Clock Reaction with a Concentration Mismatch
In a kinetic study of the reaction between A and B, a student measures initial rates at 298 K by changing one concentration at a time. The data are: Experiment 1, [A]0 = 0.020 M, [B]0 = 0.020 M, initial rate = 4.0 × 10⁻⁴ M s⁻¹; Experiment 2, [A]0 = 0.040 M, [B]0 = 0.020 M, rate = 8.0 × 10⁻⁴ M s⁻¹; Experiment 3, [A]0 = 0.020 M, [B]0 = 0.040 M, rate = 1.6 × 10⁻³ M s⁻¹. A fourth run at 308 K using the same concentrations as Experiment 1 gives a rate of 8.0 × 10⁻⁴ M s⁻¹. The student has only 30 minutes and cannot repeat all runs. The teacher asks for the rate law, the rate constant at 298 K, and whether the proposed one-step mechanism A + B → products can be correct. The student also wants to predict the effect of doubling both concentrations at 298 K.
1. Determine the order with respect to A and with respect to B, write the rate law, and calculate the rate constant at 298 K.
Read the worked answer
Compare Experiments 1 and 2: [B] is constant while [A] doubles from 0.020 to 0.040 M, and the rate doubles from 4.0 × 10⁻⁴ to 8.0 × 10⁻⁴ M s⁻¹. Since 2^m = 2, m = 1; the reaction is first order in A.
Compare Experiments 1 and 3: [A] is constant while [B] doubles from 0.020 to 0.040 M, and the rate rises from 4.0 × 10⁻⁴ to 1.6 × 10⁻³ M s⁻¹, a factor of 4. Since 2^n = 4, n = 2; the reaction is second order in B.
Rate law: rate = k[A][B]².
Using Experiment 1: k = rate/([A][B]²) = 4.0 × 10⁻⁴ / (0.020 × (0.020)²) = 4.0 × 10⁻⁴ / (8.0 × 10⁻⁶) = 50 M⁻² s⁻¹.
2. Calculate the activation energy from the 298 K and 308 K results. State any assumption you make.
Read the worked answer
At 308 K the concentrations are the same as in Experiment 1, so the rate constant is
k308 = 8.0 × 10⁻⁴ / (0.020 × (0.020)²) = 100 M⁻² s⁻¹.
Thus k308/k298 = 100/50 = 2.0. Using the Arrhenius equation in the form
ln(k2/k1) = (Ea/R)(1/T1 − 1/T2),
ln 2 = (Ea/8.314)(1/298 − 1/308).
The temperature factor is 1/298 − 1/308 = 1.089 × 10⁻⁴ K⁻¹.
Ea = (0.693 × 8.314)/(1.089 × 10⁻⁴) ≈ 5.3 × 10⁴ J mol⁻¹ ≈ 53 kJ mol⁻¹.
The assumption is that the rate law and the frequency factor remain effectively constant over this small temperature range, so the change in rate reflects the change in k.
3. Is the proposed one-step mechanism A + B → products consistent with the rate law? Predict the rate if both concentrations are doubled at 298 K.
Read the worked answer
No. A single elementary step A + B → products would have rate = k[A][B], i.e. first order in each reactant. The observed rate law is rate = k[A][B]², which is first order in A but second order in B. A one-step bimolecular mechanism cannot produce a second-order dependence on B. The mechanism must involve B in a prior or дополнительный step, for example formation of a B-containing intermediate, so the simple one-step proposal is inconsistent.
If both concentrations are doubled at 298 K:
rate ratio = (2)^1 × (2)^2 = 8.
So the rate becomes 8 × 4.0 × 10⁻⁴ = 3.2 × 10⁻³ M s⁻¹.