IB DP Chemistry Mock Exam HL Paper 1B Set 5 - 2025 Syllabus
IB DP Chemistry Mock Exam HL Paper 1B Set 5
Prepare for the IB DP Chemistry Exam with our comprehensive IB DP Chemistry Exam Mock Exam HL Paper 1B Set 5. 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
| \( T / ^\circ\text{C} \) | \( T / \text{K} \) | \( K \) |
|---|---|---|
| \( 0.0 \) | — | value from (d) |
| \( 20.0 \) | — | \( 4.74 \) |
| \( 50.0 \) | — | \( 5.76 \times 10^{-1} \) |
| \( 100.0 \) | — | \( 3.64 \times 10^{-2} \) |
▶️ Answer/Explanation
(a)(i)
Possible answers (any two precautions, each with a correct reason):
• Wear safety goggles / eye protection → to prevent corrosive \( \text{HNO}_3 \) or \( \text{NO}_2 \) from damaging the eyes.
• Work in a fume cupboard / fume hood → to avoid inhaling the toxic \( \text{NO}_2 \) gas.
(Other acceptable second precautions include using gloves or tongs to protect the skin from hot or acidic substances.)
(a)(ii)
Balanced equation:
\[ \text{Cu(s)} + 4\,\text{HNO}_3(\text{aq}) \rightarrow \text{Cu(NO}_3)_2(\text{aq}) + 2\,\text{NO}_2(\text{g}) + 2\,\text{H}_2\text{O(l)} \] Coefficients: \( 1 : 4 : 1 : 2 : 2 \).
(a)(iii)
From the balanced equation, \( 1\ \text{mol Cu} \rightarrow 2\ \text{mol NO}_2 \).
Required moles of copper:
\[ n(\text{Cu}) = \frac{n(\text{NO}_2)}{2} = \frac{0.0100}{2} = 0.00500\ \text{mol} \] Using molar mass \( M(\text{Cu}) = 63.55\ \text{g mol}^{-1} \):
\[ m(\text{Cu}) = n \times M = 0.00500 \times 63.55 = 0.3178\ \text{g} \approx 0.318\ \text{g} \] \(\boxed{0.318\ \text{g of Cu}}\)
(b)
Example pair of measurements and expected trends (any two suitable pairs):
• Measurement 1: temperature of the system → As the exothermic forward reaction \( (2\,\text{NO}_2 \rightarrow \text{N}_2\text{O}_4) \) proceeds, the temperature of the sealed system would initially increase.
• Measurement 2: pressure of the gas at constant volume → As the reaction forms fewer moles of gas (from \( 2 \) to \( 1 \) mole), the pressure is expected to decrease while the forward reaction is occurring.
(c)(i)
Place the sealed container in a thermostatically controlled water bath / hot-water bath set to \( 40^\circ\text{C} \), and monitor with a thermometer to maintain constant temperature.
(c)(ii)
Immediately recording the absorbance is unreliable because equilibrium has not yet been reached; the concentration of \( \text{NO}_2 \) (and therefore the colour/absorbance) is still changing.
(c)(iii)
Allow the system to stand at \( 40^\circ\text{C} \) until the absorbance (or colour) becomes constant, then take the reading. In practice: wait and/or repeat measurements until successive absorbance values are unchanged.
(d)
Let \( x \) be the moles of \( \text{N}_2\text{O}_4 \) formed at equilibrium.
Stoichiometry: \( 2\,\text{NO}_2 \rightleftharpoons \text{N}_2\text{O}_4 \).
Initial moles of \( \text{NO}_2 \): \( 0.0100 \).
At equilibrium: moles of \( \text{NO}_2 = 0.00732 \).
Moles of \( \text{NO}_2 \) consumed: \[ 0.0100 – 0.00732 = 0.00268\ \text{mol} \] Hence: \[ 2x = 0.00268 \Rightarrow x = 0.00134\ \text{mol of } \text{N}_2\text{O}_4 \] Volume is \( 1.00\ \text{dm}^3 \), so concentrations equal moles:
\[ [\text{NO}_2] = 0.00732\ \text{mol dm}^{-3},\quad [\text{N}_2\text{O}_4] = 0.00134\ \text{mol dm}^{-3} \] Then \[ K = \frac{[\text{N}_2\text{O}_4]}{[\text{NO}_2]^2} = \frac{0.00134}{(0.00732)^2} \approx 25.0\ \text{mol}^{-1}\text{dm}^3 \] \(\boxed{K \approx 25.0\ \text{mol}^{-1}\text{dm}^3}\)
(e)(i)
One acceptable table format is:
| Titration number | Initial burette reading / \( \text{cm}^3 \) | Final burette reading / \( \text{cm}^3 \) | Volume of \( \text{NaOH} \) added / \( \text{cm}^3 \) |
|---|---|---|---|
| 1 | |||
| 2 | |||
| 3 |
The table allows at least three titration results and records both initial and final burette readings (and hence the volume of \( \text{NaOH} \) added) with units.
(e)(ii)
Each burette reading has an uncertainty of \( \pm 0.05\ \text{cm}^3 \). For a titre calculated from initial and final readings, total absolute uncertainty is: \[ \Delta V = 0.05 + 0.05 = 0.10\ \text{cm}^3 \] Percentage uncertainty: \[ \%\,\text{uncertainty} = \frac{\Delta V}{V} \times 100 = \frac{0.10}{20.05} \times 100 \approx 0.50\% \] \(\boxed{\text{Percentage uncertainty} \approx 0.50\%}\)
(e)(iii)
Convert temperatures using \( T(\text{K}) = T(^\circ\text{C}) + 273.2 \) (to 1 decimal place as in data):
\[ \begin{aligned} 0.0^\circ\text{C} &\rightarrow 273.2\ \text{K} \\ 20.0^\circ\text{C} &\rightarrow 293.2\ \text{K} \\ 50.0^\circ\text{C} &\rightarrow 323.2\ \text{K} \\ 100.0^\circ\text{C} &\rightarrow 373.2\ \text{K} \end{aligned} \] Completed \( T/\text{K} \) column: \( 273.2,\ 293.2,\ 323.2,\ 373.2 \).
(e)(iv)
Yes, the results are consistent with the given \( \Delta H^\circ \). The reaction is exothermic (\( \Delta H^\circ = -55.3\ \text{kJ mol}^{-1} \)), so increasing temperature should favour the endothermic (reverse) reaction and thus decrease \( K \). From the table, \( K \) decreases as \( T \) increases (for example, from about \( 25.0 \) at \( 273.2\ \text{K} \) to \( 3.64 \times 10^{-2} \) at \( 373.2\ \text{K} \)), which matches this expectation.