# How exactly is activation energy defined?

In a common interpretation of the Arrhenius rate equation

$$k = A\exp\left(-\frac{E_\mathrm a}{RT}\right),$$

the activation energy $E_\mathrm a$ is understood to represent the difference in the energy of the reactants and the minimum energy required to form the activated complex or the transition state, as shown in the following diagram

But this must be wrong! As $E_\mathrm a$ is the minimum energy itself which the molecules must possess to reach the activated complex. That is, it is not the difference but the minimum energy level itself. This is the only way in which it will remain constant i.e. independent of temperature (otherwise with increase in $T$, $E_\mathrm a$ will go down) and can be used in the Arrhenius Equation where the exponential term denotes the fraction of molecules having energy higher than or equal to $E_\mathrm a$ according to Maxwell-Boltzmann distribution. Sometimes, threshold energy ($E_\mathrm t$) is defined as the minimum energy required for reaction to proceed and $E_\mathrm a = E_\mathrm t-\mathrm{KE_{average}}$. If this definition is true, then threshold energy and not activation energy should be in the Arrhenius equation. I therefore need to clarify the actual definition of activation energy $E_\mathrm a$.

• You could say Ea is the minimum energy the molecules must possess RELATIVE to the initial state. I've only seen it defined in this non-absolute fashion. – Brian Oct 13 '13 at 6:59
• @Brian But that makes $E_a$ temperature dependent since the initial state is temperature dependent. And how does that hold up against the intuition behind Arrhenius equation, the exponential factor being the fraction of molecules having energy greater than or equal to that required for reaction(Threshold energy)? – stochastic13 Oct 13 '13 at 10:03
• This is simply from wikipedia: Given the small temperature range kinetic studies occur in, it is reasonable to approximate the activation energy as being independent of the temperature. en.wikipedia.org/wiki/Arrhenius_equation – Brian Oct 13 '13 at 12:19
• Can you clarify a little your question? Others may fully understand, but for me it is not clear why you think one or other definition is wrong (or contradicting). The statement "it is not the difference but the minimum energy level itself" makes no sense to me. Also, it is not clear if you are talking about a single potential energy surface or about an ensemble of molecules - I suspect it is not clear to you, neither. – Greg Feb 5 '18 at 23:41

You are right that the energy of the initial state is a function of temperature. But in practice for this equation the energy difference $$E_\mathrm a$$ usually is fixed for an arbitrary temperature (in the middle of the diapason of experimental measurements) and assumed temperature-independent.

While this equation suggests that the activation energy is dependent on temperature, in regimes in which the Arrhenius equation is valid this is cancelled by the temperature dependence of $$k$$. Thus, $$E_\mathrm a$$ can be evaluated from the reaction rate coefficient at any temperature (within the validity of the Arrhenius equation).

As to the physical meaning of activation energy, citing Wikipedia:

Both the Arrhenius activation energy and the rate constant $$k$$ are experimentally determined, and represent macroscopic reaction-specific parameters that are not simply related to threshold energies and the success of individual collisions at the molecular level.

In other words, activation energy $$E_\mathrm a$$ is a semi-empirical quantity and is not considered to have a strictly defined physical meaning (at least on molecular level). Note that the Arrhenius equation

$$k=A\exp\left(-\frac{E_\mathrm a}{RT}\right)$$

is itself considered to be a semi-empirical relationship.

In the IUPAC Gold Book (DOI: 10.1351/goldbook.A00102), activation energy is also defined to be an empirically determined quantity, in line with the above:

An empirical parameter characterizing the exponential temperature dependence of the rate coefficient $$k$$:

$$E_\mathrm a = RT^2 \frac{\mathrm d (\ln k)}{\mathrm dT}$$