[Miffymycat]

The Equilibrium Law for aA + bB ⇌ cC + dD gives Kc = [C]c [D]d / [A]a x [ B]b at a given temperature.  Kc­ is also equal to the ratio of forward to reverse rate constants kfwd/krev.  The rate expression for the forward and reverse reactions often does not follow the stoichiometric equation, and is of the form: rate = kfwd [A]m[ B]n rather than rate = kfwd[A]a[ B]b ie where m, n often does not equal a, b.  At equilibrium, kfwd[A]a[ B]b = krev[C]c[D]d, so how is it that Kc can equal kfwd/krev … and how can the units of Kc equal those of kfwd / krev?? 

[Miffymycat]

… sorry I meant: At equilibrium kfwd[A]m[ B]n = krev[C]p[D]q ….

[Jorriss]

K= rate forward/ rate backwards only applies to an elementary reaction.

[Miffymycat]

Hmmm, thanks Jorriss, so are you saying that the expression Kc = kfwd/krev only holds true if the rate equation is the same as the stoichiometric equation in both directions?  I had assumed that Kc = kfwd/krev was a universal equality.

For reversible reactions where the rate expressions differ from the stoichiometric equation, is it therefore not meaningful to equate forward and reverse rate expressions at equilibrium, ie the ratio of rate constants has no relation to Kc or anything else?

Thanks again

[ramboacid]

If you "add" two reactions, you would multiply their equilibrium constants together to calculate the equilibrium constant for the sum of the two reactions. Therefore, you would just multiply the equilibrium constants of the elementary steps together to get the equilibrium constant for the sum of the reactions. It is still meaningful, though the caveat is that you must include the rate constants for all the elementary steps, and that the forward rate constant is the product of all the elementary forward rate constants (and same for the reverse reactions).

Also we derive the equilibrium constant from reaction rates by setting the rates of the forward an reverse reactions equal. For example

aA + bB  cC + dD

k_f[A]^a[B ]^b = K_r[C]^c[D]^d

and that can be rearranged into

K_f/K_r = ([C]^c[D]^d)/([A]^a[B ]^b) = K_c

This is how we establish the relationship between the rate constants and the equilibrium constants.

[Miffymycat]

Hi Ramboacid

Thats helpful.  So kf and kr must be the product of each step k.

You say we obtain Kc by setting f and r rates equal: kf[A]a[ B]b = Kr[C]c[D]d

but what if the rate expression is actually different to this stoichiometric form ... doesnt that matter?  Eg first order in A and zero in B?
 

[ramboacid]

The rate expressions need not match up exactly with the equilibrium constant expression because the rate expression only accounts for the steps leading up to and including the rate determining step whereas the equilibrium constant must be accountable for the sum of all the elementary reactions. If there is a zeroth order reaction with respect to one of the reactants, then that implies that that reactant is involved in elementary step(s) after the rate determining step.

[Miffymycat]

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

[ramboacid]

It's very clear you've thought this one out well, Miffymycat
This is the first time I've had to seriously examine this topic in theory, so some of my answers may be a bit weird or misguided, but I'm trying my best.
 
For your first question, the answer will seem like a cheap excuse out of the problem, but according to page 361 of Atkins' and Jones' Chemical Principles: The Quest for Insight, 4th Ed., the equilibrium constant is by definition unitless. This is so because the equilibrium constant technically uses activities instead of concentrations. If a concentration of a substance is termed [J], then the activity is termed [J]/c°, or just the concentration divided by the unit of concentration, thus cancelling out the units. Thus, if a solute is found to be at concentration 0.2 mol/L, then the activity of that solute is (0.2 mol/L)/(1 mol/L) = 0.2. Yes it sounds dumb, but the assumption that the magnitude of the concentration equals the activitity holds true only for ideal solutions. In real solutions, intermolecular forces play a role and may distort the activity of the solute from its concentration. For the dilute solutions commonly used in the lab, it is usually safe to assume that the magnitude of the concentration equals the activitity.

As for the second problem, page 540 of Atkin's and Jone's also says that if one of the reactants is in excess (as in maybe 100 times the concentration of the other reactants), then the reaction becomes a pseudo-(order minus 1) order reaction in which variations in the concentration of the excess reactant fail to effect the reaction rate, just by virtue of the fact that there's so much of it adding more doesn't help. I know you said that water "is just a reactant not reactant-solvent in excess," but that's all i can think of at the moment. I'd assume the unitlessness of activities also affects how rate constants affect equilibrium constants.

[Miffymycat]

OK thank very much and points noted.  I am going to think about it and consult some folk I know and if I come up with anything I will put it in this thread.  Thanks again!

[Miffymycat]

The simple answer is that whatever the rate expression for the forward reaction and hence kf, the reverse rate expression will always be such that kf/kr equals Kc!