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Suppose from kinetic mass balance I get the following differential equation of the molarity substances $A$,$B$, and $C$ in a reactor, with molar in- or outflow rate $\phi$, and rate constant $k$:

$$ \frac{\mathrm{d}}{\mathrm{d}t}[C] = k[A]^2[B]^3 + \phi{C_{\mathrm{in}}} - \phi{C_{\mathrm{out}}} $$

Where the dimensions convert to:

$$ [\pu{mol L-1 s-1}] = [\pu{s-1}] [\pu{mol5 L-5}] + [\pu{mol s-1}] - [\pu{mol s-1}] $$

Which can be simplified by adding the flow rate terms to:

$$ [\pu{mol L-1 s-1}] = [\pu{s-1}] [\pu{mol5 L-5}] \pm [\pu{mol s-1}] $$

By cancelling similar terms mole,seconds on both sides to:

$$[\pu{L-1}] = [\pu{mol4 L-5}] \pm 1$$

This seems that something is off about the rate constant, but also that there is no $\pu{L-1}$ term in the molar flow rate.

I don't expect exact answers to this, but could someone point me the right way, or send some clear explanation on either reaction rate constants or on kinetic mass balances? It seems I cannot find clear explanations, not even in textbooks.

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    $\begingroup$ Consider Formatting of chem expressions by using mhchem MathJax package, available in CH SE by default, involving MJ syntax \ce{} and \pu{} E.g. write $\ce{H2SO4}$ or $\ce{a A <=> p P}$ or $\pu{6.022E23 mol-1}$ to get $\ce{H2SO4}$ or $\ce{a A <=> p P}$ or $\pu{6.022E23 mol-1}$ (all eventually with double dollars in the display mode like $$\ce{H2SO4}$$. // E.g. typographic rules say unit symbols are never in italic. $\endgroup$
    – Poutnik
    Jul 10, 2022 at 11:22
  • $\begingroup$ Okay will do, but do you have an answer to my question? $\endgroup$
    – user313866
    Jul 10, 2022 at 14:57
  • $\begingroup$ I know nothing about this, but looking at the dimensions you firstly seemed to have used for concentration $mol L^{-1}$ on the LHS and simply $mol$ on the RHS, fixing that solves your second problem, and secondly by dimensionality arguments your assertion that the dimensions for k are $s^{-1}$ must be wrong. But don't ask me how to justify what the correct dimensions must be ($mol^{-4} L^4$). $\endgroup$
    – Ian Bush
    Jul 10, 2022 at 15:08
  • $\begingroup$ Yes the units of k should be mol$^{-4}$L$^4$ $\endgroup$
    – Andrew
    Jul 10, 2022 at 15:52
  • $\begingroup$ @IanBush I do not understand what you mean, the concentration is the same on both sides, except that on the LHS it is the derivative, so it is over time. $\endgroup$
    – user313866
    Jul 10, 2022 at 16:22

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