# Whether for all zero order reactions rate is independent of concentration of reactants

For example,for the reaction $\ce{A + B \longrightarrow C}$,If rate law expression is: $$\mathrm{Rate=k \times [A]^{(0.5)}\times [B]^{(-0.5)}}$$ It is a zero order reaction but whether it is independent of concentrations of reactants?

• It is not zero order; moreover, reaction order in general is not a thing at all. It is order 0.5 in A, and order -0.5 in B, period. – Ivan Neretin Sep 19 '18 at 15:05
• But overall order is sum of exponents of concentrations and here it is 0 – kanishkan senthil Sep 19 '18 at 15:18
• @kanishkansenthil Yes, and the product of exponents is -0.25, but that doesn't mean it's a number that means anything useful. – Zhe Sep 19 '18 at 15:25
• There is no such thing as overall order. – Ivan Neretin Sep 19 '18 at 15:33
• Rate=[A]^x[B]^y. [A] and [B] express the concentration of the species A and B (usually in moles per liter (molarity, M)). The exponents x and y are the partial orders of reaction for A and B and the overall reaction order is the sum of the exponents. {From Wikipedia} – kanishkan senthil Sep 19 '18 at 17:07

Your question gives a very good example of why it isn't a useful quantity: you might expect a zeroth order reaction to have no dependence on the concentration of reactants, but your reaction clearly depends on the concentration of $\ce{A} \text{ and } \ce{B}$, so the total reaction order isn't actually telling us anything about the kinetics of the reaction.