Saturday, January 24, 2015

The conformationally averaged free energy


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}} $


Post a Comment