- Give expressions for the producing
*rate*of an intermediate - Give expressions for the consumming
*rate*of an intermediate - Express concentration of intermediate in terms of concentration of reactants
- Eliminate concentrations of intermediates using concentrations of reactants.
- Derive a
**rate law**from the many elementary steps - Disscuss the derived
**rate law**.

This page gives another example to illustrate the technique of deriving
rate laws using the steady-state approximation.
The reaction considered here is between H_{2} and I_{2} gases.

**
Example
**

*Solution*

Well, this question does not have a simple answer, and there is no way to
prove one over another for its validity. Beginning chemistry students will
not be asked to propose a mechanism, but you will be asked to derive the
rate law from the proposed mechanism.

First of all, you should be able to express the rate of reaction in terms of the concentration changes,

d[HLook at the overall reaction equation again to see its relationship and the rate expressions._{2}] d[I_{2}] 1 d[HI]rate= - ----- = - ----- = --- ----- dt dt 2 dt

**Proposing a mechanism**

In order to propose a mechanism, we apply the following reasoning.
Since the bonding between I-I is weak, we expect I_{2} to dissociate
into atoms or radicals. These radicals are active, and they react with
H_{2} to produce the products. Thus we propose the three-step
mechanism:

2. 2 I(g) -

3. H

Since only step 3 gives the real products, we expect you to recognize that step 3 hints the rate law to be:

Substituting this for [I]k_{1}[I_{2}] [I]^{2}= ---------------k_{2}+k_{3}[H_{2}]

k_{1}[I_{2}]rate = k_{3}[H_{2}] --------------k_{2}+k_{3}[H_{2}]k_{1}k_{3}[H_{2}] [I_{2}] = ----------------k_{2}+k_{3}[H_{2}]

If step 3 is slow, then

Since the rate law is first order with respect to both reactant, one may argue that the rate law also support a one-step mechanism,

Suppose we use a large quantity of H_{2} compared to I_{2}, then
the change in [H_{2}] is insignificant. For example, if
[H_{2}] = 10, and [I_{2}] = 0.1 initially, [H_{2}] remains essentially
10 (9.9 with only one significant figure).
In other words, [H_{2}] hardly change when the reaction
ended. Thus,

**In an alkaline solution, peroxydisulfate oxidizes sulfite to yield sulfate,**S The following mechanism has been proposed:_{2}O_{8}^{2-}+ SO_{3}^{2-}+ 2 OH^{-}-> 3 SO_{4}^{2-}+ H_{2}O.i. S

What is the intermediate?_{2}O_{8}^{2-}+ SO_{3}^{2-}-> S_{2}O_{7}^{2-}+ SO_{4}^{2-},*k*_{1}ii. S_{2}O_{7}^{2-}+ H_{2}O -> 2 SO_{4}^{2-}+ 2 H^{+},*k*_{2}iii. H^{+}+ OH^{-}-> H_{2}O (very fast),*k*_{3}**Discussion -**

Would [S_{2}O_{7}^{2-}] change during the steady-state period?**In an alkaline solution, peroxydisulfate oxidizes sulfite to yield sulfate,**S The following mechanism has been proposed:_{2}O_{8}^{2-}+ SO_{3}^{2-}+ 2 OH^{-}-> 3 SO_{4}^{2-}+ H_{2}O.i. S

Which step would you use to derive the rate expression? Enter i, ii, or iii._{2}O_{8}^{2-}+ SO_{3}^{2-}-> S_{2}O_{7}^{2-}+ SO_{4}^{2-},*k*_{1}ii. S_{2}O_{7}^{2-}+ H_{2}O -> 2 SO_{4}^{2-}+ 2 H^{+},*k*_{2}iii. H^{+}+ OH^{-}-> H_{2}O (very fast),*k*_{3}**Discussion -**

Consider steps i and ii. Step iii is definitely inappropriate. If you consider i, the rate expression is rate =*k*_{1}[S_{2}O_{8}^{2-}] [SO_{3}^{2-}], then there is no intermediate involved. Now, write out the rate expression,

*rate = k*_{2}[S_{2}O_{8}^{2-}]**In an alkaline solution, peroxydisulfate oxidizes sulfite to yield sulfate,**S The following mechanism has been proposed:_{2}O_{8}^{2-}+ SO_{3}^{2-}+ 2 OH^{-}-> 3 SO_{4}^{2-}+ H_{2}O.i. S

Is the rate of consuming S_{2}O_{8}^{2-}+ SO_{3}^{2-}-> S_{2}O_{7}^{2-}+ SO_{4}^{2-},*k*_{1}ii. S_{2}O_{7}^{2-}+ H_{2}O -> 2 SO_{4}^{2-}+ 2 H^{+},*k*_{2}iii. H^{+}+ OH^{-}-> H_{2}O (very fast),*k*_{3}_{2}O_{7}^{2-}equal to*k*_{1}[S_{2}O_{8}^{2-}] [SO_{3}^{2-}]?**Discussion -**

The rate of consuming S_{2}O_{7}^{2-}is*k*_{2}[S_{2}O_{7}^{2-}] [H_{2}O], but [H_{2}O] is so large 56 M, that its change never detected. Thus, the rate of consuming is really k [S_{2}O_{7}^{2-}], where k =*k*_{2}[H_{2}O].**Answer***n*

**Consider...**

This is the rate of producing S_{2}O_{7}^{2-}.**In an alkaline solution, peroxydisulfate oxidizes sulfite to yield sulfate,**S The following mechanism has been proposed:_{2}O_{8}^{2-}+ SO_{3}^{2-}+ 2 OH^{-}-> 3 SO_{4}^{2-}+ H_{2}O._{2}O_{8}^{2-}+ SO_{3}^{2-}-> S_{2}O_{7}^{2-}+ SO_{4}^{2-},*k*_{1}ii. S_{2}O_{7}^{2-}+ H_{2}O -> 2 SO_{4}^{2-}+ 2 H^{+},*k*_{2}iii. H^{+}+ OH^{-}-> H_{2}O (very fast),*k*_{3}**Consider...**

Well, the expression is rather complicated. To avoid your frustration, you got the mark, and here is the expression:

rate = Have a SteadyHappyDay.*k*_{1}*k*_{2}[S_{2}O_{8}^{2-}] [SO_{3}^{2-}]