The problems are on the rate equations worksheet. Try them on your own first, then compare with the answers below.
Quick answer: A rate equation, rate = k[A]^m[B]^n, shows how concentration controls reaction speed; orders come from experiments, not from the balanced equation. These worked answers correct the 20 problems of the Chemistry 11-12 worksheet on rate equations, from rate words to spot the errors: each solution shows the formula, the substitution and the final answer with its unit.
Each answer shows how the order was deduced, the substitution into the rate equation or Arrhenius expression, and the result with its unit. Check particularly the units of k and whether you used kelvin and joules in the Arrhenius calculations.
Level: Chemistry grades 11-12 (A Level · IB · Class 11-12) · Curriculum: IB R2.2
Problem 1 Rate words
1. The change in concentration of a reactant or product per unit time.2. mol dm⁻³ s⁻¹.
3. Temperature and a catalyst change k. Changing the concentration of a reactant does not change k.
Problem 2 Reading a rate equation
1. NO second order, O₂ first order, overall third order.2. Doubling [O₂]: rate × 2. Doubling [NO]: rate × 4.
3. mol dm⁻³ s⁻¹ = k × (mol dm⁻³)³, so k has units dm⁶ mol⁻² s⁻¹.
Problem 3 Doubling and tripling
1. × 1 and × 1.2. × 2 and × 3.
3. × 4 and × 9.
4. (½)² = × 0.25.
Problem 4 Units of k
1. mol dm⁻³ s⁻¹.2. s⁻¹.
3. dm³ mol⁻¹ s⁻¹.
4. dm⁶ mol⁻² s⁻¹. Each extra order divides by another mol dm⁻³.
Problem 5 Rate against concentration
1. First order: the graph is a straight line through the origin, so rate ∝ [A].2. Gradient = 1.5 × 10⁻³ / 0.50 = 3.0 × 10⁻³ s⁻¹.
3. Rate = 3.0 × 10⁻³ × 0.35 = 1.05 × 10⁻³ mol dm⁻³ s⁻¹.
Problem 6 Using a rate equation
1. Rate = 0.030 × 0.15² = 6.75 × 10⁻⁴ mol dm⁻³ s⁻¹.2. Rate = 0.030 × 0.30² = 2.7 × 10⁻³ mol dm⁻³ s⁻¹.
3. The second is 4 times the first because the concentration doubled and the order is 2.
Problem 7 Initial rates, one reactant
1. Doubling [X] multiplies the rate by 4 and quadrupling by 16, so second order.2. Rate = k[X]².
3. k = 1.6 × 10⁻³ / 0.20² = 0.040 dm³ mol⁻¹ s⁻¹.
Problem 8 Half-life calculation
1. t½ = 0.693 / 4.6 × 10⁻³ = 151 s.2. 12.5 % is three half-lives: 3 × 150.7 = 452 s.
3. 50 % and 25 %.
Problem 9 Half-lives from a graph
1. 60 s and 60 s (0.40 at 60 s, 0.20 at 120 s).2. The half-life is constant, so the reaction is first order.
3. k = 0.693 / 60 = 1.16 × 10⁻² s⁻¹.
Problem 10 Two reactants
1. Doubling [A] gives rate × 4: second order in A. Tripling [B] gives rate × 3: first order in B.2. Rate = k[A]²[B]; overall third order.
3. k = 1.0 × 10⁻⁵ / (0.050² × 0.020) = 0.20 dm⁶ mol⁻² s⁻¹.
Problem 11 Slow step
1. Adding: 2NO₂ + NO₃ + CO → NO₃ + NO + NO₂ + CO₂; cancelling gives NO₂ + CO → NO + CO₂. The intermediate is NO₃.2. Rate = k[NO₂]².
3. CO reacts only in the fast step after the rate-determining step, so its concentration does not affect the overall rate.
Problem 12 Doubling every ten degrees
1. 3.3557 × 10⁻³ − 3.2468 × 10⁻³ = 1.0895 × 10⁻⁴ K⁻¹.2. Ea = 8.314 × ln 2 / 1.0895 × 10⁻⁴ = 52 890 J/mol = 52.9 kJ/mol.
3. Ea and A are constant over the range, and the rate constant (not the rate) doubles.
Problem 13 Effect of a catalyst
1. 25.0 kJ/mol = 25 000 J/mol.2. Factor = e^(25 000 / (8.314 × 300)) = e^10.02 = 2.25 × 10⁴.
3. A larger fraction of collisions have energy at least equal to the lower activation energy.
Problem 14 Reading an Arrhenius plot
1. Gradient = (−7.40 − (−5.00)) / (0.00340 − 0.00300) = −2.40 / 0.00040 = −6000 K.2. Ea = 6000 × 8.314 = 49 884 J/mol = 49.9 kJ/mol.
3. ln A = −5.00 + 6000 × 0.00300 = 13.0, so A = e^13 = 4.4 × 10⁵ s⁻¹.
Problem 15 A drug in the body
1. k = 0.693 / 6.0 = 0.1155 h⁻¹.2. Fraction = e^(−0.1155 × 15) = e^(−1.733) = 0.177, so 17.7 %.
3. 10 % remains: t = ln 10 / k = 2.303 / 0.1155 = 19.9 h.
Problem 16 Three reactants
1. P: doubling doubles the rate, first order. Q: tripling gives × 9, second order. R: quadrupling has no effect, zero order.2. Rate = k[P][Q]²; k = 3.0 × 10⁻⁵ / (0.10 × 0.10²) = 0.030 dm⁶ mol⁻² s⁻¹.
3. Rate = 0.030 × 0.25 × 0.15² = 1.7 × 10⁻⁴ mol dm⁻³ s⁻¹.
Problem 17 Which mechanism?
1. ICl + H₂ + HI + ICl → HI + HCl + I₂ + HCl; cancelling HI gives 2ICl + H₂ → I₂ + 2HCl.2. A: rate = k[ICl][H₂]. B: rate = k[ICl]².
3. Mechanism A fits; the intermediate is HI.
Problem 18 Rate constant at a new temperature
1. ln(k₂/k₁) = (65 000 / 8.314)(1/298 − 1/318) = 7818 × 2.1105 × 10⁻⁴ = 1.650.2. k₂ = 2.5 × 10⁻⁴ × e^1.650 = 2.5 × 10⁻⁴ × 5.207 = 1.30 × 10⁻³ s⁻¹.
3. 5.2 times faster.
Problem 19 Decomposition on a surface
1. [A] falls by 0.015 mol/dm³ every 100 s, a straight line, so zero order.2. k = 0.015 / 100 = 1.5 × 10⁻⁴ mol dm⁻³ s⁻¹.
3. t½ = [A]₀ / 2k: 0.100 / (3.0 × 10⁻⁴) = 333 s and 0.050 / (3.0 × 10⁻⁴) = 167 s. The half-life halves as the concentration halves.
Problem 20 Spot the errors
1. Orders come from experiment, not from coefficients: the rate equation cannot be written from the balanced equation.2. Units are dm³ mol⁻¹ s⁻¹; s⁻¹ is for first order.
3. A catalyst provides an alternative route with lower activation energy.
4. It is the slowest step.
