# How do reaction rates give information about mechanisms?

Dec 31, 2013

A mechanism for a reaction is a sequence of elementary steps that explains how the overall reaction proceeds.

For each elementary step, you can write a rate law. Then you put them all together to get the overall rate law for the reaction. You may propose any number of mechanisms and their corresponding rate laws.

Your task, then, is to determine how the reaction rate depends on the concentration of each component, that is, the overall rate law for the reaction. You then compare your experimental rate law with the rate laws predicted by your possible mechanisms. Your experimental rate law may agree with the rate law predicted by a proposed mechanism.

However, the fact that a mechanism explains the experimental results does not prove that the mechanism is correct. It simply says that the mechanism may be correct, because there may be other mechanisms that predict the same overall rate law.

Students in general chemistry are not generally required to propose a mechanism, but they are required to derive the rate law from a proposed mechanism and to state whether or not it agrees with their experimental data.

EXAMPLE 1

A student is studying the reaction
2NO₂ + F₂ → 2NO₂F

The student finds that (a) the rate of the reaction is directly proportional to the concentration of NO₂ and (b) the rate of the reaction is directly proportional to the concentration of F₂. Are the student’s observations consistent with the following mechanism:
i. NO₂ + F₂ → NO₂F + F (slow)
ii. NO₂ + F → NO₂F (fast)

Solution

The student found that (a) rate ∝ [NO₂] and (b) rate ∝ [F₂]. Therefore
Overall rate = -½Δ[NO₂]Δt = -Δ[F₂]/Δt = k[NO₂][F₂]

This tells us that the rate determining step probably involves the collision of an NO₂ molecule with an F₂ molecule.

In the proposed mechanism, step i is the rate-determining step. Its rate law is
rate = k[NO₂][F₂]

The student’s experimental rate law agrees with the predicted rate law for the mechanism.

EXAMPLE 2

A student finds that for the reaction
H₂ + Br₂ → 2HBr
(a) the rate of the reaction is directly proportional to the concentration of H₂ and (b) the rate of the reaction is directly proportional to the square root of the concentration of Br₂.

Are the student’s observations consistent with the following mechanism:
i. Br₂ ⇌ 2Br; k₁, k₋₁ (fast)
ii. Br + H₂ → HBr + H; k₂ (slow)
iii. H + Br₂ → HBr + Br; k₃ (fast)

SOLUTION
The student found that (a) rate ∝ [H₂] and (b) rate ∝ [Br₂]^½. Therefore
Overall rate = -Δ[H₂]Δt = -Δ[Br₂]/Δt = k[H₂][Br₂]^½

In the proposed mechanism, the rate determining step is step ii. So the rate law is
Rate = k₂[H₂][Br]

Since Br is not one of the reactants, we must derive an expression for [Br] in terms of the other concentrations. We can get this from the rapid equilibrium in step i. Since, at equilibrium. The rate of the forward reaction must equal the rate of the reverse reaction, we can write

k₁[Br₂] = k₋₁[Br]²
or
[Br] = {(k1/k-1)[Br2]}^½

Substituting this in the rate expression results in
rate = k₂[H₂] × {(k₁/k₋₁)[Br₂]}^½

Since k₁, k₋₁, and k₂ are all constants, we can combine them into one overall constant k. The predicted rate law becomes
rate = k[H₂][Br₂]½

The student’s experimental rate law agrees with the predicted rate law for the mechanism.

Hint: Whenever you see a rate law with a concentration raised to a fractional power, it often indicates that the rate determining step is preceded by a rapid equilibrium.

SUMMARY

In order to get information on mechanisms from reaction rates, you must
1. Determine the effect that the concentration of each component has on the rate.
2. Propose one or more mechanisms for the reaction.
3. Test your experimental rate law against those predicted by your proposed mechanisms.