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All my texts make it a point that in a differential rate law, the sum of the exponents on the reactant concentrations add up to the overall reaction order. However, I have yet to see any use in this. So you have two reactants, each undergoing a first-order reaction, and the overall reaction is second-order. So what? We analyze reactions for being first-order, second-order, and zero-order (at my level anyway), but that hasn't to do with the overall reaction order—just the reaction order pertaining to each individual reactant.

So my question is simple: what is the point of the overall reaction order? Why is it important?

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    $\begingroup$ In a situation where a reaction mixture is diluted by half by adding water, the rate of reaction for an overall 2nd order reaction will be a quarter, whether the rate = k[A]^2 or rate =k[A][B]. If you knew that overall a reaction was second order, but also that it was first order wrt [A], you can deduce it must also first order wrt [B] without an explicit experiment. $\endgroup$
    – Spontification
    Jan 27, 2016 at 15:13
  • $\begingroup$ It changes the mathematical form of the solution for the relationship between concentration and time, particularly for A reacting by itself compared to A + A reacting. $\endgroup$
    – Chet Miller
    Jan 27, 2016 at 15:15
  • $\begingroup$ @ChesterMiller How so? $\endgroup$
    – lightweaver
    Jan 28, 2016 at 10:46
  • $\begingroup$ Suppose that for a first order rate relation you have $\frac{dA}{dt}=-kA$, while for a 2nd order rate relation, you have $\frac{dA}{dt}=-kA^2$. Do you think the time dependent changes in concentration will be the same for both? $\endgroup$
    – Chet Miller
    Jan 31, 2016 at 1:53

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This answer is mostly a compilation of the excellent comments by Spontification and Chet Miller.

Differential rate law

The overall reaction order tells you what happens to the rate when you change all concentrations by the same factor at the same time. For example, if you decrease all concentrations by a factor 2 and the overall order is 1, the reaction will be slower by a factor of 2. If the order is 2, slower by a factor of $2^2 = 4$, if the order is 3, slower by a factor of $2^3 = 8$ and so on (overall reaction orders higher than 3 are unusual for elementary reactions). Dilution by a factor 2 would be one way of lowering concentrations; however, not all species will decrease by a factor of 2 (solvent, hydronium concentration in a buffer, any other species buffered by a fast equilibrium).

Integrated rate law

Integrated rate laws describe how the rate changes (decreases) over time. All species concentrations that are part of the differential rate law will change over time unless they are catalysts. If one species concentration is very high compared to others (e.g. the solvent) then its change will be negligible. So the integrated rate law depends on more than the differential rate law.

However, if all of the species in the differential rate law are reactants, and these reactants are present at stoichiometric ratios, the overall reaction order is directly related to the integrated rate law (e.g. overall second order reaction gives second-order integrated rate law). If not, you can get, for example, a pseudo-first order reaction even though the overall order of reaction is two.

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