AP Chemistry 5.11 Catalysis Study Notes - New Syllabus Effective fall 2024
AP Chemistry 5.11 Catalysis Study Notes.- New syllabus
AP Chemistry 5.11 Catalysis Study Notes – AP Chemistry – per latest AP Chemistry Syllabus.
LEARNING OBJECTIVE
Explain the relationship between the effect of a catalyst on a reaction and changes in the reaction mechanism.
Key Concepts:
- Catalysts and Lower Activation Energy Pathways
- Catalyst Regeneration and Constant Concentration
- Catalyst Binding and Orientation Effects (Enzyme-Type Catalysis)
- Covalent Bonding in Catalysis (Acid–Base Catalysis)
- Surface Catalysis and Adsorbed Intermediates
Catalysts and Lower Activation Energy Pathways
A catalyst is a substance that increases the rate of a chemical reaction by either providing an alternate reaction pathway with a lower activation energy or by increasing the number of effective collisions between reactant molecules. Importantly, a catalyst is not consumed in the overall reaction—it is regenerated by the end of the process.
Key Concepts:
- The presence of a catalyst lowers the required activation energy for the reaction, making it easier for reactant particles to form products.
- Because more molecules now have sufficient energy to overcome the lowered energy barrier, the reaction rate increases.
- A catalyst may also help by correctly orienting reactant molecules, increasing the frequency of effective collisions.
Graphical Representation:
- The lower peak represents the catalyzed pathway.
- The vertical distance between reactants and transition state corresponds to the activation energy.
- The overall energy change (\( \mathrm{\Delta E_{rxn}} \)) remains the same — only the energy barrier changes.
A catalyst does not alter the overall energy of reactants or products; it only provides an alternate route with a smaller energy barrier, thereby increasing the rate of reaction.
Example
The decomposition of hydrogen peroxide (\( \mathrm{H_2O_2} \)) is slow at room temperature but becomes rapid in the presence of manganese dioxide (\( \mathrm{MnO_2} \)). Explain the role of \( \mathrm{MnO_2} \) in this reaction.
▶️ Answer / Explanation
Step 1: The uncatalyzed reaction:
\( \mathrm{2H_2O_2 \rightarrow 2H_2O + O_2} \)
This reaction has a large
activation energy
, so it proceeds slowly at room temperature.
Step 2: With \( \mathrm{MnO_2} \):
\( \mathrm{MnO_2} \) provides an alternate pathway that requires less energy to reach the transition state.
It facilitates decomposition through formation of an intermediate complex on its surface.
Step 3: The overall energy change (\( \mathrm{\Delta E_{rxn}} \)) remains the same, but the rate increases significantly because more molecules can now cross the lowered energy barrier.
Result: \( \mathrm{MnO_2} \) acts as a catalyst by lowering the activation energy and increasing the number of effective collisions without being consumed in the reaction.
Catalyst Regeneration and Constant Concentration
In a reaction mechanism that includes a catalyst, the catalyst participates in one or more elementary steps, but its net concentration remains constant throughout the reaction. This happens because the catalyst is consumed in one step (often the rate-determining step) and regenerated in a later step.
Key Concepts:
- A catalyst appears in the reaction mechanism, but not in the overall balanced equation.
- It is used up in an early step and reproduced in a subsequent one.
- The catalyst’s concentration remains constant because its consumption and regeneration rates are equal at steady state.
- This allows it to facilitate multiple reaction cycles without being permanently changed.
A catalyst provides an alternate reaction path but is not a reactant or product in the overall reaction. It temporarily forms intermediates and is then restored — maintaining a constant overall amount during the reaction.
Example
Nitrogen monoxide (\( \mathrm{NO} \)) acts as a catalyst in the decomposition of ozone in the atmosphere. The mechanism is:
Step 1: \( \mathrm{NO + O_3 \rightarrow NO_2 + O_2} \) (fast)
Step 2: \( \mathrm{NO_2 + O \rightarrow NO + O_2} \) (fast)
Show that \( \mathrm{NO} \) acts as a catalyst and explain why its net concentration remains constant.
▶️ Answer / Explanation
Step 1: Identify the catalyst.
\( \mathrm{NO} \) is used in Step 1 and reappears in Step 2 — it participates in the mechanism but is regenerated.
Step 2: Determine the overall reaction.
Adding both steps and canceling intermediates gives: \( \mathrm{O_3 + O \rightarrow 2O_2} \)
The catalyst \( \mathrm{NO} \) does not appear in the overall equation, confirming it is unchanged.
Step 3: Catalyst concentration.
Because \( \mathrm{NO} \) is consumed in Step 1 and regenerated in Step 2, its overall concentration remains constant — the amount used equals the amount reformed.
Result: \( \mathrm{NO} \) serves as a catalyst that facilitates ozone decomposition by providing a lower activation energy pathway without being consumed.
Catalyst Binding and Orientation Effects (Enzyme-Type Catalysis)
Some catalysts — especially biological ones such as enzymes — speed up reactions by binding to the reactants. This binding alters the reactant orientation and provides a pathway with lower activation energy . The reactant-catalyst complex formed in this process is a new reaction intermediate, which transforms into products more easily than the uncatalyzed reactants.
- The catalyst binds to one or more reactant molecules to form a temporary intermediate complex.
- This binding can help:
- Orient reactants in a way that favors bond breaking and formation.
- Stabilize the transition state, lowering the activation energy.
- Increase the frequency of effective collisions.
- After the reaction, the catalyst separates unchanged, ready to catalyze another cycle.
Key Idea: Catalyst binding increases reaction rate by holding reactant molecules in an optimal configuration and reducing the energy needed for reaction to occur. This is a key principle behind enzyme catalysis in biological systems.
Mechanistic Representation:
\( \mathrm{Catalyst + Reactant \rightleftharpoons Catalyst\text{-}Reactant\ Complex \rightarrow Product + Catalyst} \)
- The Catalyst–Reactant Complex is a short-lived intermediate.
- The formation and decomposition of this complex occur rapidly.
Example
Explain how the enzyme catalase catalyzes the decomposition of hydrogen peroxide (\( \mathrm{H_2O_2} \)) into water and oxygen, and why this process is faster than the uncatalyzed reaction.
▶️ Answer / Explanation
Step 1: Formation of an enzyme–substrate complex.
The enzyme’s active site binds to the \( \mathrm{H_2O_2} \) molecule: \( \mathrm{E + H_2O_2 \rightleftharpoons E\text{-}H_2O_2} \)
Step 2: Breakdown of the complex.
The bound \( \mathrm{H_2O_2} \) decomposes more easily: \( \mathrm{E\text{-}H_2O_2 \rightarrow E + H_2O + \tfrac{1}{2}O_2} \)
Step 3: Catalyst regeneration.
The enzyme \( \mathrm{E} \) is released unchanged, ready to catalyze more reactions.
Step 4: Interpretation.
- The enzyme aligns \( \mathrm{H_2O_2} \) molecules properly for bond rearrangement.
- This lowers the
activation energy
by stabilizing the transition state.
- As a result, decomposition occurs rapidly at room temperature.
Result: The enzyme functions as a biological catalyst by forming a temporary bound intermediate, decreasing the activation energy , and orienting the reactant for effective decomposition.
Covalent Bonding in Catalysis (Acid–Base Catalysis)
Some catalysts accelerate a reaction by forming a temporary covalent bond with one or more reactant molecules. This process introduces a new intermediate that reacts more easily than the original species. A common example is acid–base catalysis, in which a reactant or intermediate either gains or loses a proton (\( \mathrm{H^+} \)). The formation of this new, reactive intermediate changes the reaction pathway and lowers the activation energy.
- Covalent catalysis involves formation of a temporary covalent bond between the catalyst and a reactant.
- Acid–base catalysis is a specific form where the catalyst donates or accepts a proton, forming a new intermediate.
- This new pathway has a lower activation energy because:
- Bond rearrangements occur more easily in the intermediate form.
- The transition state is stabilized by charge redistribution.
- The catalyst is regenerated after the reaction — it is not consumed permanently.
Key Idea: Acid–base catalysts work by temporarily altering the electronic structure of reactants through proton transfer, forming intermediates that require less energy to convert into products.
Example
The formation of an ester from a carboxylic acid and an alcohol is catalyzed by an acid such as \( \mathrm{H^+} \):
\( \mathrm{CH_3COOH + CH_3OH \xrightarrow{H^+} CH_3COOCH_3 + H_2O} \)
Explain how \( \mathrm{H^+} \) acts as a catalyst in this reaction.
▶️ Answer / Explanation
Step 1: Protonation of the carbonyl oxygen.
The acid donates a proton to the carbonyl oxygen of acetic acid, forming a positively charged intermediate: \( \mathrm{CH_3C(OH^+)OH} \)
Step 2: Nucleophilic attack by the alcohol.
The oxygen atom of methanol attacks the carbonyl carbon, forming a new covalent bond and creating a tetrahedral intermediate.
Step 3: Dehydration and product formation.
Water is eliminated, and the catalyst (\( \mathrm{H^+} \)) is regenerated at the end of the reaction.
Step 4: Interpretation.
- The temporary protonation step creates a more electrophilic carbon, lowering the activation energyrequired for nucleophilic attack.
- Since \( \mathrm{H^+} \) is regenerated, it is a true catalyst.
Result: The acid catalyst facilitates ester formation by temporarily bonding with the reactant, providing a lower-energy pathway for product formation.
Example
Explain how hydroxide ion (\( \mathrm{OH^-} \)) catalyzes the hydrolysis of an ester to form an alcohol and a carboxylate ion.
▶️ Answer / Explanation
Step 1: Nucleophilic attack by \( \mathrm{OH^-} \).
The hydroxide ion attacks the carbonyl carbon of the ester, forming a tetrahedral intermediate.
Step 2: Collapse of intermediate.
The intermediate collapses, releasing the alcohol and forming a carboxylate ion (\( \mathrm{RCOO^-} \)).
Step 3: Catalyst regeneration.
The \( \mathrm{OH^-} \) ion is regenerated during the final step, so its concentration remains unchanged.
Step 4: Interpretation.
- A new covalent bond is temporarily formed between the catalyst and reactant.
- The presence of \( \mathrm{OH^-} \) lowers the activation energyby stabilizing charge buildup in the transition state.
- The base acts catalytically — it facilitates reaction and is regenerated.
Result: This is an example of base catalysis, where the catalyst forms a transient covalent intermediate and lowers the activation energy of the reaction.
Surface Catalysis and Adsorbed Intermediates
In surface catalysis, reactant molecules are adsorbed (attached) onto the surface of a solid catalyst. This adsorption either involves weak physical attraction (physisorption) or strong covalent bonding (chemisorption). Once bound, the reactants undergo chemical changes more easily — forming new intermediates and lowering the activation energy required for reaction. Finally, the products detach (desorb) from the surface, regenerating the catalyst.
Key Concepts: 
- Surface catalysts provide a platform where reactants can come together in close proximity, increasing the number of effective collisions.
- The binding of reactants to the surface weakens existing bonds, making it easier for the reaction to proceed.
- Each adsorption site acts like a temporary reaction center that lowers the activation energy.
- Products detach after the reaction, leaving the catalyst surface ready for another cycle.
Surface catalysis involves adsorption, reaction, and desorption steps. The catalyst remains unchanged overall, but facilitates reaction by stabilizing intermediates on its surface and reducing the required activation energy.
Mechanistic Steps of Surface Catalysis:
- Adsorption: Reactant molecules attach to active sites on the solid catalyst surface.
- Activation: Bonds within the reactants weaken upon interaction with the surface.
- Reaction: The adsorbed species react to form new products or intermediates.
- Desorption: Products leave the surface, freeing the active sites for reuse.
Example
Describe the mechanism of hydrogenation of ethene (\( \mathrm{C_2H_4} \)) using a nickel (Ni) catalyst, and explain how the catalyst lowers the activation energy.
▶️ Answer / Explanation
Step 1: Adsorption
Both hydrogen molecules and ethene molecules are adsorbed onto the surface of nickel atoms.
Step 2: Bond Weakening
The \( \mathrm{H-H} \) bond in hydrogen and the \( \mathrm{C=C} \) double bond in ethene are weakened upon adsorption.
Step 3: Reaction
Adsorbed hydrogen atoms add to the carbon atoms of ethene, forming ethane (\( \mathrm{C_2H_6} \)).
Step 4: Desorption
The product \( \mathrm{C_2H_6} \) detaches from the nickel surface, freeing active sites for new reactant molecules.
Step 5: Interpretation
- The reaction occurs more easily on the metal surface because adsorption weakens bonds in reactants.
- This lowers the activation energy and increases reaction rate.
- The nickel surface remains unchanged and is reused.
Result: The nickel catalyst provides a surface where hydrogen and ethene interact more efficiently, producing ethane faster than in the uncatalyzed gas-phase reaction.
Example
Modern vehicles use a platinum–rhodium (Pt–Rh) catalyst in the exhaust system. Explain how this surface catalyst helps reduce pollution.
▶️ Answer / Explanation
Step 1: Adsorption of Exhaust Gases
Harmful gases like \( \mathrm{CO} \), \( \mathrm{NO} \), and \( \mathrm{C_xH_y} \) are adsorbed onto the catalyst surface.
Step 2: Surface Reactions
The following reactions occur simultaneously: \( \mathrm{2CO + O_2 \rightarrow 2CO_2} \) and \( \mathrm{2NO \rightarrow N_2 + O_2} \)
Step 3: Desorption of Products
The less harmful products \( \mathrm{CO_2} \), \( \mathrm{N_2} \), and \( \mathrm{H_2O} \) desorb from the surface.
Step 4: Catalyst Role
- Platinum and rhodium provide active sites for adsorption and reaction.
- These metals lower the activation energyof oxidation and reduction steps.
- The catalyst remains unchanged and continues to operate efficiently.
Result: The catalytic converter uses surface catalysis to transform toxic gases into harmless products through adsorption, reaction, and desorption on metal surfaces.