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Mechanisms

>> Parts of this equation/concept include:

A -	Components of Mechanisms
B -	Molecularity
C -	Rate Laws from Mechanisms >> Preceding Reversible Step

A mechanism describes the sequence of events that occur during the course of the reaction. It should also have some expression of rate (at least fast and slow). The sum of all the steps in a mechanism must result in the overall reaction. The overall mechanism must also be consistent with the rate law. If either of these criteria is not met, the mechanism is wrong.

A substance that is formed in one step of a mechanism and used up in a subsequent step is an intermediate. A substance that is used up in one step and reformed in a subsequent step is a catalyst. (Assuming the substance increases rate. It is an inhibitor if it decreases rate.) Activated complexes are something between the reactant and product of an elementary step. In an activated complex, bonds are only partially broken or formed. These do not appear in the mechanism.

Intermediates and catalysts will not appear in the overall reaction. Substances that do appear in the overall reaction are, of course, products (on the right of the arrow) or reactants (on the left of the arrow).

>> Example 1

Consider the following mechanism. Determine the overall reaction and identify reactants, products, intermediates, and catalysts.

CH₃OH + H⁺ CH₃OH₂       (slow)

CH₃OH₂ CH₃ + H₂O       (fast)

CH₃COOH H⁺ + CH₃CO₂^–        (fast)

CH₃CO₂^– + CH₃ CH₃CO₂CH₃       (fast)

Solution:

To add the reactions together, list every substance on the reactant side and every substance on the product side.

CH₃OH + H⁺ + CH₃OH₂ + CH₃COOH + CH₃CO₂^– + CH₃
                                         CH₃OH₂ + CH₃ + H₂O + H⁺ + CH₃CO₂^– + CH₃CO₂CH₃

Then cancel each substance that appears on both sides of the equation, to determine the overall reaction.

CH₃OH + CH₃COOH H₂O + CH₃CO₂CH₃

Consequently, the reactants are CH₃OH and CH₃COOH. The products are H₂O and CH₃CO₂CH₃.

Intermediates are a product before they are a reactant. Intermediates are CH₃OH₂, CH₃, and CH₃CO₂^–.

Catalysts are used up, then reformed. H⁺ is a catalyst in this reaction.

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The term molecularity describes each step of the mechanism in terms of the number of reactants. The assumption is that each elementary step is the result of a collision and that the molecularity describes the collision.

>> Example 2

Describe the molecularity of each step in the following mechanism.

CH₃OH + H⁺ CH₃OH₂       (slow)

CH₃OH₂ CH₃ + H₂O       (fast)

CH₃COOH H⁺ + CH₃CO₂^–        (fast)

CH₃CO₂^– + CH₃ CH₃CO₂CH₃       (fast)

Solution:

The first and last steps are bimolecular (two reactants). The two middle steps are unimolecular (one reactant).

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Rate is determined from the slowest step in the mechanism. For this reason the rate-determining step is the slowest step.

The rate law is based on the slow step. Rate is proportional to the product of the concentration of reactants in the slow step.

>> Example 3

What rate law is predicted by the following mechanism?

CH₃OH + H⁺ CH₃OH₂       (slow)

CH₃OH₂ CH₃ + H₂O       (fast)

CH₃COOH H⁺ + CH₃CO₂^–       (fast)

CH₃CO₂^– + CH₃ CH₃CO₂CH₃       (fast)

Solution:

The reactants are CH₃OH and H⁺ in the slow step. Since rate is proportional to the concentration of those substances, the rate law for this mechanism is

rate = k[CH₃OH][H⁺]

Note that it conforms to the general form of the rate law.

>> Preceding Reversible Step

Intermediates should not appear in the rate, although reactants, catalysts, and even products may. This will not be an issue if the first step is the slow step.

If an intermediate does act as a reactant in the slow step, its concentration should be written in terms of reactants from previous steps. This is most easily done if the previous step is reversible.

A fast, reversible step will soon reach a point where the rate of the forward reaction is the same as the rate of the reverse reaction (equilibrium). Because of this, a rate law can be written for both the forward and reverse reactions. Because the rates are equal, these two equations can be combined. Solving the combined equation for the concentration of intermediate, gives intermediate concentration in terms of reactants. This expression replaces the concentration of intermediate in the rate law.

>> Example 4

What is the rate law predicted by the following mechanism?

H₂ 2 H       (fast, reversible)

H + SO₂ HSO₂       (slow)

HSO₂ HSO + O       (fast)

HSO HS + O       (fast)

2 O O₂       (fast)

HS + H H₂S       (fast)

Solution:

Because each elementary step has its own rate constant, the easiest way to keep track of the rate constants is to use the number of the step as a subscript. (A negative number represents a reverse step.)

The slow step predicts a rate law of rate = k₂[H][SO₂]. However, H is an intermediate. The previous step is reversible. The rate predicted by the forward reaction is rate = k₁[H₂]. The rate law predicted by the reverse of the first step is rate = k_–1[H]² (Stoichiometric coefficients become exponents, since 2 H really means H + H. Since rate laws are the product of the reactants, rate = k[H][H] = k[H]².)

For the first step, the rate of the forward reaction equals the rate of the reverse reaction so

rate forward = k₁[H₂]

rate reverse = k_–1[H]²

rate forward = rate reverse

k₁[H₂] = k_–1[H]²

Solving for the intermediate,

k₁[H₂]/k_–1 = [H]²

(k₁[H₂]/k_–1)^1/2 = [H]

This expression is used to replace the intermediate in the rate expression of the slow step.

rate = k₂[H][SO₂]

rate = k₂(k₁[H₂]/k_–1)^1/2[SO₂]

Collecting all the rate constants, the rate law becomes

rate = k₂(k₁/k_–1)^1/2[H₂]^1/2[SO₂]

Fortunately, a constant times a constant, and even the square root of constants, just results in a different constant. Using k for the combined constant, the rate law (finally) is

rate = k[H]^1/2[SO₂]

Mechanisms are theoretical explanations for observed experimental results. The rate law determined from experiments with various initial concentrations or the change in concentration with time are correct. If the rate law predicted by the mechanism is not the same as the experimental rate law, the mechanism is wrong. If the rate law predicted by the mechanism matches the rate law, the mechanism might be correct.

>> Example 5

Which mechanism corresponds to the reaction SO₂ + H₂O H₂SO₃ if the rate law for the reaction is rate = k[H₂O]?

1. H₂O H + OH        (slow)

   SO₂ + OH HSO₂       (fast)

   HSO₂ + H H₂SO₃       (fast)

2. SO₂ + H₂O HSO₂ + OH       (slow)

   OH + HSO₂ H₂SO₃       (fast)

3. H₂O H + OH        (slow)

   H + SO₂ HSO₂       (fast)

   HSO₂ + OH H₂SO₃       (fast)

4. H₂O H + OH       (slow)

   OH + SO₂ HSO₂ + O       (fast)

   O + HSO₂ HSO₃       (fast)

   HSO₃ + H H₂ + SO₃       (fast)

Solution:

The rate law is determined by the slow step, which is the first step for each of these proposed mechanisms. Therefore the rate law is easily written.

1. The proposed rate law is rate = k[H₂O]. This is consistent with the experimental rate law, so this mechanism may be correct.
2. The proposed rate law is rate = k[H₂O][SO₂]. This does not match the experimental rate law, so this mechanism is incorrect.
3. The proposed rate law is rate = k[H₂O]. This is consistent with the experimental rate law, so this mechanism may be correct.
4. The proposed rate law is rate = k[H₂O], but the overall reaction does not correspond to the experimental reaction. Therefore this mechanism is incorrect.

Either mechanism a or c is correct. There is insufficient evidence to determine which. It is also possible that the correct mechanism is something completely different.

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