Question

How would you define the concept of equilibrium in terms of free energy and entropy?

Answer 1

Well, we could start from this equation:

`color(green)(DeltaG = DeltaG^@ + RTlnQ)`

where:

• `DeltaG` is the Gibbs' free energy.
• `DeltaG^@` is the Gibbs' free energy at `25^@ "C"` and `"1 bar"` (or, depending on your book, older books say `"1 atm"`, where `"1 atm = 1.01325 bar"`).
• `R` is the universal gas constant `"8.314472 J/mol"cdot"K"`.
• `T` is the temperature in `"K"`.
• `Q` is the reaction quotient, i.e. the not-yet-equilibrium constant.

When we are at equilibrium, `\mathbf(DeltaG = 0)`. Thus:

`color(blue)(DeltaG^@ = -RTlnK)`

where `Q = K` at equilibrium. `K` may or may not be `1`.

So, at equilibrium, you can use the standard Gibbs' free energy (which is tabulated in many if not all textbooks in an appendix) to solve for the equilibrium constant for the particular reaction.

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Furthermore, since `DeltaG = 0` at equilibrium, we can invoke the thermodynamic equation

`color(green)(DeltaG = DeltaH - TDeltaS)`,

where

• `DeltaH` is the enthalpy (heat flow at a constant pressure)
• `DeltaS` is the entropy (a measure of the number of ways a system can exist, i.e. loosely its "disorder"),

and recognize that now we have:

`0 = DeltaH - TDeltaS`

`TDeltaS = DeltaH`

`color(blue)(DeltaS = (DeltaH)/T)`

One example where this description of entropy at equilibrium is true is for phase changes. That is a constant-temperature phase-phase equilibrium.

So at equilibrium, you could determine the entropy from knowing the enthalpy and the current temperature.