# Is the rate of a zero order reaction always independent of the concentration of its reactants?

Consider the following reaction :

$$\ce{A + B}$$ $$\rightarrow$$ $$\ce{Products}$$

where order of the reaction w.r.t. $$\ce{A}$$ is $$-1$$ and order of the reaction w.r.t. $$\ce{B}$$ is $$+1$$. So, overall order of the reaction $$=(-1)+(1) = 0$$. So, it is a zero order reaction. My teachers told me that in a zero order reaction, the rate of the reaction is independant of concentration of the reactants. Below I have derived the integrated rate law for the above reaction. You can clearly see that in the integrated rate law there are terms of concentration of $$\ce{A}$$ and $$\ce{B}$$. Obviously, the integrated rate law is telling us that the rate of the reaction depends on the concentration of the reactants. Then why does everyone say that in a zero order reaction the rate of the reaction is independant of concentration of the reactants. Why?

• Is it possible that a reaction could be of order $-1$ with respect to a reactant ? The reaction rate should decrease when increasing the concentration of the reaction. Does it make sense ? Apr 24, 2021 at 13:12
• It may occur for complex reaction with catalysts and much more.
– Jay
Apr 24, 2021 at 13:25
• I think your teachers were talking about elementary zero order reactions. If you increase the concentration of all the reactants in a complex zero order reaction uniformly, the rate doesn't change. Apr 24, 2021 at 13:56

In this case it is important to differentiate between the overall order of the reaction and the order with respect to each species. The overall order of the reaction is zero. However, the rate law shows us that this is more complex that a simple unimolecular zero-order reaction, where rate = k.

As @M.L. pointed out, if you changed the concentration of both reactants by the same factor at the same time, then there would be no net change in the rate. However, if you change the concentration of only one reactant at a time, then the rate would change according to the order with respect to that reactant.

Let's imagine the following situation for your reaction, where the rate law is $$\ce{rate = k[A]^{-1}[B]^1}$$ and k=0.500 mol/Ls.

Trial A B Rate(mol/Ls)
1 0.10 0.10 0.50
2 0.10 0.20 1.0
3 0.20 0.10 0.25
4 0.20 0.20 0.50

You can see that when we double the concentration of B, the rate doubles. This makes sense, since the reaction is first order in B. In trial 3, we double the concentration of A and the rate decreases by half, since the reaction is -1 order in A. If we double the concentration of both reactants, as in trial 4, the resulting rate is the same as in trial 1, because the change with respect to each reactant is cancelled out.

When you look at the overall reaction, it is if you have an equal concentration of each reactant, then by the reaction progressing or adding equal amounts of both, the reaction rate won't change. For example, if you have 2M of both A and B, $$[A]^{-1}[B]^1 = 1/2*2=1$$, and increasing the concentration to 3M for both won't change the rate since $$1/3*3=1$$. The differential rate law applies to only one single species, so unless you have equimolar amounts of each reactant or make an approximation using an excess of one, you cannot apply the differential rate law.