The Initial Rates Methodology
The method of initial rates involves comparing reaction mixtures where the initial concentration of only one reactant is systematically varied while holding all other initial concentrations and temperature constant.
The Method of Initial Rates
The key principle is simple:
By keeping all concentrations constant except one, you can isolate the effect of that reactant on the rate.
Worked Example
Step-by-Step Initial Rates Problem
Experimental Data
| Exp | [A] / mol dm⁻³ | [B] / mol dm⁻³ | Initial Rate / mol dm⁻³ s⁻¹ |
|---|---|---|---|
| 1 | 0.10 | 0.10 | 2.0 × 10⁻⁴ |
| 2 | 0.20 | 0.10 | 4.0 × 10⁻⁴ |
| 3 | 0.10 | 0.20 | 8.0 × 10⁻⁴ |
Step-by-Step Solution
Finding order with respect to A (compare Exp 1 and 2):
[A] doubles (0.10 → 0.20), [B] stays constant.
Rate doubles (2.0 → 4.0): \((2)^x = 2\), so \(x = 1\) (first order in A)
Finding order with respect to B (compare Exp 1 and 3):
[B] doubles (0.10 → 0.20), [A] stays constant.
Rate quadruples (2.0 → 8.0): \((2)^y = 4\), so \(y = 2\) (second order in B)
Rate equation: Rate = k[A]¹[B]² (Overall order = 3)
Finding k
Once you know the rate equation, substitute any experiment's data to find k:
\(k = \frac{\text{Rate}}{[A][B]^2} = \frac{2.0 \times 10^{-4}}{(0.10)(0.10)^2} = \frac{2.0 \times 10^{-4}}{1.0 \times 10^{-3}} = 0.20\text{ mol}^{-2}\text{ dm}^{6}\text{ s}^{-1}\)
Systematic Pairwise Comparison
Always identify two specific experiments where only one reactant concentration changes. Clearly state your mathematical logic: "Comparing Exp 1 & Exp 2: [A] doubles while [B] is constant; Rate doubles \(\implies\) 1st order in A."
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