Is it correct to say that "Increasing the temperature will always increase the rate of any chemical reaction"?

Many articles on the internet and textbooks say yes but what about biochemical reactions that take place in our body?

To my knowledge at higher temperatures some enzymes (which are catalysts in chemical reactions in the body) become less effective which in turn slows down the reaction rate.

So is it correct to say "any chemical reaction" in the statement or do I have to think of the temperature dependence of the biochemical reactions without the enzymes?

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    $\begingroup$ Many enzymes stop working at higher temperature because the enzymes are proteins, and the structure of the protein breaks down above a certain temperature. The reaction slows down due to the loss of catalyst. Generally, the statement that 'reactions become faster at higher temperatures' indicates the ability of more molecules to overcome the activation barrier at high temperatures, rather than any catalysis. Look into Arrhenius equation. $\endgroup$ – S R Maiti Jul 23 '19 at 9:50
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    $\begingroup$ Also you can switch from spontaneous occurring to non spontaneous. Look at Gibbs free energy You can reasonably say what you refer to at least for on going single reactions. $\endgroup$ – Alchimista Jul 23 '19 at 10:39
  • $\begingroup$ @Alchimista - Spontaneous or non-spontaneous is not an issue when discussing the rate of a defined reaction step. $\endgroup$ – Karsten Theis Jul 23 '19 at 13:05
  • $\begingroup$ @Karsten. Not clear what you mean. Beside the fact that I did not refer to any step but reaction as something going from reactants to products of interest without being coupled to another reaction, again so defined. $\endgroup$ – Alchimista Jul 24 '19 at 9:13
  • $\begingroup$ My meaning was you can stop a reaction, beside enzymes denaturatiom and pot melting :) Good point is arisen by the answer of DivMit. $\endgroup$ – Alchimista Jul 24 '19 at 9:19

Well mostly it does but there can be exceptions.

Eg Consider the reaction $$\ce{2NO + O_2 -> 2NO_2}$$

It has a negative temperature dependence. The rate decrease in increasing the temperature.

There is a simple reason to why this happens. We have a RDS with an intermediate species. The rate determining step must have a positive temperature dependence. But the preceding equilibrium that produces the intermediate species could have a negative temperature dependence , that affects the overall reaction.

The Mechanism

$\ce{NO + O2 <=> ONOO}$ Fast but reversible

$\ce{ONOO + NO -> 2NO2}$ Rate determining step

If Step 1 is exothermic, as a bond is being formed between the reactants. This has a greater impact on the overall reaction and by le Chatlier's principle the equilibrium shift backward . We get a negative temperature dependence!

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    $\begingroup$ Good point, I weakened the statement in my answer accordingly. $\endgroup$ – Karsten Theis Jul 23 '19 at 19:17
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    $\begingroup$ Well, all the elementary steps get faster. Because the back reaction gets faster by a larger degree, the overall rate is lower. Nice example! $\endgroup$ – Karsten Theis Jul 23 '19 at 19:25

It is correct to say: "If nothing else changes, an increase in temperature will increase the rate of an elementary step of a chemical reaction".

All kinds of things could go wrong. If the temperature is so high that your reaction vessel melts, spilling a liquid reactant and removing it from a solid reactant, the rate will not increase. If the temperature is so high that one of the reactants undergoes a different reaction (maybe a decomposition), the rate will not increase. If the temperature is so high that your solvent turns into a plasma, the rate will not increase.

What I am saying is that enzymes are not the only aspect of a reaction that might not withstand higher temperature. On the other hand, within the range of temperatures where the enzyme has the proper conformation, the rates do go up with increasing temperature.

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    $\begingroup$ I think this is the perfect answer! $\endgroup$ – Veritas Jul 23 '19 at 18:31
  • $\begingroup$ Since you mention it, can you elaborate on the plasma case? I would think that the dependence is inverse here - a lower temperature leads to less dissociation, so the effective rate of molecule formation increases for low plasma temperatures. In fact I know this to be true in chemical equilibrium, but I've always wondered what the model for the reaction rates is. $\endgroup$ – PoorYorick Jul 23 '19 at 20:14
  • $\begingroup$ @PoorYorick Not the creation of the plasma slows down, the reaction slows down because it needs solvent but now the solvent is gone. I know nothing about plasma reactions, and I used it as an out of this world example, but I saw your question on plasma reaction, alas. $\endgroup$ – Karsten Theis Jul 23 '19 at 20:20

The statement is basically true, but only if you take into account that any reaction has two directions, and goes towards equillibrium.

Two issues still arise

  • temperature changes the point of equillibrium
  • in equillibrium, the reaction rate is, by the very definition of equillibrium, zero

Seeing that, it's better to throw away that overgeneralistic statement. The rate of elementar reaction steps indeed increases with temperature. As long as it can still work, of course, and e.g. your enzyme is still alive.


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