Reaction mechanism: Step-by-step sequence showing how reaction occurs

Why mechanisms matter:

• Explain HOW reactants become products
• Predict rate laws
• Identify intermediates
• Guide catalyst design

Most reactions are multi-step:

• Elementary steps add to give overall equation
• Each step has its own rate

Elementary Steps

Elementary step: Single molecular event

Molecularity: Number of molecules participating

Types:

• Unimolecular: A → products (rate = k[A])
• Bimolecular: A + B → products (rate = k[A][B])
• Termolecular: A + B + C → products (very rare!)

Key rule: For elementary steps ONLY, exponents = coefficients

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

Check overall: NO + Br₂ → NOBr₂ NOBr₂ + NO → 2NOBr Sum: 2NO + Br₂ → 2NOBr ✓

Check rate law:

From step 2 (RDS): Rate = k₂[NOBr₂][NO]

But NOBr₂ is intermediate - express using step 1 equilibrium:

$K_1 = \frac{[NOBr_2]}{[NO][Br_2]}$

$[NOBr_2] = K_1[NO][Br_2]$

Substitute: $\text{Rate} = k_2K_1[NO][Br_2][NO] = k_2K_1[NO]^2[Br_2]$

Let k = k₂K₁: $\text{Rate} = k[NO]^2[Br_2]$ ✓

This mechanism works!

Alternative Mechanism (Option 2):

Step 1: NO + NO → N₂O₂ (fast equilibrium) Step 2: N₂O₂ + Br₂ → 2NOBr (slow)

Check overall: 2NO → N₂O₂ N₂O₂ + Br₂ → 2NOBr Sum: 2NO + Br₂ → 2NOBr ✓

Check rate law:

From step 2: Rate = k₂[N₂O₂][Br₂]

From step 1 equilibrium: $[N_2O_2] = K_1[NO]^2$

Substitute: $\text{Rate} = k_2K_1[NO]^2[Br_2]$ ✓

This also works!

Key insight:

• Multiple mechanisms can give same rate law
• Can't prove which is correct from rate law alone
• Need additional evidence (detect intermediates, isotope studies, etc.)

From step 2 equilibrium:

$K_2 = \frac{[D]}{[B][C]}$

$[D] = K_2[B][C]$

Still have B (also intermediate!)

Step 3: Express B using equilibrium

From step 1 equilibrium:

$K_1 = \frac{[B]}{[A]}$

$[B] = K_1[A]$

Step 4: Substitute back

Substitute [B] into [D] expression:

$[D] = K_2[B][C] = K_2(K_1[A])[C] = K_1K_2[A][C]$

Step 5: Write final rate law

Substitute [D] into RDS rate law:

$\text{Rate} = k_3[D][E] = k_3(K_1K_2[A][C])[E]$

$\text{Rate} = k_3K_1K_2[A][C][E]$

Define overall rate constant:

$k_{\text{overall}} = k_3K_1K_2$

Final answer:

$\boxed{\text{Rate} = k[A][C][E]}$

where k = k₃K₁K₂

Analysis:

Reaction orders:

• First order in A
• First order in C
• First order in E
• Third order overall

Physical meaning:

• All three species participate in determining rate
• Even though RDS only involves D + E
• Pre-equilibria affect concentrations of intermediates

Verification strategy:

1. ✓ No intermediates in final rate law (B, D eliminated)
2. ✓ Only species in overall equation (A, C, E)
3. ✓ Consistent with RDS being rate-determining
4. ✓ Used equilibrium expressions properly

General approach for complex mechanisms:

1. Start with RDS rate law
2. Identify intermediates
3. Work backwards through fast equilibria
4. Substitute until only reactants remain