Reaction mechanism worksheet answers for AP Chemistry (free PDF download)

Problem 1 Molecularity and rate laws

1. Unimolecular; rate = k[O₃].
2. Bimolecular; rate = k[NO][O₃].
3. Bimolecular (two molecules collide); rate = k[NO₂]².

Problem 2 Naming the species

1. X is an intermediate (made, then used). Y is a catalyst (used, then re-formed).
2. Neither appears: both cancel.
3. The transition state.

Problem 3 Adding the steps

1. Cl + O₃ + ClO + O → ClO + O₂ + Cl + O₂. Cancel Cl and ClO: O₃ + O → 2 O₂. Atom check: 4 O on each side.
2. Catalyst Cl; intermediate ClO.

Problem 4 Reading an energy diagram

1. Eₐ(forward) = 100 − 40 = 60 kJ.
2. ΔH = 10 − 40 = −30 kJ (exothermic).
3. Eₐ(reverse) = 100 − 10 = 90 kJ.

Problem 5 True or false about catalysts

1. False: ΔH is unchanged.
2. True.
3. False: it speeds both directions and does not change the equilibrium position.
4. False: it is restored.

Problem 6 Slow step and rate law

1. A + B + C + B → C + D gives A + 2 B → D after cancelling C.
2. The intermediate is C.
3. The slow step is bimolecular, so rate = k[A][B].

Problem 7 Why heat speeds reactions

At the higher temperature, particles move faster, so collisions are more frequent. More importantly, a larger fraction of collisions have energy of at least Eₐ. Both effects increase the number of successful collisions per second, and the second effect dominates.

Problem 8 Checking a mechanism for nitric oxide

1. 2 NO + H₂ + H₂O₂ + H₂ → N₂ + H₂O₂ + 2 H₂O. Cancel H₂O₂: 2 NO + 2 H₂ → N₂ + 2 H₂O.
2. The intermediate is H₂O₂.
3. Rate = k[NO]²[H₂]: third order overall.

Problem 9 Choosing between mechanisms

1. I: rate = k[A][B]. II: rate = k[A]².
2. Mechanism I fits.
3. Several different mechanisms can predict the same rate law, and only the rate law is tested, so agreement does not prove it. Other evidence (detecting X) would be needed.

Problem 10 Fast equilibrium first

1. A + B + AB + B → AB + C gives A + 2 B → C.
2. Rate = k₂[AB][B], with [AB] = K[A][B], so rate = k₂K[A][B]².
3. k = 0.50 × 4.0 = 2.0 M⁻² s⁻¹. Rate = 2.0 × 0.10 × (0.20)² = 8.0 × 10⁻³ M/s.

Problem 11 Two-step profile

1. Step 1: 85 − 0 = 85 kJ/mol. Step 2: 60 − 30 = 30 kJ/mol.
2. Step 1, with the larger barrier.
3. ΔH = −45 − 0 = −45 kJ/mol.
4. Reverse of step 1: 85 − 30 = 55 kJ/mol.

Problem 12 Catalyst and rate constant

1. k(cat)/k = e^(25000/(8.314 × 298)) = e^(10.09) = 2.4 × 10⁴.
2. The reverse reaction is sped up by the same factor (its barrier is lowered by 25 kJ/mol too).
3. ΔH is unchanged.

Problem 13 Activation energy from a rule of thumb

1. Eₐ = R ln 2/(1/300 − 1/310) = 8.314 × 0.693/(1.075 × 10⁻⁴) = 5.36 × 10⁴ J/mol.
2. Eₐ ≈ 54 kJ/mol.

Problem 14 Predicting a rate constant

1. ln(k₂/k₁) = (80000/8.314)(1/298 − 1/318) = 9622 × 2.110 × 10⁻⁴ = 2.031, so k₂/k₁ = 7.62 and k₂ = 1.5 × 10⁻⁵ × 7.62 = 1.1 × 10⁻⁴ s⁻¹.
2. About 7.6 times faster.

Problem 15 Reading an Arrhenius plot

1. Slope = (−4.75 − (−2.15))/(0.0035 − 0.0031) = −2.60/0.0004 = −6.5 × 10³ K.
2. Eₐ = −slope × R = 6500 × 8.314 = 5.4 × 10⁴ J/mol = 54 kJ/mol.
3. ln A = −2.15 + 6500 × 0.0031 = 18.0, so A = e^18.0 = 6.6 × 10⁷ s⁻¹.

Problem 16 Catalytic decomposition of peroxide

1. H₂O₂ + I⁻ + H₂O₂ + IO⁻ → H₂O + IO⁻ + H₂O + O₂ + I⁻. Cancel I⁻ and IO⁻: 2 H₂O₂ → 2 H₂O + O₂.
2. Catalyst I⁻; intermediate IO⁻.
3. Rate = k[H₂O₂][I⁻]. Only one H₂O₂ takes part in the slow step; the second reacts after it.
4. Rate = 0.0115 × 0.50 × 0.020 = 1.2 × 10⁻⁴ M/s.

Problem 17 Designing an experiment

1. Measure initial rates at fixed [A] while changing [B] (for example 0.10 M to 0.30 M).
2. X: rate unchanged. Y: rate multiplied by the same factor as [B].
3. [B] ×3 gives rate ×3 (3.6 × 10⁻³/1.2 × 10⁻³): mechanism Y. k = 1.2 × 10⁻³/(0.10² × 0.10) = 1.2 M⁻² s⁻¹.
4. Another mechanism could give the same rate law; evidence can disprove a mechanism but only support one.

Problem 18 Food spoilage and temperature

1. ln ratio = (65000/8.314)(1/277.15 − 1/298.15) = 7818 × 2.541 × 10⁻⁴ = 1.987; ratio = 7.3.
2. Time = 4.0/7.3 = 0.55 days (about 13 hours).

Problem 19 Matching a catalysed rate

1. Factor = e^(40000/(8.314 × 300)) = e^(16.04) = 9.2 × 10⁶.
2. Equal k needs equal Eₐ/T: 100/T = 60/300, so T = 500 K (227 °C).
3. A catalyst works at low temperature, saving energy and avoiding decomposition of products, while heating by 200 K costs energy.

Problem 20 Surface catalysis and zero order

1. At high pressure all active sites on the gold are occupied, so the surface works at maximum capacity; extra gas does not raise the rate.
2. The rate rises linearly at low [N₂O] then levels off to a horizontal line.
3. t = [N₂O]₀/k = 0.012/(3.0 × 10⁻⁵) = 4.0 × 10² s.
4. As the gas is used up, the pressure falls to the low region where sites are no longer saturated, and the rate begins to depend on [N₂O].