A mechanism has several steps, usually running at very different speeds. The overall reaction cannot go faster than its slowest step, the way a line at a store moves only as fast as the slowest checkout.
Key idea
The slowest step is the rate-determining step. When it is the first step, the rate law of the whole reaction is the rate law of that step: its reactants, raised to their coefficients. Fast steps after it do not appear in the rate law.
Why it works
Products come out of the end of the mechanism only as fast as material passes through the slowest step. Steps after it are waiting for its product and finish almost at once, so changing their reactants' concentrations cannot speed anything up. Steps before it matter (the next lesson shows how), but when the slow step is first, nothing comes before it.
Worked examples
Worked example: A slow first step
Mechanism for $\mathrm{NO_2 + CO \to NO + CO_2}$:
Step 1 (slow): $\mathrm{2NO_2 \to NO_3 + NO}$
Step 2 (fast): $\mathrm{NO_3 + CO \to NO_2 + CO_2}$
What rate law does it predict?
Step 1 (slow): $\mathrm{2NO_2 \to NO_3 + NO}$
Step 2 (fast): $\mathrm{NO_3 + CO \to NO_2 + CO_2}$
What rate law does it predict?
- The slow step is the first step, and it is elementary: $\text{rate} = k[\mathrm{NO_2}]^2$.
- CO reacts only in the fast step after it, so it does not appear: the reaction is zero order in CO.
- This matches what is measured below about $500\ \text{K}$, which supports the mechanism.
Answer: $\text{rate} = k[\mathrm{NO_2}]^2$.
Worked example: A molecule falling apart first
For $$\mathrm{(CH_3)_3CBr} + \mathrm{OH^-} \to \mathrm{(CH_3)_3COH} + \mathrm{Br^-}$$:
Step 1 (slow): $\mathrm{(CH_3)_3CBr \to (CH_3)_3C^+ + Br^-}$
Step 2 (fast): $\mathrm{(CH_3)_3C^+ + OH^- \to (CH_3)_3COH}$
Predict the rate law. Does doubling $[\mathrm{OH^-}]$ change the rate?
Step 1 (slow): $\mathrm{(CH_3)_3CBr \to (CH_3)_3C^+ + Br^-}$
Step 2 (fast): $\mathrm{(CH_3)_3C^+ + OH^- \to (CH_3)_3COH}$
Predict the rate law. Does doubling $[\mathrm{OH^-}]$ change the rate?
- Slow first step, unimolecular: $\text{rate} = k[\mathrm{(CH_3)_3CBr}]$.
- $\mathrm{OH^-}$ enters only after the slow step, so doubling it does not change the rate.
Answer: $\text{rate} = k[\mathrm{(CH_3)_3CBr}]$; no change.
Worked example: A mechanism that fails the test
For $\mathrm{2A + B \to C}$ the measured rate law is $\text{rate} = k[\mathrm{A}]^2$. A student proposes:
Step 1 (slow): $\mathrm{A + B \to D}$
Step 2 (fast): $\mathrm{D + A \to C}$
Is this mechanism consistent with the data?
Step 1 (slow): $\mathrm{A + B \to D}$
Step 2 (fast): $\mathrm{D + A \to C}$
Is this mechanism consistent with the data?
- The steps add up: $\mathrm{2A + B \to C}$, with D as an intermediate.
- The slow first step predicts $\text{rate} = k[\mathrm{A}][\mathrm{B}]$: first order in B.
- The measured law is second order in A and zero order in B. The prediction disagrees, so this mechanism is ruled out, even though it adds up correctly.
Answer: No: it predicts the wrong rate law.
Common mistake
Writing the rate law from a fast step, or including a species that reacts only after the slow step. Check: every species in your rate law must take part in the slow step or in a step before it. If the slow step is not the first step, you will need the method in the next lesson.
"Slow" here means slow compared with the other steps. A rate-determining step can still finish in a fraction of a second.