That’s really helpful – you clearly have a good grip on this topic!
Can I finally put an example forward to illustrate your explanation – eg equimolar tertiary haloalkane hydrolysis (first order reaction) which I will assume has some degree of reversibility, carried out in a dipolar aprotic solvent eg propanone or DMF (ie water is just a reactant not reactant-solvent in excess):
RX + H2O
ROH + HX and so Kc = [ROH] [HX] / [RX] [H2O] ; Kc has no units
Possible forward mechanism:
RX
R+ + X- slow k1 (s-1)
R+ + H2O
ROH + X- fast k2 (mol-1 dm3)
H+ + X-
HX fast k3 (mol-1 dm3)
So k (forward) = k1 x k2 x k3 (mol-2 dm-6 s-1)
Possible reverse mechanism:
ROH
R+ + OH- slow k4 (s-1)
OH- + HBr
H2O + Br- fast k5 (mol-1 dm3 s-1)
R+ + Br-
RBr fast k6(mol-1 dm3 s-1)
And k(reverse) = k4 x k5 x k6 (mol-2 dm6 s-1)
Observed forward rate expression:
Rate = k(forward)[RX] k(forward) in s-1
Ist problem … the units of k(forward) and k1¬ x k2 x k3 disagree …
2nd issue … in order for the units of Kc and k(forward)/k(reverse) to agree, this suggests the reverse elementary steps must have the same overall combined molecularity as in this example, in order to cancel out to give no units? Seems weird as another possible mechanism could be:
ROH + HBr
R+ H2O + Br-
R+ + Br-
RBr
… which clearly gives different units for k(forward) and so Kc ≠ k(forward)/k(reverse). Any thoughts?
Units aside, my original problem, applied to this example (assuming it’s reversible) was
Rate = k(forward)(RX]
Kc = [ROH] [HX] / [RX] [H2O]
So at equilibrium, for Kc to equal k(forward)/k(reverse), this means that
Rate = k(reverse) [ROH] [HX] [H2O]-1 which seems very strange if not meaningless or impossible!!
Very grateful for your considered thoughts in due course