CChieh@UWaterloo.ca

Mechanism and Rate Law Using Steady-State Approximation

Skills to develop

  • Part II
  • Mechanism and Rate Law Using Steady-State Approximation

    Skills to develop: To propose a mechanism (future goal, not for final exam)
    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 I2 gases.

    For the reaction:

    H2(g) + I2(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[I2] 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 I2 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. I2(g) -> 2 I(g) k1 as rate constant.

    ii. 2 I(g) -> I2(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 [I2]

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

    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 is a small value. Then, k2 >> k3 [H2], and the rate law is reduced to

    rate = k [H2] [I2],

    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) + I2(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 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].

    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.

    Confidence Building Questions

    cchieh@uwaterloo.ca