in normal uncatalyst equilibrium, kforward*[S]=kreverse*[P]
if must further moved because at that instance k forward>>k forward before, but k reverse doesn't change, and [S] and [P] doesn't change
k forward increase, but k reverse increases under same degee
1 biggest doubt: how does the free energy changes overcomed?
ΔGo and RT lnQ seems not changed during equilibrium situation?
yes, at least it can't explained from this equation
So the ΔGo couldn't be used, since it's measured not adding any those enzymes their
dynamic equilibrium through enzyme
it is assumed absolutely irreversible, but actually it's not true!, a "double reaction coordinate" from forward and reverse is not exist。。。
The reverse is neglected under our steady state kinetics model but do exist
neglect because the second step's ΔG is too negative
and it's do faster than normal(catalysted)
so equilibrium never changed by any catalyst (only lower activation energy)
first law of thermal dynamics: you can't give energy from nothing
answer another similar question
why we need kinase/ phosphatase for one process
Because they are not reverse reactions。。
phosphorylation most
substrate level
ATP+moleclue---P-molecule + ADP
oxidative
ADP+pi----ATP
formula it self is crazy not spontaneous
spontaneous because mechanical energy invested
It's not chemically easily explainable
so you know why.. they are not essentially reverse processes
equilibria is determined inextricably by standard free energy change
alternative： not dynamic equilibrium through enzyme, but static equilibrium, enzyme never turned because its own machenical equilibrium
what ever which one is true, the thermal dynamic laws can't be violated
infront of laws, impossible is impossible
