The free energy averaged over $N_\mathrm{conf}$ conformations of molecule X is

$G^\circ=-RT \ln \left( \sum_I^{N_\mathrm{conf}} e^{-G^\circ_I/RT} \right)$ (1)

However, the way to compute an average property of the conformations $\left\langle x\right\rangle$ is

$\left\langle x\right\rangle=\sum_I^{N_\mathrm{conf}} p_Ix_I$ (2)

where

$p_I=\frac{e^{-G_I^\circ/RT}}{\sum_J^{N_\mathrm{conf}} e^{-G_J^\circ/RT}}$ (3)

But this equation cannot be used to compute the conformationally averaged free energy because

$G^\circ \ne \left\langle G^\circ \right\rangle$

Why not? First, let's see where the first equation comes from.

**Obtaining equation (1)**

The first equation comes from the statistical mechanical definition of the Helmholtz free energy found any any p-chem textbook

$A^\circ = -RT \ln \left( \sum_i^{\mathrm{states}} e^{-\varepsilon_i/kT} \right)=-RT \ln \left( q \right)$

The sum over microstates can be split into sums over microstates of each conformation

*I*

*$A^\circ = -RT \ln \left( \sum_I^{N\mathrm{conf}} \sum_{i\in I}^{\mathrm{states}} e^{-\varepsilon_i/kT} \right)$*

$= -RT \ln \left( \sum_I^{N\mathrm{conf}} q_I \right) = -RT \ln \left( \sum_I^{N_\mathrm{conf}} e^{-A_I^\circ/RT} \right)$

To get the corresponding expression for the Gibbs free energy [Eq (1)]:

$G^\circ = A^\circ +p^\circ V $

$=-RT \ln \left( \sum_I^{N_\mathrm{conf}} e^{-A^\circ_I/RT} \right)-RT \ln \left( e^{-p^\circ V/RT} \right)$

$=-RT \ln \left( \sum_I^{N_\mathrm{conf}} e^{-G_I^\circ/RT} \right)$

**The difference between equation (1) and (3) is the conformational entropy**

$G^\circ - \left\langle G^\circ \right\rangle=-RT \ln \left( \sum_J^{N_\mathrm{conf}} e^{-G_J^\circ/RT} \right)\sum_I^{N_\mathrm{conf}} p_I$

$+RT\sum_I^{N_\mathrm{conf}} p_I\frac{-G_I}{RT}$

$=RT \sum_I^{N_\mathrm{conf}} p_I \left( \frac{-G_I}{RT}- \ln \left( \sum_J^{N_\mathrm{conf}} e^{-G_J^\circ/RT} \right) \right)$

$=RT \sum_I^{N_\mathrm{conf}} p_I \left( \ln \left( e^{-G_I^\circ/RT} \right) - \ln \left( \sum_J^{N_\mathrm{conf}} e^{-G_J^\circ/RT} \right) \right) $

$=RT \sum_I^{N_\mathrm{conf}} p_I \ln(p_I) =-TS_{\mathrm{conf}} $