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)

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:

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

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

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.)

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:

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

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

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

RDS: step 2 ⇒ Rate = k₂[A₂][B]. From step 1: K₁ = [A₂]/[A]^2 ⇒ [A₂] = K₁[A]^2. Substitute: Rate = k₂K₁[A]^2[B] = k[A]^2[B], where k = k₂K₁.

X is a catalyst (consumed then regenerated). AX is an intermediate (formed then consumed). Overall: add steps and cancel X, AX ⇒ A + B → products. Rate law from RDS: Rate = k₂[AX][B]. From step 1: [AX] = K₁[X][A]. Thus Rate = k₂K₁[X][A][B] = k[X][A][B].

Add steps and cancel common species. Species that appear and cancel: NO (consumed in step 1, produced in step 2) cancels; O (from step 3, consumed in step 2) cancels; NO₂ (produced in step 1, consumed in step 2) cancels. Overall: O₃ → O₂. Intermediates: O and NO₂.

(a) If step 1 is slow: Rate = k[A][B]. (b) If step 2 is slow, use pre-equilibrium for step 1: [C] = K[A][B]. Rate = k₂[C][B] = k₂K[A][B]^2 = k[A][B]^2. The observed rate law changes when the RDS changes.