ChemixaLEARN CHEMISTRY

Determining a Rate Law from Experimental Data

Medium
Problem

A reaction A + B -> products is studied at constant temperature with the method of initial rates. The data are: Trial 1 — [A] = 0.10 M, [B] = 0.10 M, rate = 2.0 x 10^-3 M/s. Trial 2 — [A] = 0.20 M, [B] = 0.10 M, rate = 4.0 x 10^-3 M/s. Trial 3 — [A] = 0.10 M, [B] = 0.20 M, rate = 8.0 x 10^-3 M/s. Determine the rate law and calculate the rate constant k.

Given

  • Trial 1: [A] = 0.10 M, [B] = 0.10 M, initial rate = 2.0 x 10^-3 M/s

  • Trial 2: [A] = 0.20 M, [B] = 0.10 M, initial rate = 4.0 x 10^-3 M/s

  • Trial 3: [A] = 0.10 M, [B] = 0.20 M, initial rate = 8.0 x 10^-3 M/s

  • Temperature is held constant in all three trials.

Find

The rate law (orders m and n) and the numerical value of the rate constant k at this temperature.

Principle

Reaction orders cannot be predicted from the balanced equation; they must be determined experimentally. In the initial-rates method, two trials are compared in which only one reactant concentration changes, so the ratio of the rates equals that reactant's concentration ratio raised to its order. Once the orders are known, any single trial gives k by rearrangement.

Formula

Solution

  1. Compare trials 1 and 2, where only [A] changes: [A] doubles from 0.10 M to 0.20 M while [B] stays at 0.10 M, and the rate doubles from 2.0 x 10^-3 to 4.0 x 10^-3 M/s. The rate ratio is 2, so 2 = 2^m, giving m = 1 (first order in A).

  2. Compare trials 1 and 3, where only [B] changes: [B] doubles from 0.10 M to 0.20 M while [A] stays at 0.10 M, and the rate quadruples from 2.0 x 10^-3 to 8.0 x 10^-3 M/s. The rate ratio is 4, so 4 = 2^n, giving n = 2 (second order in B).

  3. The rate law is therefore rate = k[A][B]^2, third order overall.

  4. Substitute trial 1 values to find k: 2.0 x 10^-3 M/s = k(0.10 M)(0.10 M)^2 = k(0.10)(0.0100) M^3 = k(1.0 x 10^-3 M^3).

  5. Solve: k = (2.0 x 10^-3) / (1.0 x 10^-3) = 2.0 M^-2 s^-1.

  6. Verify with trial 3: k[A][B]^2 = 2.0 x (0.10) x (0.20)^2 = 2.0 x 0.10 x 0.0400 = 8.0 x 10^-3 M/s, which matches the measured rate exactly.

rate = k[A][B]^2 with k = 2.0 M^-2 s^-1

Unit check

Rate has units M/s, and [A][B]^2 has units M x M^2 = M^3, so k must have units (M/s)/M^3 = M^-2 s^-1. The calculated k = 2.0 M^-2 s^-1 has exactly these units, confirming dimensional consistency for a third-order reaction.

What this means physically

The reaction is third order overall: first order in A and second order in B. Doubling B quadruples the rate while doubling A only doubles it, so the reaction is much more sensitive to the concentration of B. The constant k = 2.0 M^-2 s^-1 links any pair of concentrations to their rate at this temperature.

Common mistake

Students often copy the stoichiometric coefficients from the balanced equation into the rate law. Reaction orders are measured from data, not taken from the equation — the exponents here (1 and 2) had to be found by comparing trials where only one concentration changed.