To give a rate expression from the many elementary steps

To give expressions for the rate of producing an intermediate

To give expressions for the rate of CONSUMING an intermediate

Express concentration of intermediate in terms of concentration

of reactants

REevaluate the rate expression so that only concentrations of

REACTANTS are used

DISCUSS the derived RATE LAW.

This module gives another example to illustrate the technique of
SteadyState approximation, using the reaction between H2 and I_{2} gases.

For the reaction:

H2(g) + I_{2}(g) -> 2 HI(g)

What mechanisms might be appropriate? 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 have difficulties to propose a mechanism. However, the question is asked so you think about the object of chemistry.

A more practical question is to ask you to express the rate of reaction in terms of the concentration changes, for which you should write:

d[H2] d[I_{2}] 1 d[HI]

rate = - ----- = - ----- = --- -----

dt dt 2 dt

Look at the overall reaction equation again to see its relationship and the rate expressions.

Well, 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 H2 to produce the final products. Thus we propose
the three-step mechanism:

i. I_{2}(g) -> 2 I(g) k1 as rate constant.

ii. 2 I(g) -> I_{2}(g) k2 as rate constant.

iii. H2(g) + 2 I(g) -> 2 HI(g) k3 as rate constant.

Which one would you use to write the differential rate law?

Since only step iii gives the real products, we expect you to recognize that the rate can be expressed as:

rate = k3 [H2] [I]^2

BUT this is not a proper rate law, because I is 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 SteadyState approximation by assuming the rate to produce [I] and consume [I] to be equal.

rate of producing I = 2 k1 [I_{2}]

= rate of consuming I = 2 k2 [I]^2 + 2 k3 [H2] [I]^2.

Thus, k1 [I_{2}]

[I]^2 = ---------------

k2 + k3 [H2]

Substituting this for [I]^2 into the rate expression, you have

k1 [I_{2}]

rate = k3 [H2] --------------

k2 + k3 [H2]

k1 k3 [H2] [I_{2}]

= ----------------

k2 + k3 [H2]

Discussion:

If step 3 is slow, then k3 is a small value. Then, k2 >> k3 [H2], and the rate law is reduced to

rate = k [H2] [I_{2}],

where k = k1 k3 / k2. You should work this out on paper. By reading the above derivation does not sink in your mind.

You may argue that the rate law also support a one-step mechanism

H2(g) + I_{2}(g) -> 2 HI

This is an elementary step, the same as the over all reaction.

Suppose we use a large quantity of H2 compared to I_{2}, then
the change in [H2] is insignificant. For example, if
[H2] = 10, and [I_{2}] = 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 [I_{2}].

This rate law indicates that the reaction is a PSEUDO FIRST ORDER reaction, due to large quantity of one reactant. In this case, step iii is fast, and step i is the rate determining.

**1. In an alkaline solution, peroxydisulfate oxidizes sulfite to yield sulfate, S**_{2}O_{8}(2-) + SO_{3}(2-) + 2 OH(-) -> 3 SO_{4}(2-) + H_{2}O. The following mechanism has been proposed: i S_{2}O_{8}(2-) + SO_{3}(2-) -> S_{2}O_{7}(2-) + SO_{4}(2-) k1 ii S_{2}O_{7}(2-) + H_{2}O -> 2 SO_{4}(2-) + 2 H(+) k2 iii H(+) + OH(-) -> H_{2}O (very fast) k3 What is the intermediate? Enter the formula as shown.**Hint...**

Enter one of S_{2}O_{8}(2-), SO_{3}(2-), S_{2}O_{7}(2-), SO_{4}(2-)**Answer***S*_{2}O_{7}(2-)

**Consider...**

Would [S_{2}O_{7}(2-)] change during the steady-state period?**2. In an alkaline solution, peroxydisulfate oxidizes sulfite to yield sulfate, S**_{2}O_{8}(2-) + SO_{3}(2-) + 2 OH(-) -> 3 SO_{4}(2-) + H_{2}O. The following mechanism has been proposed: i S_{2}O_{8}(2-) + SO_{3}(2-) -> S_{2}O_{7}(2-) + SO_{4}(2-) k1 ii S_{2}O_{7}(2-) + H_{2}O -> 2 SO_{4}(2-) + 2 H(+) k2 iii H(+) + OH(-) -> H_{2}O (very fast) k3 Which step would you use to derive the rate expression? Enter i, ii, or iii.**Hint...**

Consider steps i and ii. Step iii is definitely inappropriate. If you consider i, the rate expression is rate = k1 [S_{2}O_{8}(2-)] [SO_{3}(2-)], then there is no intermediate involved.**Answer***ii*

**Consider...**

Now, write out the rate expression, rate = k2 [S_{2}O_{8}(2-)]**3. In an alkaline solution, peroxydisulfate oxidizes sulfite to yield sulfate, S**_{2}O_{8}(2-) + SO_{3}(2-) + 2 OH(-) -> 3 SO_{4}(2-) + H_{2}O. The following mechanism has been proposed: i S_{2}O_{8}(2-) + SO_{3}(2-) -> S_{2}O_{7}(2-) + SO_{4}(2-) k1 ii S_{2}O_{7}(2-) + H_{2}O -> 2 SO_{4}(2-) + 2 H(+) k2 iii H(+) + OH(-) -> H_{2}O (very fast) k3 Is the rate of consuming S_{2}O_{7}(2-) equal to k1 [S_{2}O_{8}(2-)] [SO_{3}(2-)] ?**Hint...**

The rate of consuming S_{2}O_{7}(2-) is k2 [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 = k2 [H_{2}O].**Answer***n*

**Consider...**

This is the rate of producing S_{2}O_{7}(2-).**4. In an alkaline solution, peroxydisulfate oxidizes sulfite to yield sulfate, S**_{2}O_{8}(2-) + SO_{3}(2-) + 2 OH(-) -> 3 SO_{4}(2-) + H_{2}O. The following mechanism has been proposed: i S_{2}O_{8}(2-) + SO_{3}(2-) -> S_{2}O_{7}(2-) + SO_{4}(2-) k1 ii S_{2}O_{7}(2-) + H_{2}O -> 2 SO_{4}(2-) + 2 H(+) k2 iii H(+) + OH(-) -> H_{2}O (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 [S_{2}O_{8}(2-)] [SO_{3}(2-)] Have a SteadyHappyDay.