Sunday, December 16, 2012

Conformational and rotational entropy and point group symmetry

The point group affects the free energy
The use of point group symmetry in quantum chemical calculations can speed up calculations significantly, but it is often difficult to input symmetric coordinates correctly so many people opt to run calculations without symmetry (i.e. in $C_1$ symmetry) anyway.  However, the lack of symmetry changes the rotational entropy and, hence, the free energy you compute. So if you have symmetric molecules but choose to run in $C_1$ symmetry you must correct the entropies and free energies.

The rotational entropy is given by$$S_{rot}=R\ln\left(\frac{8\pi^2}{\sigma}\left(\frac{2\pi ekT}{h^2}\right)^{3/2}\sqrt{I_1I_2I_3}\right)$$ $\sigma$ is called the symmetry number and depends on the point group: for example, $\sigma=1$ for $C_1$ and $C_s$, $\sigma=n$ for $C_{nv}$ and $C_{nh}$, $\sigma= 2n$ for $D_{nh}$ and $D_{nd}$, and $\sigma= 12$ or $T_d$.  So the rotational entropy calculated with and without symmetry will differ by $R\ln(\sigma)$:$$S_{rot}=S_{rot}^{C_1}-R\ln(\sigma)$$and similarly for the free energy$$G^\circ=G^{\circ,C_1}+RT\ln(\sigma)$$
An example: $H_2O+Cl^-\rightleftharpoons HOH\cdots Cl^-$
The free energy change for this reaction is$$\Delta G^\circ=\Delta G^{\circ,C_1}+RT\ln\left(\frac{\sigma_{HOH\cdots Cl^-(C_s)}}{\sigma_{H_2O(C_{2v})}\sigma_{Cl^-(C_1)}}\right)\\ \Delta G^\circ=\Delta G^{\circ,C_1}+RT\ln\left(\frac{1}{2\times 1}\right)=\Delta G^{\circ,C_1}-RT\ln (2)$$The corresponding equilibrium constants are$$K=e^{-\Delta G^\circ/RT}=2K^{C_1}$$So, $\sigma$'s accounts for the fact that, because of the symmetry of water, there are two ways of making $HOH\cdots Cl^-$and another way to view $R\ln(2)$ is that it is the conformational entropy of the complex.

A test: $H_2O+NH_3\rightleftharpoons HOH\cdots NH_3$
For the above equilibrium what is $X$ in$$\Delta G^\circ=\Delta G^{\circ,C_1}-RT\ln (X)$$

An exception: $2H_2O\rightleftharpoons HOH\cdots OH_2$
Based on the rules outlined so far one would expect the free energy change for this equilibrium to be$$\Delta G^\circ=\Delta G^{\circ,C_1}-RT\ln (4)$$The factor of four accounts for the fact that there are four ways of making $HO^AH\cdots O^BH_2$. However, since the water molecules are identical there are actually four additional water dimer possibilities for  $HO^BH\cdots O^AH_2$, so$$\Delta G^\circ=\Delta G^{\circ,C_1}-RT\ln (8)$$In general, for $A+A\rightleftharpoons Product$ reactions the symmetry number for $A+A$ is $2\sigma_A^2$rather than $\sigma_A^2$

All these considerations also apply to activation free energies and rate constants as outlined in this excellent paper by Fernández-Ramos et al., which inspired this post.  See also this excellent paper by Gilson and Irikura.

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