This page is a workout. We take the machinery from the parent note and run it through every kind of mechanism problem an exam can hand you . Before we start, one promise: every symbol is earned. Let me re-anchor the two ideas you actually need.
Intuition What "rate" means, picturedImagine a corridor of people trying to get through a series of doors. The rate of the whole crowd getting out is set by the narrowest door — the slowest step. That narrow door is the rate-determining step (RDS) . Everything on this page is: find the narrow door, and describe the crowd waiting at it in terms of things we can actually count.
Two symbols we reuse constantly:
[ X ] k
[ X ] how much of species X ", measured in moles per litre (mol L − 1 M [ X ] k rate constant — a fixed number for a given step at a given temperature that says "how eager is this collision to succeed?" Big k k
For an elementary step aA + bB → products k [ A ] a [ B ] b number of each molecule that must collide at once — because probability of a simultaneous meeting multiplies.
Every mechanism problem is one (or a blend) of these cells. Our 9 examples below cover all of them — including a standalone Cell Z that stress-tests the formulas at their edges.
Cell
Scenario class
What makes it tricky
Example
A Slow step first , no intermediate in it
The "easy" case — read the rate off directly
Ex 1
B Slow step later , fed by a fast pre-equilibrium
Must eliminate an intermediate via K
Ex 2
C Intermediate is a reactant that was already used (denominator appears)
Fast equilibrium consuming a reactant, so it lands in the denominator
Ex 3
D Unimolecular RDS (molecularity 1)Degenerate "collision" — a single molecule falls apart
Ex 4
E Termolecular / 3-body apparent orderOrder can reach 3; is a true 3-body step realistic?
Ex 5
F No obvious slow step to Steady-State Approximation Rate set by build-up/drain balance, not one door
Ex 6
G Real-world word problem (atmospheric / catalysis)Translate prose to steps to law
Ex 7
H Exam twist : given the observed order, pick the valid mechanismReverse direction; test candidates
Ex 8
Z Degenerate/limiting checks : zero concentration, huge/tiny k Do the formulas stay sane at the edges?
Ex 9
Worked example Example 1 — the clean caseOverall: NO 2 + CO → NO + CO 2
Step 1 (slow ): NO 2 + NO 2 → NO 3 + NO = k 1 [ NO 2 ] 2
Step 2 (fast): NO 3 + CO → NO 2 + CO 2
Forecast: guess the overall rate law. Tempting answer from stoichiometry: k [ NO 2 ] [ CO ]
Step 1 — Find the narrow door. Step 1 is labelled slow , so it is the RDS.
Why this step? The whole crowd can only leave as fast as the narrowest door lets it. Step 2 is fast, so it never holds anyone up.
Step 2 — Write the RDS rate from its own stoichiometry. Two NO 2 Rate = k 1 [ NO 2 ] 2 . Why this step? Step 1 is elementary, so exponents = number of colliding molecules = 2.
Step 3 — Check for intermediates in the law. NO 3 not appear in k 1 [ NO 2 ] 2 Why this step? A finished rate law may only contain species we can measure — reactants/products.
Result: Rate = k 1 [ NO 2 ] 2 Second order in NO 2 CO — CO doesn't appear at all!
Verify: Sum the steps: 2 NO 2 + NO 3 + CO → NO 3 + NO + NO 2 + CO 2 NO 3 NO 2 NO 2 + CO → NO + CO 2 k 1 M s − 1 [ NO 2 ] 2 M 2 k 1 M − 1 s − 1
Here we meet the trick that carries the rest of the page: an intermediate sits inside the RDS and we must trade it away.
Figure s01 — a diagram of Example 2. On the left a box holds the reactants NO + Br2 k 1 k − 1 equal length connect it to a middle box holding the intermediate NOBr2 k 2 2
Worked example Example 2 — pre-equilibrium eliminationOverall: 2 NO + Br 2 → 2 NOBr
Step 1 (fast equilibrium ): NO + Br 2 ⇌ NOBr 2 k 1 k − 1
Step 2 (slow ): NOBr 2 + NO → 2 NOBr = k 2 [ NOBr 2 ] [ NO ]
Forecast: what will the order in NO
Step 1 — RDS rate. Step 2 is slow:
Rate = k 2 [ NOBr 2 ] [ NO ] . Why: the slowest door sets the pace.
Step 2 — Spot the intermediate. NOBr 2 Why: rate laws may only contain measurable reactants.
Step 3 — Use the fast equilibrium. "Fast forward and reverse" means Step 1 balances itself: forward rate = reverse rate — this is exactly the equal-length arrows in figure s01.
k 1 [ NO ] [ Br 2 ] = k − 1 [ NOBr 2 ] . Why this tool and not another? Because Step 1 is fast in both directions , it reaches equilibrium long before the slow step drains it — so the equilibrium condition (in-rate = out-rate) holds.
Step 4 — Solve for the intermediate.
[ NOBr 2 ] = k − 1 k 1 [ NO ] [ Br 2 ] . Why this step? The rate we want (k 2 [ NOBr 2 ] [ NO ] [ NOBr 2 ] [ NOBr 2 ]
Step 5 — Substitute.
Rate = k 2 ⋅ k − 1 k 1 [ NO ] [ Br 2 ] ⋅ [ NO ] = k − 1 k 1 k 2 [ NO ] 2 [ Br 2 ] . Why this step? We drop the Step-4 expression into the RDS rate in place of [ NOBr 2 ] NO Br 2 k obs = k 1 k 2 / k − 1 Rate = k obs [ NO ] 2 [ Br 2 ] .
Result: the forecast is answered — second order in NO , not first, because two NO molecules feed the path (one enters the equilibrium, one enters the slow step). Total order 2 + 1 = 3
Verify: Sum: NO + Br 2 + NOBr 2 + NO → NOBr 2 + 2 NOBr NOBr 2 2 NO + Br 2 → 2 NOBr Units of k obs rate is M s − 1 [ NO ] 2 [ Br 2 ] M 3 k obs M − 2 s − 1 k 1 k 2 / k − 1 M − 1 s − 1 ⋅ M − 1 s − 1 / s − 1 = M − 2 s − 1 k 1 = 2 , k − 1 = 4 , k 2 = 3 k obs = 2 ⋅ 3/4 = 1.5 [ NO ] = 0.1 , [ Br 2 ] = 0.2 = 1.5 ⋅ 0.01 ⋅ 0.2 = 0.003 M s − 1
Worked example Example 3 — a reactant hides in the denominatorOzone decomposition, 2 O 3 → 3 O 2
Step 1 (fast equilibrium ): O 3 ⇌ O 2 + O k 1 , k − 1
Step 2 (slow ): O + O 3 → 2 O 2 = k 2 [ O ] [ O 3 ]
Forecast: the reverse of Step 1 eats O 2 . Will [ O 2 ]
Step 1 — RDS rate: Rate = k 2 [ O ] [ O 3 ]
Step 2 — Intermediate is atomic O Eliminate it.
Step 3 — Equilibrium condition. Forward = k 1 [ O 3 ] = k − 1 [ O 2 ] [ O ] . Why: the reverse reaction is bimolecular (O 2 O [ O 2 ]
Step 4 — Solve.
[ O ] = k − 1 [ O 2 ] k 1 [ O 3 ] . Why this step? The RDS rate still hides the intermediate [ O ] [ O ] [ O 2 ] divided out — it slides into the denominator. That is precisely how a product ends up inhibiting the rate.
Step 5 — Substitute.
Rate = k 2 ⋅ k − 1 [ O 2 ] k 1 [ O 3 ] ⋅ [ O 3 ] = k − 1 k 1 k 2 [ O 2 ] [ O 3 ] 2 . Why this step? Dropping the Step-4 expression into the RDS rate evicts [ O ] O 3 O 2 k obs = k 1 k 2 / k − 1 Rate = k obs [ O 3 ] 2 [ O 2 ] − 1 .
Result: order + 2 O 3 − 1 O 2 negative order — the product O 2 inhibits the reaction, exactly what experiment shows for ozone.
Verify: Sum: O 3 + O + O 3 → O 2 + O + 2 O 2 O 2 O 3 → 3 O 2 Units of k obs rate M s − 1 k obs ⋅ M 2 / M = k obs ⋅ M k obs s − 1 k 1 k 2 / k − 1 s − 1 ⋅ M − 1 s − 1 / M − 1 s − 1 = s − 1 k 1 = 6 , k − 1 = 3 , k 2 = 2 ⇒ k obs = 4 [ O 3 ] = 0.1 , [ O 2 ] = 0.5 = 4 ⋅ 0.01/0.5 = 0.08 M s − 1
Worked example Example 4 — a single molecule falls apartOverall: N 2 O 5 → 2 NO 2 + 2 1 O 2 unimolecular :
Step 1 (slow ): N 2 O 5 → NO 2 + NO 3 = k 1 [ N 2 O 5 ]
Step 2 (fast): NO 2 + NO 3 → NO 2 + O 2 + NO
Forecast: "molecularity 1" means how many molecules must meet? What's the order?
Step 1 — RDS rate. Molecularity = 1 Rate = k 1 [ N 2 O 5 ] . Why this is the degenerate case: a "collision" with zero partners is one molecule reaching enough vibrational energy to break — see Activation Energy and Catalysts . The exponent is 1 0 [ N 2 O 5 ] number of molecules that can break per second.
Result: Rate = k 1 [ N 2 O 5 ] first order .
Verify: Units — rate M s − 1 k 1 ⋅ M k 1 s − 1 [ N 2 O 5 ] = 0 = 0
Worked example Example 5 — how high can order go?Overall: 2 NO + O 2 → 2 NO 2 termolecular step:
Step 1 (slow , 3-body): NO + NO + O 2 → 2 NO 2 = k 1 [ NO ] 2 [ O 2 ]
Forecast: what's the overall order (sum of exponents)? And is a genuine 3-body collision physically likely?
Step 1 — Read the rate. Elementary, molecularity 3:
Rate = k 1 [ NO ] 2 [ O 2 ] , order = 2 + 1 = 3.
Step 2 — Sanity on the physics. Three molecules meeting at the same instant is rare — recall from Collision Theory that even 2-body collisions are the norm and 3-body ones are roughly a thousand times less frequent. So although this gives the right law , chemists prefer the pre-equilibrium route of Example 2 (same law, no improbable triple collision).
Why mention this? An exam may accept either mechanism if both reproduce the observed order-3 law — but asks you to flag which is more realistic .
Result: Rate = k 1 [ NO ] 2 [ O 2 ]
Verify: This is exactly the k obs [ NO ] 2 [ O 2 ] same rate law , confirming a rate law can't uniquely prove a mechanism. Units of k 1 rate M s − 1 = k 1 ⋅ M 3 k 1 M − 2 s − 1 k 1 = 5 , [ NO ] = 0.2 , [ O 2 ] = 0.3 = 5 ⋅ 0.04 ⋅ 0.3 = 0.06 M s − 1
When no step is flagged "slow," the funnel has two similar doors . We use the Steady-State Approximation : a reactive intermediate is made and destroyed so fast that its amount barely changes, so its net rate of change is about zero .
Figure s02 — a graph of the intermediate concentration [ O ] d [ O ] / d t
Worked example Example 6 — steady-state derivationOverall: 2 O 3 → 3 O 2
Step 1: O 3 k 1 O 2 + O
Step − 1 O 2 + O k − 1 O 3
Step 2: O + O 3 k 2 2 O 2
Forecast: what happens to the answer when k 2 [ O 3 ] k − 1 [ O 2 ]
Step 1 — Write the net rate of the intermediate O It is made by Step 1, destroyed by Steps − 1 2 d t d [ O ] = k 1 [ O 3 ] − k − 1 [ O 2 ] [ O ] − k 2 [ O ] [ O 3 ] . Why this tool? No single door is narrow, so we can't just read off one step. Instead we track the flat top of the [ O ]
Step 2 — Set it to zero (steady state).
k 1 [ O 3 ] = [ O ] ( k − 1 [ O 2 ] + k 2 [ O 3 ] ) ⇒ [ O ] = k − 1 [ O 2 ] + k 2 [ O 3 ] k 1 [ O 3 ] . Why this step? The plateau means "made = destroyed", i.e. the whole right-hand side sums to zero. Setting it to zero turns a hard differential equation into a plain algebraic one, which we then rearrange to isolate the un-measurable [ O ]
Step 3 — Rate of product (via Step 2).
Rate = k 2 [ O ] [ O 3 ] = k − 1 [ O 2 ] + k 2 [ O 3 ] k 1 k 2 [ O 3 ] 2 . Why this step? Product O 2 k 2 [ O ] [ O 3 ] [ O ]
Step 4 — Limiting check. If k − 1 [ O 2 ] k 2 [ O 3 ] k 2 [ O 3 ] Rate → k − 1 k 1 k 2 [ O 2 ] [ O 3 ] 2 , more general law and the pre-equilibrium is its special case.
Verify: Units: numerator k 1 k 2 [ O 3 ] 2 k − 1 [ O 2 ] M s − 1 k 1 = 6 , k 2 = 2 , k − 1 = 3 , [ O 3 ] = 0.1 , [ O 2 ] = 0.5 Rate = 3 ⋅ 0.5 + 2 ⋅ 0.1 6 ⋅ 2 ⋅ 0.01 = 1.7 0.12 ≈ 0.0706 M s − 1 0.08 k 2 [ O 3 ]
Worked example Example 7 — enzyme-style bottleneck in words"An enzyme E S E S E S P E
Step 1 (fast eq): E + S ⇌ E S k 1 , k − 1
Step 2 (slow): E S k 2 E + P = k 2 [ E S ]
Forecast: with lots of enzyme and little substrate, do you expect first order in [ S ]
Step 1 — RDS rate. Rate = k 2 [ E S ]
Step 2 — E S Fast equilibrium of Step 1:
k 1 [ E ] [ S ] = k − 1 [ E S ] ⇒ [ E S ] = k − 1 k 1 [ E ] [ S ] . Why this step? We rearrange the equilibrium to isolate the un-measurable complex [ E S ]
Step 3 — Substitute.
Rate = k − 1 k 1 k 2 [ E ] [ S ] . Why this step? Dropping the Step-2 expression into the RDS rate evicts [ E S ] scarce-substrate limit the enzyme is mostly free, so [ E ] [ E ] 0 Rate ≈ k ′ [ S ] , k ′ = k − 1 k 1 k 2 [ E ] 0 .
Result: first order in substrate at low [ S ] Enzyme Kinetics (Michaelis–Menten). At high [ S ]
Verify: Low-[ S ] = 1 [ S ] = 0 k 1 = 10 , k − 1 = 2 , k 2 = 5 , [ E ] 0 = 0.01 , [ S ] = 0.1 k ′ = ( 10 ⋅ 5/2 ) ⋅ 0.01 = 0.25 = 0.25 ⋅ 0.1 = 0.025 M s − 1
Worked example Example 8 — pick the mechanism that fits the dataExperiment gives Rate = k [ H 2 ] [ ICl ] H 2 + 2 ICl → I 2 + 2 HCl
Candidate X: single step H 2 + 2 ICl → products Rate = k [ H 2 ] [ ICl ] 2 Candidate Y:
Step 1 (slow ): H 2 + ICl → HI + HCl = k 1 [ H 2 ] [ ICl ]
Step 2 (fast): HI + ICl → I 2 + HCl
Forecast: which predicted law matches the observed k [ H 2 ] [ ICl ]
Step 1 — Predict X's law. Elementary termolecular gives k [ H 2 ] [ ICl ] 2 = 2 Does not match (observed is 1).
Why reject: the exponent on ICl is wrong, and 3-body collisions are unlikely anyway.
Step 2 — Predict Y's law. RDS is Step 1, elementary, giving k 1 [ H 2 ] [ ICl ] = 1 Matches ✓.
Step 3 — Confirm Y sums correctly. H 2 + ICl + HI + ICl → HI + HCl + I 2 + HCl HI H 2 + 2 ICl → I 2 + 2 HCl
Result: Mechanism Y is accepted (fits observed law and stoichiometry; no improbable termolecular step).
Verify: Observed law order sum = 1 + 1 = 2 2 3 k 1 = 4 , [ H 2 ] = 0.5 , [ ICl ] = 0.2 = 4 ⋅ 0.5 ⋅ 0.2 = 0.4 M s − 1
Worked example Example 9 — do the formulas stay sane at the edges?Take the pre-equilibrium law from Example 2, Rate = k obs [ NO ] 2 [ Br 2 ] k obs = k 1 k 2 / k − 1
Forecast: guess which edge makes the rate blow up rather than vanish.
Step 1 — Zero reactant. Set [ NO ] = 0 Rate = k obs ⋅ 0 2 ⋅ [ Br 2 ] = 0. Why this step? If a reactant that appears in the law is absent, no path can run — the rate must be exactly zero. It is ✓.
Step 2 — Zero of a species not in the law. In Example 1, set [ CO ] = 0 = k 1 [ NO 2 ] 2 unchanged .
Why this step? Zero-order species genuinely don't gate the RDS, so removing them can't slow Step 1. This is the counter-intuitive edge students trip on.
Step 3 — Very fast reverse (k − 1 → ∞ In k obs = k 1 k 2 / k − 1 k obs → 0 ⇒ Rate → 0. Why this step? A furious reverse tears the intermediate apart before the slow step can use it — the equilibrium sits far left, starving Step 2. Vanishing rate is physically right ✓.
Step 4 — Product-inhibited law near zero product. Take Example 3, Rate = k obs [ O 3 ] 2 / [ O 2 ] [ O 2 ] → 0 Rate → + ∞. Why this step? This is the "blow-up" edge from the forecast: a denominator heading to zero sends the rate sky-high. Physically the model breaks here (you never truly have zero O 2
Verify: Edge 1 gives 0 k 1 [ NO 2 ] 2 0 lim k − 1 → ∞ k 1 k 2 / k − 1 = 0 lim [ O 2 ] → 0 + k obs [ O 3 ] 2 / [ O 2 ] = + ∞
Recall When do I use pre-equilibrium vs steady state?Use pre-equilibrium when a fast reversible step is followed by a clearly slow step (reverse far exceeds forward-drain). Use steady-state when no step is flagged slow, or to be safe — steady-state is the general result that contains pre-equilibrium as a limit.
Recall Which cell does a negative reaction order come from?Cell C ::: a product (or reactant) that appears in the reverse of a fast equilibrium lands in the denominator , giving a negative order — that species inhibits the reaction.
Mnemonic The two-move dance"Slowest sets it, intermediates get evicted." Find the RDS, write its law, then evict any intermediate using equilibrium (or steady state) until only measurable species remain.
