D30.3 Equilibrium Approximation

When the rate-determining step is not the first step, the rate law of the overall reaction can still be approximated as the rate law for the rate-determining step. However, one or more of the reactants involved in such a rate-determining step would be a reaction intermediate formed from a previous step. Hence, the rate law for the rate-determining step would include the concentration of a species that is rarely measurable. (Even if the concentration of a reaction intermediate can be measured experimentally, the accuracy is usually low because the concentration is very small.)

Experimentally determined rate laws are always expressed in terms of concentrations of reactants and/or products, for which accurate measurements are much easier to obtain. In order to compare the theoretical rate law derived from the mechanism to the experimentally determined rate law, we must express the mechanism rate law only in terms of reactant and product concentrations, that is, we cannot simply use the rate law of the rate-determining step as is (unless the rate-determining step is the first step).

In other words, it is necessary to express an intermediate’s concentration in terms of reactant and/or product concentrations. To do this, some approximation is usually needed. In one situation, where a preceding step that forms the reaction intermediate is at equilibrium, we can make use of the equilibrium approximation.

For an example, consider the following multi-step reaction:

The currently accepted mechanism for this reaction is:

Step 1:	NO(g) + Cl2(g)		NOCl2(g)	fast
Step 2:	NOCl2(g) + NO(g)		2NOCl(g)	slow
Overall:	2NO(g) + Cl2(g)	⟶	2NOCl(g)

In this mechanism, step 1 has a smaller activation energy and hence a larger rate constant than step 2. Therefore step 2 is the rate-determining step, and its rate law is:

However, this rate law involves the concentration of an intermediate, [NOCl₂], so it cannot be directly compared to experimental data. We need to express [NOCl₂] in terms of concentrations of reactants.

When NO and Cl₂ react to form NOCl₂ with rate constant k₁, there are two reactions possible for the NOCl₂:

• it can decompose back to form NO and Cl₂ with rate constant k_-1
  NOCl₂(g) NO(g) + Cl₂(g)
• it can react with NO to form two NOCl with rate constant k₂
  NOCl₂(g) + NO(g) 2 NOCl(g)

Because k₂ << k_-1, NOCl₂ molecules are much more likely to decompose than to react in step 2. When enough NOCl₂ has formed, we can approximate that the rate of its decomposition (rate_-1) is equal to the rate at which it forms (rate₁), meaning that step 1 is at equilibrium (rate₁ = rate_-1). This approximation gives us:

This relationship can be rearranged to solve for the concentration of NOCl₂:

Substituting this expression into the rate law for step 2, we have:

If we make:

then the rate law for the overall reaction becomes:

This rate law can be compared with experimental data to determine whether the proposed mechanism is a plausible one.

Activity: Rate Law from Mechanism

A fellow student says that the reaction mechanism below is a plausible mechanism for this reaction:

2 NO(g) + O₂(g)   ⟶   2 NO₂(g)          Rate = k[NO]²[O₂]

Mechanism:

Step 1 (fast):	NO(g) + O2(g)		NO3(g)
Step 2 (slow):	NO3(g) + NO(g)		2 NO2(g)

Apply the criteria for an appropriate mechanism to explain why your fellow student is correct or incorrect. Write your rationale in your notebook.

Write in your notebook, then left-click here for an explanation.

Your fellow student is correct. This is a plausible mechanism.

1. Each step in the mechanism is a bimolecular reaction, satisfying the criterion that a mechanism consists of unimolecular or bimolecular elementary reactions. The intermediate, NO₃ is known to occur in some reactions; drawing a Lewis structure reveals that NO₃ is a free radical (it has one electron less than nitrate ion, which has a complete octet on each atom). Since it is a free radical NO₃ is not extremely stable, but it is stable enough to exist as an intermediate.
2. The two reactions in the mechanism sum to the overall reaction given:
   

   Step 1:	NO(g) + O2(g)	⇌	NO3(g)
   Step 2:	NO3(g) + NO(g)	⟶	2 NO2(g)
   	NO(g) + O2(g) + NO3(g) + NO(g)	⟶	NO3(g) + 2 NO2(g)

   that is, 2 NO(g) + O₂(g)   ⟶   2 NO₂(g)

3. The mechanism predicts the experimental rate law for the reaction:
   The rate law for the slow step is the rate law for the reaction. Step 2 is the slow step, so
   Rate = k₂[NO₃][NO]

   However, NO₃ is an intermediate and its concentration is not easily measured. Therefore, we need to express the concentration of NO₃ in terms of concentrations of reactants (or other species whose concentrations are easily measured).
   Assuming that the first step is a rapid equilibrium, and the forward and reverse reaction of the first step occur at equal rates. Then

   rate_{step 1, forward} = rate_{step 1, reverse}
   k₁[NO][O₂] = k₋₁[NO₃]

   Solving for [NO₃] gives

   

   and so the predicted rate law is

   

   Because the product and quotient of constants is itself a constant, the rate law becomes

   Rate = k[NO]²[O₂]

   which agrees with the experimentally determined rate law.