The mechanism of a reaction is the sequence of the single reaction steps that sum to the overall reaction. Generally, we are presented with an overall reaction with no idea how we go from reactants to products. For example, below is an overall equation.
It is not obvious from the equation, but this reaction occurs in two steps. The proposed mechanism is below.
O + O3 (g) → 2 O2 (g)
Note, the two equations sum to the overall equation. In the first equation, an oxygen-oxygen bond in the O3 breaks leaving O2 and an oxygen atom. In the second step, the oxygen atom reacts with another O3 molecule resulting in 2 molecules of O2. Each step in a reaction mechanism is called an elementary step. An elementary step cannot be broken down into any further steps.
The oxygen atom is produced in the first step and is then consumed in the second step. The oxygen atom is an intermediate. These transient species, intermediates, are always formed in one step and then consumed in another step of the reaction mechanism. They tend to react quickly, and are high energy species. Because they are so energetic, they are seldom isolated.
An elementary step is characterized by its molecularity. Molecularity refers to the number of reactant particles involved in the reaction. For the first step of the mechanism discussed above:
the elementary step is said to be unimolecular because only one reactant particle is involved. For the second step
the reaction is bimolecular because there are two reactant particles involved. For the reaction
the reaction is termolecular because there are 3 reactant particles involved. Molecularity can be used to determine the number of steps in a reaction and the speed (kinetics) of a reaction.
The rate law for an elementary step can be determined from the stoichiometric coefficients of the elementary reaction. The reaction order is equal to the molecularity of an elementary step.
A valid reaction mechanism must meet three criteria.
2. The elementary steps must be reasonable.
3. The mechanism must correspond to the observed rate law.
The slowest step in a mechanism is the rate determining or rate limiting step. The rate law is written from the slow step in a reaction mechanism. Think of a single lane of traffic that is traveling through a tunnel. Once the cars have gone through the tunnel, the road turns into 3 lanes. The time it takes to get through the tunnel will depend on the speed the cars are traveling. If there is a car that is going very slow, then all of the cars behind that car will also have to decrease their speed. This is the same with a chemical reaction–the slowest step will determine the rate of the reaction.
The slowest step in the reaction is the rate determining step. Here, we will look at two different cases.
Case 1. If a mechanism has the first step as the slow step, we can write the rate law from that initial step.
For the overall reaction,
the mechanism is,
NO3 (g) + CO (g) → NO2 (g) + CO2 (g) (fast)
and the experimental rate law is rate = k[NO2]2.
Let’s check to see if the above mechanism is plausible. Write the rate law directly from the slow step.
Rate = k[NO2]2
This agrees with the experimental rate law. Next, check the elementary steps are reasonable. They are because both steps are bimolecular which means the reactions are likely to proceed. Finally, add the reaction mechanism steps to assure they sum to the overall equation.
\(\displaystyle \require{cancel}\cancel{NO_2}\;(g)\;+\;NO_2\;(g)\;→\;\require{cancel}\cancel{NO_3}\;(g)\;+\;NO\;(g)\)
\(\displaystyle \require{cancel}\cancel{NO_3}\;(g)\;+\;CO\;(g)\;→\;\cancel{NO_2}\;(g)\;+\;CO_2\;(g)\)
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\(\displaystyle NO_2\;(g)\;+\;CO\;(g)\;→\;NO\;(g)\;+\;CO_2\;(g)\)
We can see the elementary steps sum to the overall equation. This mechanism satisfies all three criteria and is therefore a plausible mechanism. Note, the NO3 is formed in step 1, and it is consumed in step 2. This is an intermediate. Intermediates are high energy substances with fully formed bonds that can rarely be isolated. They are formed in one step and consumed in another step.
Let’s look at another reaction mechanism for the iodide catalyzed decomposition of hydrogen peroxide with an experimental rate law of rate = k[H2O2][I–]. The reaction mechanism is:
\(\displaystyle H_2O_2\;+\;I^-\;\underrightarrow{k1}\;H_2O\;+\;IO^-\;\;\;(slow)\) \(\displaystyle H_2O_2\;+\;IO^-\;\underrightarrow{k2}\;H_2O\;+ O_2\;+\;I^-\;\;\;(fast)\)Writing the rate law from the first step:
rate = k[H2O2][I–]
This agrees with the experimental rate law. Notice the IO– ion is an intermediate because it is formed in the first step and consumed in the second step. The I – ion is a catalyst. A catalyst is a substance that will speed up a reaction by lowering the activation energy. It provides an alternate pathway for the reaction to occur. A catalyst is never consumed in a reaction. The I – ion is a reactant in the first step and is regenerated in the second step as a product.
Case 2. Below is a mechanism with a fast initial step. The rate law is written from the slow step.
Do you notice a problem with the rate law as written? It has an intermediate, and we cannot have intermediates in a rate law. We need to eliminate the intermediate with some basic algebra. Recall, for the fast equilibrium step the rate of the forward reaction is equal to the rate of the reverse reaction:
Solve the equation for the intermediate, N2O2.
This can be plugged into the rate law, rate = k[N2O2][Br2], from the slow step for the concentration of the intermediate, N2O2.
\(\displaystyle rate\;=\;\frac{k_1}{k_{_1}}[NO]^2[Br_2]\)The rate constant, k, is a constant, and a constant divided by a constant is also a constant. The rate law is
\(\displaystyle rate\;=\;k[NO]^2[Br_2]\)
In summary, when the first step is a fast equilibrium reaction, write the rate law from the slow step. Intermediates cannot be in the rate law, so use some algebra to eliminate the intermediate. As for catalysts, a catalyst can be in the rate law but it is not required to be in the rate law. It depends if the catalyst is in the rate determining step.
Exercises
Exercise 1. Indicate the molecularity of the following elementary steps.
b) NO3 (g) + CO (g) → NO2 (g) + CO2 (g)
c) NH3 (g) + HCl (g) → NH4Cl (s)
Check Solution/Answer to Exercise 1
Exercise 2. Explain the difference between a transition state and a reaction intermediate.
Check Solution/Answer to Exercise 2
Exercise 3. Consider the following mechanism.
I– (aq) + HClO (aq) → HIO (aq) + Cl– (aq) (slow)
OH– (aq) + HIO (aq) → H2O (l) + IO– (aq) (fast)
a) What is the overall equation?
b) Indicate if there are intermediates and identify them
c) What are the molecularity and rate law for each step?
d) The experimental rate law was determined to be rate = k[ClO–][I–]. Is the above mechanism a plausible mechanism? Explain.
Check Solution/Answer to Exercise 3
Exercise 4. Consider the following mechanism to answer the questions.
\(\displaystyle IO^-\;+\;H_2O\;→\;HOI\;+\;HO^-\;\;\;(fast)\)
\(\displaystyle HOI\;+\;I^-\;→\;I_2\;+\;HO^-\;\;\;(fast)\)
a) Write the rate law.
b) Write the overall reaction.
c) Identify any intermediates
Check Solution/Answer to Exercise 4
Exercise 5. Consider the following mechanism to answer the questions.
\(\displaystyle ClO\;(g)\;+\;O(g)\;→\;Cl\;(g)\;+\;O_2(g)\;\;\;(slow)\)
a) Write the overall equation.
b) Write the rate law
c) Identify intermediates and catalysts, if any.
Check Solution/Answer to Exercise 5
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