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

Mar 22, 2016

Well, we could start from this equation:

$\textcolor{g r e e n}{\Delta G = \Delta {G}^{\circ} + R T \ln Q}$

where:

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

When we are at equilibrium, $\setminus m a t h b f \left(\Delta G = 0\right)$. Thus:

$\textcolor{b l u e}{\Delta {G}^{\circ} = - R T \ln K}$

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.

Furthermore, since $\Delta G = 0$ at equilibrium, we can invoke the thermodynamic equation

$\textcolor{g r e e n}{\Delta G = \Delta H - T \Delta S}$,

where

• $\Delta H$ is the enthalpy (heat flow at a constant pressure)
• $\Delta S$ 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:

$0 = \Delta H - T \Delta S$

$T \Delta S = \Delta H$

$\textcolor{b l u e}{\Delta S = \frac{\Delta H}{T}}$

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.