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Could you please confirm or say why I am wrong?

Let us consider the steady state of the chemical reaction $A+B \leftrightarrow^{k_+}_{k_-} C$, with $k_+$ and $k_-$ the forward and backward rates.

The rate of production of $A$ is: $C k_-$.

The rate of production of $B$ is: $C k_-$.

The rate of production of $C$ is: $A B k_+$.

The system is at thermodynamic equilibrium if the detailed balanced is satisfied. In other words, we need that $Ck_-=ABk_+ \Leftrightarrow \frac{C}{AB}=\frac{k_+}{k_-}$.

Conclusion : An equilibrium state always exists, for any $k_-$, $k_+$

And then any reaction $A+B \leftrightarrow C$ can be modeled with equilibrium tools? If so, what about biological systems that are driven out of equilibrium by ATP?

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The conclusion is correct: for any system containing a single reaction, there exists an equilibrium (with detailed balance) for any values of the rate constants. I'm not sure what's meant exactly by "equilibrium tools," but it's true to say that there's no real difference in the kinetics of an equilibrium or non-equilibrium system if it only contains one reaction.

However, this is not the case in general if you have more than one reaction. For example, if you add the reaction $C \leftrightharpoons D$ to your system then what you say is still true - you can always find an equilibrium with detailed balance - but if you then add the reaction $D \leftrightharpoons A + B$ then it's no longer true. With arbitrary rate constants you will generally find a steady state where detailed balance does not hold, and instead there is a constant flow of matter from $A+B$ to $C$ to $D$ and back to $A+C$ again. (Or around the same cycle in the opposite direction.)

For biological systems driven by ATP, there are generally many reactions involved, and so the kinetics will typically be different from an equilibrium system. This allows you to have things like limit cycles as well as point attractors, for example.

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