Question

The rate law for a certain process is given by

d[I₂]/dt=k[Fe(CN)6^3-]²[I^-]²[Fe(CN)_6^4-]^-1[I₂]⁰

The overall reaction is given by

2Fe(CN)₆^3- + 2I^- = 2Fe(CN)₆^4- + I₂

_Using this information, suggest a mechanism for this process and show how it leads to the overall reaction and the rate law given. (the order of reaction with respect to one substance indicates a pre-equilibrium in process). Don't forget the intermediate.

Answer 1

2Fe(CN)₆^3- + 2I^- 2Fe(CN)₆^4- + I₂

(Fe(CN)₆^3- + e^- Fe(CN)₆^4-) x 2

2I^- I₂ + 2e^-

2Fe(CN)₆^3- + 2e^- 2Fe(CN)₆^4-

2Fe(CN)₆^3- + 2I^- 2Fe(CN)₆^4- + I₂

Rate laws must be determined experimentally; they cannot be predicted from the coefficients of balanced chemical equations. Every reaction is made up of a series of one or more elementary steps, and the rate laws and relative speeds of these steps will dictate the overall rate law. Indeed, the rate law for a reaction can be determined from its mechanism.