AP Chemistry · Free-response skill

From rate data to a rate law

The move. Use initial-rate data from a lab to find each reactant's order, write the rate law, calculate k with units, and judge whether a proposed mechanism fits the data.

How to do it

  1. For each reactant, find two trials where only its concentration changes. Write the ratio of the rates and the ratio of the concentrations, and say which power links them.
  2. Write the rate law with every order you found, then put one trial's numbers into it and solve for k. Work out the units of k from the overall order so that the rate comes out in M/s.
  3. Check k with a second trial: every trial at the same temperature should give the same k, within the precision of the data.
  4. To test a mechanism, write the rate law of its slow step from that step's coefficients, and compare it with the experimental rate law. Also check that the steps add up to the overall equation.

Practice prompt Try it

A student studies a fictional reaction, 2L(aq) + M(aq) → products, in a water bath held at 30 °C. She measures the initial rate of each trial, taking the rate as the rate at which L is used up, by following the color of the mixture with a spectrophotometer.

Initial-rate data for 2L + M → products at 30 °C

Trial[L] (M)[M] (M)Initial rate (M/s)
10.0200.0103.0 × 10⁻⁵
20.0400.0101.2 × 10⁻⁴
30.0200.0303.0 × 10⁻⁵

Answer each part below, using the data table. Show your work.

  1. Find the order of the reaction with respect to L and the order with respect to M. For each, say which trials you compared and what happened to the rate.
  2. Write the rate law, then use one trial to calculate the rate constant, k. Include its units.
  3. A classmate suggests this mechanism. Step 1 (slow): L + L → L₂. Step 2 (fast): L₂ + M → products. Say whether this mechanism agrees with your rate law, and give the reason.

Write each part, then check yourself against the checklist. Nothing you type here is saved or graded.

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Checklist

Our own checklist of what a full answer does, one line per point. Tick each one your response earns.

0 of 6 points

Model answer, annotated

(a) L: in trials 1 and 2, [M] stays at 0.010 M while [L] doubles from 0.020 to 0.040 M, and the rate rises from 3.0 × 10⁻⁵ to 1.2 × 10⁻⁴ M/s, which is 4 times larger. Since 2² = 4, the reaction is second order in L. M: in trials 1 and 3, [L] stays at 0.020 M while [M] triples from 0.010 to 0.030 M, and the rate stays at 3.0 × 10⁻⁵ M/s. Since 3⁰ = 1, it is zero order in M.

  • Point 1 Names trials 1 and 2 and shows the rate rising 4 times as [L] doubles, which means order 2.
  • Point 2 Names trials 1 and 3 and shows the rate not changing as [M] triples, which means order 0.

(b) Rate = k[L]². From trial 1: k = (3.0 × 10⁻⁵ M/s) ÷ (0.020 M)² = 0.075 M⁻¹ s⁻¹. Trial 2 gives the same k: (1.2 × 10⁻⁴) ÷ (0.040)² = 0.075 M⁻¹ s⁻¹.

  • Point 3 Uses the two orders from part a; M drops out because its order is 0.
  • Point 4 Solves for k with trial 1, checks it with trial 2, and gives second-order units.

(c) The mechanism agrees with the rate law. Step 1 is the slow step, and as an elementary step in which two L particles collide its rate law is rate = k[L]². M enters only in the fast second step, so it does not appear in the rate law, which matches the experimental rate law. The steps also add up to 2L + M → products, with L₂ cancelling as an intermediate.

  • Point 5 Gives the verdict plainly: the mechanism agrees.
  • Point 6 Writes the slow step's rate law from its coefficients and compares it with the rate law from part b.