Three Orders from Three Runs: Judging a Rate Law That Does Not Match the Equation
A student investigates the reaction A + B → P in aqueous solution at 298 K by the initial-rate method. Each run uses a freshly mixed solution, quenched after 30 s, and the product is estimated spectrophotometrically. The record book gives:
| run | [A]₀ / mol L⁻¹ | [B]₀ / mol L⁻¹ | initial rate / mol L⁻¹ s⁻¹ |
|---|---|---|---|
| 1 | 0.10 | 0.10 | 2.0 × 10⁻⁴ |
| 2 | 0.20 | 0.10 | 8.0 × 10⁻⁴ |
| 3 | 0.10 | 0.20 | 4.0 × 10⁻⁴ |
The student concludes: 'doubling [A] multiplies the rate by four and doubling [B] doubles it, so the reaction is second order overall, first order in each reactant.' A second student repeats run 1 at 308 K and obtains 4.1 × 10⁻⁴ mol L⁻¹ s⁻¹, from which the class is asked to estimate the activation energy. A third student objects that the equation says one mole of A reacts with one mole of B, 'so the rate law must be first order in each'. The report must reconcile all three statements.
1. Determine the order in A and in B, write the rate law, calculate k with units, and correct the student's conclusion.
Read the worked answer
Runs 1 and 2: [B] is fixed and [A] doubles, while the rate goes from 2.0 × 10⁻⁴ to 8.0 × 10⁻⁴, a factor of 4. So 2^m = 4 and m = 2: second order in A. Runs 1 and 3: [A] is fixed and [B] doubles, while the rate goes from 2.0 × 10⁻⁴ to 4.0 × 10⁻⁴, a factor of 2. So 2^n = 2 and n = 1: first order in B. Rate = k[A]²[B], overall order 2 + 1 = 3, not 2. The student's error is arithmetical plus conceptual: a factor of 4 in A gives second order in A, and only the sum of the exponents is the overall order. k = rate/([A]²[B]) = 2.0 × 10⁻⁴/((0.10)² × 0.10) = 2.0 × 10⁻⁴/1.0 × 10⁻³ = 0.20. Check with run 2: 8.0 × 10⁻⁴/(0.040 × 0.10) = 0.20; with run 3: 4.0 × 10⁻⁴/(0.010 × 0.20) = 0.20. Units: (mol L⁻¹ s⁻¹)/(mol L⁻¹)³ = L² mol⁻² s⁻¹.
2. With [B] held constant at 0.10 M, calculate the half-life starting from [A]₀ = 0.20 M and comment on what the answer shows.
Read the worked answer
At fixed [B] the rate law collapses to rate = k′[A]², with k′ = k[B] = 0.20 × 0.10 = 0.020 L mol⁻¹ s⁻¹. This is a pseudo-second-order reaction in A. For second order in a single reactant, t₁/₂ = 1/(k′[A]₀) = 1/(0.020 × 0.20) = 250 s. Check with the integrated form: [A] = [A]₀/(1 + k′t[A]₀) = 0.20/(1 + 0.020 × 250 × 0.20) = 0.20/2.00 = 0.10 M, which is exactly half. Good. The comment: the half-life depends on the starting concentration, so it is not a constant. From 0.20 M the first half-life is 250 s, but from 0.10 M it would be 500 s. That dependence is itself evidence against first-order behaviour in A, and it is why half-life data must be tested before an order is assumed.
3. Estimate the activation energy from the 298 K and 308 K data and evaluate the common claim that 'a 10 K rise doubles the rate'.
Read the worked answer
At the same concentrations, k ∝ rate, so k(308)/k(298) = 4.1 × 10⁻⁴/2.0 × 10⁻⁴ = 2.05. Arrhenius: ln(k₂/k₁) = (Ea/R)(1/T₁ − 1/T₂). 1/298 − 1/308 = 10/(298 × 308) = 1.0895 × 10⁻⁴ K⁻¹, and ln 2.05 = 0.718. Ea = 8.314 × 0.718/1.0895 × 10⁻⁴ = 5.97/1.0895 × 10⁻⁴ = 5.48 × 10⁴ J mol⁻¹ ≈ 54.8 kJ mol⁻¹. The claim is a coincidence here, not a rule. For Ea = 100 kJ mol⁻¹ the same 10 K rise at room temperature would multiply the rate by about 3.7, not 2; and at 1000 K the same 10 K rise would change it by only about 13%, because the fractional change in T is much smaller and the exponential is less sensitive. The 'doubling' rule works only for Ea near 50–60 kJ mol⁻¹ in the 300 K region.