13.7-13.9: Mechanisms-Catalysis

Mechanisms:

Reaction Mechanism: The set of elementary reactions whose overall effect is given by the net chemical equation. An elementary reaction is a single molecular event, such as a collision of molecules, resulting in a reaction.

Example:

Consider the reaction of nitrogen dioxide with carbon monoxide:

NO₂ (g) + CO (g) → NO (g) + CO₂ (g) (net chemical equation)
NO₂ + NO₂ → NO₃ + NO (elementary reaction)
NO₃ + CO → NO₂ + CO₂ (elementary reaction)

Note: Elementary reactions don't have phase labels. Also, NO₃ is a molecule, not the nitrate ion, and is highly unstable. It is both created and destroyed by the reaction and is called a reaction intermediate. In the last step, the CO molecule needs something desperately energetic to smack into in order to react, and the NO₃ provides that.

Following Hess's Law, obtain the overall equation by adding the elementary reactions and canceling species that occur on both sides:

NO₂ + NO₂ → NO₃ + NO (elementary reaction)
NO₃ + CO → NO₂ + CO₂ (elementary reaction)
-------------------------------------
NO₂ +NO₂ + NO₃ + CO → NO₃ + NO + NO₂ + CO₂ (overall)

Molecularity:

Molecularity: the number of molecules on the reactant side of an elementary reaction. Products are not considered in molecularity, only reactants of the elementary reaction.

Example:

Unimolecular: an elementary reaction that involves one reactant molecule.
O₂ → 2O

Bimolecular: an elementary reaction that involves two reactant molecules.
NO₂ + CO → NO + CO₂

Termolecular: an elementary reaction that involves three reactant molecules.
Br₂ +Cl₂ + C → CBr₂Cl₂ (generally exceptionally slow because highly unlikely statistically, remember Civil War colliding bullets example).

Determining the Rate Law from an Elementary Reaction:

Example:

What is the rate law for an overall reaction? In order to do this, slow step and fast step must be labeled. Need to know the slow step because it moderates the reaction; the reaction can't proceed faster than the slowest step. So the rate law for the slow step is the rate law for the overall reaction.
Note: These are both elementary reactions, the (slow step) and (fast step) imply that. Also, k is unique for each step. Once the slow step is identified, use the general form to assume the rate equation for this step:

Rate = k₁[NO₂][F₂]

Note: The components in the rate law must be part of the overall reaction. Intermediates cannot be part of the rate law.

This is pretty straightforward for reactions with an initial slow step. It's more complicated for reactions with an initial fast, equilibrium step.

Example:

Mechanism:


When we build the rate law from our slow step, we have an intermediate which needs to be eliminated:
Rate = k₂[NO₂][NO₃]

Use the fast step to find something plausible to replace [NO₃] with:

Rate_f = k₁[N₂O₅]
Rate_r = k_-1[NO₂][NO₃]

Where Rate_f is the forward rate and Rate_r is reverse rate (k₁ and k_-1 also mean forward and reverse rate). Also, at equilibrium, by definition, the forward and reverse rates are equal to each other. They are at dynamic equilibrium, where the forward reaction and the reverse reaction are happening at the same time, all the time.

So:
k₁[N₂O₅] = k_-1[NO₂][NO₃]
Solve for the component you want to eliminate, the intermediate [NO₃]:
[NO₃] = k₁[N₂O₅] / k_-1[NO₂]
Substitute this into the rate law from the slow step for [NO₃]:
Rate = k₂[NO₂] x (k₁[N₂O₅] / k_-1[NO₂])
([NO₂] will cancel)
Rate = (k₂k₁ / k_-1) [N₂O₅]
Where all k represent a constant and can be combined into a single k to get:
Rate = k[N₂O₅]

Catalysis:

Catalysts: cause a reaction to proceed faster without being consumed by the reaction. Catalysts do this by lowering the activation energy, E_a.



The solid line represents the reaction without a catalyst. It requires more energy to reach the activated state and thus has a higher E_a. The dotted line represents the reaction with the addition of a catalyst. It requires much less energy and thus has a much lower E_a.

Homogeneous Catalysis: the use of a catalyst in the same phase as the reacting species. An example would be a gas phase catalyst and gas phase reacting species; they are all in the same phase (gas).

Heterogeneous Catalysis: the use of a catalyst that exists in a different phase from the reacting species, usually a solid catalyst in contact with a gaseous or liquid solution of reactants.