Why does a small increase in temperature cause a large increase in rate of reaction?

An increase in temperature increases the average kinetic energy of the molecules, shifting the Maxwell-Boltzmann distribution to the right. This causes a significantly greater proportion of molecules to have kinetic energy equal to or greater than the activation energy (E >= Ea), leading to a much higher frequency of successful collisions.

Why is 1/t used as an approximation of the initial rate of reaction in the disappearing cross practical (RP3)?

In Required Practical 3 (sodium thiosulfate and hydrochloric acid), a fixed amount of solid sulfur precipitate is formed to obscure the cross in each run. Because the amount of product formed is constant, Rate is directly proportional to 1/time (Rate ∝ 1/t).

How does a catalyst increase reaction rate on a Maxwell-Boltzmann curve?

A catalyst does not change the shape of the Maxwell-Boltzmann curve or add thermal energy. Instead, it provides an alternative reaction pathway with a lower activation energy (Ec = Ec, increasing the frequency of successful collisions.

What safety precautions are required for the disappearing cross reaction?

Toxic sulfur dioxide (SO2) gas is produced, which is a severe respiratory irritant and asthma trigger. The reaction must be performed in a fume cupboard or well-ventilated laboratory, and flasks should be quenched immediately in a beaker of sodium carbonate solution.

AQA A-Level Chemistry Kinetics & Rates of Reaction Workbook (RP3)

Jump Directly to a Question:

Q1: Gas Syringe & Limiting Reagent (7M) Q2: Maxwell-Boltzmann & Catalyst (7M) Q3: Disappearing Cross RP3 (6M) Q4: Continuous Mass Loss Monitoring (6M) Q5: Evaluating Kinetics Misconceptions (4M)

Chemical kinetics at A-Level requires students to seamlessly combine qualitative collision theory with quantitative mathematical analysis. This revision workbook tests the core mathematical skills (MS 0.0, 1.1, 1.3, 2.2, 2.4, 3.1, 3.2) and practical apparatus techniques (AT a, b, k, l) essential for achieving an A* in AQA AS and A-Level Chemistry.

Specification & Practical Skills Matrix

Question	Syllabus Topic Reference	Practical Skills (AT)	Mathematical Skills (MS)	Marks
Q1	3.1.2 Amount of Substance & 3.1.2.2 Ideal Gas	AT a, AT k Gas collection & weighing	MS 0.0, 1.1, 2.2, 2.4	7
Q2	3.1.8.2 Maxwell-Boltzmann Distribution	—	MS 3.1, 3.2 Curve sketching	7
Q3	3.1.8.1 Collision Theory & Required Practical 3	AT b, AT l Disappearing cross & safety	MS 0.0, 1.3, 3.1 1/t rate calculations	6
Q4	3.1.2.4 Limiting Reagent & Required Practical 7	AT a, AT k Mass balance monitoring	MS 1.3, 3.1, 3.2 Graphical rate comparison	6
Q5	3.1.8.1 Factors Affecting Rates & 3.1.8.3 Catalysts	—	MS 1.1 Evaluating statements	4

Full Question Walkthroughs & Mark Schemes

Attempt each question independently under exam conditions, then toggle the examiner mark scheme to review the exact mark allocations and common student traps.

Interactive Solutions:

Question 1: Gas Collection, Stoichiometry & Ideal Gas Equation

A student reacted a 0.0830 g sample of magnesium ribbon with 25.0 cm³ of 0.500 mol dm⁻³ hydrochloric acid at 293 K. The hydrogen gas was collected in a gas syringe at a pressure of 101,000 Pa:
Mg(s) + 2HCl(aq) → MgCl₂(aq) + H₂(g) (a) Show by calculation which reactant is the limiting reagent (Mᵣ of Mg = 24.3). [2 Marks]
(b) Use the ideal gas equation (pV = nRT) to calculate the maximum volume of hydrogen gas, in cm³, that could be collected. Give your answer to 3 significant figures (R = 8.314 J K⁻¹ mol⁻¹). [4 Marks]
(c) The gas syringe has an uncertainty of ±0.5 cm³ for each reading. Calculate the percentage apparatus uncertainty in the volume of gas collected. [1 Mark]

Step-by-Step Mathematical Mark Scheme:

Part a (Limiting Reagent): n(Mg) = 0.0830 / 24.3 = 3.42 × 10⁻³ mol n(HCl) = 0.500 × (25.0 / 1000) = 0.0125 mol 1 mol Mg requires 2 mol HCl ⇒ 3.42 × 10⁻³ mol Mg requires 6.84 × 10⁻³ mol HCl. Since 0.0125 mol HCl is present, HCl is in excess, and Mg is the limiting reagent.

Part b (Ideal Gas Calculation): n(H₂) = n(Mg) = 3.4156 × 10⁻³ mol | p = 101,000 Pa | T = 293 K V = nRT / p = [(3.4156 × 10⁻³) × 8.314 × 293] / 101,000 = 8.2381 × 10⁻⁵ m³ Convert m³ → cm³ (× 10⁶): 8.2381 × 10⁻⁵ × 10⁶ = 82.4 cm³ (Accept 82.4 - 82.5)

Part c (Percentage Uncertainty): % Uncertainty = (0.5 cm³ / 82.4 cm³) × 100 = 0.607% (or 0.61%)

Note: A gas syringe only involves a single reading from 0 cm³, so do not multiply uncertainty by 2.

Question 2: Maxwell-Boltzmann Distributions & Catalyst Action

The diagram below represents the Maxwell-Boltzmann distribution of molecular energies in a gas sample at temperature T₁.

(a) Describe how to sketch a second curve representing the same sample at a higher temperature, T₂. [2 Marks]
(b) Explain, in terms of collision theory, why a small increase in temperature results in a large increase in the rate of reaction. [3 Marks]
(c) (i) Explain where to label Eₐ (uncatalysed) and E_c (catalysed) on the energy axis. [1 Mark]
(ii) Explain how a catalyst increases the rate of reaction. [1 Mark]

Step-by-Step Mark Scheme:

Part a (Drawing T₂ Curve):

M1: Peak of curve T₂ is at a lower height and shifted to the right of T₁.
M2: Starts at the origin (0, 0), crosses T₁ exactly once, and remains higher than T₁ at the high-energy tail.

Part b (Collision Theory Explanation):

M1: At higher temperature, average kinetic energy increases, shifting distribution right.
M2: A significantly greater proportion of molecules have energy ≥ Eₐ (activation energy).
M3: Consequently, there is a much higher frequency of successful collisions per unit time.

Part c (Catalyst Action):

(i) M1: E_c must be positioned to the left of Eₐ on the energy axis (E_c < Eₐ).
(ii) M2: A catalyst provides an alternative reaction pathway with a lower activation energy. A greater proportion of molecules have energy ≥ E_c, increasing the frequency of successful collisions.

Question 3: Required Practical 3 — Disappearing Cross

The reaction between sodium thiosulfate and hydrochloric acid produces a precipitate of solid sulfur:
Na₂S₂O₃(aq) + 2HCl(aq) → 2NaCl(aq) + S(s) + SO₂(g) + H₂O(l) A student measures the time (t, in seconds) for a black cross under the flask to become invisible at various temperatures:

Run 1 (20.0 °C): t = 182 s
Run 2 (30.0 °C): t = 91 s
Run 3 (40.0 °C): t = 46 s
Run 4 (50.0 °C): t = 23 s
Run 5 (60.0 °C): t = 12 s

(a) Calculate the relative rate (1/t) for each run to 3 significant figures and state the unit. [2 Marks]
(b) Explain why 1/t is a valid approximation of relative rate. [2 Marks]
(c) State one safety hazard associated with this reaction and describe how to minimise risk. [2 Marks]

Step-by-Step Mark Scheme:

Part a (Relative Rate Table): Unit: s⁻¹ Run 1 (20.0 °C): 1/182 = 5.49 × 10⁻³ s⁻¹ (0.00549) Run 2 (30.0 °C): 1/91 = 1.10 × 10⁻² s⁻¹ (0.0110) Run 3 (40.0 °C): 1/46 = 2.17 × 10⁻² s⁻¹ (0.0217) Run 4 (50.0 °C): 1/23 = 4.35 × 10⁻² s⁻¹ (0.0435) Run 5 (60.0 °C): 1/12 = 8.33 × 10⁻² s⁻¹ (0.0833)

Part b (1/t Approximation Rationale):

M1: Rate is inversely proportional to time (t) for a fixed amount of precipitate (sulfur) to form.
M2: Because the depth of solution and cross size are constant, the amount of sulfur formed to obscure the cross is identical in every run. Hence, Amount / t ∝ 1/t.

Part c (Safety & Hazard Management):

M1: Toxic sulfur dioxide (SO₂) gas is produced (toxic, respiratory irritant, asthma trigger).
M2: Perform in a fume cupboard (or well-ventilated lab) and quench reaction flasks in sodium carbonate solution immediately after the cross is obscured.

Question 4: Continuous Monitoring via Mass Loss

A student investigated the rate of reaction between calcium carbonate and dilute nitric acid:
CaCO₃(s) + 2HNO₃(aq) → Ca(NO₃)₂(aq) + CO₂(g) + H₂O(l) The student added 5.00 g of CaCO₃ chunks (Mᵣ = 100.1) to 100.0 cm³ of 0.800 mol dm⁻³ HNO₃ on a mass balance.

(a) Write the balanced ionic equation for this reaction, including state symbols. [1 Mark]
(b) Show by calculation that nitric acid is the limiting reagent. [2 Marks]
(c) Calculate the maximum theoretical mass loss recorded on the balance to 3 significant figures. [2 Marks]
(d) Describe the curve of mass vs time (Curve A), and how Curve B (using 5.00 g of powder instead of chunks) compares. [1 Mark]

Step-by-Step Mark Scheme:

Part a (Ionic Equation): CaCO₃(s) + 2H⁺(aq) → Ca²⁺(aq) + CO₂(g) + H₂O(l)

Part b (Limiting Reagent Calculation): n(CaCO₃) = 5.00 / 100.1 = 0.04995 mol n(HNO₃) = 0.800 × 0.1000 = 0.0800 mol 0.04995 mol CaCO₃ requires 0.0999 mol HNO₃. Since only 0.0800 mol HNO₃ is present, nitric acid is limiting.

Part c (Theoretical Mass Loss): Moles of CO₂ produced = n(HNO₃) / 2 = 0.0800 / 2 = 0.0400 mol Mass loss = 0.0400 mol × 44.0 g mol⁻¹ = 1.76 g

Part d (Curve Comparison):

Curve A (Chunks): Starts at initial mass, curves downwards with decreasing gradient, levels off at exactly 1.76 g lower.
Curve B (Powder): Starts at same initial mass, has a steeper initial gradient (faster rate due to greater surface area), but levels off at the exact same final mass (same moles of limiting reagent).

Question 5: Evaluation of Kinetics Misconceptions

Consider the following two student statements:

Statement 1: "Doubling the concentration of hydrochloric acid will always double the rate of its reaction with calcium carbonate, because there are twice as many acid particles per unit volume."
Evaluate this statement and explain why it is not completely correct. [2 Marks]

Statement 2: "A catalyst increases the reaction rate by transferring thermal energy to the reactants, which increases their kinetic energy."
Identify the scientific error and state the correct mechanism. [2 Marks]

Step-by-Step Mark Scheme:

Part a (Statement 1 Evaluation):

M1: Incorrect to state it will always double.
M2: The relationship depends on the reaction order with respect to H⁺. If acid is already in large excess, doubling concentration has a negligible effect, and the overall rate is also constrained by the solid calcium carbonate surface area.

Part b (Statement 2 Evaluation):

M1 (Error): A catalyst does not transfer thermal energy and does not increase molecular kinetic energy (only temperature does).
M2 (Correct Mechanism): A catalyst provides an alternative reaction pathway with a lower activation energy, so a greater fraction of existing collisions have energy ≥ E_c.

AQA A-Level Chemistry: Rates of Reaction (Kinetics) Challenge

Specification & Practical Skills Matrix

Full Question Walkthroughs & Mark Schemes

Step-by-Step Mathematical Mark Scheme:

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Mastering Rates of Reaction & Physical Chemistry