Question

Determine the rate law that is consistent with the following mechanism: Step 1: OCl− + H2O ⇄ HOCl + OH− (fast) Step 2: I− + HOCl  HOI + Cl− (slow) A) Rate = k[OCl−][H2O] B) Rate = k[I−][HOCl] C) Rate = k[OCl−][H2O][I−] D) Rate = k[OCl−][H2O]/[I−] E) Rate = k[OCl−][H2O]/[OH−] F) Rate = k[OCl−][H2O][I−]/[OH−]

Answer 1

Step 1: OCl− + H2O ⇄ HOCl + OH− (fast)

Step 2: I− + HOCl —> HOI + Cl− (slow)

Rate depends on the slowest step

So rate law can be written as:

rate = k2 [I-][HOCl]

Since [HOCl] is an intermediate, we will need to remove this

Since 1st step is an equilibrium process,

Kc = [OCl-][H2O]/[HOCl][OH-]

so,

[HOCl] = [OCl-][H2O]/Kc[OH-]

Put this in rate expression:

rate = k2 [I-][HOCl]

rate = k2 [I-][OCl-][H2O]/Kc[OH-]

Let k2/kc be k

so above expression becomes

rate = k [I-][OCl-][H2O]/[OH-]

This matches option F

Answer: F

answered by: anonymous