Home / iGCSE Chemistry 0620 Extended – 2.5 Simple molecules and covalent bonds- Exam Style Questions Paper 4

iGCSE Chemistry 0620 Extended - 2.5 Simple molecules and covalent bonds- Exam Style Questions Paper 4- New Syllabus

iGCSE Chemistry Theory (Extended) Paper 4 - All Chapters

Question

This question is about chlorine and compounds of chlorine.
(a) Gaseous chlorine reacts with gaseous phosphorus(III) chloride to form gaseous phosphorus(V) chloride. The reaction is a reversible reaction.
When the three gases are in a closed container the system reaches equilibrium.
$$Cl_{2}(g) + PCl_{3}(g) \rightleftharpoons PCl_{5}(g)$$
(i) Describe a reversible reaction at equilibrium in terms of
  • the rate of the forward reaction and the reverse reaction
  • the concentrations of reactants and products.
(ii) Complete Table 4.1 using only the words increases, decreases or no change.
(iii) When the temperature of the equilibrium mixture is increased, the equilibrium concentration of $PCl_{5}$ decreases. State what conclusion about the forward reaction can be made from this information.
(b) Ethene, $C_{2}H_{4}$, reacts with chlorine, $Cl_{2}$, to form 1,2-dichloroethane, $CH_{2}ClCH_{2}Cl$.
The equation for this reaction can be represented as shown.
Table 4.2 shows some bond energies.
Use the bond energies in Table 4.2 to calculate the enthalpy change, in kJ / mol, of the reaction.
Use the following steps.
  • Calculate the total energy needed to break the bonds in $C_{2}H_{4}$ and $Cl_{2}$.
  • Calculate the total energy released when the bonds form in $CH_{2}ClCH_{2}Cl$
  • Calculate the enthalpy change of the reaction. Your answer should include a sign.
Calculate the enthalpy change of the reaction using bond energies (C-C: $350$, C=C: $610$, C-H: $410$, Cl-Cl: $240$, C-Cl: $340 \text{ kJ/mol}$).
(c) Chlorine reacts with nitrogen to form nitrogen trichloride, $NCl_{3}$.
Complete the dot-and-cross diagram of a molecule of $NCl_{3}$.
Show outer shell electrons only.
(d) When solid magnesium carbonate, $MgCO_{3}$, is added to dilute hydrochloric acid, HCl, a chemical reaction occurs:
$$MgCO_{3}(s) + 2HCl(aq) \rightarrow MgCl_{2}(aq) + CO_{2}(g) + H_{2}O(l)$$
(i) Give two observations when solid magnesium carbonate is added to dilute hydrochloric acid.
(ii) Calculate the volume, in $cm^{3}$, of $CO_{2}(g)$ that is produced at room temperature and pressure (r.t.p.) when $50.0 \text{ cm}^{3}$ of $0.100 \text{ mol/dm}^{3}$ HCl reacts with excess $MgCO_{3}$.
Volume of $1 \text{ mol}$ of any gas is $24000 \text{ cm}^{3}$ at r.t.p.
Use the following steps.
  • Calculate the number of moles of HCl in 50.0 cm3 of 0.100 mol / dm3 HCl .
  • Deduce the number of moles of $CO_{2}(g)$ produced.
  • Calculate the volume, in cm3, of $CO_{2}(g)$ produced at r.t.p.

Most-appropriate topic codes (Cambridge IGCSE Chemistry 0620):

• Topic 6.3 — Reversible reactions and equilibrium
• Topic 5.1 — Exothermic and endothermic reactions
• Topic 2.5 — Simple molecules and covalent bonds
• Topic 3.3 — The mole and the Avogadro constant

▶️ Answer/Explanation

(a)(i)
For the correct answer:
The rate of the forward reaction and the rate of the reverse reaction are equal. The concentrations of reactants and products are no longer changing.

(a)(ii)
For the correct answer:
Pressure decreased: Effect on rate = decreases. Effect on $PCl_{5}$ conc = decreases.
Catalyst added: Effect on $PCl_{5}$ conc = no change.

Explanation: By Le Chatelier’s principle, decreasing pressure shifts the equilibrium towards the side with more moles of gas ($1 + 1 = 2$ moles on the left vs $1$ mole on the right), thereby lowering the $PCl_{5}$ concentration. Catalysts speed up both forward and reverse rates equally, merely achieving equilibrium faster without shifting its position.

(a)(iii)
For the correct answer:
The forward reaction is exothermic.

Explanation: Increasing temperature always favours the endothermic direction to absorb the excess heat. Since the product concentration decreased, the reverse reaction must be endothermic, meaning the forward reaction is exothermic.

(b)
For the correct answer:
$-180 \text{ kJ/mol}$

Calculation:
Step 1: Calculate energy required to break all bonds in the reactants (Endothermic process).
Bonds broken: $4 \times (C-H) + 1 \times (C=C) + 1 \times (Cl-Cl)$
Energy in = $4(410) + 610 + 240 = 1640 + 610 + 240 = 2490 \text{ kJ/mol}$.

Step 2: Calculate energy released forming all bonds in the products (Exothermic process).
Bonds formed: $4 \times (C-H) + 1 \times (C-C) + 2 \times (C-Cl)$
Energy out = $4(410) + 350 + 2(340) = 1640 + 350 + 680 = 2670 \text{ kJ/mol}$.

Step 3: Calculate total enthalpy change.
$\Delta H = \text{Energy required to break bonds} – \text{Energy released forming bonds}$
$\Delta H = 2490 – 2670 = -180 \text{ kJ/mol}$.

(c)
For the correct answer:
Draw $3$ dot-and-cross bonding pairs (one between the central N and each Cl), $1$ lone pair of electrons on the N atom, and $3$ lone pairs of electrons on each of the three Cl atoms.

Explanation: Nitrogen is in Group V, meaning it has $5$ valence electrons. It shares $3$ of these electrons with three Chlorine atoms to form three single covalent bonds, leaving one non-bonding pair (lone pair). Chlorine is in Group VII, meaning it has $7$ valence electrons. Each Cl shares $1$ electron with Nitrogen, leaving $6$ non-bonding electrons ($3$ lone pairs) around each Chlorine atom. This arrangement gives every atom a stable, full outer octet.

(d)(i)
For the correct answer:
1. The solid ($MgCO_{3}$) visibly disappears or dissolves.
2. Effervescence (bubbling or fizzing) is observed.

(d)(ii)
For the correct answer:
$60 \text{ cm}^{3}$

Calculation:
Step 1: Calculate the moles of limiting reactant (HCl).
$\text{Moles of HCl} = \frac{\text{Volume}}{1000} \times \text{Concentration} = \frac{50.0}{1000} \times 0.100 = 0.005 \text{ mol}$.

Step 2: Use stoichiometry to find moles of $CO_{2}$.
The balanced equation shows a $2:1$ molar ratio between HCl and $CO_{2}$.
$\text{Moles of } CO_{2} = \frac{0.005}{2} = 0.0025 \text{ mol}$.

Step 3: Convert moles of gas to volume.
$\text{Volume} = \text{Moles} \times \text{Molar Volume} = 0.0025 \times 24000 = 60 \text{ cm}^{3}$.

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