2.2.9 (HL)—Rate equations
- Syllabus
- First assessment 2025
- Objective
- 2.2.9
- Level
- HL
rate=k[A]x[B]y
Use experimental trials that vary one concentration independently. Compare the rate factor with the concentration factor to infer each exponent; the balanced overall equation alone cannot determine the rate equation.
If doubling [A] while holding [B] constant quadruples rate, the order in A is 2; if doubling [B] leaves rate unchanged, the order in B is 0, giving rate = k[A]². Choose trial pairs with only one changed concentration before combining exponents.
Worked initial-rate deduction: doubling [FeX3+] from 1.00×10−2 to 2.00×10−2moldm−3 at constant [IX−] doubles rate, so the order in FeX3+ is 1. Doubling [IX−] at constant [FeX3+] increases rate from 3.24×10−5 to 1.30×10−4moldm−3s−1, approximately fourfold, so its order is 2. Therefore v=k[FeX3+][IX−]2, third order overall.
Representative question
Two more trials ( 2 and 3 ) were carried out. The results are given below.
| Trials | Volume of 0.20 mol dm−3KI(aq)/cm3 Volume of 0.20 mol dm−3KI(aq)/cm3 | Volume of deionised water / cm3 | Volume of 3\% H2O2(aq)/cm3 Volume of 3\% H2O2(aq)/cm3 | Average rate of reaction /cm3O2(g)s−1 |
|---|---|---|---|---|
| 1 | 10.0 | 15.0 | 5.0 | |
| 2 | 10.0 | 10.0 | 10.0 | 0.0429 |
| 3 | 20.0 | 5.0 | 5.0 | 0.0451 |
Determine the rate equation for the reaction and its overall order, using your answer from (b)(i).
Rate equation:
Overall order:
Rate equation:
Rate =k[H2O2]×[KI]
Overall order:
2
Rate constant must be included.
Retrieve the route: measure a tangent rate, explain effective collisions, map rate factors, read Ea and energy profiles, evaluate mechanisms, determine molecularity and orders, calculate k, then use Arrhenius gradient and intercept for Ea and A.
Check tangent versus average slope, energy versus orientation, barrier labels, intermediate versus transition state, one-variable trial comparisons, order-dependent units, kelvin temperature and the signs of gradient and Ea.