Some help on correlating the reaction mechanism with the rate law

[dorebase2006]

Homework Statement

Consider the following mechanism:
(1) ClO[tex]^{-}[/tex] (aq) + H₂O (l) [tex]\Leftrightarrow[/tex] HClO (aq) + OH[tex]^{-}[/tex] (aq) [fast]
(2) I[tex]^{-}[/tex] (aq) + HClO (aq) [tex]\rightarrow[/tex] HIO (aq) + Cl[tex]^{-}[/tex] (aq) [slow]
(3) OH[tex]^{-}[/tex] (aq) + HIO (aq) [tex]\rightarrow[/tex] H₂O (l) + IO[tex]^{-}[/tex] (aq) [fast]

Is the mechanism consistent with the actual rate law: rate = k[ClO[tex]^{-}[/tex]][I[tex]^{-}[/tex]]

2. The attempt at a solution
I can't eliminate the intermediate OH[tex]^{-}[/tex] in my rate law. As far as I know, since the second step is slow, my rate should come from there, so rate = k[tex]_{2}[/tex] [I[tex]^{-}[/tex]][HClO]. Then I use the equilibrium (1) to express [HClO] in terms of [reactant]'s, but apparently [OH[tex]^{-}[/tex]] is there, and I don't know how to continue with this.

Another problem with which I also have the same trouble is the following proposed mechanism for the sulfonation of benzene:
(1) H₂SO₄ [tex]\rightarrow[/tex] H₃O[tex]^{+}[/tex] + HSO₄[tex]^{-}[/tex] + SO₃ [fast]
(2) SO₃ + C₆H₆ [tex]\rightarrow[/tex] H(C₆H₅[tex]^{+}[/tex])SO₃[tex]^{-}[/tex] [slow]
(3) H(C₆H₅[tex]^{+}[/tex])SO₃[tex]^{-}[/tex] + HSO₄[tex]^{-}[/tex] [tex]\rightarrow[/tex] C₆H₅SO₃[tex]^{-}[/tex] + H₂SO₄ [fast]
(4) C₆H₅SO₃[tex]^{-}[/tex] + H₃O[tex]^{+}[/tex] [tex]\rightarrow[/tex] C₆H₅SO₃H + H₂O [fast]

Thanks in advance!

 

[nate808]

After looking at the proposed mechanism, I can see that it is indeed consistent with the given rate law. Let me explain how.

First, let's look at the slow step (step 2). This step involves the reactants I^- and HClO, and the products HIO and Cl^-. According to the rate law, the rate of the reaction is dependent on the concentrations of both I^- and HClO. This matches with the slow step in the mechanism, where both reactants are involved in the formation of the products.

Now, let's consider the fast steps (steps 1, 3, and 4). These steps involve the formation of intermediates, which are then used in the slow step. In the fast step (1), we see that ClO^- and H2O react to form HClO and OH^-. This step is reversible, which means that the products can also react to form the reactants. This is important because it allows for the formation of more HClO and OH^- as the reaction progresses. This will then contribute to the concentration of HClO in the slow step.

Similarly, in the fast step (3), we see that OH^- and HIO react to form H2O and IO^-. Again, this step is reversible and will contribute to the formation of more H2O and IO^- as the reaction progresses. This will then contribute to the concentration of HIO in the slow step.

Finally, in the fast step (4), we see that H3O^+ and C6H5SO3^- react to form C6H5SO3H and H2O. This step is also reversible and will contribute to the formation of more C6H5SO3H and H2O as the reaction progresses. This will then contribute to the concentration of C6H5SO3^- in the slow step.

Overall, we can see that the proposed mechanism is consistent with the given rate law. The fast steps contribute to the formation of intermediates, which then participate in the slow step. This allows for the formation of more products and keeps the reaction going. I hope this helps to clarify any confusion you had. Keep up the good work in your scientific studies!