Quick answer: A reaction mechanism is a sequence of elementary steps whose sum is the overall equation. This AP Chemistry worksheet on reaction mechanism gives 20 problems from easy to hard: molecularity and rate laws, naming the species and adding the steps. The worked answers are on a separate page, and the worksheet is a free printable PDF.
These problems cover elementary steps, rate-determining steps, energy profiles, catalysts and the Arrhenius equation. They progress from naming species and reading diagrams to testing proposed mechanisms and handling activation-energy calculations. They follow the style of AP Chemistry Unit 5 and first-year university kinetics; give units with every numerical answer.
Level: AP Chemistry · Curriculum: AP Chem Unit 5
Problem 1 Molecularity and rate laws ★★★
Classify each elementary step and write its rate law.1. O₃ → O₂ + O
2. NO + O₃ → NO₂ + O₂
3. 2 NO₂ → N₂O₄
Problem 2 Naming the species ★★★
In a mechanism, species X is formed in step 1 and consumed in step 2, while species Y is consumed in step 1 and re-formed in step 2.1. Name each species.
2. State whether X or Y appears in the overall equation.
3. Name the species at the top of an energy-profile peak.
Problem 3 Adding the steps ★★★
Bilal in Lahore proposes: step 1, Cl + O₃ → ClO + O₂; step 2, ClO + O → Cl + O₂.1. Write the overall equation.
2. Identify the catalyst and the intermediate.
Problem 4 Reading an energy diagram ★★★
Use the profile.2. Calculate ΔH.
3. Calculate Eₐ for the reverse reaction.
Problem 5 True or false about catalysts ★★★
State whether each is true or false and correct the false ones.1. A catalyst changes ΔH.
2. A catalyst lowers the activation energy.
3. A catalyst shifts the equilibrium to the product side.
4. A catalyst is used up in the reaction.
Problem 6 Slow step and rate law ★★★
A + 2 B → D proceeds by: step 1 (slow) A + B → C; step 2 (fast) C + B → D.1. Show that the steps give the overall equation.
2. Identify the intermediate.
3. Write the predicted rate law.
Problem 7 Why heat speeds reactions ★★★
Explain, in terms of collisions, why a reaction in a pressure cooker at 120 °C is much faster than in an open pan at 100 °C.Problem 8 Checking a mechanism for nitric oxide ★★★
The reaction 2 NO + 2 H₂ → N₂ + 2 H₂O has a mechanism: step 1 (slow) 2 NO + H₂ → N₂ + H₂O₂; step 2 (fast) H₂O₂ + H₂ → 2 H₂O.1. Show that the steps add to the overall equation.
2. Identify the intermediate.
3. Predict the rate law and its overall order.
Problem 9 Choosing between mechanisms ★★★
For 2 A + B → C, experiment gives rate = k[A][B]. Mechanism I: (slow) A + B → X; (fast) X + A → C. Mechanism II: (slow) 2 A → Y; (fast) Y + B → C.1. Predict the rate law for each.
2. State which mechanism fits the data.
3. Explain why the data cannot prove that mechanism.
Problem 10 Fast equilibrium first ★★★
A + B ⇌ AB is a fast equilibrium with K = 4.0 M⁻¹, followed by a slow step AB + B → C with k₂ = 0.50 M⁻¹ s⁻¹.1. Write the overall equation.
2. Derive the rate law in terms of [A] and [B].
3. Calculate the observed rate constant and the rate when [A] = 0.10 M and [B] = 0.20 M.
Problem 11 Two-step profile ★★★
Use the energy values on the diagram.2. Identify the rate-determining step.
3. Calculate ΔH overall.
4. Calculate Eₐ for the reverse of step 1.
Problem 12 Catalyst and rate constant ★★★
A reaction has Eₐ = 75 kJ/mol. A catalyst gives a new path with Eₐ = 50 kJ/mol. Assume A is unchanged and T = 298 K.1. Calculate the factor by which k increases.
2. State what happens to the rate of the reverse reaction.
3. State what happens to ΔH.
Problem 13 Activation energy from a rule of thumb ★★★
A reaction’s rate constant doubles when the temperature rises from 300 K to 310 K.1. Calculate Eₐ.
2. Give your answer in kJ/mol.
Problem 14 Predicting a rate constant ★★★
A reaction has Eₐ = 80.0 kJ/mol and k = 1.5 × 10⁻⁵ s⁻¹ at 298 K.1. Calculate k at 318 K.
2. Calculate how many times faster the reaction runs at 318 K.
Problem 15 Reading an Arrhenius plot ★★★
The graph is for a gas-phase reaction.2. Calculate Eₐ in kJ/mol.
3. Find ln A and A.
Problem 16 Catalytic decomposition of peroxide ★★★
Iodide ions catalyse 2 H₂O₂ → 2 H₂O + O₂ by: step 1 (slow) H₂O₂ + I⁻ → H₂O + IO⁻; step 2 (fast) H₂O₂ + IO⁻ → H₂O + O₂ + I⁻.1. Show the steps add to the overall equation.
2. Identify the catalyst and the intermediate.
3. Write the rate law and explain why it is not second order in H₂O₂ even though two molecules appear in the equation.
4. Calculate the rate when k = 0.0115 M⁻¹ s⁻¹, [H₂O₂] = 0.50 M and [I⁻] = 0.020 M.
Problem 17 Designing an experiment ★★★
For 2 A + B → C, mechanism X predicts rate = k[A]² and mechanism Y predicts rate = k[A]²[B].1. Describe an experiment that distinguishes them.
2. State the result expected from each mechanism.
3. Data: [A] = 0.10 M, [B] = 0.10 M gives rate 1.2 × 10⁻³ M/s; [A] = 0.10 M, [B] = 0.30 M gives 3.6 × 10⁻³ M/s. Decide which mechanism is supported and calculate k.
4. Explain why this evidence supports but does not prove the mechanism.
Problem 18 Food spoilage and temperature ★★★
A spoilage reaction in milk has Eₐ = 65 kJ/mol.1. Calculate the ratio of rate constants at 25 °C and 4 °C.
2. Milk stays fresh for 4.0 days in a refrigerator at 4 °C. Predict how long it stays fresh at 25 °C, assuming spoilage is controlled by this reaction.
Problem 19 Matching a catalysed rate ★★★
An uncatalysed reaction has Eₐ = 100 kJ/mol; a catalyst reduces it to 60 kJ/mol. Assume the same A.1. Calculate the increase in k caused by the catalyst at 300 K.
2. Calculate the temperature at which the uncatalysed reaction has the same k as the catalysed one at 300 K.
3. Explain why using a catalyst can be preferable to heating.
Problem 20 Surface catalysis and zero order ★★★
The decomposition of N₂O on a hot gold surface is first order at low pressure and zero order at high pressure.1. Explain the zero-order behaviour using the surface.
2. Describe how the plot of rate against [N₂O] looks.
3. In the zero-order region, k = 3.0 × 10⁻⁵ M/s and [N₂O]₀ = 0.012 M. Calculate the time to finish if the order stays zero.
4. Explain why the reaction slows down at the end.
