Why does enzyme catalysis seem to violate path independence of free energy?

[samy4408]

Hello !
We all now that enzymes lower the activation energy by creating both covalent bonds and week non covalent bonds , in this book , about how the enzyme lower the activation energy using covalent bonds they said :

"Covalent interactions between
enzymes and substrates lower the activation energy (and thereby accelerate
the reaction) by providing an alternative, lower-energy reaction path"

this sentence is not compatible with a basic principle that i learned in chemistry that says "the free energy is independent of the path "
can anyone show me what is wrong with my understanding ? thanks !

 

[mjc123]

Consider two villages in neighbouring valleys. Imagine two paths between those villages, one over a mountain peak, one through a pass between the mountains. The difference in height between the start and finish points is the same for each path, but the height you have to climb on the way is not.

 

[Godot_]

The enzyme doesn't change the total energy difference between the two states, i.e. the free energy stays the same. It's merely a catalyst, providing that lower-energy reaction path.

It just lowers the activation energy's "potential hill" that has to be overcome to switch between the two states.

 

[Ygggdrasil]

Here's a nice diagram illustrating the point:

image source: https://socratic.org/questions/what-are-activation-energies

As others in the thread have said, the free energy difference between the reactants and products is still the same. However, the path between the two needs a higher activation energy without the enzyme whereas the path for the enzyme catalyzed reaction requires a smaller activation energy.

 

[samy4408]

Consider two villages in neighbouring valleys. Imagine two paths between those villages, one over a mountain peak, one through a pass between the mountains. The difference in height between the start and finish points is the same for each path, but the height you have to climb on the way is not.

Thanks a lot !