For a reaction ${I^ - } + OC{l^ - } \to I{O^ - } + C{l^ - }$ in an aqueous medium, the rate of reaction is given by $\dfrac{{d(I{O^ - })}}{{dt}} = k\dfrac{{({I^ - })(OC{l^ - })}}{{(O{H^ - })}}$ . The overall order of the reaction is ____?
A)-1
B) 0
C) 1
D) 2

Hint: The Order of Reaction refers to the power dependence of the rate on the concentration of each reactant. Thus, for a first-order reaction, the rate is dependent on the concentration of a single species.

Complete step by step answer:
 The order with respect to \[{I^ - },\;OC{l^ - }\;\& \,\;O{H^{^ - }}\;\] are 1, 1 and -1 respectively.
The overall order of the reaction is \[1 + 1 - 1 = 1\]
Therefore, overall order is =1. Option “C” is correct
As given in the rate equation, rate of reaction depends on the concentration of ${I^ - }\& \,OC{l^ - }$ , that is it depends on two reactants. But \[O{H^{^ - }}\;\]concentration is given in the denominator. Therefore, order is \[1 + 1 - 1 = 1\]
Extra points: A second-order reaction refers to one whose rate is dependent on the square of the concentration of a single reactant or the rate depends on two reactants with their power raised to unity. OR the combined first-order dependence on the concentrations of two different reactants.
$A + B \to C$
 (ii) It can be obtained by adding all the exponents of the concentration terms in the rate expression.
The order of reaction does not depend on the stoichiometric coefficients corresponding to each species in the balanced reaction.
So,Option “A” is correct.

Note:For simple one-step reactions, the order and molecularity should have the same value. Molecualrity is the number of molecules taking part in the rate determining step.
Reaction order represents the number of species whose concentration directly affects the rate of reaction. The reaction order of a chemical reaction is always defined with the help of reactant concentrations and not with product concentrations.