Non-elementary reactions: Relationship between kinetic and equilibrium constant

As far as I understand, the equation for a kinetic equilibrium is always the same, and is given by:


$$K = \frac{\text{concentration of products}^{p}} {\text{concentration of reactants}^{r}}$$


Where $p$ and $r$ are the stoichiometric coefficients.


So, for elementary reactions, it is clear that the equilibrium constant is always the same (for the same temperature $T$), notwithstanding the initial concentration. For example, for the reaction:


$$ \ce{A <=> 2B}$$


If we assume dummy kinetic rates for forward and reverse reaction:


$$ r\_f = 1·c\_\text{A} = [\ce{A}]$$


$$ r\_r = 3·c\_\text{B}^2 = 3 [\ce{B}]^2 $$


Assuming the following: $[\ce{A}]\_0 = \pu{10 kmol/m^3}$, $[\ce{B}]\_0 = \pu{0 kmol/m^3}$, $\dot{v} = \pu{1 m^3/s}$. We know that the equilibrium is reached when the rate forward equals the reverse rate


$$r\_f = r\_r$$


$$\begin{equation}[\ce{A}] = 3[\ce{B}]^2 \qquad(1)\end{equation}$$


And if we consider the mass balances:


$$ [\ce{A}] = [\ce{A}]\_0 - [\ce{B}]/2 \\ [\ce{B}] = 2 ( [\ce{A}] - [\ce{A}]\_0)$$


We can substitute these in the rate expression $(1)$ and solve the equations for the roots:


$$ [\ce{A}]\_0 - [\ce{B}]/2 = 3 [\ce{B}]^2 \\ [\ce{B}] = 1.74, [\ce{A}] = 9.13$$


And the equilibrium constant is therefore:


$$ K = \frac{1.74^2}{9.13} = 0.33$$


And if we change the initial concentration, to something like double: $[\ce{A}]\_0 = \pu{20 kmol/m^3}$, the result of the quadratic equation changes to:


$$ [\ce{B}] = 2.5, [\ce{A}] = 18.75$$


And the equilibrium constant remains the same:


$$K = \frac{2.5^2}{18.75} = 0.33$$


So everything makes sense. But I understood that this behavior should continue even when the reactions are non-elementary. If we now have the same reaction, but the kinetics are given by:


$$ \ce{A <=> 2B}$$


$$ r\_f = 1·[\ce{A}]^2$$


$$ r\_r = 3·[\ce{B}]^2$$


Now it changes. If we develop the mass balance in equilibrium:


$$ [\ce{A}]^2 = 3 [\ce{B}]^2$$


$$ ( [\ce{A}]\_0 - [\ce{B}]/2 )^2 = 3 [\ce{B}]^2$$


For an initial concentration of $[\ce{A}]\_0 = 10$:


$$[\ce{A}]\_0^2 - [\ce{A}]\_0^\vphantom{} [\ce{B}] + [\ce{B}]^2/4 = 3 [\ce{B}]^2$$


Solving the quadratic equation and mass balance:


$$ [\ce{B}] = 4.48, [\ce{A}] = 7.76$$


And the equilibrium constant:


$$ K = \frac{4.48^2}{7.76} = 2.58$$


And when we repeat the calculations for $[\ce{A}]\_0 = 20$:


$$ [\ce{B}] = 8.96, [\ce{A}] = 15.52$$


And now the equilibrium constant is:


$$ K = \frac{8.96^2}{15.52} = 5.17$$


Why does this happen? Am I calculating the equilibrium constant wrong? It should be constant for non-elementary reactions, right?




If you have a two-step reaction
$$\ce{A <=>[$k\_1$][$k\_{-1}$] I <=>[$k\_2$][$k\_{-2}$] 2B},$$
the rate of the forward reaction of the first reaction (with rate constant $k\_1$) is not equal to the rate of the reverse reaction of the second reaction (with rate constant $k\_{-2}$). Instead, for each elementary reaction, the forward and reverse rates are the same at equilibrium.


So the premise of your argument is incorrect.


It turns out that the forward rate of the first reaction is relatively easy to measure (just start with pure $\ce{A}$ and measure the initial rate), so it is easy to figure out the rate law. Similarly, you can figure out the rate law of the reverse direction of the second step. It is much harder to measure the rate laws of the intermediate reacting to form reactants, or to form products (you would have to start with pure intermediate, and where will you get that from).


However, if the steps are elementary, you know the rate law by definition. If you write down the conditions for equilibrium (all steps are at equilibrium), you will get to the equilibrium constant expression in a tedious but straight-forward way, e.g. see [this answer](https://chemistry.stackexchange.com/a/111833).


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