CIE iGCSE Co-Ordinated Science C12.5 Identification of ions and gases Exam Style Questions Paper 4
Question
Most-appropriate topic codes (Cambridge IGCSE Co-ordinated Sciences 0654):
• Topic C12.5 — Qualitative analysis / Gas tests (Part (a))
• Topic C6.2 — Rate of reaction / Catalysts (Part (b) & (c))
• Topic C5.1 — Exothermic and endothermic reactions / Energy level diagrams (Part (d)(i) & (d)(ii))
▶️ Answer/Explanation
(a) Test: Use a glowing splint. Positive result: The glowing splint relights.
(b) The catalyst decreases the activation energy (Eₐ) of the reaction.
(c) Line A shows the results using manganese(IV) oxide powder.
(d)(i) Energy level diagram for exothermic reaction:
- The reactants (2H₂O₂) line should be drawn above the products (2H₂O + O₂) line.
- An arrow should be drawn from the reactants level to the peak of the curve and labelled activation energy.
- An arrow should be drawn from the reactants level down to the products level and labelled energy change / enthalpy change.
(d)(ii) ✔ -196 kJ/mol
Question
Most-appropriate topic codes (Cambridge IGCSE Co-ordinated Sciences 0654):
• Topic C12.5 — Qualitative analysis (Part (a))
• Topic C6.2 — Rate of reaction (Part (b)(i), (b)(ii), (c))
• Topic C3.3 — The mole and the Avogadro constant (Part (b)(iii))
▶️ Answer/Explanation
(a) Test: Bubble the gas through / pass the gas into limewater.
Observation: The limewater turns milky / forms a white precipitate.
Carbon dioxide gas is commonly tested for by bubbling it through limewater (aqueous calcium hydroxide solution). When carbon dioxide reacts with limewater, it forms insoluble calcium carbonate, which appears as a white precipitate, making the solution turn milky or cloudy. The chemical equation is: \( \text{Ca(OH)}_2(\text{aq}) + \text{CO}_2(\text{g}) \rightarrow \text{CaCO}_3(\text{s}) + \text{H}_2\text{O}(\text{l}) \).
(b)(i) Any value in the inclusive range 5.6–6 minutes.
The reaction stops when no more carbon dioxide is produced. On the graph, this is when the curve becomes horizontal (plateaus). Looking at Fig. 8.1, the volume of gas stops increasing between approximately 5.6 and 6 minutes, so any value in this range is acceptable. At this point, at least one of the reactants (the acid or the calcium carbonate) has been completely used up.
(b)(ii) Average rate = 30 ÷ 2 = 15 cm³/minute.
Average rate of reaction is calculated using:
\( \text{Average rate} = \frac{\text{Total volume of gas produced}}{\text{Time taken}} \)
From the graph, at 0 minutes the volume is 0 cm³ and at 2 minutes the volume is 30 cm³. Therefore:
\( \text{Average rate} = \frac{30 – 0}{2 – 0} = \frac{30}{2} = 15 \text{ cm}^3/\text{minute} \)
(b)(iii) Moles = volume ÷ 24 = 0.050 ÷ 24 = 0.0021 mol.
To calculate the amount in moles, use the relationship:
\( \text{Moles} = \frac{\text{Volume (dm}^3\text{)}}{24} \)
First, convert 50 cm³ to dm³: 50 cm³ = 50 ÷ 1000 = 0.050 dm³.
Then: \( \text{Moles of CO}_2 = \frac{0.050}{24} = 0.002083… \approx 0.0021 \text{ mol (to 2 significant figures)} \).
(c) When the concentration of acid is increased, there are more particles per unit volume / more acid particles in the same volume. This leads to a higher frequency of collisions between the acid particles and the calcium carbonate particles, so the rate of reaction increases.
Collision theory states that for a reaction to occur, particles must collide with sufficient energy (activation energy) and in the correct orientation. When the concentration of hydrochloric acid is increased, there are more HCl particles in the same volume of solution. This means the particles are closer together, leading to:
- A greater number of particles per unit volume.
- A higher frequency of collisions between the acid particles and the calcium carbonate surface.
- More successful collisions that result in a reaction, therefore increasing the rate of reaction.