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?

  • 3
    $\begingroup$ Have you learnt about the pseudo-steady-state approximation? $\endgroup$ Sep 26, 2018 at 17:16
  • 1
    $\begingroup$ 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, 2018 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, 2018 at 8:17

1 Answer 1


You have a three step mechanism where only one step is slow and the other steps are fast.

Since the first two steps (which are, in fact equilibria) are fast, it can be assumed that each of the equilibria holds throughout the reaction. Therefore, the expression of the equilibrium constant for each step can be used: $$\ce{K_1=\frac{[I]^2}{[I2]} \,\,\, and \,\,\, K_2=\frac{[H2I]}{[H2][I]}}$$ So, $\ce{[I]=\sqrt{\ce{K_1[I_2]}}\,\,\,\,and,\,\,[H2I]=K_2[H2][I]=K2[H2]\sqrt{\ce{K1[I2]}}}$

As only the last step is slow, the rate of formation of the product (which is the rate of the overall reaction) only depends on the last step.

$$\ce{rate=\frac{1}{2}\frac{d[HI]}{dt}=k_s[H2I][I]}$$ Here, $\ce{k_s}$ is the rate constant of the last, slow step. Substituting the expressions for $\ce{[H2I]}$ and $\ce{[I]}$ in the rate law should give you an expression of the form $\ce{rate=constant\times[H2][I2]}$, which is what the given rate law is.


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