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I intend to do a kinetic study of simple alcohol catalytic dehydrogenation reactions in the gas phase. I want to start with simple power law kinetics using $K_\mathrm{eq}$ to account for the reversibility of the reactions. Would the equation below, knowing the Gibbs free energy of reaction, be valid?

$$Δ_\mathrm{r}G^\circ = -RT\ln K_\mathrm{eq}$$

I came across a paper that used this equation to calculate $K_\mathrm{eq}$ to build a kinetic model of the ethanol-to-butadiene reaction. However, other papers I read, i.e. on the methanol-to-diethyl ether reaction, have used complex equations derived from empirical observations to calculate $K_\mathrm{eq}.$ Why use one approach over the other?

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In general I would say 'no', thermodynamics only tells us about starting and ending points, thus you know the $\Delta G^\text{o}$ for the reaction but nothing about its actual mechanism, (since time does not come into thermodynamics). As $K_\mathrm{eq}$ is a ratio of rate constants these can take on many different values and still have the same ratio. To study the mechanism you will need to monitor species and intermediates vs time as the reaction proceeds.

Before doing any experiments you can only make informed guesses as to what the mechanism might be, even apparently simple reactions, such as between hydrogen and oxygen, is not simple but by experiment found to be a chain reaction.

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    $\begingroup$ May I ask you for the reason for reverting my edit? \circ in ΔG^\circ is not a typo, it's one of the two standardized ways of typesetting standard state. \Delta G^\text{o} is semantically wrong and its usage can only be justified if there is no technical means to input proper symbols. $\endgroup$ – andselisk♦ Mar 5 '20 at 14:47
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    $\begingroup$ Nothing in particular you can change it back if you wish. Whatever the standard way is, using \circ in the font used makes a rather small symbol that does seem awkward when reading. $\endgroup$ – porphyrin Mar 6 '20 at 11:31
  • $\begingroup$ I don't see how it would be incorrect to do what the asker posted, assuming the reaction is at the standard temperature. By definition, it would yield exactly Keq. The real issue at hand is that Keq is not a good metric of the kinetics of the system. $\endgroup$ – jezzo Mar 11 '20 at 17:00
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At equilibrium, the forward and reverse rates have to be equal. For mass action kinetics, this tells you that the reverse rate constant must be equal to the forward rate constant divided by the equilibrium constant.

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