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I'm currently writing about geometric numerical integration and I need an example of the following series of chemical reactions:

\begin{align} \ce{A &->[$k_1$] B} \tag{R1} \\ \ce{A + B &->[$k_2$] 2 C} \tag{R2} \end{align}

The reason I want to use this form of the reaction is because $$\frac{\mathrm d}{\mathrm dt}([\ce{A}] + [\ce{B}] + [\ce{C}]) = 0$$ and this would be an example of a linear invariant.

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  • $\begingroup$ Not exactly the same, but what about O2 -> 2 O and O2 + O -> O3 ? with d(2p(O2) + p(O) + 3p(O3))/dt=0. Or the same with molar amounts or molar fractions. Formally with molar concentrations too, but that would be little unusual for gases. But possible. $\endgroup$
    – Poutnik
    Commented Apr 5, 2022 at 9:09
  • $\begingroup$ Is that correct though? When I do my calculations I get $\frac{d[O_2]}{dt}=-k_1[O_2]-k_2[O_2][O]$, $\frac{d[O]}{dt}=2k_1[O_2]-k_2[O_2][O]$ and $\frac{d[O_3]}{dt}=k_2[O_2][O]$ and if you add these you get $\neq0$. Or am I doing something wrong? $\endgroup$
    – Erik Sahlin
    Commented Apr 5, 2022 at 9:33
  • $\begingroup$ I was not addressing the kinetics just the mass inventory in context of the invariant. As the total mass of O allotropes is constant the given derivative is zero. $\endgroup$
    – Poutnik
    Commented Apr 5, 2022 at 9:34
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    $\begingroup$ @zhe The invariant of his original supposed schema holds because of trivial total amount inventory, which is constant. $\endgroup$
    – Poutnik
    Commented Apr 5, 2022 at 15:38
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    $\begingroup$ Is there a reason you can’t have just $\ce{A -> B -> C}$? There are many examples of that scheme. $\endgroup$
    – Andrew
    Commented Apr 6, 2022 at 0:37

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