4

If we have a multi-step reaction (which involves more than one step to reach the products; a complex reaction, one may say), then is the following equation valid?

$$K_\mathrm{eq} = \frac{K_\mathrm{fwd}}{K_\mathrm{rev}}$$ $K_\mathrm{eq}$ – equilibrium constant of the reaction; $K_\mathrm{fwd}$ – rate constant of the forward reaction; $K_\mathrm{rev}$ – rate constant of the reverse reaction.

I know that it is valid for an elementary or single step reaction, where the expression for the rate involves raising the concentrations of the reactants to their coefficients in the balanced equation.

But is it so for a multi step reaction too? (Where the law of mass action is no longer valid?)

andselisk
  • 37,604
  • 14
  • 131
  • 217
Senthil Arihant
  • 141
  • 1
  • 2
  • 8
  • 2
    The case for non-elementary reactions is interesting. In fact, equilibrium constants arise from thermodynamics, with no reference to kinetics whatsoever. In the case of an elementary reaction, the strategy of equalling the forward and reverse produces the same equation, but why does or why doesn't that strategy extend to non-elementary reactions? Does microscopic reversibility guarantee the kinetics of even a very complex network of reactions generate a simple equilibrium constant between the stating materials and final products? I wonder if anyone can provide an answer touching on that. – Nicolau Saker Neto Jan 04 '18 at 07:37
  • If each 'step' in a sequence of N steps is an equilibrium and all steps are at equilibrium, there are N equilibrium constants and 2 N rate constants. Write out the differential rate law for each step (e.g. for $\ce{A<=>B}$, where $A$ does not appear in any other reaction, $\frac {d[A]}{dt}=-k_1 \cdot [A] + k_2 \cdot [B]$), and equate each to $0$. Depending on where reactants and products appear in each step, you will find whether the equation you mentioned holds or not, and in some cases you may be able to find a single expression linking the initial reactant and final product. – user6376297 Jan 04 '18 at 07:49
  • @user6376297 I don’t quite get it. Anyway, what I asked wouldn’t be valid in that case, yes? – Senthil Arihant Jan 04 '18 at 08:35
  • I edited your question a bit for clarity. Note that notation $K_\mathrm{b}$ have another meaning, where b stands for "base"; also, the term "reverse" is preferred to "backward". Notations are updated according to IUPAC recommendations. – andselisk Jan 04 '18 at 08:45
  • Related: How is it that the equilibrium constant does not depend on the mechanism?; Relationship between rate equation and equilibrium constant – andselisk Jan 04 '18 at 08:48
  • 1
    @SenthilArihant : I don't think your question can be answered as such. As I wrote, you need to specify what the individual steps are, to know which concentrations are linked at equilibrium, and how. And BTW, if your overall transformation is composed by multiple reversible steps, what do the single forward and backward rates you mentioned in your post mean? Which reaction do they refer to? You should at least have the situation completely clear in you mind before you question its details. – user6376297 Jan 04 '18 at 09:17
  • I meant the ‘K’ of the overall reaction, which occurs when I write the rate law expression for the overall reaction. – Senthil Arihant Jan 04 '18 at 09:23
  • Here’s an example: https://chemistry.stackexchange.com/q/88280/56916 Will my equation be valid here? – Senthil Arihant Jan 04 '18 at 09:25
  • @andselisk So, according to the 2nd post, unless, for a reaction, the law of mass action is valid, my equation won’t be valid, correct? – Senthil Arihant Jan 04 '18 at 09:37
  • @SenthilArihant Still, depends on what reaction you apply it to. It works fine for the brutto-reaction if you use corresponding constants for the forward and reverse brutto reactions. – andselisk Jan 04 '18 at 09:42
  • @andselisk So, what you mean is that if, by chance, the rate law expression (determined using the slowest step of the reaction mechanism) has the concentrations of the reactants raised to the powers of their stoichiometric coefficients in the balanced equation, then my equation is valid? – Senthil Arihant Jan 04 '18 at 09:47
  • Yes, I agree with that. But is that what you’re saying? – Senthil Arihant Jan 04 '18 at 09:48
  • @SenthilArihant Kinetics of that slow step is already "imprinted" in the $K_\mathrm{fwd}$ and $K_\mathrm{rev}$ values for the brutto reaction. – andselisk Jan 04 '18 at 09:53
  • @andselisk I’m not getting what you are saying. And what’s the brutto reaction? – Senthil Arihant Jan 04 '18 at 10:56
  • @SenthilArihant That's adaptation of the German word "die Bruttoreaktion", which means "overall reaction" - the one I suppose you've written the ratio for. – andselisk Jan 04 '18 at 11:07
  • Yeah, so are you saying what I think you’re saying in my 2nd comment before this one? – Senthil Arihant Jan 04 '18 at 11:53
  • And so, in this example: https://chemistry.stackexchange.com/q/88280/56916. The equation won’t be valid yes? – Senthil Arihant Jan 04 '18 at 11:54

0 Answers0