AQA AS Level CHEMISTRY Paper 2
Organic and Physical Chemistry JUNE
2025 Mark Scheme
Physical Chemistry: Energetics & Kinetics
Q1.
Define the standard enthalpy of combustion.
Answer: The enthalpy change when one mole of a substance is completely burned in excess oxygen
under standard conditions (298 K, 100 kPa), all reactants and products in their standard states.
Rationale: Key points: 1 mole, complete combustion, excess oxygen, standard conditions.
Q2.
Calculate the enthalpy change for the reaction:
CH₄(g) + 2O₂(g) → CO₂(g) + 2H₂O(l)
Given: ΔHf°[CH₄(g)] = –75 kJ mol⁻¹, ΔHf°[CO₂(g)] = –394 kJ mol⁻¹, ΔHf°[H₂O(l)] = –286 kJ mol⁻¹
Answer: ΔH° = ΣΔHf°(products) – ΣΔHf°(reactants)
= [–394 + 2(–286)] – [–75 + 0] = [–394 –572] +75 = –966 +75 = –891 kJ mol⁻¹
Rationale: Elements in standard states (O₂) have ΔHf° = 0.
Q3.
In a calorimetry experiment, 0.25 g of ethanol (Mᵣ = 46) burned and raised the temperature of 200 g of
water by 15 °C. Calculate the experimental enthalpy of combustion. (Specific heat capacity of water =
4.18 J g⁻¹ K⁻¹)
Answer: q = mcΔT = 200 × 4.18 × 15 = 12540 J = 12.54 kJ.
Moles ethanol = 0.25/46 = 0.005434 mol.
ΔH = –12..005434 = –2308 kJ mol⁻¹ (exptl).
Rationale: The negative sign because combustion is exothermic. Experimental value is less exothermic
than theoretical due to heat loss.
Q4.
Why is experimental ΔH of combustion usually less exothermic than the true value?
Answer: Heat loss to surroundings / incomplete combustion / non-standard conditions.
Rationale: Real calorimeters are not perfectly insulated.
Q5.
Define rate of reaction.
Answer: Change in concentration of reactant or product per unit time.
Rationale: Units: mol dm⁻³ s⁻¹.
,Q6.
For a reaction A + 2B → C, the rate equation is rate = k[A][B]². What is the overall order?
Answer: 3 (first order in A, second order in B).
Rationale: Orders add: 1 + 2 = 3.
Q7.
State the effect of increasing temperature on rate constant k and explain why.
Answer: k increases because more molecules have energy ≥ activation energy (larger fraction of
successful collisions).
Rationale: Arrhenius equation: k = Ae⁻Ea/RT.
Q8.
Describe a method to follow the rate of reaction producing a gas (e.g., Mg + HCl → MgCl₂ + H₂).
Answer: Measure volume of gas produced at regular intervals using a gas syringe; plot volume vs time.
Rationale: Gas syringe gives continuous, accurate data.
Q9.
What is the function of a catalyst?
Answer: Provides an alternative reaction pathway with lower activation energy, increasing rate without
being consumed.
Rationale: Does not affect ΔH or equilibrium position.
Q10.
If doubling [A] doubles rate and doubling [B] has no effect, write the rate equation.
Answer: rate = k[A][B]⁰ = k[A].
Rationale: Zero order in B, first order in A.
Organic Chemistry: Halogenoalkanes
Q11.
Explain why halogenoalkanes undergo nucleophilic substitution but alkanes do not.
Answer: The C–X bond is polar (carbon δ⁺, halogen δ⁻) due to electronegativity difference, attracting
nucleophiles. Alkanes have non-polar C–H and C–C bonds.
Rationale: Nucleophiles attack electron-deficient carbon.
Q12.
Give the mechanism for 2-bromopropane with aqueous KOH. Name the mechanism.
Answer: Mechanism: SN2 (nucleophilic substitution).
Curly arrow from OH⁻ lone pair to C δ⁺; curly arrow from C–Br bond to Br. Intermediate transition state.
Rationale: 2° halogenoalkane: both SN1 and SN2 possible, but with OH⁻ (strong nucleophile) in water,
SN2 dominates.
Q13.
Explain why 2-bromo-2-methylpropane reacts with aqueous KOH via SN1, not SN2.
Answer: 3° carbon; bulky alkyl groups hinder nucleophile attack (steric hindrance); forms stable
, carbocation intermediate.
Rationale: SN2 requires backside attack, impossible with 3°.
Q14.
Write the equation for elimination of HBr from 2-bromopentane with ethanolic KOH. Give the major
product.
Answer: CH₃CH(Br)CH₂CH₂CH₃ + KOH (ethanol, heat) → CH₃CH=CHCH₂CH₃ (pent-2-ene) + KBr + H₂O.
Rationale: Saytzeff’s rule: more substituted alkene (pent-2-ene) is major.
Q15.
Name a test to distinguish between a chloroalkane and a bromoalkane.
Answer: Add AgNO₃ in ethanol. Chloroalkane gives white precipitate (slow); bromoalkane gives cream
precipitate (faster).
Rationale: C–Br bond weaker, so Br⁻ released quicker.
Q16.
Why do halogenoalkanes react faster with CN⁻ than with H₂O?
Answer: CN⁻ is a stronger nucleophile (negative charge on carbon) and smaller than water, so better
attack.
Rationale: Nucleophilicity parallels basicity and polarisability.
Q17.
Draw the carbocation formed from 2-bromo-2-methylbutane in SN1 and give its shape and bond angle.
Answer: (CH₃)₂C⁺CH₂CH₃; trigonal planar; 120°.
Rationale: sp² hybridised carbon.
Q18.
Why are racemic mixtures produced in SN1 reactions of chiral halogenoalkanes?
Answer: Planar carbocation intermediate allows attack by nucleophile from either side with equal
probability.
Rationale: 50% R, 50% S enantiomers.
Q19.
Complete: CH₃CH₂CH₂Br + NH₃ (excess) → ?
Answer: CH₃CH₂CH₂NH₂ (propylamine) + NH₄Br.
Rationale: Excess NH₃ prevents further substitution.
Q20.
Which bond is strongest: C–F, C–Cl, C–Br, C–I? Explain.
Answer: C–F strongest because F is most electronegative and small, giving short strong bond.
Rationale: Bond strength decreases down group: F > Cl > Br > I.
Organic Analysis & Spectroscopy
Q21.
What is meant by fingerprint region in IR spectroscopy?
Organic and Physical Chemistry JUNE
2025 Mark Scheme
Physical Chemistry: Energetics & Kinetics
Q1.
Define the standard enthalpy of combustion.
Answer: The enthalpy change when one mole of a substance is completely burned in excess oxygen
under standard conditions (298 K, 100 kPa), all reactants and products in their standard states.
Rationale: Key points: 1 mole, complete combustion, excess oxygen, standard conditions.
Q2.
Calculate the enthalpy change for the reaction:
CH₄(g) + 2O₂(g) → CO₂(g) + 2H₂O(l)
Given: ΔHf°[CH₄(g)] = –75 kJ mol⁻¹, ΔHf°[CO₂(g)] = –394 kJ mol⁻¹, ΔHf°[H₂O(l)] = –286 kJ mol⁻¹
Answer: ΔH° = ΣΔHf°(products) – ΣΔHf°(reactants)
= [–394 + 2(–286)] – [–75 + 0] = [–394 –572] +75 = –966 +75 = –891 kJ mol⁻¹
Rationale: Elements in standard states (O₂) have ΔHf° = 0.
Q3.
In a calorimetry experiment, 0.25 g of ethanol (Mᵣ = 46) burned and raised the temperature of 200 g of
water by 15 °C. Calculate the experimental enthalpy of combustion. (Specific heat capacity of water =
4.18 J g⁻¹ K⁻¹)
Answer: q = mcΔT = 200 × 4.18 × 15 = 12540 J = 12.54 kJ.
Moles ethanol = 0.25/46 = 0.005434 mol.
ΔH = –12..005434 = –2308 kJ mol⁻¹ (exptl).
Rationale: The negative sign because combustion is exothermic. Experimental value is less exothermic
than theoretical due to heat loss.
Q4.
Why is experimental ΔH of combustion usually less exothermic than the true value?
Answer: Heat loss to surroundings / incomplete combustion / non-standard conditions.
Rationale: Real calorimeters are not perfectly insulated.
Q5.
Define rate of reaction.
Answer: Change in concentration of reactant or product per unit time.
Rationale: Units: mol dm⁻³ s⁻¹.
,Q6.
For a reaction A + 2B → C, the rate equation is rate = k[A][B]². What is the overall order?
Answer: 3 (first order in A, second order in B).
Rationale: Orders add: 1 + 2 = 3.
Q7.
State the effect of increasing temperature on rate constant k and explain why.
Answer: k increases because more molecules have energy ≥ activation energy (larger fraction of
successful collisions).
Rationale: Arrhenius equation: k = Ae⁻Ea/RT.
Q8.
Describe a method to follow the rate of reaction producing a gas (e.g., Mg + HCl → MgCl₂ + H₂).
Answer: Measure volume of gas produced at regular intervals using a gas syringe; plot volume vs time.
Rationale: Gas syringe gives continuous, accurate data.
Q9.
What is the function of a catalyst?
Answer: Provides an alternative reaction pathway with lower activation energy, increasing rate without
being consumed.
Rationale: Does not affect ΔH or equilibrium position.
Q10.
If doubling [A] doubles rate and doubling [B] has no effect, write the rate equation.
Answer: rate = k[A][B]⁰ = k[A].
Rationale: Zero order in B, first order in A.
Organic Chemistry: Halogenoalkanes
Q11.
Explain why halogenoalkanes undergo nucleophilic substitution but alkanes do not.
Answer: The C–X bond is polar (carbon δ⁺, halogen δ⁻) due to electronegativity difference, attracting
nucleophiles. Alkanes have non-polar C–H and C–C bonds.
Rationale: Nucleophiles attack electron-deficient carbon.
Q12.
Give the mechanism for 2-bromopropane with aqueous KOH. Name the mechanism.
Answer: Mechanism: SN2 (nucleophilic substitution).
Curly arrow from OH⁻ lone pair to C δ⁺; curly arrow from C–Br bond to Br. Intermediate transition state.
Rationale: 2° halogenoalkane: both SN1 and SN2 possible, but with OH⁻ (strong nucleophile) in water,
SN2 dominates.
Q13.
Explain why 2-bromo-2-methylpropane reacts with aqueous KOH via SN1, not SN2.
Answer: 3° carbon; bulky alkyl groups hinder nucleophile attack (steric hindrance); forms stable
, carbocation intermediate.
Rationale: SN2 requires backside attack, impossible with 3°.
Q14.
Write the equation for elimination of HBr from 2-bromopentane with ethanolic KOH. Give the major
product.
Answer: CH₃CH(Br)CH₂CH₂CH₃ + KOH (ethanol, heat) → CH₃CH=CHCH₂CH₃ (pent-2-ene) + KBr + H₂O.
Rationale: Saytzeff’s rule: more substituted alkene (pent-2-ene) is major.
Q15.
Name a test to distinguish between a chloroalkane and a bromoalkane.
Answer: Add AgNO₃ in ethanol. Chloroalkane gives white precipitate (slow); bromoalkane gives cream
precipitate (faster).
Rationale: C–Br bond weaker, so Br⁻ released quicker.
Q16.
Why do halogenoalkanes react faster with CN⁻ than with H₂O?
Answer: CN⁻ is a stronger nucleophile (negative charge on carbon) and smaller than water, so better
attack.
Rationale: Nucleophilicity parallels basicity and polarisability.
Q17.
Draw the carbocation formed from 2-bromo-2-methylbutane in SN1 and give its shape and bond angle.
Answer: (CH₃)₂C⁺CH₂CH₃; trigonal planar; 120°.
Rationale: sp² hybridised carbon.
Q18.
Why are racemic mixtures produced in SN1 reactions of chiral halogenoalkanes?
Answer: Planar carbocation intermediate allows attack by nucleophile from either side with equal
probability.
Rationale: 50% R, 50% S enantiomers.
Q19.
Complete: CH₃CH₂CH₂Br + NH₃ (excess) → ?
Answer: CH₃CH₂CH₂NH₂ (propylamine) + NH₄Br.
Rationale: Excess NH₃ prevents further substitution.
Q20.
Which bond is strongest: C–F, C–Cl, C–Br, C–I? Explain.
Answer: C–F strongest because F is most electronegative and small, giving short strong bond.
Rationale: Bond strength decreases down group: F > Cl > Br > I.
Organic Analysis & Spectroscopy
Q21.
What is meant by fingerprint region in IR spectroscopy?
