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Steady-state Approximation Part I

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

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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.


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

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.
                  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]  
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.

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