We know that an activation complex is the temporary unstable compound formed when activation energy is reached. But what dictates whether the products are indeed going to be products or if they become reactants?

Does the activation complex form when orientation is both correct and incorrect, and only forms products when the orientation is correct? Or is the decomposition of the complex purely a chance matter?

  • $\begingroup$ As far as I know, it is a matter of chance in which direction the decomposition of the activated complex goes since it symbolizes a maximum of the free energy and so a structure alteration in any direction leads to a lowering of the free energy. Maybe this answer of mine might be interesting for you. $\endgroup$
    – Philipp
    Commented Mar 3, 2014 at 18:33

3 Answers 3


Transition state theory or activated complex theory is a very advanced topic and deals with chemical reaction dynamics. It goes beyond the scope of my knowledge, however, I will attempt my best to explain with what I know.

As stated, the activated complex is a temporary unstable product. As two particles with sufficient kinetic energy collide; along the reaction path, they distort the bonds and rearrange into new atoms. The location of the transition state for the activated complex is at the highest point; or peak, of an potential energy diagram.

As stated in my general chem textbook Principles of Modern Chemistry by Oxtoby et. al, p. 685:

The activated complex is assumed to exist as if it were in equilibrium with the reactants and the theory focuses on calculating the rate at which the activated complex passes through the transition state to form products.

Statistical thermodynamics is required to calculate all motions of particles, including both rotational and vibrational energy.

In the transition state, change from products is mediated when bonds between particles or molecules become out of phase (or antisymmetrically stretched) by vibrational energies. The frequency of the molecules becoming out of phase with one another decreases as the reaction proceeds forward and the new product is transformed. The energy transferred from vibrational energy to achieve the activated complex is turned into kinetic energy.

The Eyring formula makes it possible to calculate reaction rates using transition state theory:

$$ k_r = \kappa \frac{K_\mathrm{B}T}{h}K^†$$

The kappa term is referred to as the transmission coefficent and refers to the probability that the particles will create the transition state. $K_\mathrm{B}T/h$ measures the rate at which the activated complex dissociates to form into it's products.

This equation is unwieldy to us; as you stated previously, it is extremely hard to determine equilibrium constants for the activated complex since the instability achieved at the height of the potential energy for the reaction exists for only a brief period of time.

Instead, we can relate the Eyring formula to thermodynamics and substitute $\exp (-\Delta G^†/RT)$ for $K^†$.

$$ k_r = \kappa \frac{K_\mathrm{B}T}{h}\exp (-\Delta G^†/RT)$$ $$ k_r = \kappa \frac{K_\mathrm{B}T}{h}\exp (-\Delta H^†/RT)\exp(\Delta S^†/R) $$

The equation that Arrhenius derived is:

$$k_r = A\exp (-E_a/RT)$$

Activation energy is expressed as activation enthalpy and activation entropy. The pre-exponential factor $A$ is substituted with $K_\mathrm{B}T/h$. This equation and the use of enthalpy and entropy as driving forces is used extensively in enzyme catalyzed reactions.


Collision with another molecule is purely chance matter but the orientation factor will be determined by the potential energy surfaces. If there is not a proper orientation then the it will be a surface with high energy surface but if there is a proper orientation then it will pass through the surface with lowest energy and that will dictate whether products are formed or the not


My answer may very well be reciprocated as a comment. Because I can't do that myself, I'll relay this as an answer.

The formation of the relevant activated complexes occurs only when the orientations are correct. This, though is not the case with its decomposition. According to the Hammond Postulate, the activated complexes take structures closer to the species (reactants or products) that have energy values closer to it. This way one can explain that in exothermic reactions the formation of transition state and the activated complexes is easier than in endothermic reactions. This allows for faster forward movement in reactions. The, activated complexes, thus, will now tend to move towards the state that bring it more stability. So yes, it is indeed a mater of pure chance on which side the activated complexes shift, but it is not a matter of fair chance. As such, addition or removal of heat (accordingly if it is endothermic or exothermic) helps to quicken the formation of the activated complex and pushing it towards the more stable species.

Example wise, consider the free radical bromination and the free radical chlorination of ethane. Here, the first is endothermic while the latter is exothermic, both product formation and reactant formation is easier in chlorination. But product formation is what happens in the greater proportion.


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