics states that the entropy of a system keeps increasing in an isolated system. If we now assume the universe as an isolated system with entropy S0, you can divide it internally into 2 parts: your chemical system (with entropy S) and the rest of the universe (with entropy S’).

Then in a chemical reaction/process a change of entropy will give dS0>=0 and thus dS0=dS+dS’>=0.

The heat (Q) that your chemical system extracts from the universe (for example) can change the internal energy of the rest of the universe (dU) and possibly also perform work (W). Since dS’=-Q/T, we can rewrite our first equation, via the first law of thermodynamics (Q=dU+W) as:

dS0=dS – (dU+W)/T >=0

which can be rewritten as:

dS0= (TdS-dU-W)/T >=0

in the latter we recognize the Helmholtz free energy F=U-TS. This is the energy that becomes minimal in your chemical system when it is in contact with a heat and pressure reservoir (a reasonable description for a chemical process where no particles are lost in the environment, for example through evaporation. Is there any other interaction with the environment than described here, the equation of Q changes and it will also be about a different “free energy” such as Gibbs free energy or enthalpy, which take over the role of Helmholtz free energy.)

So then we get:

dS0= (dF -W)/T >=0

Entropy increase thus means that the change in energy is greater than the work in question: -dF>=W.

If no work is now involved (W=0) then dF<=0. So we learn from this that because dS0>0 (entropy increases) that dF<0 (or the energy decreases).

With regard to your second question, it is subtly different. Entropy is often referred to as a measure of disorder, but that is a little too imprecise. It actually refers to the number of states a system can be in. So if it concerns a situation where particles can spread out then dilution causes an increase in entropy (because more states are possible), the decrease in energy is then the result of a decrease in interaction energy.

Regards,
Danny