## Freitag, 13. Dezember 2013

### Chem 001: Nernst Equation

Consider a galvanic cell composed of two $$\text{H}^{+}\left|\right.\text{H}_2$$-half cells where $$\left[\text{H}^+\right]$$ are $$0.025 \text{M}$$ and $$5 \text{M}$$ in the first and second half-cell, respectively. Calculate the electromotive force (EMF).

It's good practice to first visualize such a problem before attempting to solve it in order to gain a clear understanding of the system. The galvanic cell could look like this:

 Concentration chain of two hydrogen electrodes. The blue arc indicates the salt-bridge.The hydrogen pressures do not need to be the same (see text).
A galvanic cell with two times the same system at different electrolyte concentration is refered to as concentration chain. At the platinum electrode, the hydrogen gas is oxidized and $$\text{H}^+$$ ions enter into solution. A thermodynamic equilibrium is reached when $$\left[\text{H}^+\right]$$ is equal in both cells.

We can recognize that in electrode 1, $$\left[\text{H}^+\right]$$ is lower and so it is in electrode 1 where hydrogen is oxidized. By convention, the electrode where oxidation occurs is referred to as anode, the one where reduction occurs as kathode. The electrons thus exit the cell at electrode 1, run through the wire to electrode 2 where they enter the cell again and reduce hydrogen ions. Electrode 1 is therefore referred to as the minus pole of the cell ("where the electrons come from"). At the anode (electrode 1), the half-reaction is

$$\text{H}_2\text{(g)} \rightarrow 2\text{H}^+\text{(aq)} + 2\text{e}^-$$

In order to calculate the EMF, it is required to calculate the potential difference between both electrodes 1 and 2 compared to the standard hydrogen electrode. The EMF for the cell is then calculated as the sum of reduction potentials

$$\Delta E = E(\text{Kath.}) + E(\text{An.})$$

and this is where we have to be careful to choose the sign correspondingly to the reaction we are formulating.
For the reaction at the kathode (where reduction of $$\text{H}^+\text{(aq)}$$ occurs), the potential (compared to the normal hydrogen electrode) is given by the Nernst equation

$$E\text{(Kath.)} = E^0 - \frac{0.06}{2}\log{Q}$$

The $$2$$ in the fraction is the number of transported electrons per reduction equivalent. The $$(-)$$-sign in front of the log-expression is used when the reaction quotient $$Q$$ corresponds to a reaction formulated in direction of reduction:

$$2\text{H}^+ \text{(aq)} + 2\text{e}^- \rightarrow \text{H}_2 \text{(g)}$$

For the general reaction

$$k\cdot K + l\cdot L \rightarrow x\cdot X + y\cdot Y$$

the reaction quotient is always defined as

$$Q = \frac{a^x(X) \cdot a^y(Y)}{a^k(K) \cdot a^l(L)}$$,

where the $$a^i$$'s are activities of the species. Here, we understand the hydrogen gas has unit activity, $$a(\text{H}_2\text{(g)}) = 1$$, and so

$$E\text{(Kath.)} = E^0 - \frac{0.06}{2}\log \frac{1}{ \left[ \text{H}^+ \right] ^2 }$$

where $$\left[\text{H}^+\right] = 5 \text{M}$$ and the exponent $$2$$ is the stoichiometric coefficient. We emphasize again the minus sign in front of the $$\log$$-term used when the reaction is formulated in direction of reduction. $$E^0$$ is the electrode potential for the normal-hydrogen electrode which is zero by definition.
A similar equation is formulated for the anode half-reaction:

$$E\text{(An.)} = E^0 + \frac{0.06}{2}\log \frac{\left[ \text{H}^+ \right]^2}{1}$$

however here a $$(+)$$-sign is used in front of the $$\log$$-term because the reaction is formulated in direction of oxidation (so the oxidized species $$\left[H^+\right]$$ appears in the numerator) and $$\left[\text{H}^+\right] = 0.025 \text{M}$$.
The numeric values are $$E(\text{Kath.}) = +0.041 V$$ and $$E(\text{An.}) = -0.096 V$$. $$E(\text{An.})$$ however corresponds to an oxidation potential and so in order to calculate the EMF, we need to take its negative (such that we are adding reduction potentials). This then provides us with

$$EMF = \Delta E(\text{Kath.}) + {}^-\Delta E(\text{An.}) = 0.041 + {}^-(-0.096) = 0.137 [V]$$

The question can now be extended by asking what the EMF will be, when the hydrogen gas pressures are different from atmospheric pressure (and so the activity $$a(\text{H}_2)$$ is no longer unity):
Assume $$p(\text{H}_2) = 202.6 \text{kPa}$$ at the anode and $$p(\text{H}_2) = 10.13\text{kPa}$$ at the kathode.
Given these pressures, the activities are $$a(\text{H}_2, \text{Kath.}) = p(\text{H}_2)/101.3 \text{kPa} = 0.1$$ and $$a(\text{H}_2, \text{An.}) = p(\text{H}_2)/101.3 \text{kPa} = 2$$. By substituting these values in the Nernst equations for the respective half-cells (using the same concentrations for $$\left[H^+\right]$$), we arrive at

$$E\text{(Kath.)} = E^0 - \frac{0.06}{2}\log \frac{0.1}{ \left[ \text{H}^+ \right] ^2 } = 0.072 \left[V\right]$$

and

$$E\text{(An.)} = E^0 + \frac{0.06}{2}\log \frac{\left[ \text{H}^+ \right]^2}{2} = -0.105 \left[V\right] .$$

The EMF in this cell is therefore (after again switching the sign of the anode potential to obtain a reduction potential)

$$EMF = \Delta E(\text{Kath.}) + {}^-\Delta E(\text{An.}) = 0.071 + {}^-(-0.105) = 0.177 [V] .$$

## Mittwoch, 4. Dezember 2013

### Literature 001: Alexandrova et al., JACS, 2008, 130, 15907-15915

In "Catalytic Mechanism and Performance of Computationally Designed Enzymes for Kemp Elimination", Alexandrova et al. report calculated activities of enzymes that were previously prepared by the Baker group. The experimentally expressed enzymes are called KE07, KE10 (which contains the V131N mutation) and KE15.
To recall, the Kemp elimination is this reaction:
 http://pubs.acs.org/doi/abs/10.1021/ja804040s
From semiempirical QM/MM calculations and transition state theory, the authors obtain $$k_{cat}$$ or $$\Delta G^\ddagger$$ which they compare to experiment.
For example, for KE07 the authors report a calculated $$\Delta G^\ddagger$$ of 8.1 kcal/mol while the measured activation free energy is 17.1 kcal/mol. That appears as quite a discrepancy between predicted and measured activation energy and one might wonder how come this made it into JACS (see below). Also for KE15, the calculated activation energy doesn't really accurately predict the measured activation energy (calc. / exp. [kcal/mol] $$\Delta G^\ddagger$$: 12.3 / 17.0). For KE10, $$k_{cat}$$ wasn't measured.

However, what the authors claim is that they are not that much interested in exactly matching the measured activation energy (most likely they would use a more elaborate QM method), but much more if they can predict if the enzyme actually contributes to catalysis. I.e. they're interested in the ratio $$k_{cat}/k_{uncat}$$ where $$k_{uncat}$$ is the rate constant for the reaction in solvent using $$\text{OH}^-$$ as base. And apparently they are able to correctly predict if this ratio is larger or smaller than 1 for all enzymes:

$$k_{cat}^{KE07}/k_{uncat} = 1.6 \cdot 10^4$$

$$k_{cat}^{KE10}/k_{uncat}$$ < 3.3 (assuming $$K_M$$ to be comparable or smaller than for KE07)

$$k_{cat}^{KE15}/k_{uncat} = 1.9 \cdot 10^4$$

$$k_{cat}^{KE16}/k_{uncat} = 5.2 \cdot 10^3$$ (by 2-step mechanism and requiring to impose the protonation state of a catalytic D48 residue - why do they not estimate the $$pK_a$$ value of the residue?)

For all enzymes, catalytic activity is observed, so qualitatively they are able to tell if an enzyme will be active or not (they calculate the uncatalyzed reaction to have an activation energy of 19.8 kcal/mol and the activation energies of all enzymes is calculated to be lower than that).
Possible explanation for discrepancies provided: no backbone sampling, accuracy of SE QM (PDDG/PM3) - even if they claim this method to be suited for elimination reactions, no polarization in MM field.

Another question can be raised: how can they be sure about their prediction if they don't consider that $$K_M$$ might be different for all enzymes. So even if an enzyme might be able to catalyze the reaction, it could in principle be that it hardly binds the substrate. The way to answer this question these days is to say "The catalytic efficiency, $$k_{cat}/K_M$$, has not been computed owing to the technical challenges for $$K_M$$, which requires computation of the absolute free energy of binding for the substrate." The substrate does not dissociate from the active site in the equilibration and so the working assumption is to believe that all enzymes can bind the substrate and do so comparably well.

My impression: The paper provides elaborate discussions on observed mechanisms and explains them in minute (structural) detail. The weak points, which are very explicitly disclosed, are compensated by pointing out possible reasons and by the careful mechanistic analysis. A strong point is that the results provide suggestions for how to further improve the catalytic activity (i.e. increasing basicity of the catalytic base, optimization of substrate orientation and positioning of hydrogen bond donors should help increase activity). Furthermore, in the mindset that agreement with experimental activation energy is the key property to consider, the paper would not be considered good. But, the point is really to provide a method that allows to qualitatively tell if the enzyme is active or not, and it does so very well I think.

## Sonntag, 10. März 2013

### QM 006: INDO One-electron Integrals

In Frank, Computational Chemistry, Eq. 3.86 is
$$\langle \mu_A \left| \bf{h} \right| \mu_A \rangle = \left< \mu_A \left| -\frac{1}{2}\nabla^2 - \bf{V}_a \right| \mu_A \right> - \sum_{a \neq A}^{Nuclei} \langle \mu_A \left| \bf{V}_a \right| \mu_A \rangle$$.

Is there a reason for not writing it as
$$\langle \mu_A \left| \bf{h} \right| \mu_A \rangle = \left< \mu_A \left| -\frac{1}{2}\nabla^2 \right| \mu_A \right> - \sum_{a = A}^{Nuclei} \langle \mu_A \left| \bf{V}_a \right| \mu_A \rangle$$?

## Dienstag, 26. Februar 2013

### QM 005: Comparing semi-empirical methods

In a recent paper Liao et al. calculated the reaction profile for a newly suggested mechanism of acetylene hydratase. Conveniently, the atomic coordinates of all species along the reaction coordinate have been deposited in the supporting material.

Carrying out some SPE calculations of all species using MOPAC2012, we find the following profiles.

 Fig. 1: Calculated Reaction profiles.
In the SPE calculations, a dielectric constant of $$\epsilon=4$$ was applied. In the reference ("B3LYP" in Fig. 1), the optimization was done using the LANL2TZ(f) pseudo-potential on W, 6-311+G(d) for S and 6-31G(d,p) for the other elements. The reference SPE calculations were carried out using 6-311+G(2d,2p) on all elements but W.
Apparently, the semi-empirical methods greatly overestimate the interaction energy between W and the acetylene unit (going from 0 to 1) and underestimate the activation energy for the proton transfer in step 4.
However, without reoptimizing the transition state, PM6 is does see a very clear transition state for the hydroxylation at step 2.

Orbital diagram from supplementary material.

## Mittwoch, 16. Januar 2013

### PyMOL 007: New Feature

I recently wrote a mail to the PyMOL user mailing-list asking if PyMOL can write vector graphics based files of a session (similar to Molscript). Vector graphics are scalable, require less disc space (good for journal uploads) and look beautiful, allowing to focus attention on details that matter.

Here's the mail I wrote:

Hi PyMOL users
Can PyMOL write a vector based picture of a session? Does not require to
be very "fancy", but vector based would be cool. Something like in the
old days with molscript.

Best regards
Martin

In the following, a number of people responded suggesting SVG or EPS/PDF as file format. Within three days, Jason Vertrees of Schrödinger started a poll to check which format was most preferred by the users. Find it here:


http://pymol.org/vector_poll


It looks as if this really is a feature that could be appreciated and it would be great if it was implemented.
I guess there will be some management involved in the decision making, but in any case its another example of the great support by the PyMOL developers at Schrödinger. And sometimes this deserves mentioning.