0
$\begingroup$

I'm a biology and/or math person, not a chemistry person. I have only taken the standard sequence of undergraduate courses up to organic chemistry, and that was some time ago, so please excuse (and correct) any mistakes that I make in my terminology.


Note: For simplicity's sake, assume that all reactions occur in a fluid, and that if any solid particles form during a reaction, they are small enough (guess: <~300nm, <~2 MDa for liquid, don't know for gas) that they do not effect the fluid nature of the solution.


The typical means of expressing reactions in elementary chemistry is a chemical equation

$$\ce{c_{X_1}X_1 +\cdots + c_{X_m}X_m <=> c_{Y_1}Y_1 +\cdots+c_{Y_n}Y_n}$$

where $\rm c_Z$ is the stoichiometric coefficient of each chemical species $\rm Z$. The time-independence of such expressions subtly suggests an instantaneous conversion of reactants into products and vice-versa, so that no time passes during the rearrangement of atoms from one chemical species into another. This doesn't actually happen, of course, but for practical purposes the time spent in transition states is so small that it has no effect on the properties of a solution.

Or so I assume. But now that I'm thinking about it, I can't see any reason why the constant conversion of reactants into products back into reactants can't leave a small but non-negligible concentration of intermediates in solution at all times, especially if the products and reactants in question are only slightly more stable than the intermediate.

The presence of such intermediates could be verified by checking that the concentration of products and reactants are simultaneously lower than the expected values at equilibrium. This would indicate that the "missing" mass is bound up in some other state, which, provided that no additional [stable] species are present, can only be that of transition between reactants and products.

To clarify - the intermediates are [probably] not stable and cannot be isolated but are being produced and consumed at the same rate in the forward and reverse reactions, so that the concentration of intermediates is consistently high enough to have an observable effect on the properties of the solution.

So how about it, are there any well known mixtures which exhibit intermediates of this type?

Comment: I would expect that most if not all reactions of this type occur under constant heating.

Additional comment: I find it somewhat ironic that the addition of gratuitous "$\frac\partial{\partial t}$"s doesn't eliminate the "instant reaction" problem.

$\endgroup$
3
  • 2
    $\begingroup$ The time-independence of such expressions subtly suggests an instantaneous conversion of reactants into products and vice-versa Not at all. The reaction equation just describes equilibrium between initial and final state, with any possible time and mechanism to go from one state to the other. $\endgroup$
    – Poutnik
    Jun 17 at 3:26
  • $\begingroup$ Note that it is not an algebraic equation with applicable differentiation. Rather discrete, atom inventory equation with 2 different ways of atoms being combined to molecules or ions. Like 2 sacks with 2 apples and 1 sack with 2 pears combined to 2 sacks, 2 apples and 1 pear each. What to do there with partial derivatives? $\endgroup$
    – Poutnik
    Jun 17 at 4:58
  • 2
    $\begingroup$ What you describe is exactly the basis of the “steady-state approximation” treatment of reaction kinetics. It is well known that many reactions have intermediates that exist at close to constant low concentrations during the reaction $\endgroup$
    – Andrew
    Jun 17 at 22:38

1 Answer 1

4
$\begingroup$

The equation you give and any similar one, does not describe what happens in a reaction other that X1 moles of A react with xB of B to give this and that amount of product which may or may not be in equilibrium with reactants depending on the reaction.

What actually happens depends on the type of reaction, if it is a cis-trans isomerisation or bond breaking ($X_2 \to 2X$) then the 'intermediate' lasts for little more than a vibrational period of the bond breaking say order of 100 femtoseconds. As long as the reactant is present then some some minute fraction is as an 'intermediate', usually this is called a transition state and is fleeting.

If reaction is a bimolecular one, with practically zero activation energy, this will occur as soon as the molecules, say A and B meet, i.e. the rate is controlled by how quickly the molecules diffuse together, and again the 'intermediate' is the collision complex and will last as long as A-B exists, typically no more that nanoseconds. In other cases where there is significant activation energy the collision complex is similar in all respects except that there is not enough energy to overcome the activation barrier and so possibly a million collisions occur before one molecule reacts.

Thus there are no 'intermediates' as such in simple reactions. However, if the reaction is of the form $A\to B\to C $ then species B can be observed as it forms and decays away, you might consider this as an intermediate.

In more complex cases such as an enzyme /substrate, usually the substrate has to diffuse into the enzyme pocket and then reacts as if a bimolecular reaction with an activation energy. Usually there is always some of this complex intermediate present as (ES) for example in a Michaelis-Menten scheme. (Not a great deal is known about the actual mechanism. i.e atom positions vs time, of such reactions as time-resolved x-ray crystallography is in its infancy.)

In gas phase recombination reactions such as recombination of iodine atoms $I\cdot + I\cdot \to I_2$ and gas M is present, such as argon, $I_2$ itself or methane etc. an intermediate forms as a short lived complex with an iodine atoms $I\cdot +M \rightleftharpoons IM$ which has been detected experimentally.

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.