# General Chemistry/Reaction Mechanisms

 ← Reaction Rates · General Chemistry · Thermodynamics → Book Cover · Introduction ·  v • d • e

A step-by-step breakdown of a reaction shows its mechanism, the actual process of reactions becoming products.

Chemists often write chemical equations for reactions as a single step, which only shows the net result of a reaction. However, most chemical reactions occur in a series of steps called elementary reactions. All of these elementary reactions must add up to equal the overall balanced equation. The complete sequence of these elementary steps is called a reaction mechanism. The reaction mechanism is the step-by-step process by which reactants actually become products. It is the "how" of the reaction, whereas the overall balanced equation only shows the "what" of the reaction.

## Rate-Determining Steps

Sometimes, intermediate substances are created in the process that disappear in the end. Take the following example of a homogeneous reaction (where products and reactants are all in the same phase):

${\displaystyle {\hbox{CO}}+{\hbox{NO}}_{2}\to {\hbox{CO}}_{2}+{\hbox{NO}}}$

There are actually two reactions occurring at different speeds.

• ${\displaystyle 2{\hbox{NO}}_{2}\to {\hbox{NO}}_{3}+{\hbox{NO}}}$ (slow)
• ${\displaystyle {\hbox{NO}}_{3}+{\hbox{CO}}\to {\hbox{NO}}_{2}+{\hbox{CO}}_{2}}$ (fast)

Since the first step is the slowest, and the entire reaction must wait for it, it is known as the rate-determining step. The overall reaction rate depends almost entirely on the rate of the slowest step. The other steps are fast enough that their rate is insignificant, as they are always waiting for the slower step to complete.

Why is the first step slower? Collision theory explains whether or not particles react when they collide. The particles must collide with a minimum energy and a proper orientation if a reaction is to occur.

The minimum energy needed for a reaction to occur is its activation energy. The particles must be moving fast enough for their collision to satisfy the activation energy. Without the necessary energy, the particles will bounce off each other with no reaction. A good example of activation energy is a butane lighter, which needs a spark before the fluid burns. The spark provides enough energy to the particles to make their collisions effective.

Reactions with high activation energy will be slower than those with low activation energy. With a high activation energy, less particles are likely to generate the needed energy when they collide. Reactions that break bonds, especially double or triple covalent bonds, will have higher activation energy.

The particles must also collide with the proper orientation. For example, the reaction in the animation above shows an ammonium ion reacting with a NCO- ion. For the reaction to occur, the ammonium ion must collide with the nitrogen of the NCO-. If the ions don't collide in the right place while facing in the right direction, the reaction cannot occur.

Reactions that have very specific requirements for the orientation of the colliding particles will be much slower. Reactions that can occur without a specific orientation will happen faster.

## Determining the Validity of an Elementary Step Model

In order for a proposed elementary step equation to be valid it must fulfil these requirements.

1. The rate equation of the slow step must match the rate equation of the overall reaction.

2. The reactants and products of all the elementary steps added together must equal the one of the actual equation.

3. The rate law for the elementary step must be able to be written without the concentration of the intermediates, because the intermediates are too small to be accurately measured.

For example, if given this equation and asked to prove the rate mechanism

Equation ${\displaystyle {\hbox{2O}}_{3}\to {\hbox{3O}}_{2}}$

Proposed Mechanism

${\displaystyle {\hbox{O}}_{3}\to {\hbox{O}}_{2}+{\hbox{O}}}$ (fast)

${\displaystyle {\hbox{O}}_{3}+{\hbox{O}}\to {\hbox{2O}}_{2}}$ (slow)

${\displaystyle rate=k[O_{3}]^{2}}$

We know that when the reaction occurs backwards, the k changes from k to k^{-1}. So, we substitute in the backwards rate for O_3, and plug it into the slow step. Then, take the rate equation of the slow step. We find that it is the same as the rate law for the overall equation and meets the three criteria above, so this proposed mechanism is valid.