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The binding interaction energy during the adsorption of system B on system A can give an idea if the interaction is favorable (negative value) or unfavorable (positive value). Also, the absolute value represents the strength of the interaction.

To determine the binding energy, the procedure is to calculate the energy of both systems separately $E_A$, $E_B$ and the energy of the complex $E_{AB}$, then the binding energy can be calculated as:

$E_{bind} = E_{AB} - E_A + E_B$......... (1)

(the justification to do this is here)

Now, let suppose that I want to heat the complex to detach the system B from the surface of system A.

How to have an estimate of the temperature needed to do that? The only idea that come to my mind is to use the rule of three with the association that 300K correspond to 25meV (via $k_BT$, with $k_B$ being the Boltzmann constant).

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    $\begingroup$ Couldn't you use thermodynamics and make a van't hoff isotherm? Calculate enthalpy instead of energy and you can get the temperature dependence with respect to the equilibrium constant for the binding reaction. I don't do adsorption, but I am always very leery of using energy for anything. A specific ensemble should use the free energy to dictate equilibria. Energy is a hand wavy substitute for free energy. But like I said, I don't do adsorption. $\endgroup$ – Charlie Crown May 25 at 6:05
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    $\begingroup$ 2Charlie Crown: "Energy is a hand wavy substitute for free energy" - well said! Indeed, this setting implies energies in "vacuum". But in absolute vacuum at any finite temperature anything separated has lower free energy that anything binded. So in vacuum binding free energy is allways positive. $\endgroup$ – user36313 May 25 at 13:10
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    $\begingroup$ @CharlieCrown that's a great idea. I will take a look. $\endgroup$ – I. Camps May 25 at 13:13
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If I want to break a Li$_2$ molecule (i.e. remove atom A from atom B), one way to do it is by shining a laser on it such that the frequency ($\nu$) or energy ($h\nu$) corresponds to the difference between the energy at $v=0$ (if the molecule starts in the ground state) and the dissociation asymptote in this picture:

I think the question asks: What if we don't have a laser but we have some other way of turning up the temperature? How high would the temperature need to be, to break the molecule?.

I think that quantum mechanics tells us that even if we raise the system to a very high temperature, there is still no guarantee that the particles will separate, but we can estimate the temperature at which the dissociation happens with probability > 99%. One way to see the probability for a system with Hamiltonian $H$ and temperature $T$ to be in a particular state, is to use the formula:

\begin{equation} \rho = \frac{e^{-\beta H}}{\textrm{tr}\left(e^{-\beta H}\right)},~~~~~~~~\beta\equiv \frac{1}{k_BT}. \tag{1} \label{eq:boltzmann} \end{equation}

Here, $\rho$ is the density matrix and the diagonal elements of it give the probabilities $p_i$ of finding the system in state $|i\rangle$. There is an infinite number of possible states for this system: 10 of them are shown in the above diagram ($v=$ 0 to 9), but there's also a "continuum" of states above the dissociation asymptote, and these are the ones we want, because these are the states that correspond to atom A being removed from atom B. If we sum/integrate the probabilities $p_i$ over all states that correspond to a dissociation, we can get the total probability $P$ of successfully breaking the bond at temperature $T$. Then, all we have to do is find the value of $T$ such that $P>99\%$.

Now if we are not dealing with Li$_2$ but instead with a complicated system AB, and we want to know the probability of detaching a sub-system (A) from the rest of the complex (B), the potential energy surface may look more complicated than the one in the above figure (maybe we don't just have an internuclear distance $r$ but we also have have angles), so the Hamiltonian will be more complicated, but Eq. \eqref{eq:boltzmann} can still be used to get the probabilities of the dissociated states 🙂.

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  • $\begingroup$ I was thinking in something more easy to do in practice. As my system are intended to be used as filters, just "putting in the oven" to detach them looks a good option to recycle it. Also, each adsorbate can be different, implying different binding energies, so I think changing the oven temperature is easier than tune in the lasers. $\endgroup$ – I. Camps May 25 at 12:03
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A more approximate but sometimes easier approach than the one of Nike Dattani might be to calculate an artificial reaction coordinate of separating the two systems. This can be done manually or by employing some accelerated MD technique, e.g. metadynamics. The latter would even allow you to map the free energy along this path and thereby give you an estimate of the free energy barrier. You could then use your initial idea to directly translate this barrier to a temperature.

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  • $\begingroup$ +1. Good answer! $\endgroup$ – Nike Dattani May 25 at 7:20
  • $\begingroup$ The problem here is that, after using DFT, some covalent bonds are/can be formed. My MD jobs were only for ligand/protein complexes where the used forcefields do not treat the bond formation/breaking. I imagine that I should use some forcefields like ReaxFF, right? $\endgroup$ – I. Camps May 25 at 11:56
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    $\begingroup$ Well, if you have the means to perform a DFT based MD, that would be better I think. I'm not familiar enough with ReaxFF to judge if this kind of barrier computation would yield meaningful results. Perhaps somebody else can comment on that. Depending on your system size, CP2K with a small basis might just offer what you need. $\endgroup$ – patrickmelix May 25 at 12:50

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