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My favorite branch of chemistry is organic chemistry. Organic chemists can look at two reactants and tell you if a chemical reaction will happen. More incredibly, they can tell you what bonds break and form. Organic chemists always think about reaction mechanisms.

What is a reaction mechanism?

A chemical equation tells you the reactants and products. You know where you started and where you ended. But, what happens in the middle? A reaction mechanism describes all of a chemical reaction's steps.

Reaction mechanisms start as theoretical proposals. The chemist supports his hypothesis with experimental evidence. One option is to use spectroscopy to show the intermediates. Another type of evidence is the experimentally determined rate law. For example, if the reaction is second-order for one reactant, the mechanism must show that. Finally, the reaction mechanism must be thermodynamically favorable.

What Can a Reaction Mechanism Tell You?

In general chemistry, the reaction mechanism describes a reaction's kinetics. The mechanism tells us...

• the reaction intermediates.

• the reaction's molecularity.

• the rate-limiting step.

In organic chemistry, a reaction mechanism can tell you...

• where electrons move.

• what bonds break.

• what new bonds form.

• where atoms move.

• the molecular geometry.

Example Reaction Mechanism

Let's look at the single replacement reaction between nitrogen dioxide gas and carbon monoxide gas.

$NO_{2(g)} + CO_{(g)} \rightarrow NO_{(g)} + CO_{2(g)}$

The experimentally determined rate law is $rate=k[NO_{2(g)}]^{2}$. It is a second-order reaction for nitrogen dioxide. So, our reaction mechanism should depend on a collision between nitrogen dioxide molecules.

$Step\,1: 2NO_{2(g)} \rightarrow NO_{3(g)} + NO_{(g)}$

$Step\,2: NO_{3(g)} + CO_{(g)} \rightarrow NO_{2(g)} + CO_{2(g)}$

We can treat reaction mechanisms like addition problems. We can add the products and reactants for each step. Anything that is on both sides cancels.

$Step\,1: NO_{2(g)} + NO_{2(g)} \rightarrow NO_{3(g)} + NO_{(g)}$

$Step\,2: NO_{3(g)} + CO_{(g)} \rightarrow NO_{2(g)} + CO_{2(g)}$

$Overall: NO_{2(g)} + \cancel{NO_{2(g)}} + \cancel{NO_{3(g)}} + CO_{(g)} \rightarrow \cancel{NO_{3(g)}} + NO_{(g)} +\cancel{NO_{2(g)}} + CO_{2(g)}$

So, our reaction mechanism looks like this.

$Step\,1: NO_{2(g)} + NO_{2(g)} \rightarrow NO_{3(g)} + NO_{(g)}$

$Step\,2: NO_{3(g)} + CO_{(g)} \rightarrow NO_{2(g)} + CO_{2(g)}$

$Overall: NO_{2(g)} + CO_{(g)} \rightarrow NO_{(g)} + CO_{2(g)}$

Elementary Reactions

Each step in a reaction mechanism is called an elementary reaction. Elementary reactions are a single step. They have no intermediates! But, they do have a transition point. The energy of the transition point is the activation energy. On the energy diagram, the elementary reactions are the two hills. The first starts at the reactants and ends at the intermediate point. The second begins with the intermediates and ends with the products. The combination of the two generates the overall reaction's energy diagram.

Each elementary reaction has a rate law based on its molecularity. A reaction's molecularity is the number of molecules that collide. For example, the reaction $2NO_{2(g)} \rightarrow NO_{3(g)} + NO_{(g)}$, is bimolecular. Two $NO_{2(g)}$ molecules collide. So, we would write the rate law as $rate=k[NO_{2(g)}]^{2}$. For elementary reactions, the rate law's exponent equals the stoichiometric coefficient.

Using Elementary Reactions to Create Rate Laws (Rate Determining Steps)

Why should we care about elementary reactions? They help us determine the rate law. For example, let's say we have two elementary reactions. One is fast and the other slow. The slower reaction determines our rate law. So, we call it the rate-determining step. Let's look at two examples.

Slow First Step

Let's return to our nitrogen dioxide example.

$Step\,1: NO_{2(g)} + NO_{2(g)} \rightarrow NO_{3(g)} + NO_{(g)}$

$Step\,2: NO_{3(g)} + CO_{(g)} \rightarrow NO_{2(g)} + CO_{2(g)}$

$Overall: NO_{2(g)} + CO_{(g)} \rightarrow NO_{(g)} + CO_{2(g)}$

We can write the rate law for each elementary reaction. The $k'$ in step 2 denotes a different rate constant than step 1.

$Step 1: rate=k[NO_{2(g)}]^{2}$

$Step 2: rate=k'[NO_{3(g)}][CO_{(g)}]$

For this reaction, step one is slow. Step two is fast. So, every molecule of $NO_{3(g)}$ gets consumed immediately. The rate of the overall reaction depends on step one. Therefore, our rate law for the overall reaction is $rate=k[NO_{2(g)}]^{2}$.

Fast First Step

Sometimes a reaction has a fast initial step. It's more complicated. But we can do it!

Let's look at the synthesis reaction between nitrogen monoxide and diatomic chlorine.

$step\,1: NO_{(g)} + Cl_{2(g)} \rightleftharpoons NOCl_{2(g)}$

$step\,2: NOCl_{2(g)} + NO_{(g)} \rightarrow 2NOCl_{(g)}$

$overall: NO_{(g)} + Cl_{2(g)} \rightarrow 2 NOCl_{(g)}$

Every question with a fast initial step will be in equilibrium. So we can write two rate laws. One is for the forward reaction. The second is for the reverse reaction. For both, the rates are equal.

$rate=k_{1}[NO_{(g)}][Cl_{2(g)}]=k'_{1}[NOCl_{2(g)}]$

For step 2, the rate law is $rate=k_{2}[NOCl_{2(g)}][NO_{(g)}]$.

Since step 2 is slower, it is our rate-determining step. But, we have a problem. Step 2 depends on the intermediate, $[NOCl_{2(g)}]$. We need to solve for $[NOCl_{2(g)}]$.

$k_{1}[NO_{(g)}][Cl_{2(g)}]=k'_{1}[NOCl_{2(g)}] \rightarrow [NOCl_{2(g)}]=\frac{k_{1}}{k'_{1}}[NO_{(g)}][Cl_{2(g)}]$

Now we can plug $[NOCl_{2(g)}]$ into our step 2 rate law.

$rate=k_{2}[NOCl_{2(g)}][NO_{(g)}] \rightarrow rate=k_{2}\times\frac{k_{1}}{k'_{1}}[NO_{(g)}][Cl_{2(g)}][NO_{(g)}] \rightarrow rate=K[NO_{(g)}]^{2}[Cl_{2(g)}]$

(Video) Reaction Mechanism: Energy Diagrams, Catalysts, and Reaction Mechanisms

By Professor Dave Explains

In this video, Professor Dave explains reaction mechanism is the series of steps that occur from the reactants to the products. You can find the rate-determining step, the slowest step in a reaction mechanism, when you break down a reaction into its elementary steps. Professor Dave also points out that a specie that appears in elementary steps but not in the overall reaction is called reaction intermediate.

Reaction Mechanism: What is Reaction Mechanism?