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After thinking about it, I'm confused by what the reaction barrier of an enzyme actually means.

Imagine a reactor containing enzyme and substrate.

If the enzyme in the reactor has a reaction barrier of 15 kcal/mol,

a) then when 1 mole of product has formed, that means 15 kcal have been consumed, independent of how much enzyme is present in the reactor.

b) then when 1 mole of enzyme is present in the reactor and 1 mole of product is formed, 15 kcal have been consumed.

Is it option a), b) or otherwise?

Let us imagine two companies A and B. Both companies use very similar technical equipment to carry out a biotechnological process where a chemical reaction is catalyzed by an enzyme. Company A uses an enzyme with a reaction barrier of, say, 15 kcal/mol, while company B uses an enzyme to catalyze the same reaction but this enzyme has an activation energy of only, say 12 kcal/mol. For every Mole of product, company B saves 3 kcal worth of energy needed to drive the factory.

Does that make sense?

TMOTTM
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  • While a lower activation energy is beneficial, I don't think a difference in activation energy of 3kcal/mol will translate into 3kcal/mol energy savings. It all depends on what requirements you have for the kinetics (reaction speed) and product yield. If you want faster kinetics you will have to make more energy available, but this may also change the equilibrium conditions. When optimizing reaction conditions, you will likely have to make tradeoffs involving reaction kinetics, catalyst amount, catalyst price and reactor temperature, depending on the specific reaction and the reactor design. – jarlemag Apr 17 '14 at 19:12
  • Let's think of a first order reaction in terms of rates. Say enzyme 1 has a rate of 1000$s^{-1}$ and enzyme 2 has a rate of 2000$s^{-1}$, it's clear that enzyme 2 will only require half the time (at the same temperature) to produce 1 mole of product. Do you agree? So I only need to heat the reaction volume for half the time as I would when using enzyme 1. – TMOTTM Apr 17 '14 at 19:16
  • Yes, I agree, an enzyme giving a lower activation energy will generally give a higher reaction rate, and that will definitely affect the economics of the operation. – jarlemag Apr 17 '14 at 19:18
  • I guess then the most reliable way to do this would be to calculate the time required to produce 1 Mole of product (given the rate constants) and then calculate back to the activation free energy differences. From this also I guess one can say that the unit "kcal/mol" indeed refers to 1 Mole of product. – TMOTTM Apr 17 '14 at 19:39
  • I'm not sure I understand you correctly. Could you perhaps give a calculation example? – jarlemag Apr 17 '14 at 20:18
  • I mean just by using $k(T) = \gamma(T)\frac{k_\text{B}T}{h}e^{-\Delta G^{TS}(T)/RT}$ (see here http://www.sci.ccny.cuny.edu/~gunner/Pages-422/PDF/section.1/catalysis/cat.truhlar.pdf) and from this getting back $\Delta G$. – TMOTTM Apr 17 '14 at 23:10
  • If you already know the rate constant at a given temperature and all the other factors then yes, you can calculate the activation energy. I don't see why you would need to do that though? If you already know the rate constant, you don't really need to know the activation energy to calculate your energy needs, unless you want to use it to calculate the rate constant under different conditions. – jarlemag Apr 18 '14 at 15:42
  • Sure, the rate constant already is very informative. Still if you can assist me in this initial question: If all is equal between reactors A and B, except the activation energies, then after having produced 1 Mole of product, one reactor required dE less energy to arrive at 1 Mole of product where dE is the difference in activation energy. Do you agree? I'm assuming both reactors require the same amount of energy to be heated up to the same temperature. – TMOTTM Apr 19 '14 at 21:44
  • No, and that's what I've been getting at. As mentioned, the activation energy is only "borrowed". To find out how much energy you need to input or remove from the reactor to keep the temperature steady, you need to look at the heat of reaction. See the difference in height of the line at "reactants" and at "products" in the figure in the answer by Chris? That's what matters. If you have further questions on this, I think you may find people on the Chemistry StackExchange who can explain better. – jarlemag Apr 19 '14 at 21:57

3 Answers3

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The reaction barrier (also termed activation energy) is the energy that is needed for the reaction to take place. An enzyme is "only" a catalysator which lowers the necessary energy (and/or makes the reaction possible under the conditions) but some energy to start a chemical reaction is still needed. See this image (from here):

enter image description here

To your questions: A and B are essentially the same - but the reaction time necessary with small amounts of enzyme is much longer than when 1 mole of enzyme is present. The amount of energy needed to catalyze the production of 1 mole of product stays the same.

Chris
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The activation energy for a reaction is given for a given extent of reaction, typically per mole of product formed. In general, the amount of enzyme present in the reactor is irrelevant with respect to the activation energy. So option a) is closest to correct. However, note that the activation energy does not tell you how much energy is consumed taken up during the reaction, only how much energy is needed for the activation energy to be exceeded and the reaction to proceed. The kinetics of a reaction, is a separate issue from the thermodynamics of the reaction, which determines the equilibrium concentrations of substrate and product. What the enzyme does is lowering the activation energy, thus speeding up the kinetics of the reaction and allowing equilibrium to be reached. In practice, the activation energy in the absence of the enzyme may be so high that the reaction would not proceed at all, and thus equilibrium would not be reached.

jarlemag
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  • I agree, the activation energy would give me the amount of energy consumed during catalysis alone. Heating up the mass would be on top of that. But one can imagine that this component remains constant for different enzymes. – TMOTTM Apr 17 '14 at 18:47
  • I think there might (still) be a misunderstanding here? The heat of reaction doesn't come on top of the activation energy - they are two separate things - and a catalyzed reaction does not consume or produce more energy than an uncatalyzed reaction. Think of it like this: The activation energy is the maximum energy that is "borrowed" when the reaction happens. – jarlemag Apr 17 '14 at 18:50
  • I might have been unclear. I know that $\Delta_f H$ is an indication of how spontaneous the reaction is. With regard to the word "consume": I respectfully disagree, because to form 1 mole of product, an uncatalyzed reaction does require more energy input. – TMOTTM Apr 17 '14 at 18:57
  • Indeed, you need to have more energy available, but the energy you spend on reaching the activation energy you also get back, so I feel "consume" is the wrong word to use. I don't see much point in arguing about that though - thanks for clarifying. – jarlemag Apr 17 '14 at 19:02
  • Let me put it differently, please see edit of original post. – TMOTTM Apr 17 '14 at 19:03
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To better understand the activation energy concept my teacher had given us an excellent example. Suppose you are driving a car and want to cross a mountain. You could either go to the mountain summit and then come down or you could directly cross it through a tunnel. What enzyme does is to construct a tunnel for the reaction to proceed.

biogirl
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  • I dont think I have a problem with understanding the concept of a potential energy surface, in fact I would be a bit careful with that analogy. But it is a good starting point. Rather than a tunnel, I would say the enzyme finds a rout which doesn't reach that high up as the summit. Speaking of tunnels can be misleading when talking about enzymes, so one has to be careful about the terms one uses. – TMOTTM Apr 17 '14 at 18:37
  • @TMOTTM yes I agree. Tunnel is a bit too much. – biogirl Apr 17 '14 at 18:38