Using initial rates to find a rate law and propose a mechanism
The oxidation of bromide by bromate(V) in acidic solution is studied at 25 °C: BrO3−(aq) + 5Br−(aq) + 6H+(aq) → 3Br2(aq) + 3H2O(l). A spectrophotometer is used to follow the absorbance of bromine at 400 nm, and initial rates are obtained from the tangent to the absorbance–time curve at t = 0. The table gives the initial concentrations and initial rates for four experiments. A student claims that because the stoichiometric equation contains five bromide ions and six hydrogen ions, the rate law must have the same exponents. Another student argues that the rate law can only be found experimentally. The teacher asks you to resolve the disagreement using the data, to calculate the rate constant, and to suggest a rate-determining step. The temperature is held constant at 25 °C and the ionic strength is approximately constant.
| Experiment | [BrO3−] / mol dm−3 | [Br−] / mol dm−3 | [H+] / mol dm−3 | Initial rate / mol dm−3 s−1 |
|---|---|---|---|---|
| 1 | 0.10 | 0.10 | 0.10 | 1.2 × 10−4 |
| 2 | 0.20 | 0.10 | 0.10 | 2.4 × 10−4 |
| 3 | 0.10 | 0.20 | 0.10 | 2.4 × 10−4 |
| 4 | 0.10 | 0.10 | 0.20 | 4.8 × 10−4 |
1. Deduce the order of reaction with respect to each reactant and write the rate equation. Show clearly which experiments you compare.
Read the worked answer
Compare experiments in which only one concentration changes.
Order with respect to BrO3−: compare experiments 1 and 2. [Br−] and [H+] are constant; [BrO3−] doubles from 0.10 to 0.20 mol dm−3. The rate doubles from 1.2 × 10−4 to 2.4 × 10−4 mol dm−3 s−1. Doubling the concentration doubles the rate, so the reaction is first order in BrO3−.
Order with respect to Br−: compare experiments 1 and 3. [BrO3−] and [H+] are constant; [Br−] doubles and the rate doubles. The reaction is first order in Br−.
Order with respect to H+: compare experiments 1 and 4. [BrO3−] and [Br−] are constant; [H+] doubles from 0.10 to 0.20 mol dm−3. The rate increases from 1.2 × 10−4 to 4.8 × 10−4, a factor of four. Doubling [H+] quadruples the rate, so the reaction is second order in H+.
Rate equation:
rate = k[BrO3−][Br−][H+]^2
The overall order is 1 + 1 + 2 = 4. The student who reads the exponents from the stoichiometric equation is wrong: the equation gives the overall stoichiometry, not the mechanism.
2. Calculate the rate constant k from experiment 1 and give its units.
Read the worked answer
Use the rate equation and the data from experiment 1:
rate = k[BrO3−][Br−][H+]^2
1.2 × 10−4 = k × (0.10) × (0.10) × (0.10)^2
1.2 × 10−4 = k × 1.0 × 10−4
k = 1.2
Units:
rate has units mol dm−3 s−1. The concentration product has units (mol dm−3)^4 because the overall order is 4.
units of k = (mol dm−3 s−1) / (mol dm−3)^4 = mol−3 dm9 s−1
So k = 1.2 mol−3 dm9 s−1 at 25 °C. The large positive powers of dm3 are normal for a fourth-order rate law and simply reflect the units needed to make the rate come out in mol dm−3 s−1.
3. Propose a rate-determining step consistent with the rate equation. Explain why the stoichiometric equation alone cannot predict the rate law.
Read the worked answer
The rate law shows that the activated complex of the rate-determining step contains one BrO3−, one Br− and two H+ ions. A possible slow step is:
BrO3− + Br− + 2H+ → HBrO2 + HOBr (slow)
This is followed by faster steps in which the intermediate bromine-containing species react with more Br− and H+ to give Br2 and water. The slow step is consistent with the experimental rate law because its stoichiometry matches the species whose concentrations appear in the rate equation.
However, a rate law is not a proof of a mechanism. Other mechanisms, including ones with fast pre-equilibria before the slow step, can give the same rate equation. The proposed step should be regarded as consistent with the data, not uniquely proven by it.
The stoichiometric equation 5Br− + BrO3− + 6H+ → 3Br2 + 3H2O tells us the overall ratio in which reactants are consumed. It does not tell us which collisions or bond-breaking steps are slow. The rate law is determined by the slow step and by any pre-equilibria, so it must be found experimentally. This is why the exponents in the rate equation are not simply copied from the balanced equation.