Exam paper: written questions
Answer all questions in the spaces provided or on paper, then check against the mark scheme. Total: 70 marks.
Question 1: Reaction Kinetics and Collision Theory
12 marks(a) State what is meant by the term activation energy of a reaction. [1]
(b) In a reaction vessel, nitrogen dioxide reacts with carbon monoxide at 300 °C:
NO₂(g) + CO(g) → NO(g) + CO₂(g)
On axes labeled with energy (E) on the horizontal axis and fraction of molecules on the vertical axis, describe the essential features of a Maxwell-Boltzmann distribution curve for this gas mixture at temperature T1 = 300 °C. State where the curve starts, its shape, and the relative positions of the most probable energy (Em) and the activation energy (Ea). [3]
(c) Describe two differences between the distribution curve at 300 °C (T1) and a second curve drawn for the same mixture at a higher temperature T2 = 400 °C. [2]
(d) Explain, with reference to collision theory and the Maxwell-Boltzmann distribution, why increasing the temperature from 300 °C to 400 °C results in a significant increase in reaction rate. [3]
(e) In a separate experiment, the decomposition of aqueous hydrogen peroxide is catalysed by adding solid manganese(IV) oxide:
2H₂O₂(aq) → 2H₂O(l) + O₂(g)
Explain how a catalyst increases the rate of reaction. Do not use the term "collision frequency" in your answer. [3]
Show Mark Scheme
(a) [1 mark]
- M1: Minimum kinetic energy colliding particles must possess for a collision to result in a chemical reaction [1].
- Accept: Minimum energy needed to start a reaction.
- Ignore: Simple "energy needed for molecules to collide" without specifying reaction or successful collision.
(b) [3 marks]
- M1: Curve starts at the origin (0, 0) and approaches the horizontal axis at high energy but does not touch or cross it [1].
- M2: Curve is asymmetrical, with the peak skewed towards the left [1].
- M3: Em is positioned directly below the peak/apex, and Ea is located on the high-energy tail to the right of Em [1].
(c) [2 marks]
- M1: The peak at T2 is lower and shifted to the right of the peak at T1 [1].
- M2: The T2 curve crosses the T1 curve only once and stays above T1 at higher energies (with approximately equal total area) [1].
(d) [3 marks]
- M1: At higher temperature (T2), the distribution shifts to higher energies, so a significantly larger fraction of molecules possess kinetic energy E ≥ Ea (represented by the area under the curve to the right of Ea) [1].
- M2: This produces a higher frequency of successful collisions (collisions with energy ≥ Ea) [1].
- M3: Therefore, the rate of reaction increases significantly [1].
- Ignore: Simply stating "more collisions occur" without specifying frequency of collisions or successful collisions.
(e) [3 marks]
- M1: The catalyst provides an alternative reaction pathway / route [1].
- M2: This alternative pathway has a lower activation energy (Ea) [1].
- M3: A greater fraction of molecules have kinetic energy greater than or equal to this lower activation energy, increasing reaction rate [1].
- Do not accept: "Catalyst lowers activation energy of existing pathway" without stating an alternative route.
Question 2: Chemical Equilibria and Equilibrium Constant Kc
12 marks(a) State Le Chatelier's principle. [2]
(b) A liquid mixture of ethanoic acid, ethanol, ethyl ethanoate, and water reaches dynamic equilibrium at 25 °C:
CH₃COOH(l) + C₂H₅OH(l) ⇌ CH₃COOC₂H₅(l) + H₂O(l)
Write an expression for the equilibrium constant Kc for this reaction. [1]
(c) A chemist placed 1.50 mol of ethanoic acid and 1.50 mol of ethanol into a sealed flask. The reaction mixture was kept at 25 °C in a total volume of 2.00 dm³ until equilibrium was established.
At equilibrium, titration showed that 0.50 mol of ethanoic acid remained unreacted.
Calculate the equilibrium concentration of each of the four components in the flask, calculate the value of Kc at 25 °C, and state the units, if any, for Kc. Show all your working. [4]
(d) The enthalpy change (ΔH) for the forward esterification reaction in (b) is −21 kJ mol⁻¹.
Explain the effect of increasing the temperature on:
1. The equilibrium yield of ethyl ethanoate.
2. The value of the equilibrium constant Kc. [3]
(e) Phosphorus pentachloride decomposes in a closed vessel according to the gaseous equilibrium:
PCl₅(g) ⇌ PCl₃(g) + Cl₂(g) ΔH = +88 kJ mol⁻¹
Explain why increasing the total pressure of the container at constant temperature decreases the equilibrium yield of phosphorus trichloride (PCl₃). [2]
Show Mark Scheme
(a) [2 marks]
- M1: When a system in dynamic equilibrium is subjected to a change in conditions [1]...
- M2: ...the position of equilibrium shifts to oppose that change [1].
(b) [1 mark]
- M1: Kc = [CH₃COOC₂H₅][H₂O] / ([CH₃COOH][C₂H₅OH]) [1].
- Accept: Displayed formulas or names in square brackets.
- Do not accept: Round brackets or missing brackets.
(c) [4 marks]
- M1: Equilibrium moles deduction:
Acid reacted = 1.50 - 0.50 = 1.00 mol.
Ethanol eqm = 1.50 - 1.00 = 0.50 mol.
Ethyl ethanoate eqm = 1.00 mol; Water eqm = 1.00 mol [1]. - M2: Equilibrium concentrations (moles divided by 2.00 dm³):
[CH₃COOH] = 0.50 / 2.00 = 0.25 mol dm⁻³.
[C₂H₅OH] = 0.50 / 2.00 = 0.25 mol dm⁻³.
[CH₃COOC₂H₅] = 1.00 / 2.00 = 0.50 mol dm⁻³.
[H₂O] = 1.00 / 2.00 = 0.50 mol dm⁻³ [1]. - M3: Value of Kc:
Kc = (0.50 × 0.50) / (0.25 × 0.25) = 0.25 / 0.0625 = 4.0 (or 4) [1]. - M4: Units: None / no units / dimensionless [1].
- Note: Allow full ECF for arithmetic errors in M1. If candidate omits volume division but arrives at Kc = 4.0 (since volume terms cancel), award M1, M3, and M4 (max 3/4 marks).
(d) [3 marks]
- M1: The forward reaction is exothermic (ΔH < 0), so increasing temperature shifts equilibrium in the endothermic reverse direction to absorb heat [1].
- M2: The equilibrium yield of ethyl ethanoate decreases [1].
- M3: The value of Kc decreases because product concentrations decrease while reactant concentrations increase [1].
(e) [2 marks]
- M1: There is 1 mole of gas on the left side and 2 moles of gas on the right side (1 PCl₃ + 1 Cl₂) [1].
- M2: Increasing pressure causes the equilibrium to shift to the left (the side with fewer gas moles) to oppose the increase in pressure, decreasing the yield of PCl₃ [1].
Question 3: Alkanes and Halogenoalkanes
11 marks(a) When butane (C₄H₁₀) undergoes incomplete combustion in a limited supply of oxygen, carbon monoxide and water are formed. Write a balanced chemical equation for this reaction. Include state symbols. [1]
(b) Under ultraviolet (UV) radiation, butane reacts with bromine by a free-radical substitution mechanism to produce 1-bromobutane.
Write balanced equations for the initiation step and the two propagation steps that lead to the formation of 1-bromobutane. Show the unpaired electron on each radical with a dot (•). [4]
(c) Outline the mechanism for the reaction of 1-bromobutane with warm aqueous sodium hydroxide to form butan-1-ol.
Show relevant dipoles (δ+, δ−), lone pairs of electrons, and curly arrows showing the movement of electron pairs. [4]
(d) When 1-bromobutane is heated under reflux with potassium hydroxide dissolved in ethanol instead of water, an elimination reaction occurs.
State the role of the hydroxide ion in this elimination reaction and give the systematic IUPAC name of the organic product formed. [2]
Show Mark Scheme
(a) [1 mark]
- M1: C₄H₁₀(g) + 4.5O₂(g) → 4CO(g) + 5H₂O(l)
OR 2C₄H₁₀(g) + 9O₂(g) → 8CO(g) + 10H₂O(l) [1]. - Accept: H₂O(g). Correct state symbols are mandatory.
(b) [4 marks]
- M1: Initiation equation: Br₂ → 2Br• [1].
- M2: Condition: Ultraviolet (UV) light specified on arrow or explicitly stated [1].
- M3: Propagation step 1: C₄H₁₀ + Br• → C₄H₉• + HBr [1].
- M4: Propagation step 2: C₄H₉• + Br₂ → C₄H₉Br + Br• [1].
- Criteria: Radical dots must be accurately positioned on the bromine atom and on the carbon atom of the alkyl radical.
(c) [4 marks]
- M1: Correct partial charges on 1-bromobutane: Cδ+ bonded to Brδ− [1].
- M2: Curly arrow originating from a lone pair of electrons on the oxygen of :OH⁻ pointing directly to the δ+ carbon atom of the C-Br bond [1].
- M3: Curly arrow originating from the C-Br bond pointing directly to the bromine atom [1].
- M4: Correct products: butan-1-ol (CH₃CH₂CH₂CH₂OH) and bromide ion (:Br⁻) [1].
- Penalty: Deduct 1 mark if curly arrow starts on the negative charge rather than the lone pair on :OH⁻.
(d) [2 marks]
- M1: Role of hydroxide ion: Base (or proton acceptor) [1].
- M2: Organic product: But-1-ene [1].
- Do not accept: "Nucleophile" for M1. Locant is required for M2.
Question 4: Alkenes, Stereoisomerism, and Electrophilic Addition
12 marks(a) Give the systematic IUPAC name for the alkene: CH₃-CH₂-CH=CH-CH₃. [1]
(b) Define the term stereoisomerism and explain the two structural features that enable the alkene in (a) to exist as a pair of geometric (E/Z) isomers. [3]
(c) Propene (CH₃CH=CH₂) reacts with hydrogen bromide (HBr) at room temperature to form 2-bromopropane as the major product.
Name the mechanism and outline its steps. Show all dipoles, lone pairs, carbocation intermediate structure, and curly arrows. [5]
(d) Explain, in terms of carbocation stability, why 2-bromopropane is formed in higher yield than 1-bromopropane in the reaction in (c). [3]
Show Mark Scheme
(a) [1 mark]
- M1: Pent-2-ene [1]. (Allow 2-pentene; reject pentene without locant).
(b) [3 marks]
- M1: Definition: Stereoisomers have the same structural formula but a different arrangement of their atoms in space [1].
- M2: Feature 1: Restricted rotation about the planar C=C double bond due to the presence of the π-bond [1].
- M3: Feature 2: Each carbon atom of the C=C double bond is bonded to two different atoms or groups (-H and -CH₃ on one carbon; -H and -CH₂CH₃ on the other carbon) [1].
(c) [5 marks]
- M1: Mechanism name: Electrophilic addition [1].
- M2: Curly arrow originating from the electron-rich C=C double bond pointing to the H atom of H-Br [1].
- M3: Dipoles shown on Hδ+-Brδ− and curly arrow from the H-Br bond pointing to the bromine atom [1].
- M4: Correct structure of the secondary carbocation intermediate: CH₃-CH⁺-CH₃ [1].
- M5: Curly arrow originating from a lone pair of electrons on :Br⁻ pointing to the positively charged carbon atom (C⁺) [1].
- Penalty: If primary carbocation intermediate (CH₃CH₂CH₂⁺) is drawn instead of secondary, deduct 2 marks (maximum 3/5).
(d) [3 marks]
- M1: 2-bromopropane forms via a secondary carbocation intermediate (CH₃-CH⁺-CH₃) [1].
- M2: 1-bromopropane forms via a primary carbocation intermediate (CH₃CH₂CH₂⁺) [1].
- M3: A secondary carbocation is more stable than a primary carbocation because it has two electron-releasing alkyl (methyl) groups that disperse the positive charge by the positive inductive effect, lowering the activation energy for its formation [1].
Question 5: Alcohols and Organic Synthesis
11 marks(a) Propan-1-ol can be oxidised to propanal or to propanoic acid using acidified potassium dichromate(VI). Explain how the method used decides which product is obtained. [3]
(b) In a laboratory preparation, ethanol is oxidised to ethanal (CH₃CHO) using simple distillation apparatus. State the reagents used as the oxidising agent and describe the colour change observed during the reaction. [2]
(c) Write a balanced chemical equation for the complete oxidation of ethanol to ethanoic acid under reflux. Use [O] to represent the oxidising agent. [2]
(d) Cyclohexanol can be converted into cyclohexene by an acid-catalysed dehydration (elimination) reaction (Required Practical 4):
C₆H₁₁OH → C₆H₁₀ + H₂O
Describe the experimental procedure for this preparation, including:
• The catalyst used and initial heating setup.
• How crude cyclohexene is separated from the reaction mixture.
• How the organic product is washed and dried to obtain pure cyclohexene. [4]
Show Mark Scheme
(a) [3 marks]
- M1: For propanal, use excess alcohol (limited oxidising agent) and distil the aldehyde off as it forms [1].
- M2: For propanoic acid, use excess acidified potassium dichromate(VI) and heat under reflux [1].
- M3: Distilling removes the aldehyde (lower boiling point, no hydrogen bonding between its molecules) before it can be oxidised further, while reflux keeps it in contact with the oxidising agent [1].
(b) [2 marks]
- M1: Reagents: Acidified potassium dichromate(VI) (or K₂Cr₂O₇ and dilute H₂SO₄) [1]. (Both dichromate and sulfuric acid are required).
- M2: Colour change: From Orange to Green [1].
(c) [2 marks]
- M1: C₂H₅OH → CH₃COOH (correct organic formulas) [1].
- M2: Balanced with + 2[O] and + H₂O:
C₂H₅OH + 2[O] → CH₃COOH + H₂O [1].
(d) [4 marks]
- M1: Heat cyclohexanol with concentrated sulfuric acid (H₂SO₄) or concentrated phosphoric acid (H₃PO₄) in a round-bottom flask [1].
- M2: Distil the reaction mixture and collect the distillate boiling below 85 °C in a cooled receiver flask [1].
- M3: Transfer distillate to a separating funnel, wash with saturated sodium chloride solution / water to remove acid impurities, allow layers to separate, and discard the lower aqueous layer [1].
- M4: Run the organic layer into a conical flask and add an anhydrous drying agent (e.g. anhydrous CaCl₂ or anhydrous MgSO₄) until the liquid becomes completely clear/transparent [1].
Question 6: Organic Analysis and Infrared Spectroscopy
12 marks(a) A laboratory technician provides four unlabelled bottles containing separate organic liquids:
• Bottle P: Propan-1-ol
• Bottle Q: Propanal
• Bottle R: Propanoic acid
• Bottle S: Cyclohexene
Describe chemical tests, with reagents and expected observations, that would allow a student to identify each of the four liquids unambiguously. [6]
(b) An infrared spectrum is recorded for one of the four bottles. It displays:
• A very broad absorption peak in the region 2500-3000 cm⁻¹.
• A strong, sharp absorption peak at 1715 cm⁻¹.
Identify which of the four bottles (P, Q, R, or S) corresponds to this spectrum. Identify the bond and functional group responsible for each of the two absorption peaks, and explain why the other three compounds are ruled out. [4]
(c) State what is meant by the "fingerprint region" of an infrared spectrum and explain how it can be used to confirm the precise identity of an organic compound. [2]
Show Mark Scheme
(a) [6 marks]
- M1 (Bottle S: Cyclohexene): Add bromine water (Br₂(aq)); observation: decolourises from orange/brown to colourless [1].
- M2 (Bottle R: Propanoic acid): Add sodium carbonate (Na₂CO₃) or sodium hydrogencarbonate (NaHCO₃); observation: effervescence / bubbles of carbon dioxide gas [1].
- M3 (Bottle Q: Propanal): Warm with Tollens' reagent; observation: silver mirror formed on glass wall (OR warm with Fehling's solution; observation: blue solution forms brick-red precipitate) [1].
- M4 (Bottle P: Propan-1-ol): Warm with acidified potassium dichromate(VI) (K₂Cr₂O₇ / H₂SO₄); observation: solution turns from orange to green [1].
- M5 & M6 (Logical scheme): Full discrimination achieved with clear, non-ambiguous stepwise logic [2]. (1 mark for partial scheme separating at least 3 liquids; 2 marks for complete discrimination of all 4).
(b) [4 marks]
- M1: Compound identified: Bottle R (Propanoic acid) [1].
- M2: Very broad absorption between 2500-3000 cm⁻¹ corresponds to the O-H bond of a carboxylic acid [1].
- M3: Strong, sharp absorption at 1715 cm⁻¹ corresponds to the C=O carbonyl bond of a carboxylic acid [1].
- M4: Deduction by elimination: Propanal (Q) contains C=O but lacks carboxylic O-H; Propan-1-ol (P) contains alcohol O-H (3230-3550 cm⁻¹) but lacks C=O; Cyclohexene (S) contains C=C (1620-1680 cm⁻¹) but lacks both C=O and O-H [1].
(c) [2 marks]
- M1: The region below 1500 cm⁻¹ contains a complex, unique pattern of absorption peaks specific to that particular molecule [1].
- M2: The spectrum of the unknown compound is compared with a computerised library / database of known spectra; an exact match confirms its identity [1].
Extra practice (topics not in the mock paper)
These questions test Unit 2 topics that the mock paper above does not cover. They do not count towards the 70 marks.
Question 7: Epoxyethane
4 marks(a) Write an equation for the reaction. [1]
(b) Explain why epoxyethane is very reactive. [2]
(c) Give one use of the product. [1]
Show Mark Scheme
(a) [1 mark]
- M1: C₂H₄O + H₂O → HOCH₂CH₂OH [1].
(b) [2 marks]
- M1: Ring strain: the bond angles in the ring are about 60° instead of 109.5° [1].
- M2: The C-O bonds are weak and polar, so the δ+ carbon is attacked by nucleophiles [1].
(c) [1 mark]
- M1: Antifreeze, or making polyesters [1].
Question 8: E-Z Isomers and the CIP Priority Rules
3 marksUse the Cahn-Ingold-Prelog rules to explain why the isomer of CH₃C(Cl)=CHCH₃ that has the Cl atom and the CH₃ group on the other carbon of the C=C on the same side is the Z isomer. [3]
Show Mark Scheme
- M1: On C2, Cl has a higher atomic number than C, so Cl has the higher priority [1].
- M2: On C3, C (of CH₃) has a higher atomic number than H, so CH₃ has priority [1].
- M3: The two higher-priority groups are on the same side, so the isomer is Z [1].
Question 9: Removing Sulfur Dioxide from Flue Gases
2 marksExplain why calcium oxide can be used to remove sulfur dioxide from flue gases. Write an equation. [2]
Show Mark Scheme
- M1: SO₂ is acidic and CaO is basic, so they react in a neutralisation [1].
- M2: CaO + SO₂ → CaSO₃ [1].
Question 10: Concentration and Rate of Reaction
2 marksExplain how increasing the concentration of a reactant in solution increases the rate of reaction. [2]
Show Mark Scheme
- M1: There are more particles in the same volume, so collisions are more frequent [1].
- M2: So there are more successful collisions (with energy ≥ Ea) per unit time [1].
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