# Reaction mechanisms

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##### Intros
###### Lessons
1. How do reactants become products?
2. Steps in chemical reactions.
3. Why one reaction step matters most.
4. Drawing activated complexes and transition states.
5. Writing elementary steps and overall reactions.
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##### Examples
###### Lessons
1. Identify products and intermediates in a multi-step chemical reaction.
The reactions below are some elementary steps in the free-radical substitution reaction between methane and chlorine.

Cl2 $\,$$\,$ 2Cl·
CH4 + Cl· $\,$$\,$ CH3· + HCl
CH3· + Cl2 $\,$$\,$ CH3Cl + Cl·
2Cl· $\,$$\,$ Cl2
1. Identify the reactants, products and intermediates in this series of reactions.
2. Write the rate law for a balanced chemical equation.
The reaction below is the rate determining step of a chemical process.

2A + 3B $\,$$\,$ C + 2D

1. Write the rate law expression for this reaction. What is the order with respect to B?
2. In an experiment to measure the rate of reaction, by how much would you expect the rate increase if the concentration of B was doubled?
###### Topic Notes

In this lesson, we will learn:

• The definition of reaction mechanism, transition state and intermediate.
• How to interpret reaction mechanisms in steps and recognize the rate-determining step (RDS) in a chemical process.
• To understand rate laws as expressions of the rate-determining step.

Notes:

In this lesson, we will learn:

• Collision theory states that the reactant particles must collide with sufficient energy and correct orientation. What about with complicated reactions?
• If a chemical reaction involves 5, 10, or even more reactant molecules converting to different products, what is the chance that all of them will collide in the same space at the same time? It’s extremely unlikely.
However, complex reactions do happen at respectable rates. This suggests that reactions do not happen all in one step – reactions often happen in a sequence of steps which, combined, form the overall reaction. The specific steps in how reactants form the products is called the reaction mechanism.

• Just as the reaction rate for an overall reaction can be measured, the rate of individual or elementary steps in reactions can be studied and measured.
• Even though chemical reactions can have many steps, there is normally one elementary step that is orders of magnitude slower (hundreds or thousands of times) than the other steps. This is the rate-determining step that practically sets the overall rate because this step is the rate bottleneck of the rest of the reaction.
• Studying individual steps of a reaction is very complicated, but the concentration of reactants, intermediates (substances formed and used up entirely during the reaction) and products can be measured. This helps identify the rate determining step (RDS).
• Because the RDS is so much slower than the other steps, this is the only step that ‘bottlenecks’ reactants, leading to a noticeable concentration of these reactants if the reaction is analyzed while taking place.
• In the same way as the RDS bottlenecks, the other steps have relatively no influence – increasing concentration of reactants in the non-rate-determining steps doesn’t affect the overall rate. The rate is solely determined by the concentration of reactants in the rate-determining step (RDS).

• To write an overall reaction equation, all the elementary steps must be combined and the intermediates cancelled out from both sides.
• EXAMPLE: Consider the reaction below which has two elementary steps:
Step one: $\mathrm{A \to B + C}$
Step two: $\mathrm{A + B \to 2C}$
In the above reaction, C is called an intermediate because it is formed entirely and used up entirely during the reaction.
• It is only an intermediate if every single molecule of the substance is produced and then used up in the middle of the reaction. There will be no trace of it in the overall reactants or products.
Intermediates and elementary steps must still balance when combining to get an overall equation – use this and the fact that intermediates entirely cancel out to balance these equations!
The overall equation for the example above would be worked out like this:
Step one: $\mathrm{A \to B + C}$
Step two: $\mathrm{A + B \to 2C}$
Combined: $\mathrm{2A + B \to B + 3C}$
Cancel out: $\mathrm{2A + \cancel{B} \to \cancel{B} + 3C}$
Overall equation: $\mathrm{2A \to 3C}$

• The elementary steps of the reaction show the mechanism - practice looking at the difference between overall equations and elementary steps. The overall equation and elementary steps are often contradictory:
• The two elementary steps of the example show a molecule of reactant A decomposes to form product C and intermediate B, which then reacts with another reactant A molecule to form two molecules of product C.
• However, the overall equation suggests two molecules of reactant A react together to form three molecules of product C.
If you need to describe the mechanism, you must discuss the elementary steps!

• Each elementary step can be treated as an individual process in terms of the activated complex that the molecules make when they collide.
• If you need to draw an 'activated complex' for an elementary step of a reaction then use dashed lines to represent bonds in the middle of being broken and formed.

• When drawing energy diagrams of reaction mechanisms, show each step individually – each elementary step will have a flat section indicating the intermediate(s), and an activation energy barrier.
• Each activation energy 'hump' is the boundary between the chemical substances changing, and one step of the reaction moving into the next. The number of 'humps' should be the same as the number of reaction step.

• As said above, when finding reaction rate there is usually only one elementary step that is important because one is much slower than all the other steps – the rate-determining step (RDS). The rate determining step is used to derive the rate law.
• The rate law is a mathematical expression that relates rate of reaction to the concentration of reactants in the rate determining step. Generally:
For the reaction with rate determining step: $xA + yB \to products$

$Rate = k[A]^x[B]^y$

Where $k = rate\;constant$
x = rate order with respect to A
y = rate order with respect to B

As the equation shows, the rate is found by multiplying a rate constant k (a constant with a unique value for every chemical reaction) by the concentration of the reactants ([A] and [B]) in the rate-determining step raised to the power of their molar ratio (x and y respectively).

• There are some types of reactions where the rate law is easy to find – simple reactions involving only one (unimolecular) or two (bimolecular) reactant molecules.
• In unimolecular reactions (most decomposition reactions), the reaction is wholly dependent on one molecule and the breaking of the bonds. The rate law for unimolecular reactions (whether an elementary or overall step) is always:
$A \to products$

$Rate = k[A]$