# Can energy savings be estimated from activation energy differences?

Imagine two companies X and Y. Both companies use very similar technical equipment to carry out a biotechnological process where a chemical reaction is catalyzed by an enzyme. Company X uses an enzyme with a reaction barrier of, say, 15 kcal/mol, while company Y uses a slightly different enzyme to catalyze the same reaction but this enzyme has an activation energy of only 12 kcal/mol. The reaction energies in the two reactors are, obviously, the same since both companies do $\ce{A} \rightarrow \ce{B}$. Still, as I see it, company X has to input 3 kcal per mole of product more into activating the enzyme.

Does company Y therefore require 3 kcal less energy to produce 1 mole of product?

The statement of this questions mixes a number of issues, so there is, in my opinion, no definite answer to this question.

First of all, one must take care that activation energies will have an influence on the kinetics (rate constant) of the reaction that is catalyzed. In a very general point of view, the overall energetic requirement for a chemical reaction does not directly depend on the activation energy (for instance the equilibrium constant is not changed).

Thus if two separate processes are considered, a different activation energy will mostly mean a reaction rate that will be slower, so overall a process that might take longer but not necessarily need a higher energy input.

Now if one would add to this problem a constraint that would be that the reaction rates of bith processes should be similar, the main method to achieve such a result would be to change the temperature at which the process takes place. This is related to the activation energy through the Arrhenius equation:

$$k = A e^{-\frac{E_{a}}{RT}}$$

If one assumes that the pre-exponential factor is the same for both processes (in effect meaning that the transition states in the catalyzed reaction have similar constraints) then one can reach some conclusions on both processes.

For instance, the reaction rates at constant temperature can be calculated. At 298 K, 3 kcal/mol difference would mean a ratio $k_{1}/k_{2}$ of about 150, so the process with lower activation energy would proceed 150 times faster than the other one. An other option would then be to increase the temperature of the slower process. In such a case, the temperature would depend on the ratio of the activation energies (and not the difference) as :

$$T_{2} = \frac{E_{a2}}{E_{a1}} T_{1}$$

If one takes the values of the qustion and considers a process that would take place at 298 K, one would need to heat at 373 K to reach the same rate constant. From this, one could consider the additional energy input required to heat the reaction medium up to this temperature.