Simple rate equations, orders of reaction and rate constants — practice questions

a) Deducing orders:

• For propanone [CH₃COCH₃]: Compare experiments 1 and 2. [CH₃COCH₃] doubles (0.40 → 0.80), while [I₂] and [H⁺] are constant. The rate also doubles (2.4 x 10⁻⁵ → 4.8 x 10⁻⁵). Since rate ∝ [CH₃COCH₃]¹, the reaction is first order with respect to propanone.

• For iodine [I₂]: Compare experiments 1 and 3. [I₂] doubles (0.02 → 0.04), while [CH₃COCH₃] and [H⁺] are constant. The rate does not change (2.4 x 10⁻⁵). Since rate is unaffected by [I₂], the reaction is zero order with respect to iodine.

• For H⁺ ions: Compare experiments 1 and 4. [H⁺] doubles (0.20 → 0.40), while [CH₃COCH₃] and [I₂] are constant. The rate doubles (2.4 x 10⁻⁵ → 4.8 x 10⁻⁵). Since rate ∝ [H⁺]¹, the reaction is first order with respect to H⁺ ions.

b) Rate equation:

Rate = k[CH₃COCH₃]¹[I₂]⁰[H⁺]¹ Since [I₂]⁰ = 1, the simplified rate equation is: Rate = k[CH₃COCH₃][H⁺]

c) Calculating k:

Rearrange the rate equation: k = Rate / ([CH₃COCH₃][H⁺]) Use data from any experiment, e.g., Experiment 1: k = (2.4 x 10⁻⁵ mol dm⁻³ s⁻¹) / (0.40 mol dm⁻³ × 0.20 mol dm⁻³) k = (2.4 x 10⁻⁵) / (0.08) k = 3.0 x 10⁻⁴

To find the units: units of k = (mol dm⁻³ s⁻¹) / (mol dm⁻³ × mol dm⁻³) units of k = s⁻¹ / (mol dm⁻³) units of k = dm³ mol⁻¹ s⁻¹

a) Calculate half-life:

For a first-order reaction, we use the formula: $t_{1/2} = \frac{\ln(2)}{k}$ Substitute the value of k: $t_{1/2} = \frac{\ln(2)}{7.30 \times 10^{-4} \text{ s}^{-1}}$ $t_{1/2} = \frac{0.693}{7.30 \times 10^{-4} \text{ s}^{-1}}$ $t_{1/2} = 949.5$ s To 3 significant figures, $t_{1/2} = 950$ s.

b) Concentration after three half-lives:

Initial concentration = 1.60 mol dm⁻³ After 1 half-life: Concentration = 1.60 / 2 = 0.80 mol dm⁻³ After 2 half-lives: Concentration = 0.80 / 2 = 0.40 mol dm⁻³ After 3 half-lives: Concentration = 0.40 / 2 = 0.20 mol dm⁻³

Alternatively, Concentration = Initial Concentration / $2^n$, where n is the number of half-lives. Concentration = 1.60 / $2^3$ = 1.60 / 8 = 0.20 mol dm⁻³.