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# Deriving Rate Laws Using the Steady-State Approximation - part II

### Skills to develop

• 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.

### Deriving Rate Law Using Steady-State Approximation - part II

This page gives another example to illustrate the technique of deriving rate laws using the steady-state approximation. The reaction considered here is between H2 and I2 gases.

Example

For the reaction: H2(g) + I2(g) -> 2 HI(g) what mechanisms might be appropriate? Derive a rate law from the proposed mechanism.

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[H2]     d[I2]     1  d[HI]
rate = - ----- = - -----  = --- -----
dt        dt       2   dt
```
Look at the overall reaction equation again to see its relationship and the rate expressions.

Proposing a mechanism

In order to propose a mechanism, we apply the following reasoning. Since the bonding between I-I is weak, we expect I2 to dissociate into atoms or radicals. These radicals are active, and they react with H2 to produce the products. Thus we propose the three-step mechanism:

1. I2(g)   -k1->   2 I(g)
2. 2 I(g)   -k2->   I2(g)
3. H2(g) + 2 I(g)   -k3->   2 HI(g)
Which step would you use to write the differential rate law?

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

rate = k3 [H2] [I]2
But this is not a proper rate law, because I is an intermediate, not a reactant. So, you have to express [I] or [I]2 in terms of the concentration of reactants. To do this, we use the steady-state approximation and write out the following relationships: rate of producing I = 2 k1 [I2]
rate of consuming I = 2 k2 [I]2 + 2 k3 [H2] [I]2.
producing rate of I = consuming rate of I.
Thus,
```                  k1 [I2]
[I]2  =  ---------------
k2  +  k3 [H2]
```
Substituting this for [I]2 into the rate expression, you have
```                         k1 [I2]
rate = k3 [H2]  --------------
k2  +  k3 [H2]

k1 k3 [H2] [I2]
= ----------------
k2  +  k3 [H2]
```
Discussion:
If step 3 is slow, then k3 and k2 >> k3 [H2]. The rate law is reduced to rate = k [H2] [I2], where k = k1 k3 / k2. Work this out on paper yourself, reading the above derivation does not lead to learning.

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,

H2(g) + I2(g) -> 2 HI This elementary step is the same as the over all reaction.

Suppose we use a large quantity of H2 compared to I2, then the change in [H2] is insignificant. For example, if [H2] = 10, and [I2] = 0.1 initially, [H2] remains essentially 10 (9.9 with only one significant figure). In other words, [H2] hardly change when the reaction ended. Thus,

k3 [H2] >> k3 and the rate law becomes: rate = k1 [I2]. Thus, the reaction is a pseudo first order reaction, due to large quantity of one reactant. The results suggests iii a fast step (due to large quantity of H2, and i the rate determining step.

### Confidence Building Questions

• In an alkaline solution, peroxydisulfate oxidizes sulfite to yield sulfate, S2O82- + SO32- + 2 OH- -> 3 SO42- + H2O. The following mechanism has been proposed:
```  i. S2O82- + SO32- -> S2O72- + SO42-, k1
ii. S2O72- + H2O -> 2 SO42- + 2 H+,  k2
iii. H+ + OH- -> H2O  (very fast),   k3
```
What is the intermediate?

Discussion -
Would [S2O72-] change during the steady-state period?

• In an alkaline solution, peroxydisulfate oxidizes sulfite to yield sulfate, S2O82- + SO32- + 2 OH- -> 3 SO42- + H2O. The following mechanism has been proposed:
```  i. S2O82- + SO32- -> S2O72- + SO42-, k1
ii. S2O72- + H2O -> 2 SO42- + 2 H+,  k2
iii. H+ + OH- -> H2O  (very fast),   k3
```
Which step would you use to derive the rate expression? Enter i, ii, or iii.

Discussion -
Consider steps i and ii. Step iii is definitely inappropriate. If you consider i, the rate expression is rate = k1 [S2O82-] [SO32-], then there is no intermediate involved. Now, write out the rate expression,
rate = k2 [S2O82-]

• In an alkaline solution, peroxydisulfate oxidizes sulfite to yield sulfate, S2O82- + SO32- + 2 OH- -> 3 SO42- + H2O. The following mechanism has been proposed:
```  i. S2O82- + SO32- -> S2O72- + SO42-, k1
ii. S2O72- + H2O -> 2 SO42- + 2 H+,  k2
iii. H+ + OH- -> H2O  (very fast),   k3
```
Is the rate of consuming S2O72- equal to k1 [S2O82-] [SO32-]?

Discussion -
The rate of consuming S2O72- is k2 [S2O72-] [H2O], but [H2O] is so large 56 M, that its change never detected. Thus, the rate of consuming is really k [S2O72-], where k = k2 [H2O].

Consider...
This is the rate of producing S2O72-.

• In an alkaline solution, peroxydisulfate oxidizes sulfite to yield sulfate, S2O82- + SO32- + 2 OH- -> 3 SO42- + H2O. The following mechanism has been proposed:
```  i. S2O82- + SO32- -> S2O72- + SO42-, k1
ii. S2O72- + H2O -> 2 SO42- + 2 H+,  k2
iii. H+ + OH- -> H2O  (very fast),   k3
```
What is the rate law for this reaction derived from the mechanism?

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

rate = k1 k2 [S2O82-] [SO32-] Have a SteadyHappyDay.