IB DP Chemistry Mock Exam SL Paper 1B Set 4 - 2025 Syllabus
IB DP Chemistry Mock Exam SL Paper 1B Set 4
Prepare for the IB DP Chemistry Exam with our comprehensive IB DP Chemistry Exam Mock Exam SL Paper 1B Set 4. Test your knowledge and understanding of key concepts with challenging questions covering all essential topics. Identify areas for improvement and boost your confidence for the real exam
Question
___Cu (s) + ___HNO₃ (aq) → ___Cu(NO₃)₂ (aq) + ___NO₂ (g) + ___H₂O (l)
2NO₂ (g) ⇌ N₂O₄ (g) ∆\(H^⦵\) = –55.3 kJ mol⁻¹
Propose two different measurements, excluding colour observation, that could be used to follow the progress of this reaction over time, and describe the expected trend for each.
(i) Suggest a straightforward method to maintain this constant temperature in a school laboratory.
(ii) The equilibrium concentration of NO₂ was tracked using a colorimeter. A student took an absorbance reading immediately after starting. Explain why this reading might be unreliable.
(iii) Suggest how the issue identified in part (c)(ii) could be resolved.
Given that \( K = \frac{[\text{N}_2\text{O}_4]}{[\text{NO}_2]^2} \), calculate the equilibrium constant, K, at 0 °C. Express your answer to two significant figures.
2NO₂(g) + H₂O(l) → HNO₃(aq) + HNO₂(aq)
The resulting solution was diluted to 250.0 cm³. Then, 25.0 cm³ portions of this solution were titrated against a 0.0500 mol dm⁻³ standard solution of sodium hydroxide, NaOH.
(i) Identify the most precise piece of equipment for:
• Transferring 25.0 cm³ of the solution for titration.
• Adding the NaOH solution during the titration.
(ii) Design a table suitable for recording the titration results. Include headers for variables, units, and other relevant information, but leave the data cells empty.
(iii) The burette used for the NaOH titration had an uncertainty of ±0.05 cm³. The average titre volume was 20.05 cm³. Determine the percentage uncertainty in this volume.
2NO₂(g) ⇌ N₂O₄(g) ∆\(H^⦵\) = –55.3 kJ mol⁻¹
| T (°C) | T (K) | K |
|---|---|---|
| 0.0 | Value from (d) | |
| 20.0 | 4.74 | |
| 50.0 | 5.76 × 10⁻¹ | |
| 100.0 | 3.64 × 10⁻² |
(v) Using the completed table, determine whether the results are consistent with the given standard enthalpy change, ∆\(H^⦵\).
▶️ Answer/Explanation
(a) (i)
Precaution 1: Use a fume cupboard.
Reason: To prevent inhalation of toxic NO₂ gas.
Precaution 2: Wear safety goggles.
Reason: To protect eyes from corrosive acid splashes or glass fragments.
\(\boxed{\text{Fume cupboard}}\) & \(\boxed{\text{Safety goggles}}\)
(a) (ii)
The balanced equation is:
Cu (s) + 4HNO₃ (aq) → Cu(NO₃)₂ (aq) + 2NO₂ (g) + 2H₂O (l)
\(\boxed{1, 4, 1, 2, 2}\)
(a) (iii)
From the equation, 1 mol Cu → 2 mol NO₂.
Moles of Cu needed = \( \frac{0.0100}{2} = 0.00500 \) mol.
Molar mass of Cu = 63.55 g mol⁻¹.
Mass = \( 0.00500 \times 63.55 = 0.318 \) g.
\(\boxed{0.318}\)
(b)
Measurement 1: Temperature.
Expected result: Increases initially (exothermic forward reaction).
Measurement 2: Total pressure (at constant volume).
Expected result: Decreases (fewer gas moles in forward direction).
\(\boxed{\text{Temperature}}\) & \(\boxed{\text{Pressure}}\)
(c) (i)
Use a thermostatically controlled water bath.
\(\boxed{\text{Water bath}}\)
(c) (ii)
The system has not reached equilibrium; concentrations are still changing.
\(\boxed{\text{Equilibrium not reached}}\)
(c) (iii)
Wait until the absorbance reading becomes constant before recording.
\(\boxed{\text{Wait for constant absorbance}}\)
(d)
\( [\text{N}_2\text{O}_4] = \frac{0.00134}{1} = 0.00134 \) mol dm⁻³
\( [\text{NO}_2] = \frac{0.00732}{1} = 0.00732 \) mol dm⁻³
\( K = \frac{0.00134}{(0.00732)^2} = \frac{0.00134}{0.0000536} \approx 25 \)
\(\boxed{25}\)
(e) (i)
• Transferring 25.0 cm³: Volumetric pipette.
• Adding NaOH: Burette.
\(\boxed{\text{Pipette}}\) & \(\boxed{\text{Burette}}\)
(e) (ii)
Example table:
| Titration | Initial Burette Reading / cm³ | Final Burette Reading / cm³ | Volume of NaOH / cm³ |
|---|---|---|---|
| 1 | |||
| 2 | |||
| 3 |
\(\boxed{\text{Table with headers}}\)
(e) (iii)
% uncertainty = \( \frac{0.05}{20.05} \times 100 \approx 0.25\% \)
\(\boxed{0.25}\)
(f) (iv)
T(K) = T(°C) + 273.15
0.0 °C → 273.2 K
20.0 °C → 293.2 K
50.0 °C → 323.2 K
100.0 °C → 373.2 K
\(\boxed{273.2, 293.2, 323.2, 373.2}\)
(f) (v)
Yes, the results are consistent. As temperature increases, K decreases, confirming the reaction is exothermic (∆H⦵ < 0).
\(\boxed{\text{Yes}}\)
