2023 Cambridge IGCSE Chemistry (0620) Practice Paper with Answers

Silicon (Si) is a Group IV metalloid that forms a giant covalent structure similar to diamond, where each silicon atom is tetrahedrally bonded to four other silicon atoms.

Si

Aluminium (Al) forms aluminium oxide (\text{Al}_2\text{O}_3), which is amphoteric as it reacts with both acids and bases.

Al

Phosphorus (P) is in Group V (Group 15) of the Periodic Table, so it has 5 electrons in its outer shell.

P

Sodium (Na) is an alkali metal (Group I) that reacts vigorously with water and oxygen in the air, so it must be stored under oil.

Na

Chlorine (Cl) is a halogen (Group VII) that exists as a diatomic gas (\text{Cl}_2) at room temperature and pressure.

Cl

Argon (Ar) is a noble gas (Group VIII/0) with a full valence shell (8 outer-shell electrons), meaning it is stable and exists as monatomic atoms.

Ar

(a)
- Anode: Bubbles of a pale green/yellow-green gas (chlorine) are observed.
- Cathode: A pink/brown solid (copper metal) is deposited on the electrode.

(b)
- Anode half-equation: \( 2\text{Cl}^-\text{(aq)} \rightarrow \text{Cl}_2\text{(g)} + 2\text{e}^- \) (or \( 2\text{Cl}^-\text{(aq)} - 2\text{e}^- \rightarrow \text{Cl}_2\text{(g)} \))
- Cathode half-equation: \( \text{Cu}^{2+}\text{(aq)} + 2\text{e}^- \rightarrow \text{Cu(s)} \)

(c) The blue colour of the electrolyte fades and eventually becomes colourless. This is because the blue copper(II) ions, \( \text{Cu}^{2+} \), are discharged and removed from the solution to form copper metal at the cathode.

(d)
(i) \( \text{Moles of Cu} = \frac{0.16\text{ g}}{64\text{ g/mol}} = 0.0025\text{ mol} \)
(ii) According to the stoichiometry of the electron transfer:
\( 1\text{ mol of Cu} \) requires \( 2\text{ mol of e}^- \), and \( 1\text{ mol of Cl}_2 \) produces \( 2\text{ mol of e}^- \).
Therefore, the mole ratio of \( \text{Cu} : \text{Cl}_2 \) is \( 1 : 1 \).
\( \text{Moles of Cl}_2 = 0.0025\text{ mol} \).
(iii) \( \text{Volume of Cl}_2 = 0.0025\text{ mol} \times 24\text{ dm}^3\text{/mol} = 0.06\text{ dm}^3 \)
In \( \text{cm}^3 \): \( 0.06 \times 1000 = 60\text{ cm}^3 \).

(a) Anode: bubbles / effervescence / fizzing of green-yellow gas [1]
Cathode: pink / brown / reddish-brown solid [1] (Reject: 'copper' on its own as it is an inference, must be an observation).

(b) Anode equation: \( 2\text{Cl}^- \rightarrow \text{Cl}_2 + 2\text{e}^- \) [1], correct state symbols: \( \text{Cl}^-\text{(aq)} \) and \( \text{Cl}_2\text{(g)} \) [1].
Cathode equation: \( \text{Cu}^{2+} + 2\text{e}^- \rightarrow \text{Cu} \) [1], correct state symbols: \( \text{Cu}^{2+}\text{(aq)} \) and \( \text{Cu(s)} \) [1].

(c) Fades / turns colourless [1], because \( \text{Cu}^{2+} \) ions are removed / concentration of copper ions decreases [1].

(d) (i) \( 0.0025 \) / \( 2.5 \times 10^{-3} \) (mol) [1]
(ii) \( 0.0025 \) (mol) / allow error carried forward from (d)(i) [1]
(iii) \( 0.06\text{ dm}^3 \) or \( 60\text{ cm}^3 \) [1], unit must match value [1].

(a) Place the conical flask containing dilute hydrochloric acid on a digital mass balance. Add the marble chips, insert a cotton wool plug in the neck of the flask, and start a timer. Record the mass at regular intervals. The cotton wool plug allows carbon dioxide gas to escape while preventing any acid spray/droplets from being lost, which would lead to an inaccurate mass loss reading.

(b) (i) The sketch should show:
- Y-axis labelled 'mass of flask and contents' (or 'mass') and X-axis labelled 'time'.
- Both curves starting at the same initial mass and falling to the same final horizontal asymptote (since marble chips are in excess, the limiting reactant, acid, determines the total mass loss).
- The curve for \( 40^\circ\text{C} \) is steeper initially and levels off sooner than the curve for \( 20^\circ\text{C} \).

(ii) At \( 40^\circ\text{C} \), the particles have more kinetic energy and move faster. Consequently, the collision frequency of reactant particles increases. More importantly, a much larger proportion of the colliding particles possess energy greater than or equal to the activation energy (\( E_a \)). This leads to a higher frequency of successful/effective collisions.

(c)
- The initial rate of reaction increases (doubles) because there are more reactant particles per unit volume, which increases the frequency of collisions between reactant particles.
- The final total mass of carbon dioxide produced doubles. This is because doubling the concentration of the acid while keeping the volume constant doubles the number of moles of the limiting reactant (\( \text{HCl} \)), producing twice as many moles of product.

(a) Flask placed on a mass balance/balance [1]; cotton wool allows gas/\( \text{CO}_2 \) to escape [1]; cotton wool prevents loss of acid spray/mist [1].

(b) (i) Correct axes (mass on y, time on x) and both curves decreasing to the same final mass [1]; \( 40^\circ\text{C} \) curve is steeper/levels off earlier than \( 20^\circ\text{C} \) [1]; both curves clearly labelled [1].

(ii) Particles gain kinetic energy / move faster [1]; higher frequency of collisions [1]; more particles have energy greater than/equal to the activation energy [1]; higher frequency of successful/fruitful collisions [1].

(c) Initial rate increases / is higher [1]; total mass of carbon dioxide increases / doubles [1]; because there are more moles of acid reacting / double the moles of \( \text{HCl} \) [1].

(a) Suitable soluble salts are barium nitrate, \( \text{Ba(NO}_3\text{)}_2 \) (or barium chloride, \( \text{BaCl}_2 \)) and sodium sulfate, \( \text{Na}_2\text{SO}_4 \) (or potassium sulfate, \( \text{K}_2\text{SO}_4 \), or dilute sulfuric acid, \( \text{H}_2\text{SO}_4 \)).

(b) Using barium nitrate and sodium sulfate:
\( \text{Ba(NO}_3\text{)}_2\text{(aq)} + \text{Na}_2\text{SO}_4\text{(aq)} \rightarrow \text{BaSO}_4\text{(s)} + 2\text{NaNO}_3\text{(aq)} \)

(c)
1. Filter the mixture to separate the insoluble barium sulfate precipitate (residue) from the soluble sodium nitrate solution (filtrate).
2. Wash the residue (barium sulfate) with distilled/deionised water to remove any remaining soluble impurity ions.
3. Dry the pure barium sulfate precipitate in a warm oven (or leave it to dry between sheets of filter paper).

(d) Barium sulfate is highly insoluble in water and stomach acid. Therefore, it does not dissolve or release toxic \( \text{Ba}^{2+} \) ions into the body / bloodstream.

(e) Ionic equation: \( \text{Ba}^{2+}\text{(aq)} + \text{SO}_4^{2-}\text{(aq)} \rightarrow \text{BaSO}_4\text{(s)} \)

(a) Barium nitrate / barium chloride [1] and sodium sulfate / potassium sulfate / ammonium sulfate / sulfuric acid [1]. (Reject: insoluble sulfates or barium carbonate/oxide/hydroxide if not specified as soluble).

(b) Correct formulae of reactants and products [1]; balanced equation [1]; correct state symbols: (aq) for soluble reactants/products, (s) for \( \text{BaSO}_4 \) [1].

(c) Filter / filtration to obtain residue [1]; wash residue with distilled/deionised water [1]; dry residue [1]; description of drying (e.g., in oven, using filter paper, desiccator) [1].

(d) It is insoluble / does not dissolve in body fluids/blood [1].

(e) \( \text{Ba}^{2+} \) and \( \text{SO}_4^{2-} \) on left and \( \text{BaSO}_4 \) on right [1]; state symbols \( \text{(aq)} + \text{(aq)} \rightarrow \text{(s)} \) [1].

(a)
(i) The functional group is the carboxylic acid group / carboxyl group (\( \text{-COOH} \)).
(ii) Displayed formula of butanoic acid:
```
H H H O
| | | //
H - C - C - C - C
| | | \
H H H O - H
```
(Alternatively, methylpropanoic acid displayed formula is acceptable).
(iii) Word equation:
\( \text{butanoic acid} + \text{sodium carbonate} \rightarrow \text{sodium butanoate} + \text{carbon dioxide} + \text{water} \)
(Accept methylpropanoic acid / sodium methylpropanoate).

(b)
(i) Concentrated sulfuric acid.
(ii) Ethanol and ethanoic acid (to make ethyl ethanoate). Other acceptable pairs include:
- Methanol and propanoic acid (to make methyl propanoate).
- Propan-1-ol (or propan-2-ol) and methanoic acid (to make propyl methanoate / isopropyl methanoate).
(iii) Displayed formula of ethyl ethanoate:
```
H O H H
| // | |
H - C - C - O - C - C - H
| | |
H H H
```
The ester linkage is the \( \text{-C(=O)-O-} \) group and must be circled.
(iv) Ethyl ethanoate (or systematic name corresponding to the drawn structure in b(iii)).

(a) (i) Carboxylic acid group / carboxyl group / \( \text{-COOH} \) [1].
(ii) Correct skeleton of 4 carbons [1]; correct functional group with \( \text{C=O} \) and \( \text{O-H} \) shown fully displayed [1].
(iii) Carboxylic acid + sodium carbonate \( \rightarrow \) sodium salt + carbon dioxide + water [2] (1 mark for correct reactants, 1 mark for correct products. Allow error carried forward from the specific acid named).

(b) (i) (Concentrated) sulfuric acid [1] (Reject: dilute sulfuric acid, hydrochloric acid).
(ii) Suitable alcohol [1] and carboxylic acid [1] that have a total of 4 carbons (e.g. ethanol and ethanoic acid).
(iii) Correct displayed formula of a 4-carbon ester [2] (1 mark for correct ester structure, 1 mark for showing all bonds including O-H if any, though none in esters, check C-H, C-O, C=O); correct circle around the ester linkage (\( \text{-COO-} \)) [1].
(iv) Correct systematic IUPAC name matching structure in (b)(iii) [2] (1 mark for the alkyl part, e.g. 'ethyl', 1 mark for the carboxylate part, e.g. 'ethanoate').

(a) Increasing the pressure shifts the position of equilibrium to the right (towards the products side). This is because the forward reaction results in a decrease in the number of moles of gas (3 moles of gaseous reactants on the left react to form 1 mole of gaseous product on the right). Shifting to the right opposes the increase in pressure.

(b)
(i) A higher temperature decreases the equilibrium yield of methanol. This is because the forward reaction is exothermic (releases heat, \( \Delta H \) is negative). According to Le Chatelier's principle, increasing the temperature favours the endothermic reverse reaction to absorb the added thermal energy.
(ii) A lower temperature is not used because it would dramatically decrease the rate of the reaction. At very low temperatures, the reaction would take too long to reach equilibrium, making the process industrially unprofitable.

(c)
(i) The catalyst increases the rate of the forward reaction (and the reverse reaction equally).
(ii) The catalyst has no effect on the position of equilibrium.

(d) The reactants have a higher energy level than the products. The reaction is exothermic.

(a) Equilibrium shifts to the right [1]; because there are fewer moles of gas on the right / 3 moles on left and 1 mole on right [1]; this decreases the pressure [1].

(b) (i) Yield of methanol decreases [1]; because the forward reaction is exothermic [1]; system shifts to the left / favours endothermic reaction to absorb heat [1].
(ii) Rate of reaction would be too slow [1]; fewer successful collisions per unit time / less kinetic energy of particles [1].

(c) (i) Increases / speeds up [1].
(ii) No effect / unchanged [1].

(d) Reactants have higher energy than products [1]; reaction is exothermic [1].

(a) Isotopes are atoms of the same element (same number of protons / atomic number) with different numbers of neutrons (different mass / nucleon numbers).

(b) Phosphorus has atomic number 15.
- For \( ^{31}\text{P} \) atom:
- Protons = 15
- Neutrons = \( 31 - 15 = 16 \)
- Electrons = 15
- For \( ^{32}\text{P}^{3-} \) ion:
- Protons = 15
- Neutrons = \( 32 - 15 = 17 \)
- Electrons = \( 15 + 3 = 18 \)

(c)
(i) Covalent bonding.
(ii) Phosphorus is in Group V (5 outer electrons) and Chlorine is in Group VII (7 outer electrons). In \( \text{PCl}_3 \):
- The central P atom shares one pair of electrons with each of the three Cl atoms (three P-Cl single covalent bonds).
- The P atom has 2 non-bonding outer electrons left (one lone pair).
- Each Cl atom has 6 non-bonding outer electrons left (three lone pairs).

(d) Phosphorus trichloride has a simple molecular structure. There are weak intermolecular forces (or weak forces of attraction between molecules) which require very little thermal energy to overcome.

(a) Atoms of the same element / same number of protons [1]; different numbers of neutrons [1] (Reject: 'elements' instead of 'atoms').

(b)
- \( ^{31}\text{P} \) atom row: 15 protons, 16 neutrons, 15 electrons [2] (all correct = 2 marks, 1 or 2 correct = 1 mark).
- \( ^{32}\text{P}^{3-} \) ion row: 15 protons, 17 neutrons, 18 electrons [2] (all correct = 2 marks, 1 or 2 correct = 1 mark).

(c)
(i) Covalent [1].
(ii) Three shared pairs of electrons between P and each Cl [1]; lone pair (2 electrons) on the P atom [1]; three lone pairs (6 electrons) on each Cl atom [1].

(d) Simple molecular structure / simple covalent structure [1]; weak intermolecular forces / weak forces between molecules [1] (Reject: 'weak covalent bonds'); requires little energy to overcome/break these forces [1].