Reaction Mechanisms and Intermediates

Understand multi-step mechanisms, rate-determining steps, intermediates, and how mechanisms relate to observed rate laws.

Reaction Mechanisms and Intermediates

What is a Mechanism?

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: Keq=[C][A][B]K_{eq} = \frac{[C]}{[A][B]} [C]=Keq[A][B][C] = K_{eq}[A][B]

Rate law from RDS: Rate=k2[C][D]=k2Keq[A][B][D]\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

📚 Practice Problems

1Problem 1easy

❓ Question:

Given mechanism: (1) NO₂ + NO₂ → NO₃ + NO (slow), (2) NO₃ + CO → NO₂ + CO₂ (fast). Find: (a) overall equation, (b) intermediates, (c) rate law.

💡 Show Solution

Given mechanism:

  1. NO₂ + NO₂ → NO₃ + NO (slow)
  2. NO₃ + CO → NO₂ + CO₂ (fast)

(a) Overall equation

Add the two steps:

Step 1: NO₂ + NO₂ → NO₃ + NO Step 2: NO₃ + CO → NO₂ + CO₂

Sum: NO₂ + NO₂ + NO₃ + CO → NO₃ + NO + NO₂ + CO₂

Cancel species on both sides:

  • 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)

2Problem 2medium

❓ Question:

The reaction 2NO + Br₂ → 2NOBr has rate law: Rate = k[NO]²[Br₂]. Propose a mechanism consistent with this rate law.

💡 Show Solution

Given:

  • Overall: 2NO + Br₂ → 2NOBr
  • Observed rate law: Rate = k[NO]²[Br₂]

Need: Mechanism that gives this rate law

Strategy:

  • Rate law is third order: second in NO, first in Br₂
  • Suggests RDS involves 2 NO and 1 Br₂
  • Or RDS after fast equilibrium

Proposed Mechanism (Option 1):

Step 1: NO + Br₂ → NOBr₂ (fast equilibrium) Step 2: NOBr₂ + NO → 2NOBr (slow)

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:

K1=[NOBr2][NO][Br2]K_1 = \frac{[NOBr_2]}{[NO][Br_2]}

[NOBr2]=K1[NO][Br2][NOBr_2] = K_1[NO][Br_2]

Substitute: Rate=k2K1[NO][Br2][NO]=k2K1[NO]2[Br2]\text{Rate} = k_2K_1[NO][Br_2][NO] = k_2K_1[NO]^2[Br_2]

Let k = k₂K₁: Rate=k[NO]2[Br2]\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: [N2O2]=K1[NO]2[N_2O_2] = K_1[NO]^2

Substitute: Rate=k2K1[NO]2[Br2]\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.)

3Problem 3hard

❓ Question:

For mechanism: (1) A ⇌ B (fast, K₁), (2) B + C ⇌ D (fast, K₂), (3) D + E → F (slow, k₃). Derive the rate law in terms of A, C, and E only.

💡 Show Solution

Given mechanism:

  1. A ⇌ B (fast equilibrium, K₁)
  2. B + C ⇌ D (fast equilibrium, K₂)
  3. D + E → F (slow, k₃)

Goal: Rate law in terms of A, C, E (not intermediates B or D)

Step 1: Identify RDS

Step 3 is slow → rate-determining step

Initial rate law: Rate = k₃[D][E]

Problem: D is intermediate (can't appear in final rate law)

Step 2: Express D using equilibrium

From step 2 equilibrium:

K2=[D][B][C]K_2 = \frac{[D]}{[B][C]}

[D]=K2[B][C][D] = K_2[B][C]

Still have B (also intermediate!)

Step 3: Express B using equilibrium

From step 1 equilibrium:

K1=[B][A]K_1 = \frac{[B]}{[A]}

[B]=K1[A][B] = K_1[A]

Step 4: Substitute back

Substitute [B] into [D] expression:

[D]=K2[B][C]=K2(K1[A])[C]=K1K2[A][C][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:

Rate=k3[D][E]=k3(K1K2[A][C])[E]\text{Rate} = k_3[D][E] = k_3(K_1K_2[A][C])[E]

Rate=k3K1K2[A][C][E]\text{Rate} = k_3K_1K_2[A][C][E]

Define overall rate constant:

koverall=k3K1K2k_{\text{overall}} = k_3K_1K_2

Final answer:

Rate=k[A][C][E]\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
🎴

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