The reaction:OCl⊖+I- ⟶OH-OI⊖+Cl⊖takes place in the following steps:i. OCl⊖+H2O⇋k1k2HOCl+OH        (fast)ii. I⊖+HOCl⟶k3HOI+Cl⊖            (slow)iii. O⊖H+HOIk1′k2′H2O+OI⊖        (fast)The rate of consumption of I in the following equation is

Step (ii) is the rate-determining step.∴ Rate =k3I⊖[HOCl](HOCI) produced in step (i) is consumed in step (ii), so it is a reactive intermediate. Similarly, (HOI) produced in step (ii) is consumed in step (iii), so it is also a reactive intermediate. The concentration of both reactive species is determined fromtheir equilibrium constant value.∴ from step (i)k1k2=[HOCl][OH]OCl⊖⇒[HOCl]=k1k2OCl⊖[OH]   .. . (2) Substitute the value of [HOCI] from Eq. (2) in Eq. (1), Rate =k1k3k2OCl⊖I⊖[O⊝H]Hence answer is (3).