In A-level chemistry we are taught that:

To every reaction: $$\sum_{i=1}^n c_iA_i\to\sum_{j=1}^m d_jB_j$$There is a rate equation: $$r=\kappa[A_1]^{\alpha_1}[A_2]^{\alpha_2}\cdots[A_n]^{\alpha_n}$$With the $(\alpha_i)$ called the partial orders and $\sum_{i=1}^n\alpha_i$ the overall order, $\kappa$ some constant.

Since the case of multiple reactants is confusing (e.g. I have not been able to get a clear answer of how 'rate' is defined in general) I'll stick to a reaction with only one reactant, $A\to bB+cC+dD+\cdots$.

We can then explicitly solve the differential equations, for $[A]_t$ treated as a function in time $t\ge0$, with $[A]_0>0$: $$\frac{\mathrm{d}[A]_t}{\mathrm{d}t}=\kappa\cdot[A]_t^\alpha\implies\kappa t+C=\begin{cases}\frac{1}{1-\alpha}[A]_t^{1-\alpha}&\alpha\neq0\\\ln[A]_t&\alpha=0\end{cases}$$After some shuffling, you get: $$[A]_t=\begin{cases}\{(1-\alpha)(\kappa\cdot t+C)\}^{\frac{1}{1-\alpha}}&\alpha\neq1\\Ce^{\kappa t}&\alpha=1\end{cases}$$Where $C$ is some constant in both cases. If $\alpha=1$, we get $C=[A]_t$: otherwise, $C=\frac{1}{1-\alpha}[A]_0^{1-\alpha}$.

I played with this a little, plotting concentration-time reaction curves on Desmos, and I quickly noticed a significant difference between the cases $\alpha<1$ and $\alpha\ge1$: if the order $\alpha$ is less than $1$, the model predicts the reaction terminates in finite time, at $t=-C/\kappa$. Else, the reaction never terminates!

Intuitively, I don't think a chemical reaction can ever fully terminate, so I'm inclined to believe the following conjecture:

In a reaction with only one reactant, the order of reaction must be greater than or equal to $1$.

More generally, I suppose:

In any reaction for which the rate equation holds, the overall order must be greater than or equal to $1$.

Does that make sense? I imagine I'm quite wrong, since I'm doing this on the back of relatively little chemical knowledge. In particular, it's hard to picture what solving the differential equations looks like for multiple reactants.

  • 1
    $\begingroup$ Aside of photochemical reactions of zeroth order, consider complex, multistep reactions with reaction loopbacks. Things about their order can go rather wild. Even for the simple looking $\ce{Br2 + H2 -> 2 HBr}$. $\endgroup$
    – Poutnik
    Dec 6, 2022 at 17:16
  • $\begingroup$ see coursehero.com/file/21287225/Chemical-Kinetics-3 $\endgroup$
    – Poutnik
    Dec 6, 2022 at 17:22
  • 5
    $\begingroup$ Long story short, if the reaction is truly elementary, then all orders would be positive integers; and if it isn't, then its order is just an approximation that won't hold all the way down to zero concentrations. $\endgroup$ Dec 6, 2022 at 17:33
  • $\begingroup$ :( Even or all this nice formatting, I'm still tempted to downvote. It maybe could be a duplicate target, if well answered, but the topic's being asked so many times before... $\endgroup$
    – Mithoron
    Dec 6, 2022 at 21:26
  • 1
    $\begingroup$ @NilayGhosh I often self-answer over at MSE. However, I don't have the authority to self-answer here, since I am not even close to knowing the details properly. $\endgroup$
    – FShrike
    Dec 7, 2022 at 17:37


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