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Consider a simple chemical reaction of an enzyme that binds to a substrate to create a complex

$$E+S\longleftrightarrow ES$$

the forward/backward reactions occur with rates $k_f,k_d$

If we consider conservation of material $[E]+[ES]=E_{tot}$ and $[S]+[ES]=S_{tot}$, we can obtain all the information in the following ODE

$$\frac{d[E]}{dt}=-k_f[E][S]+k_r[ES]$$

which in equilibrium defines the dissociation constant of the reaction

$$K_d=\frac{[E][S]}{[ES]}$$

The dissociation constant is related to the temperature, but in all the books and info online the derivation is only made through the gibbs free energy and enthalpic terms

$$\Delta G=-RT ln(K_d)$$

I wish to derive the relation for the Helmholtz free energy at constant volume, from physical thermodynamic principles that consider the binding and unbinding energy of the molecules, and will be written in terms of $k_b$ and not $R$ (this is a solution not and ideal gas!).

I would like at the end to obtain a term that looks like

$$K_d=e^{-\frac{\Delta F}{k_bT}}$$

is there any brave physicist who is willing to take this challenge? or chemical problems are too much for physicists to handle...

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  • $\begingroup$ Are you sure you mean at constant pressure? It is very odd to have a physical system where both the pressure and volume are held constant. Normally one is allowed to vary in order to keep the other fixed. If that is really what you want then you will most likely need to use some very odd potential, rather than the Helmholtz free energy. $\endgroup$ – By Symmetry Feb 6 at 15:15
  • $\begingroup$ Alternatively, if you are assuming your solution to be more or less incompressable, so that constant pressure and constant volume are more or less interchangable, why do you care about whether you are using the Helmholtz or Gibbs free energy? $\endgroup$ – By Symmetry Feb 6 at 15:17
  • $\begingroup$ @BySymmetry, corrected: at constant volume, you can go ahead and make the derivation. $\endgroup$ – jarhead Feb 6 at 15:34
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I’ll discuss your question in the context of the kinetics of a gas-phase reaction at constant volume. Liquid-phase reactions are generally run at constant pressure, largely as a matter of experimental convenience, but the work done against pressure and energy stolen from the solvent are confusing complications that don’t invalidate thermodynamic arguments about minimizing free energy, which finesse the kinetics.

Your question writes the rate constant for dissociation ${{k}_{r}}$ in one place and ${{k}_{d}}$ in another, but they are the same thing. I’ll write ${{k}_{assoc}}$ and ${{k}_{dissoc}}$ for clarity.

The equilibrium constant ${{K}_{d}}\equiv [E][S]/[ES]$ should obviously equal the ratio ${{k}_{dissoc}}/{{k}_{assoc}}$. In gas phase, dissociation cannot happen unless dissociation is exothermic, and ${{k}_{dissoc}}$ will be independent of temperature (unless rotational state matters). By contrast, ${{k}_{assoc}}$ is sensitive to temperature because the colliding molecules must have sufficient kinetic energy in the CM frame to stick, and that’s where the Boltzmann factor $\exp (-\Delta U/{{k}_{B}}T)$ comes in. (There is no guarantee that the colliding molecules will stick, even if they have enough energy.) The general principle of detailed balance can be proven in terms of QM.

The free energy $\Delta F$ includes $-T\Delta S$, which you may calculate using the semi-classical Sackur-Tetrode equation. The bottom line will indeed be either ${{K}_{d}}=\exp (-\Delta F/{{k}_{B}}T)$ or $\exp (-\Delta F/RT)$, depending on whether you state $\Delta F$ per molecule or per mole. For a $1\to 2$ body dissociation, the equilibrium constant will turn out $\propto {{h}^{-3}}$.

Don’t be fooled by the name gas constant. The value of R was first measured in dilute gases, but it’s not specific to gases. It is nothing but the universal conversion factor between temperature and energy per mole.

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An equilibrium state is an equilibrium state. The system does not care which properties we are "holding" constant and which properties are "floating" at constant values in response. You can use the regular formula with $\Delta G$ at equilibrium even when $V$ is held constant instead of $P$.

If you know $\Delta F$ instead of $\Delta G$, then under your constant-volume constraint you can say

\begin{align} \Delta G &= \Delta (F +PV) \\ &= \Delta F + V \Delta P \end{align} so to proceed given only $\Delta F$, you'd need to calculate $\Delta P$ (the rate at which system pressure changes with reaction progress, at equilibrium at the prescribed $T$ and $V$). You would need some sort of equation(s) of state to do this properly.

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Sorry in advance for my english !

The problem of your derivation with the kinetic constants is that you are making a particular hypothesis about the mechanism of the reaction: simple mechanism with first order in the reactive and in the products. This is rarely true. The thermodynamic relation is more general.

About the Boltzmann constant : the free energy is a molar quantity. You just have to define a "molecular free energy" ${{G}_{\text{molecular}}}=G/N$ with $N$ the Avogadro number. ($R={{k}_{B}}N$)

And it have been said in the comments that for a liquid solution, the dependence of the free Gibbs enegy with pressure can often be neglected. (At constant pressure or at constant volume don't change a lot the conditions).

But even if you want to take into account the change in volume, the precedent answer has also explained that the law of mass action remain valid : there is only one equilibrium condition.

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