What is the extent of say, this reaction?

$\ce{H_2CO_3 + HOCO_2^- <=> HOCO_2^- + H_2CO_3}$

K for the reaction appears to equal 1, and given the stoichiometry of the reaction, this implies that the reaction reaches an extent of 50% at equilibrium.

However, is this actually true?

  • $\begingroup$ Yes, K=1, but why do you say "this implies that the reaction reaches an extent of 50% at equilibrium."? K=1 doesn't imply anything like that. $\endgroup$ – Greg Jun 16 '14 at 2:41
  • $\begingroup$ Why not? For this reaction Keq = [product]^2/[reactant]^2 = 1, and if you take the square root of Keq, you get 1 = [product]/[reactant]. Only way for that to be true is for [product]=[reactant]. Now I do understand that for non-symmetric reactions K = 1 does not imply 50%. $\endgroup$ – Dissenter Jun 16 '14 at 5:33
  • $\begingroup$ I am sorry, but since [R1] = [P1] and [R2]=P2] the actual equqtion is K = [P1][P2] / ([P1][P2]) = 1, and it is true for any P1 and P2. You already rewrote this equation assuming that P1=P2, yet not assuming that the concentration of a material is same with itself. $\endgroup$ – Greg Jun 17 '14 at 20:33

The reaction you've shown is termed an "identity reaction", the products are the same as the reactants so $\ce{K_{eq}}$ must equal one. In the particular case you've drawn the reaction involves a simple proton transfer and will likely be diffusion controlled. Other examples of identity reactions are shown below. That they are actually occurring can be demonstrated using isotopically labeled substrates. The last one is one of my favorites, the degenerate rearrangement of bullvalene into itself via a Cope rearrangement. My drawing of bullvalene doesn't do it justice, just google bullvalene and prepare to be fascinated. If you number each of the ten carbon atoms in bullvalene you'll find that there are over 1.2 million possible arrangements. Sequential Cope rearrangements will convert each of these arragments into every other possible rearrangement. Connectivity in bullvalene is lost, each carbon atom is rapidly bound to every other carbon atom. At a bit over 100 °C, both the proton and carbon-13 NMR of bullvalene show only 1 sharp peak.

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Thermodynamics only applies to statistical behaviour over macroscopic quantities, rather than individual proton shuffles. When considered on this scale and with this level of abstraction, no reaction has taken place and so this isn't a meaningful question.

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  • $\begingroup$ Excuse me, I don't follow what you mean by no reaction has taken place. $\endgroup$ – Dissenter Jun 15 '14 at 21:06
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    $\begingroup$ @Dissenter, the individual molecules consumed and formed are identical, the bonds broken and formed are identical (both intra- and intermolecularly), hence there's no change in enthalpy. There's also no net change in concentration of any species, and no change in physical states specified, so there's no change in entropy. $\Delta H$ and $\Delta S$ are both zero, so $\Delta G$ is zero. Since $K_{eq} = e^{\frac{-\Delta G}{RT}} = 0$ the reaction is at equilibrium, forward and backward reaction rates are equal, and there is no discernible macroscopic change. $\endgroup$ – Greg E. Jun 15 '14 at 21:28
  • $\begingroup$ @Dissenter, sorry, that should read $K_{eq} = 1$, not zero; apologies for the typo. $\endgroup$ – Greg E. Jun 15 '14 at 21:35

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