A-Level Chemistry — Practice Paper 1

Organic mechanisms, equilibria, kinetics, thermodynamics and electrode potentials. Mechanism questions are marked on the arrows, so the arrow conventions are made explicit throughout.

How to use this: attempt the mechanisms with a pen before reading. Curly arrows are the one part of A-Level Chemistry you cannot learn by reading — the marks are for drawing them from the right place to the right place.

Section A — Physical chemistry

1. Define the term enthalpy change of formation (2 marks)

The enthalpy change when one mole of a compound is formed from its elements in their standard states under standard conditions (100 kPa, stated temperature, usually 298 K).

Where the marks sit: 1 for one mole of the compound, 1 for elements in their standard states. Omitting "one mole" is the usual loss.

Consequence worth knowing: $\Delta H_f^{\ominus}$ of any element in its standard state is zero by definition. Several Hess-cycle questions depend on spotting this.

2. For the equilibrium below, state and explain the effect of increasing the pressure (3 marks)

$$\ce{N2(g) + 3H2(g) <=> 2NH3(g)} \qquad \Delta H = -92\ \text{kJ mol}^{-1}$$

1. There are 4 moles of gas on the left and 2 on the right
2. Increasing pressure shifts the equilibrium towards the side with fewer moles of gas
3. So the position of equilibrium moves to the right, increasing the yield of ammonia

Where the marks sit: 1 for comparing moles of gas, 1 for the direction of shift, 1 for stating the effect on yield.

Common mistake: saying the equilibrium constant $K_p$ increases. $K_p$ depends only on temperature. Pressure changes the position of equilibrium, not the constant — this distinction is examined nearly every year.

3. Calculate the pH of a $0.050\ \text{mol dm}^{-3}$ solution of a strong acid, HCl (2 marks)

HCl is strong ⟹ fully dissociated ⟹ [H⁺] = 0.050 mol dm⁻³

pH = -log₁₀[H⁺]
   = -log₁₀(0.050)
   = 1.30   (2 d.p.)

Where the marks sit: 1 for $[\ce{H+}] = 0.050$, 1 for the pH to 2 decimal places.

Convention: pH is quoted to 2 decimal places at A-Level. Writing 1.3 can lose the accuracy mark.

4. The rate equation for a reaction is $\text{rate} = k[\ce{A}]^2[\ce{B}]$. State the overall order and the units of $k$ (3 marks)

• Overall order = $2 + 1 =$ 3rd order — 1 mark

rate       = k [A]² [B]

mol dm⁻³ s⁻¹ = k × (mol dm⁻³)² × (mol dm⁻³)
             = k × (mol dm⁻³)³

            mol dm⁻³ s⁻¹
k  =  ------------------------  =  mol⁻² dm⁶ s⁻¹
         (mol dm⁻³)³

• Units of $k$ = mol⁻² dm⁶ s⁻¹ — 2 marks

Where the marks sit: 1 for the order, 2 for correctly derived units.

Do not memorise the units — derive them from the rate equation each time. The order changes between questions and a memorised set will be wrong.

Section B — Organic chemistry

5. Draw the mechanism for the nucleophilic substitution of bromoethane by hydroxide ions (4 marks)

        H   H                              H   H
        |   |                              |   |
HO⁻ ⟶  C — C — Br            HO — C — C   +   Br⁻
        |   |     ⟍                        |   |
        H   H      ⟍ (arrow from C–Br      H   H
                      bond to Br)

The four marks:
1. Lone pair shown on the hydroxide ion, with the curly arrow starting from the lone pair (not from the negative charge, and not from nowhere)
2. Arrow points to the carbon atom bonded to bromine
3. A second curly arrow from the C–Br bond to the bromine atom
4. Correct products: ethanol and Br⁻

Where these marks are lost: an arrow starting at the minus sign rather than the lone pair, or an arrow that stops at the C–Br bond rather than at the carbon. Curly arrows represent movement of an electron pair, so each must start at a lone pair or a bond and end where that pair goes.

Mechanism type: $S_N2$ for a primary halogenoalkane — one concerted step.

6. Explain why tertiary halogenoalkanes undergo nucleophilic substitution faster than primary ones (3 marks)

1. Tertiary halogenoalkanes react by an $S_N1$ mechanism, forming a carbocation intermediate
2. The tertiary carbocation is stabilised by the positive inductive effect of three alkyl groups donating electron density
3. This lowers the activation energy for the rate-determining step, so the reaction is faster

Where the marks sit: 1 for naming $S_N1$ / carbocation, 1 for inductive stabilisation by alkyl groups, 1 for linking to activation energy.

7. A compound has the molecular formula $\ce{C3H6O}$ and shows a strong infrared absorption at $1715\ \text{cm}^{-1}$. Its $^1$H NMR spectrum shows a single peak. Identify the compound and justify (3 marks)

• The absorption at $1715\ \text{cm}^{-1}$ indicates a C=O (carbonyl) group — 1 mark
• A single NMR peak means all six hydrogens are in the same chemical environment — 1 mark
• The compound is propanone, $\ce{CH3COCH3}$ — 1 mark

Why not propanal? Propanal ($\ce{CH3CH2CHO}$) also has a C=O, but its hydrogens sit in three different environments, so it would show three peaks. The single peak is what discriminates.

8. State what is meant by an electrophile, and give an example (2 marks)

An electron-pair acceptor — a species attracted to a region of high electron density.

Example: $\ce{NO2+}$ (nitronium ion), or $\ce{Br^{\delta+}}$ in a polarised bromine molecule.

Where the marks sit: 1 for electron pair acceptor, 1 for a valid example. "Positively charged species" is not sufficient — some electrophiles are neutral but polarised.

9. Using the standard electrode potentials below, deduce whether the reaction is feasible (3 marks)

$$\ce{Zn^2+ + 2e- <=> Zn} \qquad E^{\ominus} = -0.76\ \text{V}$$
$$\ce{Cu^2+ + 2e- <=> Cu} \qquad E^{\ominus} = +0.34\ \text{V}$$

Proposed reaction: $\ce{Zn + Cu^2+ -> Zn^2+ + Cu}$

The more positive electrode potential is reduced;
the more negative is oxidised.

Cu²⁺ + 2e⁻ → Cu      (reduction, +0.34 V)
Zn → Zn²⁺ + 2e⁻      (oxidation, reversed: +0.76 V)

E°cell = E°(reduced) - E°(oxidised)
       = +0.34 - (-0.76)
       = +1.10 V

Since $E^{\ominus}_{cell}$ is positive, the reaction is feasible.

Where the marks sit: 1 for identifying which species is oxidised and which reduced, 1 for the calculation, 1 for the conclusion linked to the sign.

Important caveat, and a common discussion mark: a positive $E^{\ominus}_{cell}$ shows the reaction is thermodynamically feasible, not that it will happen at an observable rate. A large activation energy can make a feasible reaction immeasurably slow.

Where A-Level Chemistry marks are actually lost

1. Curly arrows drawn carelessly. Start at a lone pair or a bond; end where the electron pair goes. An arrow from a charge sign scores nothing even if the products are right.
2. Confusing position of equilibrium with $K$. Only temperature changes $K$.
3. Memorised units for $k$. Derive them; they change with the order.
4. Feasible taken to mean fast. Thermodynamics and kinetics answer different questions.

Where to go next

• GCSE Biology Practice Paper 1 — for the extended-response technique, which transfers
• KCSE Chemistry Practice Paper 1 — foundational mole and redox drilling

Ask StudyAI to generate more mechanism questions — it will mark your arrows and tell you which one started in the wrong place.