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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?

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    $\begingroup$ For non-elementary reactions, all bets are generally off. Kinetics of consuming reactants or producing products is not generally directly related to the powered multiplication of their concentrations (or equivalent quantities). There is multiple forward and backward reactions and their respective rate constants. $\endgroup$
    – Poutnik
    Commented Jul 21, 2023 at 15:45
  • $\begingroup$ See e.g. chemistry.stackexchange.com/q/81063/35806 $\endgroup$
    – Poutnik
    Commented Jul 21, 2023 at 15:54
  • $\begingroup$ Nitpick: IUPAC's Green Book discerns number concentration $C_\mathrm{B} = N_\mathrm{B}/V$ in units like $\pu{m^{-3}}$ from amount concentration $c_\mathrm{B} = n_\mathrm{B}/V = [\ce{B}]$ in units like $\pu{mol m^{-3}}$ (though mol/L is more practical and hence more frequently seen). And SI units typically are not set italics. See mhchem you can use for the body of questions, answers and comments here on chemistry.se (don't use it on titles of questions). $\endgroup$
    – Buttonwood
    Commented Jul 21, 2023 at 16:41
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    $\begingroup$ I don't understand your notation, if $r_f$ is a rate then $k_f=C_A$ does not make sense, $k_f$ is, I assume, a rate constant, with units 1/time for first order reaction, and $C_A$ is concentration. The rate is $d[concentration]/dt=...$. $\endgroup$
    – porphyrin
    Commented Jul 21, 2023 at 17:04
  • $\begingroup$ While the post now is set with mhchem, there obviously a hump in the second equation. That is, (amount) concentrations run in dimensions like $\pu{mol/L}$, but lower case $k$, the rate constant has (at least) a time related contribution like $\ce{s^{-1}}$. By dimension analysis, the use of equality signs however requires the left and the right of this sign to express the same number (perhaps differently expressed, as we have e.g. prefixes like kilo, milli), as well as the dimension. (I didn't introduce this by the use of mhchem.) $\endgroup$
    – Buttonwood
    Commented Jul 21, 2023 at 17:11

1 Answer 1

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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.

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  • $\begingroup$ But, isn't it possible that you have a power law reaction that reproduces correctly the experimental kinetic, and that it is not an elementary reaction? For example: N2 + 3H2 <-> 2NH3 The elementary kinetic law would be "r = kPN2PH2^3 - kPNH3^2", but what if the exponents found through experiments are different? $\endgroup$
    – Daniel V.
    Commented Jul 22, 2023 at 8:38
  • $\begingroup$ @DanielV. You would expect them to be different. The reaction is typically catalyzed (getting all four reactants to collide in mid-air with sufficient kinetic energy is practically impossible). Nonetheless, when you do the tedious math, you can derive the equilibrium constant expression from the rate laws of elementary steps, and it looks the same no matter what the steps may be ("catalysts change the kinetics but not the equilibrium constant of a reaction"). $\endgroup$
    – Karsten
    Commented Jul 22, 2023 at 8:51
  • $\begingroup$ And If I wanted just the number of the constant (Because I am going to fix the temperature), how would I obtain it, when it changes with the initial concentration? Because then the division changes as well, like in the example I wrote, right? $\endgroup$
    – Daniel V.
    Commented Jul 22, 2023 at 8:58
  • $\begingroup$ @DanielV. The value of the equilibrium constant does not change with initial concentration. It just looks like that because your strategy to calculate it has a flaw. For multi-step reactions, you could get the equilibrium constant by measuring equilibrium concentrations, or you could measure the standard Gibbs energy of reaction (via electrode potentials or enthalpy and entropy measurements) or calculate it from thermodynamic data (via Gibbs energy of formation). $\endgroup$
    – Karsten
    Commented Jul 22, 2023 at 9:02
  • $\begingroup$ I see. Then in an experimental setting, how would the "global" equilibrium constant of the example reaction be calculated? Or is it impossible without knowing the multi-step reactions? (I'm gonna mark your answer as the answer already, since you already helped me a lot and it is the correct answer, I'm sure, I'm just having difficulty grasping it) $\endgroup$
    – Daniel V.
    Commented Jul 22, 2023 at 9:08

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