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The problem is from Principles of general chemistry, 2nd edition by Silverberg, chapter 16 problem no.81:

"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: $rate = k\ce{[H2][I2]}$. The long accepted mechanism had a single bimolecular step; that is, the overall reaction was thought to be elementary:

$$\ce{H2_{(g)} + I2_{(g)} -> 2HI_{(g)}}$$

In the 1960s, however, spectroscopic evidence showed the presence of free I atoms during the reaction. Kineticists have since proposed a three-step mechanism:

$$\quad \ce{ I2_{(g)} <=> 2I_{(g)} \quad\quad[fast]}\tag {i }$$ $$\ce{H2_{(g)} + I_{(g)} <=> H2I_{(g)} [fast]\tag{ii}}$$ $$\ce{ H2I_{(g)} + I_{(g)} -> 2HI_{(g)} [slow]}\tag{iii}$$

Show that this mechanism is consistent with the rate law."

I cannot understand what the problem asks. I think, this reaction is consistent with the rate law by the third reaction, $\ce{H2I + I->2HI}$, which is the rate-determining step. Is my guess correct?

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    $\begingroup$ Have you learnt about the pseudo-steady-state approximation? $\endgroup$ – a-cyclohexane-molecule Sep 26 '18 at 17:16
  • $\begingroup$ You should express the rate law in terms of reactants if you can. Neither $\ce{H2I}$ nor $\ce{I}$ are reactants in the original reaction equation. $\endgroup$ – Zhe Sep 26 '18 at 18:06
  • $\begingroup$ write down the rate of production of product say $k_3[H_2I][I]$ this means that you have to find $[H_2I]$ and $ [I]$. The rate equation for the intermediate $[H_2I]$ is $d[H_2I]/dt= \cdots$ (it three terms) and set this to zero and find $[H_2I]$. All that is left is to find $[I]$ from the first equation. $\endgroup$ – porphyrin Sep 28 '18 at 8:17

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