Intermediates vs Catalysts

Intermediate:

• Formed in one step, consumed in later step
• Appears in mechanism but NOT overall equation
• Short-lived species

Catalyst:

• Consumed in one step, regenerated in later step
• Appears in mechanism but NOT overall equation
• Speeds reaction, not consumed overall

Product:

• Formed and remains
• Appears in overall equation

Rate-Determining Step

Rate-determining step (RDS): Slowest step in mechanism

Analogy: Traffic bottleneck - slowest lane determines overall flow

Key points:

• Controls overall reaction rate
• Rate law comes from RDS (with modifications)
• Like slowest step on assembly line

If RDS is first step:

• Overall rate law = rate law of first step
• Straightforward

If RDS is later step:

• May involve intermediates
• Must substitute from fast equilibrium

Deriving Rate Laws from Mechanisms

Case 1: RDS is First Step

Mechanism:

1. A + B → C (slow)
2. C + D → E (fast)

Overall: A + B + D → E

Rate law: Rate = k[A][B] (from slow step)

Case 2: RDS is Second Step

Mechanism:

1. A + B ⇌ C (fast equilibrium)
2. C + D → E (slow)

Overall: A + B + D → E

Can't use C in rate law (intermediate!)

From equilibrium: $K_{eq} = \frac{[C]}{[A][B]}$ $[C] = K_{eq}[A][B]$

Rate law from RDS: $\text{Rate} = k_2[C][D] = k_2K_{eq}[A][B][D]$

Overall: Rate = k[A][B][D] where k = k₂K_{eq}

Writing Valid Mechanisms

Requirements:

1. Sum to overall equation - elementary steps must add up
2. Rate law must match - mechanism must give observed rate law
3. Reasonable steps - mostly uni- or bimolecular
4. Energy profile - intermediates at local minima

Note: Can't prove mechanism, only support or disprove

Common Mechanism Patterns

Two-Step with Fast Equilibrium

Example: 2NO + O₂ → 2NO₂

Mechanism:

1. NO + NO ⇌ N₂O₂ (fast equilibrium)
2. N₂O₂ + O₂ → 2NO₂ (slow)

Rate = k[NO]²[O₂] (matches experimental!)

Pre-Equilibrium Approximation

When first step is fast equilibrium:

• Forward and reverse rates equal
• Can use K_{eq} to express intermediate concentration
• Substitute into RDS rate law

Steady-State Approximation

For complex mechanisms:

• Intermediate concentration stays roughly constant
• Production rate ≈ consumption rate
• Advanced technique

Catalysis Mechanisms

Homogeneous Catalyst

Example: I⁻ catalyzing H₂O₂ decomposition

1. H₂O₂ + I⁻ → H₂O + IO⁻ (slow)
2. H₂O₂ + IO⁻ → H₂O + O₂ + I⁻ (fast)

Overall: 2H₂O₂ → 2H₂O + O₂

I⁻ consumed then regenerated → catalyst

Heterogeneous Catalyst

Surface mechanism (typical):

1. Reactant adsorbs onto surface
2. Reaction occurs on surface (lower Ea)
3. Product desorbs from surface

Example: Hydrogenation with Pt catalyst

Analyzing Mechanisms - Practice Strategy

Given mechanism, can:

1. Identify intermediates (formed then consumed)
2. Identify catalysts (consumed then reformed)
3. Write overall equation (cancel intermediates)
4. Determine rate law (from RDS + fast equilibrium)

Given rate law, can:

1. Propose mechanism consistent with rate law
2. Multiple mechanisms may fit - can't prove which is correct

• NO₃ appears on both sides → cancel
• One NO₂ appears on both sides → cancel (2 on left, 1 on right leaves 1 on left)

Overall: NO₂ + CO → NO + CO₂

(b) Identify intermediates

Intermediate: Formed in one step, consumed in another

NO₃:

• Formed in step 1 (product)
• Consumed in step 2 (reactant)
• NO₃ is intermediate

Check other species:

• NO₂: reactant (overall)
• CO: reactant (overall)
• NO: product (overall)
• CO₂: product (overall)

Answer: NO₃ is the only intermediate

(c) Rate law

Rate-determining step (RDS): Step 1 (slow)

For elementary step: exponents = coefficients

Step 1: NO₂ + NO₂ → NO₃ + NO

Rate law: Rate = k[NO₂]²

Answer: Rate = k[NO₂]²

Note:

• Step 1 is RDS and first step → rate law directly from it
• No intermediates in RDS → no substitution needed
• CO doesn't appear because step 2 is fast (not rate-determining)