Unit 5: Topic 9 - Steady-State Approximation

Question 1

What approximations and assumptions can be made when determining the rate law of a reaction when the first step to the reaction mechanism is not the rate-determining step?

Answer 1

In a rate mechanism, the rate-determining step can be used to determine the rate law of the original reaction. Sometimes, doing this will involve intermediates, a species that transiently exists throughout the reaction but is not a reactant or a product. When the reaction involves intermediates, the concentration of this intermediate will remain constant at some stage of the reaction. This is the steady-state approximation. Therefore, if we can find two representations for the concentration of the intermediate, we can set them equal to each other under the steady-state approximation.

Question 2

Derive the rate law for the reaction 2N2O5 –> 4NO2 + O2. The three-step reaction mechanism is shown here.

Answer 2

First use stoichiometry: the rate law is equal to the production of oxygen: rate = d[O2]/dt.
NO and NO3 are the intermediates. Let us consider nitric oxide first.
From step 2, the production rate of NO is k2[NO3][NO2].
From step 3, the consumption rate of NO is k3[NO3][NO].
Steady-state assumption: k2[NO3][NO2] = k3[NO3][NO], so [NO] = k2[NO2]/k3.
The production rate of NO3 is kf[N2O5]. The consumption rate is k2[NO3][NO2] + k3[NO3][NO] + kb[NO3][NO2], as we sum up the contributions from each step. We can equate these two expressions.
Now, using step two, we have d[O2]/dt = k2[NO3][NO2].
Substituting everything, we will eventually arrive at d[O2]/dt = k[N2O5], k = kfk2/(kb + 2k2). This is the experimental rate law.

Note: students must work this out by themselves to understand the concept and the math.