Non-Elementary Reaction with Elementary Rate Law

There are many non-elementary reactions following the elementary rate law. How can someone spot them? This is an application of PSSH (Pseudo-Steady-State Hypothesis).

For example:

(CH₃)₂O à CH₄ + H₂ + CO

We can say this is a reaction follows the form of

AàP

r_p = -k[A]

The mechanism of this reaction consists of 3 elementary reactions.

1. Activation of A: A + M àA* + M          (K1 as rate constant)

2. Deactivation of A*: A* + M àP + M     (K2 as rate constant)

3. Decomposition of A*: A*àP                  (K3 as rate constant)

Here: M is inert species that does not react at all. A* is denoted as the active intermediate of A. Its presence is due to collision of A molecules. When collision occurs, kinetic energy of one A molecule is transferred to internal rotational and vibrational energies of the other A molecule, so it is activated and being highly reactive. 

From 1. r_1A* = k₁[A][M]

From 2. r_2A* = -k₂[A*][M]

From 3. r_3A* = -k₃[A*]

             r_p = k₃[A*]

Now we will apply PSSH, which states that the net rate of formation of an active intermediate is zero.

r_A* = r_{1A* +} r_{2A* +} r_3A*

r_A* = k₁[A][M] - k₂[A*][M] - k₃[A*]

Apply PSSH, r_A* = k₁[A][M] - k₂[A*][M] - k₃[A*] = 0

Solve for [A*]

Since r_p = k₃[A*], by substitute this, we get the following.

Because concentration of M is constant, we say the following.

 So, r_A = -r_p = k[A]

This shows the reaction follows first order rate law, and it is elementary. But, again, the reaction is not an elementary reaction. However, it is a series of elementary reactions.

PS: In a rate law, if there is a concentration in the denominator, it is probably the species that is colliding with the active intermediate. If there is a constant in the denominator, that probably implies there is a reaction step which is the decomposition of the active intermediate. If there is a concentration in the numerator, that probably says there is a step to produce the active intermediate.