In my reaction engineering textbook (Levenspiel), it is stated that reactions with high activation energies are very temperature sensitive, while reactions with low activation energies are relatively temperature-insensitive. This principle was then applied to a homework problem of ours where for two reaction pathways in competition, at high temperature, the reaction rate of the higher activation energy pathway would dominate, with the opposite being true at low temperatures.

Could someone explain why the high activation pathway dominates at high temperatures? In my mind, I still expect the rate of the low activation energy pathway to dominate due to path of least resistance principle.

  • $\begingroup$ Were you given a graphic along with that question? If so, attaching a picture to your question would give a good starting point for discussion, though this is not absolutely necessary. $\endgroup$
    – airhuff
    Commented Feb 19, 2017 at 19:52
  • 1
    $\begingroup$ You are correct, the higher activation energy reaction will never dominate, it will only become more competitive at higher temperatures. $\endgroup$
    – ron
    Commented Feb 19, 2017 at 21:50

1 Answer 1


In chemicals reactions that have more than one product, being one of them the kinetic product and the other the thermodynamic product, the temperature dependency is a relevant factor. For example, the kinetic vs. thermodynamic control can manipulate the regiochemistry of the alkylation of enolates. In other words, you have a competition between reactivity and stability.

Reactions are usefully to describe it in terms of potential energy diagrams, $\Delta G$ vs. reaction coordinate ($\xi$), where a one-step reaction have a global maximum corresponding to a transition state leading to the product and the free energy of this transition state, relative to the reactant, determine only the rate of reaction. The enthalpy part of $\Delta G$ deal with the stability of the molecule or intermediate but this doesn't give information about the rate of reaction.

Absolute rate theory (or transition state theory), by Eyring, have an expression for the rate constant $k$:

$$k=\frac{\kappa k_{B}T}{h}e^{\Delta S^{\ddagger}/R}e^{-\Delta H^{\ddagger}/RT}$$

where $\kappa$ is the transmission coefficient, $k_{B}$ the Boltzmann's constant, $h$ the Planck's constant, $R$ is the gas constant and $T$ is the absolute temperature. Here, the term $\frac{\kappa k_{B}T}{h}e^{\Delta S^{\ddagger}/R}$ varies slightly with temperature compared to the enthalpic term. Arrhenius equation also reflects the temperature dependence of reactions: $$k=Ae^{-E_{a}/RT}$$ For remark, both relationships are temperature dependent.

So, if you have a reaction under kinetic control, the ratio of the products is determined by the relative energies of the transitions states. Whereas under thermodynamic control, the ratio of the products is determined by the relative energies of the products. This leads to the question if there are any relationship between $\Delta G$ or $\Delta G^{\ddagger}$, where you might search about Bell-Evans-Polanyi formula, Hammond's postulate and the Marcus equation.

Suppose you have a reaction $$\text{A}\leftarrow\text{R}\rightarrow\text{B}$$ with the energy profile: enter image description here The product A have a lower $E_{a}$ or $\Delta G^{\ddagger}$ (kinetic product), whereas B have a higher $E_{a}$ but a lower $\Delta G$. The potential energy profile generally is temperature independent, i.e. the barrier and the heat of reaction are determined by the molecular structure. When you raise the temperature, the kinetic energy of the molecules is raised and this helps the molecules to pass the potential energy barrier more easily (you can check it by a Maxwell-Boltzmann distribution for a molecule changing the temperature and fixing an energy corresponding to $E_{a}$), being obvious that if the barrier is smaller will be more easy to pass.

If the temperature is high enough to make the molecules pass over all the barriers on the energy profile, the molecules will choose the pathway with the most stable product. All of this is determined by whether the products are or aren't able to equilibrate under the reaction conditions (search for the principle of microscopic reversibility). So, at lowers temperature you will disfavor the equilibrium and only get the kinetic product.


Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge you have read our privacy policy.

Not the answer you're looking for? Browse other questions tagged or ask your own question.