3
$\begingroup$

It is often claimed that a (normal) catalyst catalyzes the forward and the backward reaction to the same extent, so the equilibrium position is not shifted.

I would like to ask why must the forward reaction and the backward reaction follow the same pathway?

I could imagine a mechanism for this reversible reaction from $A$ to $D$:

$$\ce{A <=> D}$$ $$\ce{A -> B<=> D}$$ $$\ce{A <=> C <- D}$$

So that the forward reaction and the backward reaction must follow different pathways, as one step of both pathways are irreversible.

Is this scenario impossible in reality? If so, why? If not, is there any example in real life?

$\endgroup$
  • 1
    $\begingroup$ So you think if you remove the catalyst for the B-D step, you'd have an endless energy source? Fat chance. ;-) There is no such thing as an "irreversible" reaction, especially not two that are in an equillibrium. $\endgroup$ – Karl Sep 4 '16 at 12:44
0
$\begingroup$

A catalyst cannot change the equillibrium of a two component reaction, because that'd always violate the principle of conservation of energy.

Also there is no such thing as an irreversible reaction. If the backward reaction is very unfavourable energy-wise, the catalyst won't change that, but you cannot have two highly exothermic reactions in your funny circle anyway. Where should the energy come from?

|improve this answer|||||
$\endgroup$
  • 1
    $\begingroup$ Thank you. I did not know that there is no such thing as irreversible reaction until I saw this page. $\endgroup$ – DHMO Sep 4 '16 at 13:54
2
$\begingroup$

A reaction could proceed by many different pathways, for example an extreme way would be splitting the reactants into atoms and combining them to form products, but this is essentially impossible, the energy required would be huge and how would the atoms reassemble just into the product required?

In any reaction the molecules follow the lowest energy pathway, which is to say that the lowest activation barrier between reactants and products will be crossed in preference to any other pathway. The same argument applies to a catalysed as to a 'normal' reaction even though a catalysed reaction proceeds by a different mechanism. The lowest energy pathway also ensures in some general way that close to minimum number of bonds are broken in the reaction.

The reason for this is that the activation energy (barrier height) is usually far higher than average thermal energy and many collisions happen between reactants before by chance there is enough energy to cross the barrier. The probability $p$ of having sufficient energy $E$ is given by the Boltzmann exponential distribution, $p \approx exp(-E/(RT))$. This is a rapidly falling distribution as $E$ increases thus the minimum energy pathway is generally hugely favoured over others.

In biochemical reactions it is possible for the same molecules to be synthesised by different pathways, involving several steps, but the same arguments apply to each stage in the reaction within the enzymes used.
Synthetic chemists will have many examples of the same products being formed from the same starting molecules and progressing by different sequences of pathways in a series of $separate$ reactions. However, the yields will be generally be different; one of the aims of synthetic chemistry is often to optimise yields, but I take the question to ask that the reaction is not that of a separate sequence of steps but in one process.

|improve this answer|||||
$\endgroup$

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.