Trying to elucidate a reaction pathway, I calculated the energies for different transition states possible for that reaction (computational chem. DFT). I found activation free energies ranging from 29 to 39 kcal/mol, and wanted to understand if all those transition states are possible in the reaction conditions that were used in previous papers. That is, given that this reaction is known to occur at around 355 K, is it possible to overcome those free energy barriers of 29 to 39 kcal/mol?

I know I can calculate the rate constant for a given temperature using the Eyring equation but I'm not sure how to relate the rate constant to the reaction "feasibility".


2 Answers 2


You could convert the rate constant($k$) to half-life($t_{1/2}$) which would give you an idea of the time scale required for the reaction to finish at a certain temperature.

The equation to obtain half-life from $k$ is different depending on the order of the reaction. For an unimolecular reaction($\ce{A->P}$), it is: $$t_{1/2}=\frac{\text{ln 2}}{k}$$

For a bimolecular reaction ($\ce{2A->P}$), $$t_{1/2}=\frac{1}{k[\text{A}]_0}$$ where $[\text{A}]_0$ is the concentration of A at the beginning of the reaction. See this page for more.

The half-life($t_{1/2}$) is defined as the time required for the initial concentration of the reactant to be reduced to half of its initial value.

The following table is from Shriver and Atkins' Inorganic Chemistry, and it shows the representative time scales for various ligand exchange reactions in metal complexes. (You can get an idea of the feasibility of the reaction from the timescales, and that is applicable to all types of reactions)

timescales for various chemical and physical processes

Anything with a half-life of less than a minute is fast. Half-lives in the range of minutes to hours are moderately fast reactions. Hours to days is slow. Reactions with half-lives more than weeks is intractable (without catalyst/heating).


  1. Shriver and Atkins' Inorganic Chemistry, P. Atkins, T. Overton, J. Rourke, M. Weller, F. Armstrong, M. Hagerman, 5th ed., 2010, W.H. Freeman and Company, p509

From the Eyring equation, we can simply calculate the $k$ value for it.

\begin{align} k &= \frac{k_\mathrm{b} T}{h}\exp\left(\frac{-\Delta G^\ddagger}{RT}\right)\\ k_\mathrm{b} &= \pu{1.38E-9 J K^-1}\\ T &= \pu{355 K}\\ h &= \pu{6.626E-34 J s}\\ R &= \pu{8.3145 J K^-1 mol^-1} \end{align}

Here, I am assuming the average of $\pu{29 kcal/mol}$ and $\pu{39 kcal/mol}$ to be $\Delta G^\ddagger = \pu{34 kcal mol^-1} = \pu{142256 J mol^-1}$

If we plug in the values into the equation, we get a $k$ value equal to about $\pu{8.64 x 10^-9 s^-1}$ which is quite small.

If we take for example a Gibbs free energy of activation of $\pu{23 kcal mol^-1}$, that would equate to a half-life of 2 hours at room temperature based on $t_{1/2} = \frac{\ln2}{k}$ for first-ordered reactions. If we use the same formula for the $k$ we calculated at $\pu{355 K}$ for $\Delta G^\ddagger = \pu{34 kcal mol^-1}$, we get a half-life of $\pu{8.02 x 10^7 s}$ which is about $2.54$ years. Therefore, the reaction is likely not going to proceed quickly at $\Delta G^\ddagger = \pu{34 kcal mol^-1}$ and can be considered unfeasible as the reaction goes extremely slowly.

If we instead take the lower end with $\Delta G^\ddagger \pu{= 29 kcal mol^-1}$, the half-life is less than a day at $\pu{355 K}$ which is reasonable. However, anything after $\Delta G^\ddagger = \pu{30 kcal mol^-1}$ are essentially unfeasible at $\pu{355 K}$. Even with $\Delta G^\ddagger = \pu{30 kcal mol^-1}$ at $\pu{355 K}$, the half-life is over 3 days and it will only increase exponentially as $\Delta G^\ddagger$ increases.

Note that $\Delta G^\ddagger$ is temperature dependant and the calculations above are done assuming that $\Delta G^\ddagger$ is constant at the given temperatures. Therefore, the theoretical values I calculated may not perfectly match the experimental results of the reaction pathway you are studying. However, it should hopefully give a general idea of the rate and the feasibility of reactions for the range of $\Delta G^\ddagger$ values you derived.

  • $\begingroup$ G_a is temperature dependent; it'll rise with rising temperature. Other reactions possibly take over in the meantime. Anything > 100 kJ/mol and STP needs a bit more persuasion. As for DFT, without calibration, up to 80 kJ/mol gives you a guesstimate but nothing more. $\endgroup$ Commented Jul 7, 2021 at 23:56
  • $\begingroup$ @M.L. Note that you are actually dealing with the change in the Gibbs free energy ($\Delta G$) not the absolute value of Gibbs free energy ($G$) here. $\endgroup$
    – S R Maiti
    Commented Jul 8, 2021 at 2:56
  • $\begingroup$ @Martin-マーチン Yes, I forgot to take that into account. I'm not an expert in this area but what do you think I could say to fix my answer? Should I say that we assume delta G is independent of temperature? $\endgroup$
    – M.L
    Commented Jul 8, 2021 at 6:03
  • $\begingroup$ I've linked a question (above) that covers some more ground, have a look at that maybe it inspires you to improve/include some points. I think making sure that a reader will see the approximate nature of what you have written is important. The numbers at the end are just too tempting if taken at face value. Other than that, I think your answer is still fine; especially the part with the "hundred year" half-life. $\endgroup$ Commented Jul 9, 2021 at 15:39

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