DAT General Chemistry: Intermediates Versus Catalysts in a Mechanism

DAT High-Yield Concept

Reaction Intermediates: Formed in an early elementary step and consumed in a later step. They appear as a product first, then a reactant.
Catalysts: Consumed in an early elementary step and regenerated in a later step. They appear as a reactant first, then a product.
Net Reaction Equation: Neither intermediates nor catalysts appear in the overall balanced chemical equation.
Mechanism Validation: The sum of all elementary steps must equal the overall reaction after cancelling species common to both sides.

Reaction Intermediates in Elementary Steps

A reaction mechanism describes the step-by-step sequence of elementary reactions leading to overall chemical transformation. A reaction intermediate is a chemical species produced in one elementary step and subsequently consumed in a later step.

Key Definition: An intermediate is a transient species that exists only during the reaction sequence. It is generated as a product and consumed as a reactant, leaving no net presence in the overall stoichiometry.

Worked Example: Identifying Intermediates

Problem: Consider the two-step mechanism for ozone decomposition:

Step 1: $\ce{O3(g) -> O2(g) + O(g)}$ Step 2: $\ce{O3(g) + O(g) -> 2O2(g)}$ Identify the reaction intermediate.

Reasoning: Atomic oxygen, $\ce{O(g)}$ , is produced on the right side in Step 1 and consumed on the left side in Step 2.

Answer: $\ce{O(g)}$ is the reaction intermediate.

Common Exam Trap: Students often mistake intermediates for final products because they see them on the right side of Step 1. Always inspect subsequent steps to check if the species is consumed.

Distinguishing Intermediates from Catalysts

Both intermediates and catalysts are absent from the overall balanced equation, but their order of appearance in the mechanism differs:

Reaction Intermediate: Formed first (Product) $\rightarrow$ Consumed second (Reactant).
Catalyst: Consumed first (Reactant) $\rightarrow$ Regenerated second (Product).

An energy profile diagram demonstrating the effect of a catalyst for the generic exothermic reaction of X + Y →Z. The catalyst offers an alternate reaction pathway (shown in red) where the rate determining step has a smaller ΔG≠. The relative thermodynamic stabilities remain the same. (Source: Smokefoot · CC0 · Wikimedia Commons)

DAT Mnemonic:
• Intermediate = Inside the reaction (Product first

\rightarrow

Reactant second)
• Catalyst = Consumed first (Reactant first

\rightarrow

Product second)

Property	Reaction Intermediate	Catalyst
Order of Appearance	Product in early step $\rightarrow$ Reactant in later step	Reactant in early step $\rightarrow$ Product in later step
Overall Consumption	Formed then destroyed (net change = 0)	Consumed then regenerated (net change = 0)
Net Chemical Equation	Does NOT appear	Does NOT appear

Worked Example: Classifying Species in a Multi-Step Mechanism

Problem: Identify all catalysts and intermediates in the following 3-step aqueous mechanism:

Step 1: $\ce{Ce^{4+}(aq) + Mn^{2+}(aq) -> Ce^{3+}(aq) + Mn^{3+}(aq)}$ Step 2: $\ce{Ce^{4+}(aq) + Mn^{3+}(aq) -> Ce^{3+}(aq) + Mn^{4+}(aq)}$ Step 3: $\ce{Mn^{4+}(aq) + Tl^{+}(aq) -> Mn^{2+}(aq) + Tl^{3+}(aq)}$

Reasoning:

$\ce{Mn^{2+}(aq)}$ is consumed in Step 1 and regenerated in Step 3 $\rightarrow$ Catalyst.
$\ce{Mn^{3+}(aq)}$ is produced in Step 1 and consumed in Step 2 $\rightarrow$ Intermediate.
$\ce{Mn^{4+}(aq)}$ is produced in Step 2 and consumed in Step 3 $\rightarrow$ Intermediate.

Answer: Catalyst = $\ce{Mn^{2+}(aq)}$ ; Intermediates = $\ce{Mn^{3+}(aq)}$ and $\ce{Mn^{4+}(aq)}$ .

Summing Elementary Steps to the Overall Balanced Equation

For a proposed mechanism to be valid, the elementary steps must sum to yield the overall balanced reaction equation, conserving both mass and charge.

Worked Example: Verification and Summation

Problem: Sum the elementary steps to find the overall balanced equation:

Step 1: $\ce{Cl(g) + O3(g) -> ClO(g) + O2(g)}$ Step 2: $\ce{ClO(g) + O3(g) -> Cl(g) + 2O2(g)}$

Reasoning:

Combine all reactants and products: $\ce{Cl(g) + O3(g) + ClO(g) + O3(g) -> ClO(g) + O2(g) + Cl(g) + 2O2(g)}$
Cancel species present on both sides:
- Cancel catalyst $\ce{Cl(g)}$ from both sides.
- Cancel intermediate $\ce{ClO(g)}$ from both sides.
Combine like terms: $\ce{2O3(g) -> 3O2(g)}$

Answer: $\ce{2O3(g) -> 3O2(g)}$ . Both sides contain 6 oxygen atoms and a net charge of 0.

Common Exam Trap: Leaving catalysts or intermediates in the net chemical equation. Ensure all non-reactants/non-products cancel out completely.

Critical Thinking Questions

If a species appears as a reactant in Step 1 and is never regenerated in subsequent steps, is it a catalyst, an intermediate, or a reactant?
How would adding a homogeneous catalyst change the appearance of intermediates on a potential energy profile diagram?
Why can a rate law derived from an elementary step include an intermediate, and why must that intermediate be substituted out for the final overall rate law expression?

References

Sources returned by Google Search grounding during lesson verification.

3.6: Day 23- Reaction Mechanisms - Chemistry LibreTexts — chem.libretexts.org
Reaction mechanism and rate law | Khan Academy — khanacademy.org

Intermediates Versus Catalysts in a Mechanism

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