Velocity law for non-elementary reactions

As explained in the text Law of speed of chemical reactions, the equation used to represent the law of the speed of a reaction is given by the product of the constant characteristic of the reaction at a certain temperature and the concentrations of the reactants raised to their respective exponents: v = k. [THE]^α. [B]^β.

See an example:

2NO_(g) → N₂O_2(g)

The equation for the speed of this reaction is given by: v = k. [AT THE]².

Does this mean that in all cases the exponent of the concentration of the reactant will be exactly equal to its coefficient in the reaction?

Do not. This only happened in this case because it is an elementary reaction, that is, it is a reaction that takes place in a single step, without intermediate compounds. In cases where the reaction is not elemental, the exponents must be determined experimentally.But how is this done? And how is it possible to know if the reaction is elementary or not?

Well, let's consider another reaction:

CO + NO₂ → CO₂ + NO

Let's say a scientist performed this reaction several times, changing the concentration of the reactants in different ways, but keeping the temperature constant. He obtained the following data:

Note that from the first to the second step, he doubled the CO concentration, which did not change the reaction rate.

Therefore, the exponent of this substance is zero. Since any number raised to zero equals 1, CO does not participate in the reaction rate equation.

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Now, see that from the 2nd experiment to the 3rd the NO concentration doubled₂, which caused the reaction speed to quadruple.

Thus, the exponent of the concentration of this substance in the equation for the rate of reactions is equal to 2 (4/2).

In this way, we find out what the equation for the speed of this reaction is: v = k. [AT THE₂]².

Note that in this case the exponent in the equation was not equal to the coefficient in the reaction. Therefore, we can conclude that this reaction is not elementary. After experimentally verifying the law of speed, the scientist should then suggest a mechanism that explained this reaction, that is, it should propose a set of steps consistent with the experimental data of this process.

The following mechanism was proposed:

Stage 1 (slow):  AT THE_2(g) + NO_2(g) → NO_3(g) + NO_(g)
Step 2 (quick):AT THE_3(g) + CO_(g) → CO_2(g) + NO_2(g)

Global equation:CO + NO₂ → CO₂ + NO

See that the law of experimental speed coincides with the slowest step:

v_global = v_{slow step}

k. [AT THE₂]² = k. [AT THE₂]. [AT THE₂]

This shows us that, in any mechanism, the stage that determines the rate of development of a reaction will always be the slow step, that is, the rate of development of the global reaction will be proportional only to the concentrations of the reagents that participated in the slow step.

It is important to correctly determine these exponents because they are the ones that will indicate the order of the reaction.

By Jennifer Fogaça
Graduated in Chemistry

Would you like to reference this text in a school or academic work? Look:

FOGAÇA, Jennifer Rocha Vargas. "Law of speed for non-elementary reactions"; Brazil School. Available in: https://brasilescola.uol.com.br/quimica/lei-velocidade-para-reacoes-nao-elementares.htm. Accessed on June 27, 2021.