Even when a mechanism is consistent with the rate law, later work may show it to be incorrect. For example, the reaction between hydrogen and iodine has this rate law: Rate = k[H₂][I₂]. The long-accepted mechanism had a single bimolecular step; that is, the overall reaction was thought to be elementary:H₂(g) + I₂(g) → 2 HI(g) In the 1960s, however, spectroscopic evidence showed the presence of free I atoms during the reaction. Kineticists have since proposed a three-step mechanism:Show that this mechanism is consistent with the rate law.

Since each of the equilibria holds throughout the reaction, it may be assumed that the first two steps—which are in reality equilibria—occur quickly.

So, the following equation for the equilibrium constant at each phase is appropriate:

[tex]K_{1} =\frac{[I]^{2} }{[I_{2}] }[/tex] and [tex]K_{2} =\frac{[H_{2} I]}{[H_{2}][I] }[/tex]

[I] = [tex]\sqrt{[K_{1}] }[I_{2} ][/tex] and [H₂I] = [tex]K_{2}[/tex][H₂][I] = [tex]K_{2} [H_{2} ]\sqrt{K_{1} } [I_{2} ][/tex]

The pace of production of the product (which is the rate of the total reaction) only depends on the final phase because only that step is sluggish.

Rate = [tex]\frac{1}{2} \frac{d[HI]}{dt} =k_{s}[H_{2} I] [I][/tex]

The rate constant [tex]k_{s}[/tex] of the final slow step is used here.

Hence, rate = constant x [H₂][I₂] is rate law.

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