SME
General Chemistry II · Problem type
Rate law from a mechanism
Use this for “The slow step is NO₂ + NO₂ → NO₃ + NO, then NO₃ + CO → NO₂ + CO₂ is fast. What rate law fits?”
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Which rate law does each mechanism give?
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The slow step sets the rate: its reactants, counted by their coefficients, give the rate law.
Assumptions
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The slowest step limits the rate, so the rate law is the rate law of that step alone.
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An elementary step’s rate law comes from its reactants: each one’s order is its coefficient in that step.
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Every slow step here comes first, so its reactants are all starting materials, never intermediates.
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With real species, the first reactant named in the slow step plays A and the second plays B.
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Related lessons
Kinetics
Main lesson
A second-order reaction
Use this for “A second-order reaction has k = 0.500 M⁻¹s⁻¹. Starting at 0.100 M, find the concentration after 60 s and the first half-life.”
A zero-order reaction
Use this for “A zero-order reaction has k = 2.00 × 10⁻³ M/s. Starting at 0.500 M, find the concentration after 100 s, the half-life and when A runs out.”
Order and rate constant from initial rates
Use this for “With [B] held at 0.100 M, raising [A] from 0.100 M to 0.200 M raises the initial rate from 2.00 × 10⁻³ to 8.00 × 10⁻³ M/s. The reaction is first order in B. Find the order in A and k.”
Activation energy from two temperatures
Use this for “A rate constant doubles, from 1.00 × 10⁻³ s⁻¹ at 25 °C to 2.00 × 10⁻³ s⁻¹ at 35 °C. Find the activation energy and the frequency factor A.”