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.
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 dtLook 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:
Since only step 3 gives the real products, we expect you to recognize that step 3 hints the rate law to be:
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:
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 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,
i. S2O82- + SO32- -> S2O72- + SO42-, k1 ii. S2O72- + H2O -> 2 SO42- + 2 H+, k2 iii. H+ + OH- -> H2O (very fast), k3What is the intermediate?
Would [S2O72-] change during the steady-state period?
i. S2O82- + SO32- -> S2O72- + SO42-, k1 ii. S2O72- + H2O -> 2 SO42- + 2 H+, k2 iii. H+ + OH- -> H2O (very fast), k3Which step would you use to derive the rate expression? Enter i, ii, or iii.
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-]
i. S2O82- + SO32- -> S2O72- + SO42-, k1 ii. S2O72- + H2O -> 2 SO42- + 2 H+, k2 iii. H+ + OH- -> H2O (very fast), k3Is the rate of consuming S2O72- equal to k1 [S2O82-] [SO32-]?
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].
This is the rate of producing S2O72-.
Well, the expression is rather complicated. To avoid your frustration, you got the mark, and here is the expression: