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AQA A LEVEL CHEMISTRY PAPER MARK SCHEME MS

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AQA A LEVEL CHEMISTRY PAPER MARK SCHEME MS

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Uploaded on
October 12, 2025
Number of pages
21
Written in
2025/2026
Type
Exam (elaborations)
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  • aqa a level chemistry

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AQA A LEVEL CHEMISTRY PAPER MARK
SCHEME MS
 Course
 AQA A LEVEL CHEMISTRY PAPER 3 7405-3 2021 Ms

1. Thermodynamics – Enthalpy of Formation

Question:
Define the standard enthalpy of formation of a compound and write the equation for the
formation of methane, CH₄(g).

Answer:
Definition: The enthalpy change when 1 mole of a compound is formed from its constituent
elements in their standard states under standard conditions (298 K, 100 kPa).
Equation:
C(s) + 2H₂(g) → CH₄(g)

Rationale:
The standard states must be used (graphite for carbon, H₂ gas for hydrogen). Formation always
produces one mole of product.



2. Equilibrium – Le Chatelier’s Principle

Question:
For the reaction:
N₂(g) + 3H₂(g) ⇌ 2NH₃(g), ΔH = -92 kJ mol⁻¹
Explain and predict the effect on yield of ammonia when temperature is increased.

Answer:
The yield of ammonia decreases when temperature increases.

Rationale:
The forward reaction is exothermic. Increasing temperature shifts equilibrium to the
endothermic direction (left), favoring decomposition of NH₃.



3. Atomic Structure

Question:
Explain why the first ionisation energy of oxygen is lower than that of nitrogen.

,Answer:
In oxygen, the 2p electrons start to pair up, causing increased electron–electron repulsion.
Therefore, less energy is required to remove an electron compared to nitrogen, where all 2p
electrons are unpaired.

Rationale:
Although oxygen has a higher nuclear charge, the repulsion in the paired orbital outweighs the
attraction.



4. Redox and Electrochemistry

Question:
Given the half-equations:
Cu²⁺ + 2e⁻ → Cu (E° = +0.34 V)
Zn²⁺ + 2e⁻ → Zn (E° = -0.76 V)
Calculate the standard cell potential for the cell and identify the direction of electron flow.

Answer:
E°cell = 0.34 - (-0.76) = +1.10 V
Electrons flow from Zn → Cu.

Rationale:
Zinc has a more negative E°, so it is oxidised at the anode; copper is reduced at the cathode.



5. Kinetics

Question:
In a reaction between A and B, the rate equation is: rate = k[A]²[B].
If the concentration of A is doubled and B is halved, by what factor does the rate change?

Answer:
Rate factor = (2²) × (½) = 4 × ½ = 2
So, rate doubles.

Rationale:
Rate depends quadratically on [A] and linearly on [B]. Adjusting both must be combined
multiplicatively.



6. Inorganic Chemistry – Group 2

, Question:
Explain why the thermal stability of Group 2 carbonates increases down the group.

Answer:
As ionic radius increases down the group, polarising power decreases, so the carbonate ion is
less distorted, making decomposition more difficult.

Rationale:
Smaller cations (e.g., Mg²⁺) distort CO₃²⁻ more strongly, weakening C–O bonds and lowering
stability.



7. Organic Chemistry – Mechanisms

Question:
Describe the mechanism for the reaction of 2-bromopropane with hydroxide ions in ethanol.

Answer:
Mechanism: Nucleophilic substitution (S_N2).
Step:
OH⁻ attacks the carbon bonded to Br from the opposite side → transition state → Br⁻ leaves →
propan-2-ol formed.

Rationale:
Ethanol is a polar aprotic solvent favoring the S_N2 mechanism for secondary halogenoalkanes.



8. Organic Analysis

Question:
How can you distinguish between an aldehyde and a ketone using chemical tests?

Answer:
Use Tollens’ reagent (ammoniacal silver nitrate).

 Aldehyde: silver mirror forms.
 Ketone: no reaction.

Rationale:
Aldehydes are oxidised to carboxylic acids; ketones cannot be further oxidised under these
conditions.

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