An original Thinka practice paper modelled on the structure and difficulty of the Jun 2022 Pearson Edexcel A Level Chemistry (9CH0) paper. Not affiliated with or reproduced from Pearson.
Paper 1: Advanced Inorganic and Physical Chemistry
Answer all questions. Show all your working in calculations and include units where appropriate.
33 Question · 99.5 marks
Question 1 · Multiple Choice
1 marks
A buffer solution is prepared by mixing \(50.0\text{ cm}^3\) of \(0.100\text{ mol dm}^{-3}\) propanoic acid (\(K_a = 1.35 \times 10^{-5}\text{ mol dm}^{-3}\)) with \(25.0\text{ cm}^3\) of \(0.120\text{ mol dm}^{-3}\) sodium hydroxide. What is the pH of the resulting buffer solution at \(298\text{ K}\)?
A.4.69
B.4.87
C.5.05
D.5.23
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Worked solution
First, calculate the initial moles of propanoic acid (HA) and sodium hydroxide (NaOH): Moles of HA = \(0.0500\text{ dm}^3 \times 0.100\text{ mol dm}^{-3} = 0.00500\text{ mol}\); Moles of NaOH = \(0.0250\text{ dm}^3 \times 0.120\text{ mol dm}^{-3} = 0.00300\text{ mol}\). The NaOH reacts completely with HA to produce A^- and water: Moles of A^- formed = \(0.00300\text{ mol}\); Moles of HA remaining = \(0.00500 - 0.00300 = 0.00200\text{ mol}\). Next, calculate the pKa of propanoic acid: pKa = \(-\log_{10}(1.35 \times 10^{-5}) = 4.87\). Using the Henderson-Hasselbalch equation: pH = \(\text{p}K_a + \log_{10}\left(\frac{[\text{A}^-]}{[\text{HA}]}\right) = 4.87 + \log_{10}\left(\frac{0.00300}{0.00200}\right) = 4.87 + 0.18 = 5.05\).
Marking scheme
1 mark for calculating the correct pH of 5.05 (Option C). Reject all other options.
Question 2 · Multiple Choice
1 marks
An element \(X\) in Period 3 of the Periodic Table has the following successive ionization energies in \(\text{kJ mol}^{-1}\): \(IE_1 = 1000\), \(IE_2 = 2250\), \(IE_3 = 3360\), \(IE_4 = 4560\), \(IE_5 = 7010\), \(IE_6 = 8500\), \(IE_7 = 27100\), \(IE_8 = 31700\). In which group of the Periodic Table is element \(X\)?
A.Group 14
B.Group 15
C.Group 16
D.Group 17
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Worked solution
Analyze the successive ionization energies to find the largest jump. The energy required increases steadily from the 1st to the 6th ionization energy (1000 to 8500 \(\text{kJ mol}^{-1}\)). There is an extremely large jump between the 6th and 7th ionization energies (from 8500 to 27100 \(\text{kJ mol}^{-1}\)). This indicates that the 7th electron is removed from an inner core shell, which means the element has 6 valence electrons in its outer shell. Therefore, it belongs to Group 16 (Group 6).
Marking scheme
1 mark for identifying Group 16 as the correct group (Option C).
Question 3 · Multiple Choice
1 marks
How many stereoisomers exist for the octahedral complex ion \([\text{Co}(\text{en})_2\text{Cl}_2]^+\) (where \(\text{en}\) represents the bidentate ligand ethane-1,2-diamine)?
A.2
B.3
C.4
D.6
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Worked solution
The complex exhibits geometric isomerism with cis and trans forms. The trans-isomer has a plane of symmetry and is achiral, so it exists as only 1 stereoisomer. The cis-isomer lacks a plane of symmetry and is chiral, existing as a pair of non-superimposable optical isomers (2 enantiomers). Therefore, there is a total of 1 (trans) + 2 (cis-enantiomers) = 3 stereoisomers.
Marking scheme
1 mark for identifying 3 stereoisomers (Option B).
Question 4 · Multiple Choice
1 marks
Use the following data to calculate the experimental lattice energy, \(\Delta_{\text{latt}}H^\ominus\), of calcium chloride, \(\text{CaCl}_2(\text{s})\) in \(\text{kJ mol}^{-1}\): \(\Delta_fH^\ominus(\text{CaCl}_2(\text{s})) = -796\text{ kJ mol}^{-1}\), \(\Delta_{\text{at}}H^\ominus(\text{Ca}(\text{s})) = +178\text{ kJ mol}^{-1}\), \(1^{\text{st}}\text{ IE}(\text{Ca}) = +590\text{ kJ mol}^{-1}\), \(2^{\text{nd}}\text{ IE}(\text{Ca}) = +1145\text{ kJ mol}^{-1}\), \(\Delta_{\text{at}}H^\ominus(\text{Cl}_2(\text{g})) = +122\text{ kJ mol}^{-1}\) (enthalpy of atomization per mole of \(\text{Cl}(\text{g})\)), \(1^{\text{st}}\text{ EA}(\text{Cl}) = -349\text{ kJ mol}^{-1}\).
A.-1557
B.-2133
C.-2255
D.-2377
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1 mark for the correct calculation of -2255 kJ mol^-1 (Option C).
Question 5 · Multiple Choice
1 marks
A vessel is filled with \(1.00\text{ mol}\) of \(\text{PCl}_5(\text{g})\) and allowed to reach equilibrium at a constant temperature and a total pressure of \(2.00\text{ atm}\). \(\text{PCl}_5(\text{g}) \rightleftharpoons \text{PCl}_3(\text{g}) + \text{Cl}_2(\text{g})\). At equilibrium, the mixture contains \(0.40\text{ mol}\) of \(\text{Cl}_2(\text{g})\). What is the value of the equilibrium constant \(K_p\) in \(\text{atm}\)?
A.0.20
B.0.38
C.0.53
D.0.80
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Worked solution
From the stoichiometry of the reaction, at equilibrium: Moles of \(\text{Cl}_2\) = \(0.40\text{ mol}\), Moles of \(\text{PCl}_3\) = \(0.40\text{ mol}\), Moles of \(\text{PCl}_5\) = \(1.00 - 0.40 = 0.60\text{ mol}\). Total moles at equilibrium = \(0.60 + 0.40 + 0.40 = 1.40\text{ mol}\). Partial pressures are: \(p(\text{PCl}_5) = \frac{0.60}{1.40} \times 2.00 = 0.857\text{ atm}\), \(p(\text{PCl}_3) = \frac{0.40}{1.40} \times 2.00 = 0.571\text{ atm}\), \(p(\text{Cl}_2) = \frac{0.40}{1.40} \times 2.00 = 0.571\text{ atm}\). Substituting into the \(K_p\) expression: \(K_p = \frac{p(\text{PCl}_3) \times p(\text{Cl}_2)}{p(\text{PCl}_5)} = \frac{0.571 \times 0.571}{0.857} = 0.38\text{ atm}\).
Marking scheme
1 mark for the correct calculation of Kp as 0.38 atm (Option B).
Question 6 · Multiple Choice
1 marks
Which of the following statements correctly explains the trend in thermal stability of the Group 2 nitrates down the group?
A.The cationic radius decreases, increasing the polarization of the nitrate ion and making the compound more thermally stable.
B.The cationic radius increases, decreasing the polarization of the nitrate ion and making the compound more thermally stable.
C.The cationic radius increases, increasing the polarization of the nitrate ion and making the compound less thermally stable.
D.The cationic radius decreases, decreasing the polarization of the nitrate ion and making the compound less thermally stable.
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Worked solution
As you descend Group 2, the cationic radius of the metal ion increases while the charge remains constant at 2+. This leads to a decrease in charge density and polarizing power. Consequently, the larger cation polarizes the large, electron-clouded nitrate anion to a lesser extent, requiring more thermal energy to decompose the compound. Thus, thermal stability increases down the group.
Marking scheme
1 mark for identifying the correct explanation in terms of increasing cationic radius and decreasing polarization (Option B).
Question 7 · Multiple Choice
1 marks
Chlorine reacts with hot, concentrated aqueous sodium hydroxide to form sodium chloride, sodium chlorate(V), and water. What are the oxidation states of chlorine in the products of this disproportionation reaction?
A.-1 and +1
B.-1 and +3
C.-1 and +5
D.-1 and +7
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Worked solution
The chemical equation for the reaction of chlorine with hot, concentrated sodium hydroxide is: \(3\text{Cl}_2(\text{g}) + 6\text{NaOH}(\text{aq}) \to 5\text{NaCl}(\text{aq}) + \text{NaClO}_3(\text{aq}) + 3\text{H}_2\text{O}(\text{l})\). In sodium chloride (NaCl), chlorine is in the -1 oxidation state. In sodium chlorate(V) (\(\text{NaClO}_3\)), chlorine is in the +5 oxidation state.
Marking scheme
1 mark for identifying the oxidation states of chlorine in NaCl and NaClO3 as -1 and +5 respectively (Option C).
Question 8 · Multiple Choice
1 marks
Which of the following molecules or ions has a non-linear shape with a bond angle of approximately \(104.5^\circ\)?
A.CO2
B.SO2
C.NO2^+
D.NH2^-
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Worked solution
According to VSEPR theory, \(\text{NH}_2^-\)\ has 2 bonding pairs and 2 lone pairs on the central nitrogen atom (isoelectronic with \(\text{H}_2\text{O}\)). The 4 electron pairs arrange tetrahedrally to minimize repulsion, but the presence of 2 lone pairs compresses the bond angle from the tetrahedral angle of \(109.5^\circ\) to approximately \(104.5^\circ\), resulting in a non-linear (bent) shape.
Marking scheme
1 mark for identifying NH2^- as the correct ion (Option D).
Question 9 · Multiple Choice
1 marks
A buffer solution is prepared by mixing \(50.0\text{ cm}^3\) of \(0.100\text{ mol dm}^{-3}\) propanoic acid (\(K_a = 1.35 \times 10^{-5}\text{ mol dm}^{-3}\)) with \(25.0\text{ cm}^3\) of \(0.100\text{ mol dm}^{-3}\) sodium hydroxide.
What is the pH of this buffer solution at \(298\text{ K}\)?
A.4.57
B.4.87
C.5.17
D.9.13
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2. Calculate moles of \(\text{NaOH}\) added: \(\text{moles of NaOH} = 0.0250\text{ dm}^3 \times 0.100\text{ mol dm}^{-3} = 0.00250\text{ mol}\)
3. The reaction that occurs is: \(\text{CH}_3\text{CH}_2\text{COOH} + \text{NaOH} \rightarrow \text{CH}_3\text{CH}_2\text{COONa} + \text{H}_2\text{O}\)
4. Calculate the remaining moles of propanoic acid and the moles of propanoate ions formed: \(\text{moles of propanoic acid remaining} = 0.00500 - 0.00250 = 0.00250\text{ mol}\) \(\text{moles of propanoate ions formed} = 0.00250\text{ mol}\)
5. Since the moles of the weak acid and its conjugate base are equal, this is the half-neutralisation point: \([\text{HA}] = [\text{A}^-]\)
- **[1 mark]** for the correct answer (B). - **Correct method**: Recognizing the mixture forms a buffer at the half-neutralisation point where \(\text{pH} = \text{p}K_a\), leading to \(\text{pH} = -\log_{10}(1.35 \times 10^{-5}) = 4.87\).
Question 10 · Multiple Choice
1 marks
Which of the following gaseous ions has the same number of unpaired d-electrons as a gaseous \(\text{Co}^{2+}\) ion?
A.\(\text{Cr}^{3+}\)
B.\(\text{Fe}^{3+}\)
C.\(\text{Ni}^{2+}\)
D.\(\text{Cu}^{2+}\)
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Worked solution
1. Determine the electron configuration of a gaseous \(\text{Co}^{2+}\) ion: - Cobalt (\(\text{Co}\)) has atomic number 27: \([\text{Ar}] 3\text{d}^7 4\text{s}^2\). - \(\text{Co}^{2+}\) has the configuration \([\text{Ar}] 3\text{d}^7\). - The seven d-electrons fill the five d-orbitals according to Hund's rule: three orbitals contain single unpaired electrons, and two orbitals contain paired electrons. Thus, there are **3 unpaired electrons**.
2. Determine the electron configurations and number of unpaired electrons for each option: - **\(\text{Cr}^{3+}\)**: Chromium (\(\text{Cr}\)) is \([\text{Ar}] 3\text{d}^5 4\text{s}^1\). \(\text{Cr}^{3+}\) is \([\text{Ar}] 3\text{d}^3\). The three d-electrons occupy separate orbitals singly, giving **3 unpaired electrons**. - **\(\text{Fe}^{3+}\)**: Iron (\(\text{Fe}\)) is \([\text{Ar}] 3\text{d}^6 4\text{s}^2\). \(\text{Fe}^{3+}\) is \([\text{Ar}] 3\text{d}^5\), giving **5 unpaired electrons**. - **\(\text{Ni}^{2+}\)**: Nickel (\(\text{Ni}\)) is \([\text{Ar}] 3\text{d}^8 4\text{s}^2\). \(\text{Ni}^{2+}\) is \([\text{Ar}] 3\text{d}^8\). Eight d-electrons occupy the five orbitals: three pairs and two single electrons, giving **2 unpaired electrons**. - **\(\text{Cu}^{2+}\)**: Copper (\(\text{Cu}\)) is \([\text{Ar}] 3\text{d}^{10} 4\text{s}^1\). \(\text{Cu}^{2+}\) is \([\text{Ar}] 3\text{d}^9\). Nine d-electrons occupy the five orbitals: four pairs and one single electron, giving **1 unpaired electron**.
Therefore, \(\text{Cr}^{3+}\) has the same number of unpaired d-electrons (3) as \(\text{Co}^{2+}\).
Marking scheme
- **[1 mark]** for the correct answer (A). - **Correct method**: Identifying that \(\text{Co}^{2+}\) has a \(3\text{d}^7\) configuration with 3 unpaired electrons and matching it with \(\text{Cr}^{3+}\) which has a \(3\text{d}^3\) configuration with 3 unpaired electrons.
Question 11 · Short Answer
2.5 marks
A buffer solution is prepared by mixing \(20.0\text{ cm}^3\) of \(0.150\text{ mol dm}^{-3}\) methanoic acid, \(\text{HCOOH}\), with \(10.0\text{ cm}^3\) of \(0.100\text{ mol dm}^{-3}\) sodium hydroxide, \(\text{NaOH}\). Calculate the pH of the resulting buffer solution at 298 K. [\(K_a\text{ for HCOOH} = 1.6 \times 10^{-4}\text{ mol dm}^{-3}\text{ at 298 K}\)]
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Worked solution
First, calculate the initial number of moles of methanoic acid and sodium hydroxide: \(n(\text{HCOOH}) = 0.0200\text{ dm}^3 \times 0.150\text{ mol dm}^{-3} = 3.00 \times 10^{-3}\text{ mol}\) \(n(\text{NaOH}) = 0.0100\text{ dm}^3 \times 0.100\text{ mol dm}^{-3} = 1.00 \times 10^{-3}\text{ mol}\)
Upon mixing, \(\text{NaOH}\) reacts completely with \(\text{HCOOH}\) according to the equation: \(\text{HCOOH} + \text{NaOH} \rightarrow \text{HCOONa} + \text{H}_2\text{O}\)
Calculate the moles of species in the buffer solution at equilibrium: \(n(\text{HCOO}^-) = 1.00 \times 10^{-3}\text{ mol}\) \(n(\text{HCOOH})_{\text{remaining}} = 3.00 \times 10^{-3} - 1.00 \times 10^{-3} = 2.00 \times 10^{-3}\text{ mol}\)
Using the acid dissociation constant expression: \(K_a = \frac{[\text{H}^+][\text{HCOO}^-]}{[\text{HCOOH}]}\) Rearranging to solve for \([\text{H}^+]\): \([\text{H}^+] = K_a \times \frac{n(\text{HCOOH})}{n(\text{HCOO}^-)}\) \([\text{H}^+] = 1.6 \times 10^{-4} \times \frac{2.00 \times 10^{-3}}{1.00 \times 10^{-3}} = 3.20 \times 10^{-4}\text{ mol dm}^{-3}\)
Calculate the pH: \(\text{pH} = -\log_{10}(3.20 \times 10^{-4}) = 3.49\)
Marking scheme
- **1 Mark**: Calculate the moles of \(\text{HCOOH}\) remaining (\(2.00 \times 10^{-3}\text{ mol}\)) and \(\text{HCOO}^-\) formed (\(1.00 \times 10^{-3}\text{ mol}\)). - **1 Mark**: Correctly substitute these values into the \(K_a\) expression to find \([\text{H}^+] = 3.20 \times 10^{-4}\text{ mol dm}^{-3}\). - **0.5 Marks**: Correctly calculate pH to 2 decimal places (\(3.49\)). (Accept \(3.5\) if correct working shown but penalise for incorrect rounding of intermediate steps).
Question 12 · Short Answer
2.5 marks
An anhydrous chloride of chromium, \(\text{CrCl}_x\), has a mass of \(0.125\text{ g}\). It was dissolved in water and reacted with an excess of silver nitrate solution, producing \(0.339\text{ g}\) of dry silver chloride precipitate, \(\text{AgCl}\). Deduce the value of \(x\) to find the formula of the chromium chloride. [Molar masses: \(\text{Cr} = 52.0\text{ g mol}^{-1}\), \(\text{Ag} = 107.9\text{ g mol}^{-1}\), \(\text{Cl} = 35.5\text{ g mol}^{-1}\)]
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Worked solution
1. Calculate the moles of \(\text{AgCl}\) precipitate: \(n(\text{AgCl}) = \frac{0.339}{107.9 + 35.5} = \frac{0.339}{143.4} = 2.364 \times 10^{-3}\text{ mol}\) Since \(1\text{ mol}\) of \(\text{AgCl}\) contains \(1\text{ mol}\) of \(\text{Cl}^-\), \(n(\text{Cl}^-) = 2.364 \times 10^{-3}\text{ mol}\).
2. Calculate the mass of chlorine in the sample: \(m(\text{Cl}) = 2.364 \times 10^{-3}\text{ mol} \times 35.5\text{ g mol}^{-1} = 0.0839\text{ g}\)
3. Calculate the mass of chromium in the sample: \(m(\text{Cr}) = 0.125\text{ g} - 0.0839\text{ g} = 0.0411\text{ g}\)
4. Calculate the moles of chromium in the sample: \(n(\text{Cr}) = \frac{0.0411}{52.0} = 7.904 \times 10^{-4}\text{ mol}\)
5. Determine the ratio of \(\text{Cl} : \text{Cr}\): \(\text{Ratio} = \frac{2.364 \times 10^{-3}}{7.904 \times 10^{-4}} = 2.99 \approx 3\)
Therefore, \(x = 3\) and the formula is \(\text{CrCl}_3\).
Marking scheme
- **1 Mark**: Correctly calculate the moles of \(\text{AgCl}\) (\(2.36 \times 10^{-3}\text{ mol}\)) and the mass of chloride (\(0.0839\text{ g}\)). - **1 Mark**: Deduce the mass of chromium (\(0.0411\text{ g}\)) and calculate the moles of chromium (\(7.90 \times 10^{-4}\text{ mol}\)). - **0.5 Marks**: Calculate the molar ratio to obtain the whole number \(x = 3\) (or state formula as \(\text{CrCl}_3\)).
Question 13 · Short Answer
2.5 marks
Use the following data to calculate the first electron affinity of fluorine, in \(\text{kJ mol}^{-1}\): - Standard enthalpy of formation of \(\text{NaF(s)}\) = \(-574\text{ kJ mol}^{-1}\) - Enthalpy of atomisation of sodium, \(\text{Na(s)} \rightarrow \text{Na(g)}\) = \(+107\text{ kJ mol}^{-1}\) - First ionisation energy of sodium = \(+496\text{ kJ mol}^{-1}\) - Enthalpy of atomisation of fluorine, \(\frac{1}{2}\text{F}_2\text{(g)} \rightarrow \text{F(g)}\) = \(+79\text{ kJ mol}^{-1}\) - Lattice energy of \(\text{NaF(s)}\) = \(-930\text{ kJ mol}^{-1}\)
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Worked solution
Using a Born-Haber cycle, the standard enthalpy of formation is equal to the sum of the enthalpy changes of the steps: \(\Delta H_f^{\theta} = \Delta H_{at}^{\theta}(\text{Na}) + \text{IE}_1(\text{Na}) + \Delta H_{at}^{\theta}(\text{F}) + \text{EA}_1(\text{F}) + \Delta H_{latt}^{\theta}(\text{NaF})\)
Substitute the given values into the equation: \(-574 = +107 + 496 + 79 + \text{EA}_1(\text{F}) - 930\) \(-574 = -248 + \text{EA}_1(\text{F})\) \(\text{EA}_1(\text{F}) = -574 - (-248) = -326\text{ kJ mol}^{-1}\)
Marking scheme
- **1 Mark**: Correct algebraic Born-Haber cycle expression relating the given enthalpy changes. - **1 Mark**: Correct substitution of values into the expression. - **0.5 Marks**: Correct final answer of \(-326\text{ kJ mol}^{-1}\) (including the negative sign and unit).
Question 14 · Short Answer
2.5 marks
A \(2.97\text{ g}\) sample of anhydrous magnesium nitrate, \(\text{Mg(NO}_3)_2\), is heated until it completely decomposes according to the following equation:
Calculate the total volume of gas, in \(\text{dm}^3\), produced at room temperature and pressure (RTP). [Molar volume of gas at RTP = \(24.0\text{ dm}^3\text{ mol}^{-1}\); Molar mass of \(\text{Mg(NO}_3)_2 = 148.3\text{ g mol}^{-1}\)]
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Worked solution
1. Calculate the number of moles of \(\text{Mg(NO}_3)_2\): \(n(\text{Mg(NO}_3)_2) = \frac{2.97\text{ g}}{148.3\text{ g mol}^{-1}} = 0.02003\text{ mol}\)
2. Determine the total moles of gas produced: From the stoichiometry, \(2\text{ moles}\) of \(\text{Mg(NO}_3)_2\) produce \(5\text{ moles}\) of gas (\(4\text{ mol of NO}_2 + 1\text{ mol of O}_2\)). \(n(\text{total gas}) = 0.02003\text{ mol} \times \frac{5}{2} = 0.05008\text{ mol}\)
3. Calculate the total gas volume at RTP: \(V = 0.05008\text{ mol} \times 24.0\text{ dm}^3\text{ mol}^{-1} = 1.20\text{ dm}^3\) (to 3 significant figures)
Marking scheme
- **1 Mark**: Calculate the moles of \(\text{Mg(NO}_3)_2\) correctly as \(0.0200\text{ mol}\). - **1 Mark**: Use the stoichiometric ratio (2.5 moles of gas per mole of reactant) to find the total moles of gas as \(0.0500\text{ mol}\). - **0.5 Marks**: Calculate the total gas volume at RTP as \(1.20\text{ dm}^3\) (allow error carried forward from incorrect moles if method is clear).
Question 15 · Short Answer
2.5 marks
Explain the differences in the bond angles of ammonia (\(\text{NH}_3\)) and the amide ion (\(\text{NH}_2^-\)), stating the shape and approximate bond angle in each species.
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Worked solution
- Ammonia (\(\text{NH}_3\)) has a trigonal pyramidal shape with a bond angle of approximately \(107^\circ\) (accept \(106.5^\circ - 107.5^\circ\)). It has 3 bonding pairs and 1 lone pair of electrons. - The amide ion (\(\text{NH}_2^-\)) has a bent / non-linear shape with a bond angle of approximately \(104.5^\circ\) (accept \(104^\circ - 105^\circ\)). It has 2 bonding pairs and 2 lone pairs of electrons. - Lone pairs repel other electron pairs more strongly than bonding pairs. The presence of an extra lone pair in \(\text{NH}_2^-\)- compared to \(\text{NH}_3\) causes greater repulsion, which pushes the N-H bonds closer together, resulting in a smaller bond angle.
Marking scheme
- **1 Mark**: State the correct shape and approximate bond angle for both \(\text{NH}_3\) (trigonal pyramidal, \(107^\circ\)) and \(\text{NH}_2^-\)/amide ion (bent/non-linear, \(104.5^\circ\)). - **1 Mark**: State that \(\text{NH}_3\) has 3 bonding pairs and 1 lone pair, whereas \(\text{NH}_2^-\) has 2 bonding pairs and 2 lone pairs. - **0.5 Marks**: Explain that lone pair-lone pair repulsion is greater than lone pair-bonding pair (or bonding pair-bonding pair) repulsion, hence reducing the bond angle in \(\text{NH}_2^-\).
Question 16 · Short Answer
2.5 marks
In a closed vessel, the following equilibrium is established at a specific temperature:
At equilibrium, the partial pressure of \(\text{PCl}_5\) is \(0.240\text{ atm}\), and the partial pressure of \(\text{Cl}_2\) is \(0.360\text{ atm}\). Assuming the mixture was initially formed from pure \(\text{PCl}_5\), calculate the equilibrium constant, \(K_p\), for this reaction, including its units.
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Worked solution
Since the equilibrium mixture was formed from pure \(\text{PCl}_5\), the mole ratio of the produced gases is \(1:1\). Therefore, the equilibrium partial pressure of \(\text{PCl}_3\) must be equal to the partial pressure of \(\text{Cl}_2\): \(p(\text{PCl}_3) = p(\text{Cl}_2) = 0.360\text{ atm}\)
The expression for \(K_p\) is: \(K_p = \frac{p(\text{PCl}_3) \times p(\text{Cl}_2)}{p(\text{PCl}_5)}\)
Substitute the values into the expression: \(K_p = \frac{0.360\text{ atm} \times 0.360\text{ atm}}{0.240\text{ atm}} = 0.540\text{ atm}\)
Marking scheme
- **1 Mark**: State or use the fact that the partial pressure of \(\text{PCl}_3\) is equal to that of \(\text{Cl}_2\) (\(0.360\text{ atm}\)). - **1 Mark**: Write the correct \(K_p\) expression and substitute the values to obtain \(0.540\) (or \(0.54\)). - **0.5 Marks**: Give the correct unit of \(\text{atm}\).
Question 17 · Short Answer
2.5 marks
A student titrates \(25.0\text{ cm}^3\) of an acidified solution containing \(\text{Fe}^{2+}\) ions with \(0.0200\text{ mol dm}^{-3}\) potassium manganate(VII), \(\text{KMnO}_4\). The end-point is reached when \(18.50\text{ cm}^3\) of the \(\text{KMnO}_4\) solution has been added. Calculate the concentration of the \(\text{Fe}^{2+}\) ions in the original solution, to 3 significant figures.
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Worked solution
1. Write the balanced redox equation for the titration: \(\text{MnO}_4^- + 5\text{Fe}^{2+} + 8\text{H}^+ \rightarrow \text{Mn}^{2+} + 5\text{Fe}^{3+} + 4\text{H}_2\text{O}\) This shows a reaction ratio of \(1\text{ mol of MnO}_4^-\text{ to } 5\text{ mol of Fe}^{2+}\).
2. Calculate the moles of manganate(VII) used: \(n(\text{MnO}_4^-) = 0.01850\text{ dm}^3 \times 0.0200\text{ mol dm}^{-3} = 3.70 \times 10^{-4}\text{ mol}\)
3. Calculate the moles of \(\text{Fe}^{2+}\) in the \(25.0\text{ cm}^3\) sample: \(n(\text{Fe}^{2+}) = 5 \times 3.70 \times 10^{-4}\text{ mol} = 1.85 \times 10^{-3}\text{ mol}\)
4. Calculate the concentration of \(\text{Fe}^{2+}\): \([\text{Fe}^{2+}] = \frac{1.85 \times 10^{-3}\text{ mol}}{0.0250\text{ dm}^3} = 0.0740\text{ mol dm}^{-3}\)
Marking scheme
- **1 Mark**: Identify the 1:5 stoichiometric ratio of \(\text{MnO}_4^-\text{ to } \text{Fe}^{2+}\). - **1 Mark**: Calculate the moles of \(\text{MnO}_4^-\) as \(3.70 \times 10^{-4}\text{ mol}\) and moles of \(\text{Fe}^{2+}\) as \(1.85 \times 10^{-3}\text{ mol}\). - **0.5 Marks**: Calculate the final concentration of \(\text{Fe}^{2+}\) as \(0.0740\text{ mol dm}^{-3}\) (must be to 3 significant figures and include units).
Question 18 · Short Answer
2.5 marks
The table below shows the successive ionisation energies, in \(\text{kJ mol}^{-1}\), for a Period 3 element, \(X\): - 1st: \(578\) - 2nd: \(1817\) - 3rd: \(2745\) - 4th: \(11578\) - 5th: \(14831\)
Identify element \(X\) and explain your choice by referring to the data.
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Worked solution
1. Identify the element: Element \(X\) is Aluminium (\(\text{Al}\)).
2. Explain using the data: There is a very large increase (or 'jump') between the 3rd and the 4th ionisation energies (from \(2745\) to \(11578\text{ kJ mol}^{-1}\)). This indicates that the 4th electron is being removed from an inner electron shell (which is closer to the nucleus and experiences less shielding and stronger electrostatic attraction). Therefore, there are three electrons in the outermost shell, placing it in Group 3. Since the element is in Period 3, it must be aluminium.
Marking scheme
- **1 Mark**: Correctly identify the element as Aluminium (or \(\text{Al}\)). - **1 Mark**: Note the large jump/increase between the 3rd and 4th ionisation energies. - **0.5 Marks**: Explain that this jump shows the 4th electron is being removed from an inner shell (or that there are 3 valence electrons in the outer shell).
Question 19 · Short Answer
2.5 marks
A buffer solution is prepared by mixing \(50.0\text{ cm}^3\) of \(0.150\text{ mol dm}^{-3}\) propanoic acid (\(K_a = 1.35 \times 10^{-5}\text{ mol dm}^{-3}\)) and \(25.0\text{ cm}^3\) of \(0.200\text{ mol dm}^{-3}\) sodium propanoate at \(298\text{ K}\). Calculate the pH of this buffer solution, giving your answer to two decimal places.
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Worked solution
1. Calculate the moles of propanoic acid (HA) and propanoate ions (\(\text{A}^-\)): Moles of HA = \(0.0500\text{ dm}^3 \times 0.150\text{ mol dm}^{-3} = 7.50 \times 10^{-3}\text{ mol}\) Moles of \(\text{A}^-\)\ = \(0.0250\text{ dm}^3 \times 0.200\text{ mol dm}^{-3} = 5.00 \times 10^{-3}\text{ mol}\)
1 mark: Calculation of moles of both HA (\(7.50 \times 10^{-3}\text{ mol}\)) and \(\text{A}^-\)\ (\(5.00 \times 10^{-3}\text{ mol}\)). 1 mark: Calculation of \([\text{H}^+] = 2.025 \times 10^{-5}\text{ mol dm}^{-3}\) (or correct substitution into Henderson-Hasselbalch equation). 0.5 marks: Final pH of 4.69 (must be strictly 2 decimal places).
Question 20 · Short Answer
2.5 marks
Explain, in terms of the size and charge of the cations, why magnesium carbonate decomposes at a lower temperature than barium carbonate.
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Worked solution
Magnesium (\(\text{Mg}^{2+}\)) and barium (\(\text{Ba}^{2+}\)) ions both have a \(+2\) charge. However, the \(\text{Mg}^{2+}\) ion is smaller than the \(\text{Ba}^{2+}\) ion. Because of its smaller ionic radius, the \(\text{Mg}^{2+}\) ion has a higher charge density. This enables it to more strongly polarise (distort) the electron cloud of the carbonate (\(\text{CO}_3^{2-}\)) ion. This polarisation weakens the covalent carbon-oxygen bonds within the carbonate ion, lowering the thermal energy required to decompose it into magnesium oxide and carbon dioxide.
Marking scheme
1 mark: Identifies that \(\text{Mg}^{2+}\) has a smaller ionic radius / higher charge density than \(\text{Ba}^{2+}\) (with same charge). 1 mark: Explains that \(\text{Mg}^{2+}\) polarises or distorts the electron cloud of the carbonate ion more strongly. 0.5 marks: Explains that this weakens the covalent C-O bond within the carbonate ion (requiring less energy/temperature to break).
Question 21 · Short Answer
2.5 marks
A transition metal complex ion has the formula \([\text{Co}(\text{NH}_3)_4(\text{H}_2\text{O})\text{Cl}]^{n+}\). The cobalt ion has an oxidation state of \(+3\). State: (i) the coordination number of the cobalt ion, (ii) the value of \(n\), and (iii) the shape of this complex ion.
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Worked solution
(i) The coordination number is the number of coordinate bonds formed to the central metal ion. There are four \(\text{NH}_3\) ligands, one \(\text{H}_2\text{O}\) ligand, and one \(\text{Cl}^-\)\ ligand, all of which are monodentate, giving a coordination number of 6. (ii) Cobalt has an oxidation state of \(+3\). \(\text{NH}_3\) and \(\text{H}_2\text{O}\) are neutral ligands (charge = 0), and \(\text{Cl}^-\)\ has a charge of \(-1\). Therefore, the overall charge of the complex ion is \(+3 + 0 + 0 - 1 = +2\), so \(n = 2\). (iii) A transition metal complex with a coordination number of 6 adopts an octahedral geometry.
Marking scheme
1 mark: State coordination number is 6. 1 mark: State value of \(n = 2\) (accept \(2+\) or \(+2\)). 0.5 marks: State shape is octahedral.
Question 22 · Short Answer
2.5 marks
Calculate the standard entropy change of system (\(\Delta S^\ominus_{\text{sys}}\)) at \(298\text{ K}\) for the following reaction: \(\text{N}_2\text{(g)} + 3\text{H}_2\text{(g)} \rightarrow 2\text{NH}_3\text{(g)}\)
Given the standard entropies (\(S^\ominus\)) at \(298\text{ K}\): \(\text{N}_2\text{(g)} = 191.6\text{ J K}^{-1}\text{ mol}^{-1}\) \(\text{H}_2\text{(g)} = 130.6\text{ J K}^{-1}\text{ mol}^{-1}\) \(\text{NH}_3\text{(g)} = 192.3\text{ J K}^{-1}\text{ mol}^{-1}\)
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1 mark: Correct substitution of standard entropy values into the equation: \([2 \times 192.3] - [191.6 + (3 \times 130.6)]\). 1 mark: Correct numerical value of \(-198.8\). 0.5 marks: Correct unit of \(\text{J K}^{-1}\text{ mol}^{-1}\) associated with the answer (must have negative sign).
Question 23 · Short Answer
2.5 marks
Sulfur tetrafluoride, \(\text{SF}_4\), is a highly reactive species. Determine the number of bonding pairs and lone pairs of electrons surrounding the central sulfur atom. Use this information to predict the shape and the approximate equatorial \(\text{F-S-F}\) bond angle in \(\text{SF}_4\).
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Worked solution
Sulfur is in Group 6 and has 6 outer-shell electrons. It forms four single covalent bonds with the four fluorine atoms, which uses 4 electrons, leaving 2 non-bonding electrons (1 lone pair). Thus, there are 4 bonding pairs and 1 lone pair around the central sulfur atom. This gives a total of 5 electron pairs, which arrange themselves in a trigonal bipyramidal base. Since there is one lone pair (which occupies an equatorial position to minimise repulsion), the shape is described as seesaw (or sawhorse). The strong repulsion from the lone pair compresses the equatorial bond angle from the ideal \(120^\circ\) to less than \(120^\circ\) (typically around \(102^\circ\)).
Marking scheme
1 mark: Correctly identifies 4 bonding pairs and 1 lone pair of electrons on sulfur. 1 mark: Predicts the shape is 'seesaw' (or 'sawhorse'). 0.5 marks: States that the equatorial angle is less than \(120^\circ\) (accept any value in the range \(101^\circ\) to \(105^\circ\) or less than \(120^\circ\)).
Question 24 · Short Answer
2.5 marks
A sample of an unknown volatile liquid with a mass of \(0.231\text{ g}\) was vaporised completely in a gas syringe at a temperature of \(97^\circ\text{C}\) and a pressure of \(1.00 \times 10^5\text{ Pa}\). The volume of the gas produced was \(74.0\text{ cm}^3\). Calculate the molar mass of the liquid. [Gas constant, \(R = 8.31\text{ J K}^{-1}\text{ mol}^{-1}\)]
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Worked solution
1. Convert all units to SI units: \(T = 97 + 273 = 370\text{ K}\) \(V = 74.0\text{ cm}^3 = 7.40 \times 10^{-5}\text{ m}^3\) \(p = 1.00 \times 10^5\text{ Pa}\)
2. Use the ideal gas equation \(pV = nRT\) to find the amount in moles (\(n\)): \(n = \frac{pV}{RT} = \frac{1.00 \times 10^5 \times 7.40 \times 10^{-5}}{8.31 \times 370}\) \(n = \frac{7.40}{3074.7} = 2.4067 \times 10^{-3}\text{ mol}\)
3. Calculate the molar mass (\(M\)): \(M = \frac{\text{mass}}{n} = \frac{0.231}{2.4067 \times 10^{-3}} = 95.98\text{ g mol}^{-1} \approx 96.0\text{ g mol}^{-1}\)
Marking scheme
1 mark: Converts units correctly to \(370\text{ K}\) and \(7.40 \times 10^{-5}\text{ m}^3\). 1 mark: Rearranges the ideal gas equation to solve for \(n\) and calculates \(n = 2.41 \times 10^{-3}\text{ mol}\). 0.5 marks: Calculates molar mass of \(96.0\text{ g mol}^{-1}\) (accept range \(95.9\) to \(96.1\) with or without units).
Question 25 · Short Answer
2.5 marks
A standard electrochemical cell is constructed from two standard half-cells:
(i) Write the conventional cell representation for this operating cell under standard conditions. (ii) Calculate the standard cell potential, \(E^\ominus_{\text{cell}}\), and state which electrode is positive.
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Worked solution
(i) The \(\text{Ag}^+/\text{Ag}\) half-cell has a more positive standard electrode potential (\(+0.80\text{ V}\)) than the \(\text{Fe}^{3+}/\text{Fe}^{2+}\) half-cell (\(+0.77\text{ V}\)). Therefore, silver ions are reduced at the cathode, and iron(II) ions are oxidised at the anode. Since the \(\text{Fe}^{3+}/\text{Fe}^{2+}\) half-cell contains only aqueous ions, an inert platinum electrode is required. The conventional cell representation is: \(\text{Pt(s)} \mid \text{Fe}^{2+}\text{(aq)}, \text{Fe}^{3+}\text{(aq)} \parallel \text{Ag}^+\text{(aq)} \mid \text{Ag(s)}\)
(ii) Calculate \(E^\ominus_{\text{cell}}\): \(E^\ominus_{\text{cell}} = E^\ominus_{\text{reduction}} - E^\ominus_{\text{oxidation}} = +0.80 - (+0.77) = +0.03\text{ V}\) Since reduction occurs at the silver electrode, it is the positive electrode (cathode).
Marking scheme
1 mark: Correct conventional cell diagram: \(\text{Pt(s)} \mid \text{Fe}^{2+}\text{(aq)}, \text{Fe}^{3+}\text{(aq)} \parallel \text{Ag}^+\text{(aq)} \mid \text{Ag(s)}\) (including platinum electrode and correct ordering). 1 mark: Calculation of standard cell potential, \(E^\ominus_{\text{cell}} = +0.03\text{ V}\) (must show sign and unit). 0.5 marks: States that the silver (\(\text{Ag}\)) electrode is the positive electrode.
Question 26 · Structured
6.5 marks
A buffer solution is prepared by mixing 50.0 cm3 of 0.200 mol dm-3 lactic acid (CH3CH(OH)COOH, Ka = 1.38 x 10^-4 mol dm-3) and 50.0 cm3 of 0.300 mol dm-3 sodium lactate. Calculate the pH of this buffer solution after the addition of 2.00 cm3 of 0.100 mol dm-3 hydrochloric acid. Show your working to 3 significant figures.
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Worked solution
First, calculate the initial moles of lactic acid (HA) and lactate ions (A-): n(HA) = (50.0 / 1000) * 0.200 = 0.0100 mol. n(A-) = (50.0 / 1000) * 0.300 = 0.0150 mol. Next, calculate the moles of H+ ions added from the HCl: n(H+) = (2.00 / 1000) * 0.100 = 0.00020 mol. The added strong acid reacts completely with the lactate ions: A- + H+ -> HA. This changes the mole quantities as follows: New n(HA) = 0.0100 + 0.00020 = 0.0102 mol. New n(A-) = 0.0150 - 0.00020 = 0.0148 mol. Using the Henderson-Hasselbalch equation: pH = pKa + log10([A-]/[HA]), where pKa = -log10(1.38 x 10^-4) = 3.860. pH = 3.860 + log10(0.0148 / 0.0102) = 3.860 + 0.162 = 4.022. Rounded to 3 significant figures, the pH is 4.02.
Marking scheme
M1: Moles of HA initially = 0.0100 mol (1 mark). M2: Moles of A- initially = 0.0150 mol (1 mark). M3: Moles of H+ added = 0.00020 mol (0.5 marks). M4: New moles of HA = 0.0102 mol and new moles of A- = 0.0148 mol (1 mark). M5: Correct rearrangement of Ka expression or use of Henderson-Hasselbalch equation (1 mark). M6: Final answer of pH = 4.02 (allow 4.02 to 4.03 depending on intermediate rounding) (2 marks).
Question 27 · Structured
6.5 marks
A sample of hydrated ammonium iron(II) sulfate, (NH4)2Fe(SO4)2 . xH2O, with a mass of 1.48 g was dissolved in excess dilute sulfuric acid and made up to 250.0 cm3 in a volumetric flask. A 25.0 cm3 portion of this solution required 15.10 cm3 of 0.00500 mol dm-3 potassium manganate(VII) solution for complete oxidation. Calculate the value of x in the formula of the hydrated salt. [Molar masses in g mol-1: N = 14.0, H = 1.0, Fe = 55.8, S = 32.1, O = 16.0]
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Worked solution
First, find the moles of manganate(VII) ions used: n(MnO4-) = (15.10 / 1000) * 0.00500 = 7.55 x 10^-5 mol. The equation for the redox titration is MnO4- + 5Fe2+ + 8H+ -> Mn2+ + 5Fe3+ + 4H2O. Therefore, the ratio of MnO4- : Fe2+ is 1 : 5. Moles of Fe2+ in the 25.0 cm3 aliquot = 5 * 7.55 x 10^-5 = 3.775 x 10^-4 mol. Moles of Fe2+ in the 250.0 cm3 volumetric flask = 10 * 3.775 x 10^-4 = 3.775 x 10^-3 mol. This is equal to the total moles of (NH4)2Fe(SO4)2 . xH2O in the 1.48 g sample. The molar mass of the hydrated salt is M = mass / moles = 1.48 / (3.775 x 10^-3) = 392.1 g mol-1. The molar mass of anhydrous (NH4)2Fe(SO4)2 = 2 * (14.0 + 4.0) + 55.8 + 2 * (32.1 + 64.0) = 36.0 + 55.8 + 192.2 = 284.0 g mol-1. Mass of water of crystallisation per mole = 392.1 - 284.0 = 108.1 g mol-1. Therefore, x = 108.1 / 18.0 = 6.01, which rounds to the nearest integer 6.
Marking scheme
M1: Moles of MnO4- = 7.55 x 10^-5 mol (1 mark). M2: Moles of Fe2+ in 25.0 cm3 = 3.775 x 10^-4 mol using the 1:5 ratio (1 mark). M3: Moles of Fe2+ in 250.0 cm3 = 3.775 x 10^-3 mol (0.5 marks). M4: Molar mass of the hydrated salt = 392.1 g mol-1 (1 mark). M5: Molar mass of anhydrous salt = 284.0 g mol-1 (1 mark). M6: Correct subtraction and determination of x = 6 with clear working (2 marks).
Question 28 · Structured
6.5 marks
Using the following data, calculate the standard lattice energy (lattice enthalpy of formation) of calcium chloride, CaCl2(s). Show all working. Data: Standard enthalpy change of formation of CaCl2(s) = -795.8 kJ mol-1; Standard enthalpy change of atomisation of calcium = +178.2 kJ mol-1; First ionisation energy of calcium = +590.0 kJ mol-1; Second ionisation energy of calcium = +1145.0 kJ mol-1; Standard enthalpy change of atomisation of chlorine = +121.7 kJ mol-1; First electron affinity of chlorine = -349.0 kJ mol-1. Then, explain why the experimental lattice energy of calcium iodide is less exothermic than expected from a theoretical purely ionic model.
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Worked solution
The Born-Haber cycle equation is: Delta Hf = Delta Hat[Ca] + IE1[Ca] + IE2[Ca] + 2 * Delta Hat[Cl] + 2 * EA1[Cl] + Delta HL. Substituting the values: -795.8 = 178.2 + 590.0 + 1145.0 + 2 * (121.7) + 2 * (-349.0) + Delta HL. -795.8 = 178.2 + 590.0 + 1145.0 + 243.4 - 698.0 + Delta HL. -795.8 = 1458.6 + Delta HL. Delta HL = -795.8 - 1458.6 = -2254.4 kJ mol-1. For the second part: The iodide ion has a large ionic radius and is easily polarized. The Ca2+ cation has a high charge density and polarizes the electron cloud of the iodide ion, which introduces some covalent character. This additional covalent bonding makes the experimental lattice energy more exothermic than the theoretical value calculated using a purely ionic model.
Marking scheme
M1: Multiplies Cl atomisation enthalpy by 2 (+243.4 kJ mol-1) (1 mark). M2: Multiplies Cl electron affinity by 2 (-698.0 kJ mol-1) (1 mark). M3: Correct algebraic expression for the cycle (1 mark). M4: Correct calculation of Delta HL = -2254.4 kJ mol-1 with negative sign and units (1.5 marks). M5: Explains that the iodide ion is large and highly polarizable by the calcium cation (1 mark). M6: Explains that polarization leads to covalent character, making the experimental lattice energy more exothermic than the theoretical model (1 mark).
Question 29 · Structured
6.5 marks
Explain the trend in the thermal stability of Group 2 carbonates down the group, from magnesium carbonate to barium carbonate, in terms of cation size, charge density, and polarization.
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Worked solution
As you go down Group 2 from magnesium to barium: 1. The ionic radius (size) of the Group 2 cation increases. 2. The charge of the cation remains constant at +2, meaning the charge density of the cation decreases down the group. 3. Consequently, the polarizing power of the cation on the carbonate anion decreases down the group. 4. A more polarizing cation (like Mg2+) distorts the electron cloud of the carbonate ion more heavily, weakening the carbon-oxygen double bonds within the carbonate ion and facilitating its decomposition into an oxide and carbon dioxide. 5. Since Ba2+ is much larger with a lower charge density, it causes minimal polarization of the carbonate ion. Therefore, more thermal energy (a higher temperature) is required to decompose the carbonate down the group, meaning thermal stability increases.
Marking scheme
M1: State that cation ionic radius / size increases down the group (1 mark). M2: State that the 2+ charge remains constant, so charge density decreases (1 mark). M3: State that the polarizing power of the cation decreases down the group (1 mark). M4: Describe the polarization of the carbonate ion (its electron cloud is distorted) (1 mark). M5: Explain that this polarization weakens the C-O bond within the carbonate ion (1.5 marks). M6: Conclude that more thermal energy / higher temperature is needed to break down the carbonates down the group (1 mark).
Question 30 · Structured
6.5 marks
Predict and compare the shapes and bond angles of the phosphorus trichloride molecule (PCl3) and the tetrachlorophosphate(V) ion (PCl4+). Additionally, explain why the boiling temperature of phosphorus trichloride (PCl3) is higher than that of nitrogen trichloride (NCl3) despite nitrogen being more electronegative than phosphorus.
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Worked solution
For PCl3: The phosphorus atom has 5 valence electrons and forms 3 single covalent bonds with chlorine atoms, leaving 1 lone pair. According to VSEPR theory, these 4 electron pairs arrange themselves tetrahedrally to minimize repulsion, but the presence of the lone pair results in a trigonal pyramidal shape with a bond angle of approximately 107 degrees. For PCl4+: The phosphorus atom loses one electron to form a P+ species which has 4 valence electrons. It forms 4 bonding pairs with chlorine atoms and has 0 lone pairs. The shape is tetrahedral with a bond angle of 109.5 degrees. Boiling temperature comparison: Both PCl3 and NCl3 are polar molecules with permanent dipole-dipole interactions and London forces. Phosphorus is in Period 3 and is a larger atom with more electrons than nitrogen in Period 2. Therefore, PCl3 has more total electrons than NCl3, leading to stronger, more easily polarized London forces which require more energy to overcome, resulting in a higher boiling temperature.
Marking scheme
M1: Identifies 3 bonding pairs and 1 lone pair for PCl3, giving a trigonal pyramidal shape (1 mark). M2: Identifies a bond angle of approx 107 degrees (accept 100-108) for PCl3 (0.5 marks). M3: Identifies 4 bonding pairs and 0 lone pairs for PCl4+, giving a tetrahedral shape (1 mark). M4: Identifies a bond angle of 109.5 degrees for PCl4+ (0.5 marks). M5: States that both compounds have London forces and permanent dipole-dipole forces (0.5 marks). M6: Explains that PCl3 has more electrons than NCl3 (1 mark). M7: Concludes that the London forces are stronger in PCl3 and require more energy to overcome (2 marks).
Question 31 · Structured
6.5 marks
An oxide of nitrogen is found to contain 30.43% nitrogen and 69.57% oxygen by mass. (a) Determine the empirical formula of this oxide. (b) Under conditions of 101 kPa and 298 K, a 1.15 g sample of this gaseous oxide occupies a volume of 306 cm3. Using the ideal gas equation (PV = nRT), calculate the molar mass of the oxide to 3 significant figures and deduce its molecular formula. [R = 8.31 J K-1 mol-1, molar masses: N = 14.0 g mol-1, O = 16.0 g mol-1]
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Worked solution
(a) Moles of N = 30.43 / 14.0 = 2.174 mol. Moles of O = 69.57 / 16.0 = 4.348 mol. Divide by the smaller number: N = 2.174 / 2.174 = 1, O = 4.348 / 2.174 = 2. Empirical formula is NO2. (b) Convert units: P = 101,000 Pa, V = 306 cm3 = 306 x 10^-6 m3 = 3.06 x 10^-4 m3, T = 298 K. Calculate moles using PV = nRT: n = PV / RT = (101,000 * 3.06 x 10^-4) / (8.31 * 298) = 30.906 / 2476.38 = 0.01248 mol. Calculate molar mass: M = mass / moles = 1.15 / 0.01248 = 92.1 g mol-1. The empirical formula mass of NO2 is 14.0 + 2*(16.0) = 46.0 g mol-1. Since 92.1 / 46.0 approx 2, the molecular formula is (NO2)2 = N2O4.
Marking scheme
M1: Calculates correct mole ratio for N and O (1 mark). M2: Deduces correct empirical formula of NO2 (1 mark). M3: Converts V to m3 and P to Pa (1 mark). M4: Correct substitution of values into PV = nRT to find moles of gas = 0.0125 mol (1 mark). M5: Calculates molar mass of 92.1 g mol-1 (allow 92.0 - 93.0) (1 mark). M6: Deduces molecular formula is N2O4 with showing that molecular mass is double the empirical mass (1.5 marks).
Question 32 · Structured
6.5 marks
When chlorine gas is reacted with hot concentrated sodium hydroxide solution, a disproportionation reaction occurs, forming sodium chloride, sodium chlorate(V), and water. (a) Define the term 'disproportionation'. (b) Write the half-equations for the oxidation and reduction processes, and use them to construct the overall balanced ionic equation for this reaction.
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Worked solution
(a) Disproportionation is a reaction in which an element in a single chemical species is simultaneously oxidized and reduced. (b) Reduction half-equation: Cl2 + 2e- -> 2Cl-. Oxidation half-equation (forming chlorate(V) in alkaline conditions): Cl2 + 12OH- -> 2ClO3- + 6H2O + 10e-. To balance the electrons, multiply the reduction half-equation by 5: 5Cl2 + 10e- -> 10Cl-. Add the two equations: 6Cl2 + 12OH- -> 10Cl- + 2ClO3- + 6H2O. Dividing all coefficients by 2 gives the simplest ionic equation: 3Cl2 + 6OH- -> 5Cl- + ClO3- + 3H2O.
Marking scheme
M1: Defines disproportionation as simultaneous oxidation and reduction of the same element in a single species (1.5 marks). M2: Correct reduction half-equation: Cl2 + 2e- -> 2Cl- (1 mark). M3: Correct oxidation half-equation with correct species and balanced electrons: Cl2 + 12OH- -> 2ClO3- + 6H2O + 10e- (2 marks). M4: Correctly multiplies and combines to get a balanced overall equation: 6Cl2 + 12OH- -> 10Cl- + 2ClO3- + 6H2O (1 mark). M5: Simplifies to 3Cl2 + 6OH- -> 5Cl- + ClO3- + 3H2O (1 mark).
Question 33 · Structured
6.5 marks
A mixture containing 2.00 mol of SO2(g) and 1.00 mol of O2(g) is allowed to reach equilibrium in a sealed container at a constant temperature: 2SO2(g) + O2(g) <=> 2SO3(g). At equilibrium, the total pressure of the system is 3.00 atm and the mixture is found to contain 0.400 mol of SO2. Calculate the equilibrium constant, Kp, for this reaction at this temperature. Include units in your answer.
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Worked solution
Set up an ICE table: Initial moles: SO2 = 2.00, O2 = 1.00, SO3 = 0. Change in moles: Since equilibrium moles of SO2 = 0.400, change in SO2 = -1.60 mol. According to the stoichiometry, change in O2 = -0.800 mol and change in SO3 = +1.60 mol. Equilibrium moles: SO2 = 0.400 mol, O2 = 1.00 - 0.800 = 0.200 mol, SO3 = 1.60 mol. Total moles at equilibrium = 0.400 + 0.200 + 1.60 = 2.20 mol. Calculate partial pressures: p(SO2) = (0.400 / 2.20) * 3.00 atm = 0.5455 atm. p(O2) = (0.200 / 2.20) * 3.00 atm = 0.2727 atm. p(SO3) = (1.60 / 2.20) * 3.00 atm = 2.1818 atm. Write Kp expression: Kp = [p(SO3)]^2 / ([p(SO2)]^2 * p(O2)). Substitute values: Kp = (2.1818)^2 / ((0.5455)^2 * 0.2727) = 4.760 / (0.2976 * 0.2727) = 4.760 / 0.08115 = 58.66 atm^-1. Rounding to 3 significant figures gives Kp = 58.7 atm^-1.
Marking scheme
M1: Calculates equilibrium moles of SO3 = 1.60 mol and O2 = 0.200 mol (1.5 marks). M2: Calculates total equilibrium moles = 2.20 mol (0.5 marks). M3: Calculates correct partial pressures for all three gases (1.5 marks). M4: Writes correct Kp expression (1 mark). M5: Correctly calculates numerical value of 58.7 (accept 58.6 to 59.0 depending on rounding) (1 mark). M6: States correct units of atm^-1 (1 mark).
Paper 2: Advanced Organic and Physical Chemistry
Answer all questions. Show all your working in calculations and include units where appropriate.
34 Question · 102 marks
Question 1 · multiple-choice
1 marks
For a reaction between reactants P and Q, the rate equation is: \(\text{rate} = k[\text{P}]^2[\text{Q}]\). In a series of experiments, the initial concentration of P is doubled, and the initial concentration of Q is halved. What is the overall effect on the initial rate of the reaction?
A.The rate is halved.
B.The rate remains unchanged.
C.The rate is doubled.
D.The rate is quadrupled.
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Worked solution
Initial rate is proportional to \([\text{P}]^2[\text{Q}]\). If \([\text{P}]\) is doubled (factor of 2) and \([\text{Q}]\) is halved (factor of 0.5), the new rate is proportional to \(2^2 \times 0.5 = 4 \times 0.5 = 2\) times the original rate. Therefore, the rate is doubled.
Marking scheme
[1] c - The rate is doubled. Correctly identifying that doubling P increases rate by 4, and halving Q decreases rate by 0.5, giving a net factor of 2.
Question 2 · multiple-choice
1 marks
Which carbonyl compound reacts with hydrogen cyanide, \(\text{HCN}\), in the presence of potassium cyanide, \(\text{KCN}\), to form a product that exists as a racemic mixture?
A.Propanone
B.Methanal
C.Propanal
D.Pentan-3-one
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Worked solution
Nucleophilic addition of \(\text{CN}^-\) to propanal (\(\text{CH}_3\text{CH}_2\text{CHO}\)) forms 2-hydroxybutanenitrile (\(\text{CH}_3\text{CH}_2\text{CH(OH)CN}\)), which has a chiral carbon. Since the reactant carbonyl group is planar, attack from either side is equally likely, leading to a racemic mixture. Propanone, methanal, and pentan-3-one produce symmetrical molecules that are not chiral.
Marking scheme
[1] c - Propanal. Identification of propanal as the only precursor that yields a chiral center upon nucleophilic addition.
Question 3 · multiple-choice
1 marks
In the electrophilic substitution reaction of benzene to form nitrobenzene, what is the role of the concentrated sulfuric acid?
A.To act as a catalyst and protonate nitric acid to generate the \(\text{NO}_2^+\) electrophile.
B.To act as a reducing agent to convert nitric acid into nitrogen dioxide.
C.To act as an oxidising agent to open the delocalised pi system of benzene.
D.To act as a nucleophile that attacks the benzene ring.
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Worked solution
Sulfuric acid is stronger than nitric acid and protonates it: \(\text{HNO}_3 + 2\text{H}_2\text{SO}_4 \rightleftharpoons \text{NO}_2^+ + \text{H}_3\text{O}^+ + 2\text{HSO}_4^-\). It acts as a catalyst by regenerating later in the mechanism when a proton is lost from the intermediate.
Marking scheme
[1] a - Catalyst and proton donor. Correct role of sulfuric acid in generating the nitronium ion.
Question 4 · multiple-choice
1 marks
How many peaks are observed in the carbon-13 (\(^{13}\text{C}\)) NMR spectrum of methylbenzene?
A.4
B.5
C.6
D.7
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Worked solution
Methylbenzene consists of a methyl carbon (1 environment) and a benzene ring. Due to a plane of symmetry along the C1-C4 axis, the ring has 4 carbon environments: C1 (attached to methyl), C2/C6 (ortho), C3/C5 (meta), and C4 (para). Total environments = 1 + 4 = 5 peaks.
Marking scheme
[1] b - 5 peaks. One peak for the methyl carbon and four peaks for the ring carbons.
Question 5 · multiple-choice
1 marks
When propene is reacted with bromine water, \(\text{Br}_2\text{(aq)}\), the major product obtained is 1-bromopropan-2-ol. Which statement correctly explains why this is the major product?
A.Water molecules are stronger electrophiles than bromine molecules.
B.Water is present in much higher concentration than bromide ions and attacks the more stable partial carbocation site on the intermediate.
C.The bromide ion is sterically hindered from attacking the secondary carbocation.
D.The bromonium ion intermediate is highly unstable and decomposes to form propene and oxygen.
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Worked solution
Bromine reacts first to form a cyclic bromonium ion. Since the reaction takes place in aqueous solution, \(\text{H}_2\text{O}\) molecules are in much greater abundance than \(\text{Br}^-\). The secondary carbon holds a larger partial positive charge, attracting the nucleophilic water molecule to form 1-bromopropan-2-ol as the major product.
Marking scheme
[1] b - Correct explanation based on water concentration and regiochemical preference of nucleophilic attack.
Question 6 · multiple-choice
1 marks
Which of the following isomeric organic compounds has the highest boiling point?
A.Butan-1-ol
B.Butanal
C.Butanone
D.Ethoxyethane
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Worked solution
Butan-1-ol has a highly polar O-H bond, enabling the formation of intermolecular hydrogen bonds, which are stronger than the permanent dipole-dipole interactions in butanal and butanone, and the weaker dipole-dipole/London forces in ethoxyethane. More energy is required to overcome these forces, resulting in the highest boiling point.
Marking scheme
[1] a - Butan-1-ol. Identification of hydrogen bonding as the strongest intermolecular force among the choices.
Question 7 · multiple-choice
1 marks
A student reacts 9.20 g of ethanol (\(M_r = 46.0\)) with excess ethanoic acid to produce ethyl ethanoate (\(M_r = 88.0\)) in a condensation reaction. If the percentage yield of the reaction is 60.0%, what mass of ethyl ethanoate is obtained?
A.5.28 g
B.10.56 g
C.17.60 g
D.29.33 g
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Worked solution
Moles of ethanol = \(9.20 / 46.0 = 0.200\text{ mol}\). Theoretical yield of ethyl ethanoate = \(0.200\text{ mol}\). With 60.0% yield, actual moles = \(0.200 \times 0.600 = 0.120\text{ mol}\). Mass of ethyl ethanoate = \(0.120 \times 88.0 = 10.56\text{ g}\).
Marking scheme
[1] b - 10.56 g. Correctly calculating moles of ethanol, applying the 60% yield factor, and multiplying by the molar mass of ethyl ethanoate.
Question 8 · multiple-choice
1 marks
An organic compound X is known to be either propanone, propan-1-ol, or propanoic acid. The infrared spectrum of X shows a broad absorption band at \(3300\text{ cm}^{-1}\) but does not show any absorption at \(1700\text{ cm}^{-1}\). Identify X.
A.Propanone
B.Propan-1-ol
C.Propanoic acid
D.A mixture of propanone and propanoic acid
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Worked solution
The absorption at \(3300\text{ cm}^{-1}\) corresponds to the \(\text{O}-\text{H}\) stretching vibration of an alcohol. The absence of a strong absorption at \(1700\text{ cm}^{-1}\) indicates that no carbonyl group (\(\text{C}=\text{O}\)) is present. Thus, the compound must be propan-1-ol, as propanoic acid would show both bands and propanone would show only the carbonyl band.
Marking scheme
[1] b - Propan-1-ol. Correct interpretation of both the presence of the O-H band and the absence of the C=O band.
Question 9 · Multiple Choice
1 marks
The rate constant, \(k\), of a reaction was determined at several different temperatures. A graph of \(\ln k\) against \(\frac{1}{T}\) (where \(T\) is measured in \(\text{K}\)) was plotted and found to be a straight line with a gradient of \(-9.50 \times 10^3\text{ K}\). What is the activation energy, \(E_a\), of this reaction? [Gas constant \(R = 8.31\text{ J K}^{-1}\text{ mol}^{-1}\)]
A.\(+78.9\text{ kJ mol}^{-1}\)
B.\(-78.9\text{ kJ mol}^{-1}\)
C.\(+1.14\text{ kJ mol}^{-1}\)
D.\(+7.89 \times 10^4\text{ kJ mol}^{-1}\)
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Worked solution
According to the logarithmic form of the Arrhenius equation: \(\ln k = -\frac{E_a}{R}\left(\frac{1}{T}\right) + \ln A\). Therefore, a plot of \(\ln k\) against \(\frac{1}{T}\) yields a straight line with a gradient of \(-\frac{E_a}{R}\). Substituting the given gradient value: \(-9.50 \times 10^3\text{ K} = -\frac{E_a}{8.31\text{ J K}^{-1}\text{ mol}^{-1}}\). Solving for \(E_a\): \(E_a = 9.50 \times 10^3 \times 8.31 = 78945\text{ J mol}^{-1}\). To convert this value to \(\text{kJ mol}^{-1}\), we divide by 1000: \(E_a = \frac{78945}{1000} = +78.9\text{ kJ mol}^{-1}\).
Marking scheme
1 mark: Correct option chosen (A). Reject alternative calculations: B has the incorrect sign (activation energy must be positive), C is calculated using division by R instead of multiplication, and D has incorrect units (not converted to kJ).
Question 10 · Multiple Choice
1 marks
Which of the following describes the reaction that occurs when phenol is treated with excess bromine water at room temperature?
A.Electrophilic substitution occurs to form a white precipitate of 2,4,6-tribromophenol.
B.Electrophilic addition occurs to form a white precipitate of 2,4,6-tribromophenol.
C.Electrophilic substitution occurs to form a pale yellow solution of 4-bromophenol.
D.Nucleophilic substitution occurs to form a white precipitate of 2,4,6-tribromophenol.
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Worked solution
Phenol reacts readily with bromine water at room temperature without the need for a halogen carrier catalyst. This is because the lone pair of electrons on the oxygen atom of the hydroxyl group is partially delocalised into the benzene ring, increasing the electron density and making the ring more susceptible to electrophilic attack. The reaction is an electrophilic substitution reaction, and it produces a white precipitate of 2,4,6-tribromophenol while decolourising the orange bromine water.
Marking scheme
1 mark: Correct option chosen (A). B is incorrect because the mechanism is electrophilic substitution, not addition. C is incorrect because the major product is 2,4,6-tribromophenol, not 4-bromophenol, and it forms a precipitate rather than a solution. D is incorrect because the mechanism is electrophilic, not nucleophilic.
Question 11 · Short Answer
2.5 marks
A reaction has rate constants \(k_1 = 3.40 \times 10^{-4} \text{ s}^{-1}\) at \(298 \text{ K}\) and \(k_2 = 1.25 \times 10^{-2} \text{ s}^{-1}\) at \(348 \text{ K}\). Calculate the activation energy (\(E_a\)) for this reaction in \( \text{kJ mol}^{-1} \). (Gas constant \(R = 8.31 \text{ J K}^{-1} \text{ mol}^{-1}\)). Give your answer to 3 significant figures.
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Worked solution
We use the Arrhenius equation in logarithmic form: \(\ln\left(\frac{k_2}{k_1}\right) = -\frac{E_a}{R} \left(\frac{1}{T_2} - \frac{1}{T_1}\right) = \frac{E_a}{R} \left(\frac{1}{T_1} - \frac{1}{T_2}\right)\)
1. Calculate the LHS of the equation: \(\ln\left(\frac{1.25 \times 10^{-2}}{3.40 \times 10^{-4}}\right) = \ln(36.765) = 3.6045\)
Converting to \(\text{kJ mol}^{-1}\) and rounding to 3 significant figures: \(E_a = 62.1 \text{ kJ mol}^{-1}\) (or \(62.2 \text{ kJ mol}^{-1}\) if using a more precise value for \(R\) or rounding steps differently).
Marking scheme
- 1 mark: Correct calculation of the temperature reciprocal term \(4.82 \times 10^{-4} \text{ K}^{-1}\) and logarithmic ratio term \(3.60\). - 1 mark: Correct calculation of \(E_a\) in \(\text{J mol}^{-1}\) (allow value between \(62100\) and \(62200\)). - 0.5 mark: Correct conversion to \(\text{kJ mol}^{-1}\) and rounding to 3 significant figures: \(62.1\) or \(62.2\).
Question 12 · Short Answer
2.5 marks
Identify the organic product formed when propanoyl chloride reacts with ethylamine. Write the IUPAC name of the organic product and name the mechanism of this reaction.
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Worked solution
When an acyl chloride (propanoyl chloride) reacts with a primary amine (ethylamine), an N-substituted amide is produced along with hydrogen chloride (which then reacts with excess amine to form a salt). The product is \(N\)-ethylpropanamide. The mechanism of this reaction is nucleophilic addition-elimination.
Marking scheme
- 1.5 marks: For the correct IUPAC name of the product: N-ethylpropanamide (accept N-ethyl propanamide, reject ethylpropanamide without N-). - 1 mark: For identifying the mechanism as nucleophilic addition-elimination (reject nucleophilic substitution).
Question 13 · Short Answer
2.5 marks
A Grignard reagent, ethylmagnesium bromide, reacts with a carbonyl compound followed by hydrolysis to form 3-methylpentan-3-ol. Deduce the IUPAC name and the structural formula of this carbonyl compound.
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Worked solution
3-methylpentan-3-ol is a tertiary alcohol with the structure \(\text{CH}_3\text{CH}_2\text{C(OH)(CH}_3)\text{CH}_2\text{CH}_3\). The Grignard reagent, ethylmagnesium bromide, supplies one of the ethyl groups (\(-\text{CH}_2\text{CH}_3\)). Removing this group leaves the precursor carbonyl compound, which must contain 4 carbon atoms with the carbonyl group at carbon-2: \(\text{CH}_3\text{COCH}_2\text{CH}_3\). The IUPAC name is butanone (or butan-2-one).
Marking scheme
- 1.5 marks: Correct IUPAC name: butanone or butan-2-one. - 1 mark: Correct structural formula of the carbonyl compound: \(\text{CH}_3\text{COCH}_2\text{CH}_3\) or equivalent displayed/skeletal formula.
Question 14 · Short Answer
2.5 marks
An organic compound has the molecular formula \(\text{C}_5\text{H}_{10}\text{O}_2\) and exhibits only three peaks in its \(^{13}\text{C}\) NMR spectrum. One of these peaks is in the region \(\delta = 160\text{--}185\text{ ppm}\). Identify the functional group responsible for this peak, and deduce the IUPAC name of the compound.
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Worked solution
The molecular formula \(\text{C}_5\text{H}_{10}\text{O}_2\) represents a carboxylic acid or an ester. The peak in the \(\delta = 160\text{--}185\text{ ppm}\) range confirms the presence of a carbonyl group in an ester or carboxylic acid. Since there are only three carbon environments (peaks) for a five-carbon compound, there must be high symmetry. The only carboxylic acid or ester with \(\text{C}_5\text{H}_{10}\text{O}_2\) that has 3 peaks is 2,2-dimethylpropanoic acid, \((\text{CH}_3)_3\text{CCOOH}\). Its environments are: the three equivalent methyl groups, the quaternary central carbon, and the carboxyl carbon.
Marking scheme
- 1 mark: Identifying the functional group as a carboxylic acid / ester / carboxyl / \(\text{-COOH}\) group. - 1.5 marks: Correct IUPAC name: 2,2-dimethylpropanoic acid (accept correct skeletal/structural formula instead).
Question 15 · Short Answer
2.5 marks
When a single enantiomer of 2-bromobutane reacts with warm aqueous sodium hydroxide, the organic product, butan-2-ol, is formed as a single enantiomer with inversion of configuration. State the mechanism of this reaction and explain why the configuration of the chiral centre is inverted.
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Worked solution
The reaction proceeds via an \(\text{S}_\text{N}2\) mechanism. Because it is a one-step mechanism, the hydroxide nucleophile must attack the chiral carbon atom from the side opposite to the leaving group (the bromine atom) to minimise electrostatic repulsion. This 'backside attack' causes the other three groups attached to the chiral carbon to 'flip' over (inversion of configuration) as the C-OH bond forms and the C-Br bond breaks.
Marking scheme
- 1 mark: Stating that the mechanism is \(\text{S}_\text{N}2\) (accept nucleophilic substitution, bimolecular). - 1.5 marks: Explaining the inversion: the hydroxide nucleophile attacks the carbon from the opposite side to the leaving group / C-Br bond (1 mark) to minimise repulsion / because the bromine atom blocks frontside attack (0.5 mark).
Question 16 · Short Answer
2.5 marks
Ethyl benzoate (\(\text{C}_9\text{H}_{10}\text{O}_2\), molar mass \(= 150.2 \text{ g mol}^{-1}\)) is hydrolysed using excess sodium hydroxide to form benzoic acid (\(\text{C}_7\text{H}_6\text{O}_2\), molar mass \(= 122.1 \text{ g mol}^{-1}\)) after acidification. Starting with \(4.51 \text{ g}\) of ethyl benzoate, a student obtains \(2.75 \text{ g}\) of pure benzoic acid. Calculate the percentage yield of benzoic acid. Give your answer to 3 significant figures.
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- 1 mark: Correct calculation of moles of ethyl benzoate (\(0.0300 \text{ mol}\)) or theoretical mass of benzoic acid (\(3.67 \text{ g}\)). - 1 mark: Correct calculation of percentage yield (\(75.0\%\)). - 0.5 mark: Giving the final answer to 3 significant figures (\(75.0\%\)).
Question 17 · Short Answer
2.5 marks
In the mass spectrum of 2-chloropropane, two major molecular ion peaks are observed at \(m/z = 78\) and \(m/z = 80\). State the isotope of chlorine responsible for each of these peaks, and give the theoretical ratio of the abundance of the \(m/z = 78\) peak to the \(m/z = 80\) peak.
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Worked solution
The molecular formula of 2-chloropropane is \(\text{C}_3\text{H}_7\text{Cl}\). The organic fragment \(\text{C}_3\text{H}_7\) has a mass of 43. - For the peak at \(m/z = 78\): \(78 - 43 = 35\), indicating the presence of the \(^{35}\text{Cl}\) isotope. - For the peak at \(m/z = 80\): \(80 - 43 = 37\), indicating the presence of the \(^{37}\text{Cl}\) isotope. Since the natural abundance ratio of \(^{35}\text{Cl}\) to \(^{37}\text{Cl}\) is approximately 3:1, the peak heights at \(m/z = 78\) and \(m/z = 80\) will also be in a 3:1 ratio.
Marking scheme
- 1.5 marks: Correct identification of the isotopes: \(^{35}\text{Cl}\) (or Cl-35) for the \(m/z = 78\) peak and \(^{37}\text{Cl}\) (or Cl-37) for the \(m/z = 80\) peak. - 1 mark: Correct theoretical ratio of 3:1 (accept 3 to 1; reject 1:3).
Question 18 · Short Answer
2.5 marks
Explain, in terms of intermolecular forces, why propan-1-ol (boiling point \(97^\circ\text{C}\)) has a significantly higher boiling point than propanal (boiling point \(49^\circ\text{C}\)), despite both compounds having similar molar masses.
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Worked solution
Both molecules have similar molar masses, meaning their London forces are similar in strength. However, propan-1-ol contains a highly polar \(\text{O-H}\) bond which allows its molecules to form strong intermolecular hydrogen bonds. Propanal contains a polar \(\text{C=O}\) bond but cannot form hydrogen bonds because it lacks a hydrogen atom bonded to an oxygen; it only forms weaker permanent dipole-dipole forces between its molecules. Hydrogen bonds require significantly more thermal energy to overcome, resulting in a higher boiling point for propan-1-ol.
Marking scheme
- 1 mark: Stating that propan-1-ol molecules form hydrogen bonds (due to the polar O-H group). - 1 mark: Stating that propanal molecules only form permanent dipole-dipole forces / cannot form hydrogen bonds. - 0.5 mark: Explaining that hydrogen bonds are stronger and require more energy to break than permanent dipole-dipole forces.
Question 19 · Short Answer
2.5 marks
The rate constant, \(k\), for a certain organic decomposition reaction was measured at two different temperatures. At \(300\text{ K}\), \(k_1 = 3.20 \times 10^{-4}\text{ s}^{-1}\), and at \(340\text{ K}\), \(k_2 = 1.50 \times 10^{-2}\text{ s}^{-1}\). Calculate the activation energy, \(E_a\), for this reaction in \\text{kJ mol}^{-1}\). The gas constant, \(R = 8.31\text{ J K}^{-1}\text{ mol}^{-1}\).
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Worked solution
We use the Arrhenius equation in the form: \(\ln\left(\frac{k_2}{k_1}\right) = -\frac{E_a}{R} \left(\frac{1}{T_2} - \frac{1}{T_1}\right)\)
To three significant figures, \(E_a = 81.5\text{ kJ mol}^{-1}\).
Marking scheme
- 1.0 mark: Correct calculation of \(\ln(k_2/k_1)\) (3.85) and the reciprocal temperature difference (\(-3.92 \times 10^{-4}\)). - 1.0 mark: Correct calculation of \(E_a\) in \(\text{J mol}^{-1}\) (81500 to 81600). - 0.5 mark: Correct conversion to \(\text{kJ mol}^{-1}\) and quoted to 3 significant figures with units (81.5 \(\text{kJ mol}^{-1}\)).
Question 20 · Short Answer
2.5 marks
Explain why the nucleophilic substitution of optically active 2-bromobutane with aqueous sodium hydroxide proceeds via an \(S_N1\) mechanism to form a racemic mixture, whereas an \(S_N2\) mechanism would produce a single optical isomer.
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Worked solution
In an \(S_N1\) mechanism, the rate-determining step involves the loss of the bromide leaving group to form a carbocation intermediate. The carbocation has a trigonal planar geometry around the positively charged carbon atom. The nucleophile (\(\text{OH}^-\)) has an equal probability of attacking this flat intermediate from either side (above or below the plane). This results in equal amounts of the two enantiomers, forming a racemic mixture. In contrast, the \(S_N2\) mechanism is a single-step process where the nucleophile attacks from the side directly opposite to the leaving group (backside attack), leading to complete inversion of configuration at the chiral centre, thus producing a single optical isomer.
Marking scheme
- 1.0 mark: Stating that the \(S_N1\) intermediate is a planar carbocation (or trigonal planar around the positive carbon). - 1.0 mark: Explaining that the nucleophile can attack with equal probability from either side / above or below the plane, producing equal quantities of both enantiomers (racemic mixture). - 0.5 mark: Explaining that \(S_N2\) involves one-step backside attack resulting in inversion of configuration to yield a single optical isomer.
Question 21 · Short Answer
2.5 marks
State the reagents and reaction conditions required to convert nitrobenzene to phenylamine in a laboratory synthesis.
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Worked solution
To reduce nitrobenzene to phenylamine, the nitrobenzene is first reacted with tin (\(\text{Sn}\)) and concentrated hydrochloric acid (\(\text{HCl}\)) under reflux. This reaction produces the phenylammonium salt. To obtain the free amine, aqueous sodium hydroxide (\(\text{NaOH}\)) is subsequently added to neutralise the acid and liberate the phenylamine.
Marking scheme
- 1.0 mark: Tin (\(\text{Sn}\)) and concentrated hydrochloric acid (\(\text{HCl}\)). (Accept Sn and conc. HCl; reject dilute HCl). - 1.0 mark: Subsequent addition of sodium hydroxide (\(\text{NaOH}\)) / alkali to liberate the amine. - 0.5 mark: Heat under reflux.
Question 22 · Short Answer
2.5 marks
State the number of peaks observed in the \(^{13}\text{C}\) NMR spectrum of isopropyl methanoate, \(\text{HCOOCH(CH}_3)_2\), and explain your answer by identifying the different carbon environments.
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Worked solution
There are 3 peaks in the \(^{13}\text{C}\) NMR spectrum of isopropyl methanoate. This is because there are three distinct carbon environments: 1. The ester carbonyl carbon (\(\text{H}\underline{\text{C}}\text{OO}-\)). 2. The central carbon of the isopropyl group (\(-\underline{\text{C}}\text{H}(\text{CH}_3)_2\)). 3. The two methyl carbons (\(-\text{CH}(\underline{\text{C}}\text{H}_3)_2\)), which are chemically equivalent due to symmetry and thus contribute to a single peak.
Marking scheme
- 1.0 mark: Deducing that there are 3 peaks in the spectrum. - 1.0 mark: Identifying that the two methyl (\(-\text{CH}_3\)) groups are chemically equivalent / in the same environment. - 0.5 mark: Stating that the other two carbons (carbonyl carbon and \(\text{CH}\) carbon) are in distinct environments.
Question 23 · Short Answer
2.5 marks
A student synthesises 1-bromobutane by reacting \(15.0\text{ g}\) of butan-1-ol with an excess of sodium bromide and sulfuric acid. After purification, \(19.5\text{ g}\) of 1-bromobutane is obtained. Calculate the percentage yield of 1-bromobutane. Give your answer to 3 significant figures. [Molar masses: butan-1-ol = \(74.0\text{ g mol}^{-1}\); 1-bromobutane = \(136.9\text{ g mol}^{-1}\)]
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Worked solution
1. Calculate the amount in moles of butan-1-ol used: \(n(\text{butan-1-ol}) = \frac{15.0}{74.0} = 0.2027\text{ mol}\)
2. Determine the theoretical mass of 1-bromobutane (1:1 stoichiometry): \(n(\text{1-bromobutane}) = 0.2027\text{ mol}\) \(\text{Theoretical mass} = 0.2027 \times 136.9 = 27.75\text{ g}\)
Deduce the rate equation and calculate the rate constant, \(k\), including its units.
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Worked solution
1. Find orders of reaction: - From Run 1 and Run 2, doubling \([A]\) (while keeping \([B]\) constant) increases the rate by a factor of \(\frac{8.0 \times 10^{-4}}{2.0 \times 10^{-4}} = 4\). Therefore, the reaction is second order with respect to \(A\). - From Run 2 and Run 3, doubling \([B]\) (while keeping \([A]\) constant) causes no change in the rate. Therefore, the reaction is zero order with respect to \(B\). - Rate equation: \(\text{Rate} = k[A]^2\)
2. Calculate the rate constant, \(k\): Using Run 1 data: \(2.0 \times 10^{-4} = k(0.10)^2\) \(k = \frac{2.0 \times 10^{-4}}{0.010} = 2.0 \times 10^{-2}\text{ dm}^3\text{ mol}^{-1}\text{ s}^{-1}\) (or \(0.020\)).
3. Units: \(\text{units of } k = \frac{\text{mol dm}^{-3}\text{ s}^{-1}}{(\text{mol dm}^{-3})^2} = \text{dm}^3\text{ mol}^{-1}\text{ s}^{-1}\).
Arrange the following compounds in order of increasing boiling point: propan-1-ol, propanal, butane. Explain the trend in boiling points based on the intermolecular forces present.
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Worked solution
1. Order of increasing boiling point: butane < propanal < propan-1-ol
2. Explanation: - Butane is non-polar and experiences only weak London forces between its molecules, requiring the least energy to overcome. - Propanal is a polar molecule containing a carbonyl (\(\text{C=O}\)) group. It experiences permanent dipole-dipole interactions in addition to London forces, which are stronger and require more energy to break. - Propan-1-ol has a polar hydroxyl (\(-\text{OH}\)) group capable of forming intermolecular hydrogen bonds, which are significantly stronger than permanent dipole-dipole interactions, requiring the most energy to break.
Marking scheme
- 1.0 mark: Correct order: butane < propanal < propan-1-ol. - 1.0 mark: Correctly identifying the strongest intermolecular force in each: butane has London forces, propanal has permanent dipole-dipole, propan-1-ol has hydrogen bonding. - 0.5 mark: Explaining that hydrogen bonds are stronger than permanent dipole-dipole interactions (and London forces), thus requiring more thermal energy to overcome.
Question 26 · Short Answer
2.5 marks
An unknown organic compound, \(X\), containing carbon, hydrogen, and oxygen, has a molecular ion peak at \(m/z = 58\) in its mass spectrum. The infrared spectrum of \(X\) shows a strong absorption peak at \(1715\text{ cm}^{-1}\) but no absorption peak between \(3200\text{ and } 3600\text{ cm}^{-1}\). Identify the functional group present, determine the molecular formula of \(X\), and draw or state a possible structural formula for \(X\).
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Worked solution
1. Functional group identification: - The strong absorption at \(1715\text{ cm}^{-1}\) indicates a carbonyl group (\(\text{C=O}\)), characteristic of aldehydes or ketones. - The absence of a broad peak between \(3200\text{ and } 3600\text{ cm}^{-1}\) rules out an alcohol (\(\text{O-H}\)) group.
2. Molecular formula determination: - The molecular ion peak is at \(m/z = 58\), so \(M_r = 58\). - Subtracting the mass of one oxygen atom (16.0) leaves a hydrocarbon backbone mass of \(58 - 16 = 42\). - A hydrocarbon combination with mass 42 is \(\text{C}_3\text{H}_6\) (\(3 \times 12 + 6 = 42\)). - Therefore, the molecular formula of \(X\) is \(\text{C}_3\text{H}_6\text{O}\).
3. Structure: - The possible structures for an acyclic carbonyl compound with formula \(\text{C}_3\text{H}_6\text{O}\) are propanone (\(\text{CH}_3\text{COCH}_3\)) or propanal (\(\text{CH}_3\text{CH}_2\text{CHO}\)).
Marking scheme
- 1.0 mark: Identifying the functional group as a carbonyl (aldehyde or ketone) due to the peak at \(1715\text{ cm}^{-1}\). - 1.0 mark: Deducing the correct molecular formula (\(\text{C}_3\text{H}_6\text{O}\)) using the molecular ion peak. - 0.5 mark: Proposing a correct structural formula: either propanone (\(\text{CH}_3\text{COCH}_3\)) or propanal (\(\text{CH}_3\text{CH}_2\text{CHO}\)).
Question 27 · Structured
6.5 marks
A student investigates the rate of the dehydrochlorination of a chloroalkane at different temperatures. The rate constant, \(k\), was determined at two different temperatures: - At \(350\text{ K}\), \(k_1 = 3.20 \times 10^{-4}\text{ s}^{-1}\) - At \(410\text{ K}\), \(k_2 = 2.45 \times 10^{-2}\text{ s}^{-1}\)
Calculate the activation energy, \(E_a\), for this reaction in \\text{kJ mol}^{-1}\). The gas constant, \(R = 8.31\text{ J K}^{-1}\text{ mol}^{-1}\). Show all your working.
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Worked solution
We use the Arrhenius equation in logarithmic form: \(\ln\left(\frac{k_2}{k_1}\right) = \frac{E_a}{R} \left(\frac{1}{T_1} - \frac{1}{T_2}\right)\)
1. Calculate the left side of the equation: \(\ln\left(\frac{2.45 \times 10^{-2}}{3.20 \times 10^{-4}}\right) = \ln(76.5625) = 4.3381\)
- **1 Mark**: Correct substitution of all experimental values into the Arrhenius equation. - **1 Mark**: Correct calculation of \(\ln(k_2/k_1)\) as 4.34. - **1 Mark**: Correct calculation of \((1/T_1 - 1/T_2)\) as \(4.18 \times 10^{-4}\text{ K}^{-1}\). - **1.5 Marks**: Correct algebraic rearrangement of the Arrhenius equation to isolate \(E_a\). - **1 Mark**: Calculation of \(E_a\) in joules as \(86200 - 86300\text{ J mol}^{-1}\). - **1 Mark**: Conversion to \(86.2\text{ kJ mol}^{-1}\) (allow 3 sig figs: 86.2, reject 86 or 86.22) with correct unit. [Accept answers using \(\ln(k_1/k_2) = -\frac{E_a}{R}(\frac{1}{T_1}-\frac{1}{T_2})\)].
Question 28 · Structured
6.5 marks
The reaction of both butanal and butanone with HCN in the presence of KCN produces mixtures of optical isomers.
Explain, in terms of the structure of the carbonyl group, why both reactions produce a racemic mixture. Compare the relative reactivity of butanal and butanone towards this nucleophilic addition, justifying your answer.
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Worked solution
The carbonyl carbon atom in both butanal and butanone has a planar shape (trigonal planar geometry) with respect to the \(sp^2\) hybridised carbon and its three attached groups. Therefore, the nucleophilic cyanide ion (\(\text{CN}^-\)) can attack the planar carbonyl carbon with equal probability from either above or below the plane. This yields an equimolar (50:50) mixture of the two optical isomers (enantiomers), which is a racemic mixture.
Butanal is more reactive than butanone. This is because: 1. **Steric Hindrance**: Butanal has only one bulky alkyl group (propyl) attached to the carbonyl group, whereas butanone has two alkyl groups (ethyl and methyl). This makes the carbonyl carbon in butanone more crowded and harder for the nucleophile to approach. 2. **Inductive Effect**: Alkyl groups are electron-donating. Butanone has two electron-donating alkyl groups which reduce the partial positive charge (\(\delta^+\)) on the carbonyl carbon more than the single alkyl group in butanal, making it less electrophilic and less susceptible to nucleophilic attack.
Marking scheme
- **1 Mark**: Identifies the carbonyl group / carbonyl carbon as planar. - **1.5 Marks**: Explains that the nucleophile (\(\text{CN}^-\)) can attack from above or below the plane with equal probability, resulting in equal amounts of enantiomers (racemic mixture). - **1 Mark**: Correctly states that butanal is more reactive than butanone. - **1.5 Marks**: Explains the difference in reactivity in terms of steric hindrance (butanal has less hindrance/crowding than butanone). - **1.5 Marks**: Explains the difference in reactivity in terms of the inductive effect (alkyl groups are electron-donating, reducing the \(\delta^+\) charge on the carbonyl carbon in butanone more than in butanal).
Question 29 · Structured
6.5 marks
An unknown organic compound Y has the molecular formula \(\text{C}_4\text{H}_8\text{O}_2\). In its \(^1\text{H}\) NMR spectrum, the following signals are recorded: - Singlet at \(\delta = 2.05\text{ ppm}\) (relative peak area = 3) - Triplet at \(\delta = 1.25\text{ ppm}\) (relative peak area = 3) - Quartet at \(\delta = 4.12\text{ ppm}\) (relative peak area = 2)
In its mass spectrum, a major fragment peak appears at \(m/z = 43\).
Deduce the structural formula of Y. Show how all the NMR signals and the mass spectrum peak support your proposed structure.
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Worked solution
1. **Structure Deduction**: The molecular formula is \(\text{C}_4\text{H}_8\text{O}_2\). The presence of two oxygens suggests an ester or a carboxylic acid. Since there is no singlet representing an \(\text{O-H}\) proton (usually \(\delta = 10-12\text{ ppm}\)), it is likely an ester.
2. **NMR Analysis**: - The triplet at \(\delta = 1.25\text{ ppm}\) (3H) and quartet at \(\delta = 4.12\text{ ppm}\) (2H) show a coupling pattern characteristic of an ethyl group, \(-\text{CH}_2\text{CH}_3\), because \(n+1\) splitting shows the adjacent carbon has 3 and 2 hydrogens respectively. - The chemical shift of the quartet at \(\delta = 4.12\text{ ppm}\) is relatively high, indicating that the \(-\text{CH}_2-\) protons are adjacent to an electronegative oxygen atom, confirming the structural fragment \(-\text{O}-\text{CH}_2\text{CH}_3\). - The singlet at \(\delta = 2.05\text{ ppm}\) (3H) has no adjacent protons, indicating a isolated methyl group, \(\text{CH}_3-\). Its chemical shift is characteristic of protons adjacent to a carbonyl group, \(\text{CH}_3\text{C}=\text{O}\).
Combining these fragments gives ethyl ethanoate: \(\text{CH}_3\text{COOCH}_2\text{CH}_3\).
3. **Mass Spectrometry Analysis**: - The peak at \(m/z = 43\) corresponds to the acylium fragment ion: \([\text{CH}_3\text{CO}]^+\). (\(15 + 12 + 16 = 43\)).
Marking scheme
- **1 Mark**: Correctly identifies Y as ethyl ethanoate (draws or writes structural formula \(\text{CH}_3\text{COOCH}_2\text{CH}_3\)). - **1.5 Marks**: Explains the spin-spin coupling of the triplet at \(\delta = 1.25\text{ ppm}\) and quartet at \(\delta = 4.12\text{ ppm}\) as indicating an ethyl group (\(-\text{CH}_2\text{CH}_3\)). - **1 Mark**: Explains that the chemical shift of the quartet (\(4.12\text{ ppm}\)) shows the \(-\text{CH}_2-\) group is directly bonded to an oxygen atom (\(-\text{O}-\text{CH}_2-\)). - **1 Mark**: Explains that the singlet at \(\delta = 2.05\text{ ppm}\) represents a isolated \(\text{CH}_3\) group adjacent to a carbonyl group (\(\text{CH}_3\text{C}=\text{O}\)). - **2 Marks**: Identifies the mass spec peak at \(m/z = 43\) as the \([\text{CH}_3\text{CO}]^+\) acylium ion fragment. (Must include the positive charge on the formula to gain the final mark; reject neutral fragments).
Question 30 · Structured
6.5 marks
Outline a multi-step synthesis to prepare 4-(dimethylamino)azobenzene starting from phenylamine (aniline). State the reagents, necessary reaction conditions, and the structure of the intermediate compound.
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Worked solution
Step 1: **Diazotisation** - Reagents: Sodium nitrate(III) (\(\text{NaNO}_2\)) and concentrated hydrochloric acid (\(\text{HCl}\)) (to form nitrous acid, \(\text{HNO}_2\) in situ). - Reaction conditions: Keep the temperature below \(10^\circ\text{C}\) (typically between \(0^\circ\text{C}\) and \(5^\circ\text{C}\)). - Intermediate: Benzenediazonium chloride, \(\text{C}_6\text{H}_5\text{N}_2^+\text{Cl}^-\). - Explanation of temperature: If the temperature rises above \(10^\circ\text{C}\), the diazonium salt decomposes rapidly to phenol and nitrogen gas.
- **1 Mark**: Reagents for Step 1: \(\text{NaNO}_2\) and \(\text{HCl}\) (or \(\text{HNO}_2\)). - **1 Mark**: Conditions for Step 1: Temperature between \(0^\circ\text{C}\) and \(10^\circ\text{C}\). - **1 Mark**: Correct structure or formula of benzenediazonium chloride (\(\text{C}_6\text{H}_5\text{N}_2^+\text{Cl}^-\)). - **1 Mark**: Explains why temperature must be low (to prevent decomposition of the diazonium salt to phenol). - **1 Mark**: Reagent for Step 2: N,N-dimethylaniline. - **1.5 Marks**: Reaction conditions for Step 2: Cold/iced, followed by neutralisation or under alkaline/weakly acidic conditions to facilitate coupling.
Question 31 · Structured
6.5 marks
Explain how you would use Infrared (IR) spectroscopy and Mass Spectrometry (MS) to distinguish between three structural isomers with the molecular formula \(\text{C}_3\text{H}_6\text{O}\): propanal, propanone, and prop-2-en-1-ol. Give characteristic absorption values and structural fragments where appropriate.
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Worked solution
1. **IR Spectroscopy identification**: - **Prop-2-en-1-ol**: Shows a broad, strong absorption band at \(3200-3600\text{ cm}^{-1}\) due to the alcohol \(\text{O-H}\) bond stretch. It also shows a sharp absorption at \(1620-1680\text{ cm}^{-1}\) due to the alkene \(\text{C=C}\) bond stretch. It lacks any carbonyl \(\text{C=O}\) stretch. - **Propanal** and **Propanone**: Both show a strong, sharp absorption band at \(1675-1750\text{ cm}^{-1}\) due to the carbonyl \(\text{C=O}\) stretch, and neither has an \(\text{O-H}\) stretch. - To distinguish **propanal** from propanone, look for the aldehyde \(\text{C-H}\) stretch which appears as two weak, sharp bands at \(2700-2900\text{ cm}^{-1}\) in propanal, but is completely absent in propanone.
2. **Mass Spectrometry distinction**: - While all three have the same molecular ion peak at \(m/z = 58\), their fragmentation patterns will differ. - **Propanone** undergoes fragmentation to give a very stable acylium ion peak at \(m/z = 43\) (due to \([\text{CH}_3\text{CO}]^+\)) and a methyl fragment at \(m/z = 15\) (\([\text{CH}_3]^+\)). - **Propanal** will show a prominent peak at \(m/z = 29\) due to the formyl/ethyl ions (\([\text{CHO}]^+\) or \([\text{C}_2\text{H}_5]^+\)).
Marking scheme
- **1.5 Marks**: Identifies prop-2-en-1-ol using IR: presence of broad \(\text{O-H}\) band (\(3200-3600\text{ cm}^{-1}\)) AND \(\text{C=C}\) band (\(1620-1680\text{ cm}^{-1}\)). - **1 Mark**: Identifies carbonyl compounds (propanal and propanone) by presence of \(\text{C=O}\) band (\(1675-1750\text{ cm}^{-1}\)). - **1 Mark**: Distinguishes propanal from propanone using IR: presence of aldehyde \(\text{C-H}\) stretches at \(2700-2900\text{ cm}^{-1}\). - **1.5 Marks**: Uses MS to identify propanal: presence of peak at \(m/z = 29\) representing \([\text{CHO}]^+\) or \([\text{C}_2\text{H}_5]^+\). - **1.5 Marks**: Uses MS to identify propanone: presence of peak at \(m/z = 43\) representing \([\text{CH}_3\text{CO}]^+\).
Question 32 · Structured
6.5 marks
A organic ester contains only carbon, hydrogen, and oxygen. A \(1.50\text{ g}\) sample of this liquid ester was burned completely in excess oxygen.
The combustion products were \(3.00\text{ g}\) of carbon dioxide and \(1.23\text{ g}\) of water.
Calculate the empirical formula of the ester. Show all steps in your working.
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Worked solution
1. **Find mass of Carbon (C) in the sample**: \(\text{Mass of C} = 3.00\text{ g CO}_2 \times \frac{12.0}{44.0} = 0.8182\text{ g}\)
2. **Find mass of Hydrogen (H) in the sample**: \(\text{Mass of H} = 1.23\text{ g H}_2\text{O} \times \frac{2.0}{18.0} = 0.1367\text{ g}\)
3. **Find mass of Oxygen (O) in the sample**: \(\text{Mass of O} = 1.50\text{ g} - (0.8182\text{ g} + 0.1367\text{ g}) = 0.5451\text{ g}\)
4. **Calculate moles of each element**: - \(\text{Moles of C} = \frac{0.8182}{12.0} = 0.06818\text{ mol}\) - \(\text{Moles of H} = \frac{0.1367}{1.0} = 0.1367\text{ mol}\) - \(\text{Moles of O} = \frac{0.5451}{16.0} = 0.03407\text{ mol}\)
5. **Find the simplest whole number molar ratio**: Divide each by the smallest number of moles (0.03407): - \(\text{C} = \frac{0.06818}{0.03407} = 2.00\) - \(\text{H} = \frac{0.1367}{0.03407} = 4.01 \approx 4\) - \(\text{O} = \frac{0.03407}{0.03407} = 1\)
Therefore, the empirical formula is \(\text{C}_2\text{H}_4\text{O}\).
Marking scheme
- **1.5 Marks**: Correctly calculates the masses of C (\(0.818\text{ g}\)) and H (\(0.137\text{ g}\)). [Accept 1.5 marks for both, 1 mark for only one correct]. - **1 Mark**: Correctly subtracts these masses from \(1.50\text{ g}\) to find the mass of O (\(0.545\text{ g}\)). - **1.5 Marks**: Converts masses of C, H, and O into moles correctly (\(\text{C} = 0.0682\), \(\text{H} = 0.137\), \(\text{O} = 0.0341\)). - **1.5 Marks**: Shows the ratio division by \(0.0341\) to get a ratio of approximately \(2:4:1\). - **1 Mark**: States the correct empirical formula as \(\text{C}_2\text{H}_4\text{O}\).
Question 33 · Structured
6.5 marks
A student compared the relative rates of hydrolysis of 1-chlorobutane, 1-bromobutane, and 1-iodobutane by reacting each compound with aqueous silver nitrate in ethanol solvent at \(50^\circ\text{C}\).
State the relative order of rates of hydrolysis of these three halogenoalkanes. Explain this order, writing relevant ionic equations, and state whether bond enthalpy or bond polarity is the deciding factor in determining the rate of reaction. Explain the role of the ethanol.
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Worked solution
The relative order of the rates of hydrolysis (from fastest to slowest) is: 1-iodobutane > 1-bromobutane > 1-chlorobutane.
**Explanation**: - The hydrolysis reaction involves breaking the carbon-halogen (\(\text{C-X}\)) bond. - The carbon-halogen bond enthalpies decrease down Group 7: \(\text{C-Cl} > \text{C-Br} > \text{C-I}\). - The \(\text{C-I}\) bond has the lowest bond enthalpy (is the weakest bond) and thus requires the least energy to break, resulting in the fastest rate of hydrolysis. - Even though the \(\text{C-Cl}\) bond is the most polar due to chlorine's higher electronegativity, **bond enthalpy** is the dominant factor determining the reaction rate here, not bond polarity.
**Role of Ethanol**: - Halogenoalkanes are insoluble in water, but soluble in organic solvents. Silver nitrate is soluble in water. Ethanol acts as a **mutual solvent** (cosolvent), allowing the halogenoalkane and the aqueous silver nitrate reactant to mix into a single phase so they can react.
- **1 Mark**: Correctly states the rate order: 1-iodobutane is fastest, 1-chlorobutane is slowest. - **1 Mark**: Explains that ethanol acts as a mutual solvent to allow immiscible reactants (halogenoalkane and aqueous silver nitrate) to mix. - **1.5 Marks**: Writes both correct equations: the hydrolysis reaction equation (producing \(\text{X}^-\)) AND the ionic equation for precipitate formation (\(\text{Ag}^+ + \text{X}^- \rightarrow \text{AgX}\)). - **1 Mark**: Explicitly states that bond enthalpy is the dominant/deciding factor (not bond polarity). - **2 Marks**: Explains that down Group 7, the \(\text{C-X}\) bond strength/enthalpy decreases (or \(\text{C-I}\) is the weakest bond), so the \(\text{C-I}\) bond is broken most easily.
Question 34 · Structured
6.5 marks
Ethanol (\(\text{CH}_3\text{CH}_2\text{OH}\)) has a boiling point of \(78^\circ\text{C}\), whereas its structural isomer dimethyl ether (\(\text{CH}_3\text{OCH}_3\)) has a boiling point of \(-24^\circ\text{C}\).
Explain this difference in boiling points in terms of intermolecular forces. In addition, discuss the solubility of both isomers in water, explaining how each is able to interact with water molecules.
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Worked solution
1. **Boiling Point Difference**: - Ethanol contains a highly polar hydrogen-oxygen (\(\text{O-H}\)) bond, which enables strong **hydrogen bonding** to occur between its molecules. - Dimethyl ether is polar but does not contain a hydrogen atom directly bonded to an oxygen (no \(\text{O-H}\) bond). Therefore, the intermolecular forces between dimethyl ether molecules are limited to weaker **permanent dipole-dipole forces** and **London (dispersion) forces**. - Because hydrogen bonds are significantly stronger than dipole-dipole and London forces, much more energy is required to separate ethanol molecules, leading to its much higher boiling point.
2. **Solubility in Water**: - Both ethanol and dimethyl ether are highly soluble in water. - Ethanol can form **hydrogen bonds** with water molecules using both its polar \(\text{O-H}\) group (acting as a hydrogen donor and acceptor). - Dimethyl ether can also form **hydrogen bonds** with water. Although it has no \(\text{O-H}\) bond itself to act as a hydrogen donor, the oxygen atom in dimethyl ether has lone pairs of electrons that can accept hydrogen bonds from the highly polar \(\delta^+\) hydrogen atoms of water molecules.
Marking scheme
- **1 Mark**: Identifies that ethanol molecules form hydrogen bonds between themselves. - **1 Mark**: Identifies that dimethyl ether molecules form permanent dipole-dipole forces and London forces (but NO hydrogen bonds between themselves). - **1 Mark**: Explains that hydrogen bonds are stronger than dipole-dipole/London forces, hence more energy is needed to boil ethanol. - **1 Mark**: States that both ethanol and dimethyl ether are soluble in water. - **1.25 Marks**: Explains ethanol's solubility via hydrogen bonding between its \(\text{O-H}\) group and water. - **1.25 Marks**: Explains dimethyl ether's solubility via hydrogen bonding between the lone pairs on its ether oxygen and the hydrogen of water.
Paper 3: General and Practical Principles in Chemistry
Answer all questions. Write your answers in the spaces provided. Show all working.
25 Question · 120 marks
Question 1 · Multiple Choice
2 marks
A student heats a sample of hydrated calcium sulfate, \(\text{CaSO}_4 \cdot x\text{H}_2\text{O}\), in a crucible to determine the value of \(x\).
Experimental data: - Mass of empty crucible = \(18.20\text{ g}\) - Mass of crucible + hydrated salt = \(21.64\text{ g}\) - Mass of crucible + anhydrous salt after heating to constant mass = \(20.92\text{ g}\)
What is the value of \(x\) based on this experimental data? (Molar masses: \(\text{CaSO}_4 = 136.1\text{ g mol}^{-1}\), \(\text{H}_2\text{O} = 18.0\text{ g mol}^{-1}\))
A.0.5
B.1
C.2
D.5
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Worked solution
1. Calculate the mass of the anhydrous salt: \(20.92\text{ g} - 18.20\text{ g} = 2.72\text{ g}\).
2. Calculate the mass of water lost: \(21.64\text{ g} - 20.92\text{ g} = 0.72\text{ g}\).
- 1 mark for calculating the moles of anhydrous salt (0.0200 mol) and water (0.0400 mol). - 1 mark for calculating the correct value of x (2).
Question 2 · Multiple Choice
2 marks
During the purification of crude cyclohexene prepared from cyclohexanol, the crude liquid is shaken with saturated sodium chloride solution (brine) in a separating funnel. What is the primary purpose of this step?
A.To protonate unreacted cyclohexanol to make it more soluble in the aqueous phase.
B.To catalyse the dehydration of any remaining cyclohexanol.
C.To decrease the solubility of cyclohexene in the aqueous layer, helping the organic layer to separate more cleanly.
D.To dry the cyclohexene by removing all traces of dissolved water.
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Worked solution
Shaking the organic product with saturated sodium chloride solution (brine) is a 'salting out' technique. It decreases the solubility of the organic product (cyclohexene) in the aqueous phase, encouraging a sharper phase boundary and preventing emulsion formation, thereby maximising the recovery of the organic layer.
Marking scheme
- 1 mark for identifying that brine decreases the solubility of the organic product in water. - 1 mark for identifying that this improves separation/layer boundary clarity.
Question 3 · Multiple Choice
2 marks
A student investigates the kinetics of a clock reaction at different temperatures. A plot of \(\ln(\text{rate})\) against \(\frac{1}{T}\) (where temperature \(T\) is in Kelvin) yields a straight line with a gradient of \(-6120\text{ K}\).
What is the activation energy, \(E_a\), for this reaction? \((R = 8.31\text{ J K}^{-1}\text{ mol}^{-1})\)
A.+50.9 kJ mol^-1
B.-50.9 kJ mol^-1
C.+0.736 kJ mol^-1
D.+736 kJ mol^-1
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Worked solution
Using the logarithmic form of the Arrhenius equation: \(\ln(k) = -\frac{E_a}{R}\left(\frac{1}{T}\right) + \ln(A)\)
Since rate is proportional to the rate constant \(k\), the gradient of a plot of \(\ln(\text{rate})\) versus \(\frac{1}{T}\) is equal to \(-\frac{E_a}{R}\).
- 1 mark for correctly setting up the Arrhenius gradient relationship: \(E_a = -\text{gradient} \times R\). - 1 mark for correct calculation to 3 significant figures and standard unit conversion (\(+50.9\text{ kJ mol}^{-1}\)).
Question 4 · Multiple Choice
2 marks
A student wants to measure the standard electrode potential of the \(\text{Fe}^{3+}(\text{aq}) / \text{Fe}^{2+}(\text{aq})\) half-cell. Which of the following describes the correct experimental setup for this half-cell?
A.An iron metal electrode immersed in a solution containing both \(\text{Fe}^{3+}(\text{aq})\) and \(\text{Fe}^{2+}(\text{aq})\) ions.
B.A platinum electrode immersed in a solution containing both \(\text{Fe}^{3+}(\text{aq})\) and \(\text{Fe}^{2+}(\text{aq})\) ions, each at a concentration of \(1.0\text{ mol dm}^{-3}\).
C.A platinum electrode immersed in a solution containing \(1.0\text{ mol dm}^{-3}\text{ Fe}^{3+}(\text{aq})\) only, connected to an iron half-cell.
D.An iron metal electrode immersed in a solution containing \(1.0\text{ mol dm}^{-3}\text{ Fe}^{2+}(\text{aq})\) only, at \(0\text{ }^\circ\text{C}\).
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Worked solution
For a half-cell containing two aqueous ions of the same element in different oxidation states, an inert platinum electrode is used to allow electron transfer. Under standard conditions, both ions must have a concentration of \(1.0\text{ mol dm}^{-3}\) at \(298\text{ K}\).
Marking scheme
- 1 mark for identifying the use of a platinum electrode. - 1 mark for specifying standard concentration conditions (\(1.0\text{ mol dm}^{-3}\) for both ions).
Question 5 · Multiple Choice
2 marks
In an experiment to determine the molar volume of a gas, a student reacts a known mass of magnesium ribbon with excess hydrochloric acid and collects the hydrogen gas in a gas syringe.
If a small amount of hydrogen gas escapes from the delivery tube before the stopper is fully inserted, how will this affect the calculated molar volume of the gas?
A.The calculated molar volume will be higher because the volume reading is lower.
B.The calculated molar volume will be lower because the volume reading is lower.
C.The calculated molar volume will be unchanged because the moles of magnesium used remain constant.
D.The calculated molar volume will be higher because the rate of reaction is unaffected.
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Worked solution
Molar volume of a gas is calculated using the formula: \(V_m = \frac{\text{Volume of gas collected}}{\text{Moles of gas produced}}\)
If hydrogen gas escapes, the measured volume of gas collected will be smaller than it should be. Since the mass of magnesium (and thus the theoretical moles of gas produced) remains unchanged, the calculated molar volume will be lower than the true value.
Marking scheme
- 1 mark for identifying that the measured volume of gas decreases. - 1 mark for explaining that this leads to a lower calculated molar volume.
Question 6 · Multiple Choice
2 marks
An unknown transition metal compound \(\text{Y}\) is dissolved in water to form a pale blue solution. When aqueous ammonia is added dropwise, a pale blue precipitate is formed. Upon adding excess aqueous ammonia, this precipitate dissolves to form a deep blue solution.
Identify the formula of the complex ion present in the final deep blue solution.
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Worked solution
The pale blue solution contains hydrated copper(II) ions, \([\text{Cu}(\text{H}_2\text{O})_6]^{2+}\). When dropwise ammonia is added, hydroxide ions from the partial ionisation of ammonia precipitate copper(II) hydroxide, \(\text{Cu}(\text{OH})_2(\text{H}_2\text{O})_4\) (a pale blue precipitate). With excess ammonia, ligand exchange occurs to form the tetraamminediaphoracopper(II) complex, \([\text{Cu}(\text{NH}_3)_4(\text{H}_2\text{O})_2]^{2+}\), which is a deep blue solution.
Marking scheme
- 1 mark for identifying the transition metal as copper(II). - 1 mark for identifying the correct complex formula with four ammonia and two water ligands.
Question 7 · Multiple Choice
2 marks
In a calorimetry experiment, \(50.0\text{ cm}^3\) of \(1.00\text{ mol dm}^{-3}\text{ HCl}\) is mixed with \(50.0\text{ cm}^3\) of \(1.00\text{ mol dm}^{-3}\text{ NaOH}\) in a polystyrene cup. The temperature of the mixture rises by \(6.5\text{ }^\circ\text{C}\).
What is the enthalpy change of neutralisation, \(\Delta H_{\text{neut}}\), in \(\text{kJ mol}^{-1}\)? (Assume the density of the solution is \(1.00\text{ g cm}^{-3}\) and its specific heat capacity is \(4.18\text{ J g}^{-1}\text{ K}^{-1}\).)
A.-54.3 kJ mol^-1
B.-108.7 kJ mol^-1
C.+54.3 kJ mol^-1
D.-27.2 kJ mol^-1
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Worked solution
1. Calculate energy released (\(q\)): \(q = m c \Delta T\) \(m = 50.0\text{ cm}^3 + 50.0\text{ cm}^3 = 100.0\text{ g}\) \(q = 100.0\text{ g} \times 4.18\text{ J g}^{-1}\text{ K}^{-1} \times 6.5\text{ K} = 2717\text{ J} = 2.717\text{ kJ}\)
2. Calculate moles of reaction (water formed): \(n(\text{H}_2\text{O}) = c \times V = 1.00\text{ mol dm}^{-3} \times 0.0500\text{ dm}^3 = 0.0500\text{ mol}\)
- 1 mark for correctly calculating the heat energy released (\(2.717\text{ kJ}\)). - 1 mark for calculating the molar enthalpy change with a negative sign (\(-54.3\text{ kJ mol}^{-1}\)).
Question 8 · Multiple Choice
2 marks
A student tests an aqueous solution containing halide ions. On adding dilute nitric acid followed by silver nitrate solution, a cream precipitate is formed.
Which reagent should the student add next, and what observation would confirm the identity of the halide ion?
A.Dilute aqueous ammonia; the cream precipitate dissolves.
B.Concentrated aqueous ammonia; the cream precipitate remains insoluble.
C.Concentrated aqueous ammonia; the cream precipitate dissolves.
D.Dilute aqueous ammonia; the cream precipitate turns yellow.
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Worked solution
A cream precipitate with silver nitrate indicates the presence of bromide ions (\(\text{AgBr}\)). Silver bromide is insoluble in dilute aqueous ammonia but dissolves in concentrated aqueous ammonia. This distinguishes it from silver chloride (which dissolves in dilute ammonia) and silver iodide (which is insoluble in concentrated ammonia).
Marking scheme
- 1 mark for identifying bromide as the halide that forms a cream precipitate. - 1 mark for identifying that \(\text{AgBr}\) dissolves in concentrated ammonia but not dilute ammonia.
Question 9 · multiple-choice
2 marks
A student carries out a titration using a Class B burette which has an uncertainty of \(\pm 0.05 \text{ cm}^3\) per reading. The initial burette reading is \(0.15 \text{ cm}^3\) and the final reading is \(24.35 \text{ cm}^3\). What is the percentage uncertainty in the calculated titre volume?
A.\(0.21\%\)
B.\(0.41\%\)
C.\(0.45\%\)
D.\(0.83\%\)
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Worked solution
The uncertainty of a single burette reading is \(\pm 0.05 \text{ cm}^3\). A titration requires two readings (initial and final), so the total absolute uncertainty is \(2 \times 0.05 = 0.10 \text{ cm}^3\). The titre volume is \(24.35 - 0.15 = 24.20 \text{ cm}^3\). The percentage uncertainty is calculated as: \((0.10 / 24.20) \times 100 = 0.413\%\), which rounds to \(0.41\%\).
Marking scheme
[1 mark] for calculating the correct titre volume of 24.20 cm3 and doubling the uncertainty to 0.10 cm3. [1 mark] for calculating the final percentage uncertainty of 0.41%.
Question 10 · multiple-choice
2 marks
During the preparation and purification of a liquid organic product, cyclohexene, a student uses anhydrous calcium chloride to dry the organic layer. Which of the following is the correct observation when the organic liquid is fully dry, and what is the correct subsequent step?
A.The liquid goes from cloudy to clear; then decant or filter the liquid.
B.The drying agent clumps together; then carry out fractional distillation.
C.The liquid goes from clear to cloudy; then wash with sodium hydrogencarbonate.
D.The drying agent dissolves completely; then carry out simple distillation.
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Worked solution
When an organic liquid containing suspended water droplets is dried using an anhydrous salt, the cloudy emulsion becomes clear. Once dried, the solid drying agent is removed from the product by either filtration or decantation before any subsequent distillation or analysis.
Marking scheme
[1 mark] for identifying that the liquid changes from cloudy to clear when dry. [1 mark] for identifying that the liquid is separated from the solid drying agent by decantation or filtration.
Question 11 · multiple-choice
2 marks
In an iodine clock reaction used to study chemical kinetics, which of the following describes the role of sodium thiosulfate (\(\text{Na}_2\text{S}_2\text{O}_3\)) and starch indicator?
A.Thiosulfate reacts instantly with starch; starch acts as a catalyst to speed up iodine production.
B.Thiosulfate limits the concentration of iodine by converting it back to iodide; starch signals when all thiosulfate has been consumed.
C.Thiosulfate acts as a catalyst to generate iodide ions; starch turns blue-black in the presence of iodide.
D.Thiosulfate oxidizes starch to form a blue-black complex; starch acts as a pH buffer.
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Worked solution
In an iodine clock reaction, a small, fixed amount of sodium thiosulfate is added. It immediately reduces any iodine (\(\text{I}_2\)) produced back to iodide (\(\text{I}^-\)). Once all the thiosulfate is consumed, the iodine concentration increases rapidly and reacts with the starch indicator to produce a blue-black starch-iodine complex.
Marking scheme
[1 mark] for explaining that thiosulfate reacts with iodine to keep its concentration at zero initially. [1 mark] for explaining that starch indicates the endpoint once all thiosulfate is consumed.
Question 12 · multiple-choice
2 marks
A student heated a \(2.50\text{ g}\) sample of an anhydrous Group 2 metal carbonate, \(\text{MCO}_3\), until constant mass to determine the identity of \(\text{M}\). The final mass of the Group 2 metal oxide (\(\text{MO}\)) obtained was \(1.40\text{ g}\). Identify the metal \(\text{M}\). (Atomic masses: \(\text{C} = 12.0\), \(\text{O} = 16.0\), \(\text{Mg} = 24.3\), \(\text{Ca} = 40.1\), \(\text{Sr} = 87.6\), \(\text{Ba} = 137.3\))
A.Magnesium (\(\text{Mg}\))
B.Calcium (\(\text{Ca}\))
C.Strontium (\(\text{Sr}\))
D.Barium (\(\text{Ba}\))
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Worked solution
The thermal decomposition equation is: \(\text{MCO}_3 \rightarrow \text{MO} + \text{CO}_2\). The mass loss is due to carbon dioxide: \(2.50 - 1.40 = 1.10\text{ g}\). Moles of \(\text{CO}_2 = 1.10 / 44.0 = 0.0250\text{ mol}\). Since the stoichiometry is 1:1, moles of \(\text{MCO}_3 = 0.0250\text{ mol}\). Molar mass of \(\text{MCO}_3 = 2.50 / 0.0250 = 100\text{ g mol}^{-1}\). \(\text{Ar}\) of \(\text{M} = 100 - (12.0 + 3 \times 16.0) = 40.0\text{ g mol}^{-1}\). This corresponds to Calcium.
Marking scheme
[1 mark] for determining the moles of carbon dioxide lost as 0.0250 mol. [1 mark] for calculating the molar mass of the carbonate (100 g/mol) and identifying the metal as calcium.
Question 13 · multiple-choice
2 marks
A student uses a colorimeter to determine the concentration of copper(II) ions in an aqueous solution. Which filter color should be selected for the colorimeter, and what is the chemical justification?
A.Blue filter, because copper(II) solutions appear blue and therefore absorb blue light most strongly.
B.Red or orange filter, because copper(II) solutions appear blue and absorb light of the complementary color.
C.Green filter, because green light is transmitted through the blue-green solution.
D.Violet filter, because copper(II) ions have d-d transitions that specifically absorb high-energy violet light.
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Worked solution
A colorimeter should use a filter that represents the complementary color of the solution being analyzed to ensure maximum absorbance of light. Since aqueous copper(II) ions form a blue solution, they absorb light in the orange/red region of the spectrum, meaning a red or orange filter should be chosen.
Marking scheme
[1 mark] for identifying red/orange as the correct filter color. [1 mark] for justifying that the complementary color to blue is absorbed most strongly.
Question 14 · multiple-choice
2 marks
A student runs a thin-layer chromatography (TLC) experiment to separate a mixture of amino acids. The solvent front travels \(8.0\text{ cm}\) and one of the amino acid spots (X) travels \(3.6\text{ cm\). Which of the following shows the correct \(R_f\) value for spot X and the correct method for locating the spots?
A.\(R_f = 0.45\); locate the spots by spraying with ninhydrin reagent.
B.\(R_f = 2.22\); locate the spots by spraying with ninhydrin reagent.
C.\(R_f = 0.45\); locate the spots by spraying with Tollens' reagent.
D.\(R_f = 2.22\); locate the spots by placing the plate under a UV lamp.
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Worked solution
The \(R_f\) value is calculated by dividing the distance traveled by the spot by the distance traveled by the solvent front: \(R_f = 3.6 / 8.0 = 0.45\). Since amino acids are colorless, they must be visualized using a locating agent. Spraying with ninhydrin reagent and heating develops a purple/brown color for the spots.
Marking scheme
[1 mark] for calculating the Rf value as 0.45. [1 mark] for identifying ninhydrin as the appropriate locating agent for amino acids.
Question 15 · multiple-choice
2 marks
A bomb calorimeter is calibrated by burning \(0.500\text{ g}\) of benzoic acid (molar mass = \(122.1\text{ g mol}^{-1}\); enthalpy of combustion \(\Delta_c H = -3227\text{ kJ mol}^{-1}\)). The temperature of the calorimeter increases by \(2.45\text{ K}\). What is the heat capacity of the calorimeter in \(\text{kJ K}^{-1}\)?
A.2.20
B.5.39
C.13.2
D.26.4
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Worked solution
First, calculate the moles of benzoic acid: \(0.500 / 122.1 = 0.004095\text{ mol}\). The heat released is \(q = 0.004095\text{ mol} \times 3227\text{ kJ mol}^{-1} = 13.214\text{ kJ}\). The heat capacity of the calorimeter is \(C = q / \Delta T = 13.214 / 2.45 = 5.39\text{ kJ K}^{-1}\).
Marking scheme
[1 mark] for calculating the heat released (13.2 kJ) from the combustion of benzoic acid. [1 mark] for dividing the heat by the temperature change to find the heat capacity (5.39 kJ/K).
Question 16 · structured
9 marks
A student weighs out \(1.624\text{ g}\) of hydrated copper(II) malonate, \(\text{Cu}(C_3H_2O_4) \cdot xH_2O\). The salt is dissolved in deionised water and made up to \(250.0\text{ cm}^3\) in a volumetric flask. A \(25.0\text{ cm}^3\) aliquot of this solution is transferred to a conical flask, and an excess of potassium iodide, \(\text{KI}\), is added. The liberated iodine is titrated with \(0.0400\text{ mol dm}^{-3}\) sodium thiosulfate solution, \(\text{Na}_2\text{S}_2\text{O}_3\), using starch indicator. The mean titre is \(18.50\text{ cm}^3\). The equations for the reactions are: \(2\text{Cu}^{2+}(aq) + 4\text{I}^{-}(aq) \rightarrow 2\text{CuI}(s) + \text{I}_2(aq)\) and \(\text{I}_2(aq) + 2\text{S}_2\text{O}_3^{2-}(aq) \rightarrow 2\text{I}^{-}(aq) + \text{S}_4\text{O}_6^{2-}(aq)\). (a) Describe the colour change at the end-point of this titration when starch is used. (b) Explain why starch is only added near the end-point, rather than at the start. (c) Calculate the moles of \(\text{Cu}^{2+}\) in the \(25.0\text{ cm}^3\) aliquot and hence the concentration of \(\text{Cu}^{2+}\) in the original volumetric flask. (d) Calculate the value of \(x\) in \(\text{Cu}(C_3H_2O_4) \cdot xH_2O\) to the nearest whole number. [Molar masses: \(\text{Cu} = 63.5\), \(\text{C} = 12.0\), \(\text{H} = 1.0\), \(\text{O} = 16.0\)]
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Worked solution
(a) The colour change at the end-point is from blue-black to off-white / light brown precipitate (due to the presence of solid \(\text{CuI}\)). (b) Starch is added near the end-point because if added at the start, the concentration of iodine is high, and the starch-iodine complex forms irreversibly, preventing a sharp end-point. (c) Moles of \(\text{S}_2\text{O}_3^{2-}\) in titration = \(0.0400 \times \frac{18.50}{1000} = 7.40 \times 10^{-4}\text{ mol}\). Since \(1\text{ mol } \text{Cu}^{2+} \equiv 1\text{ mol } \text{S}_2\text{O}_3^{2-}\), moles of \(\text{Cu}^{2+}\) in the \(25.0\text{ cm}^3\) aliquot = \(7.40 \times 10^{-4}\text{ mol}\). Concentration of \(\text{Cu}^{2+}\) in volumetric flask = \(\frac{7.40 \times 10^{-4}}{0.0250} = 0.0296\text{ mol dm}^{-3}\). (d) Total moles in \(250.0\text{ cm}^3\) = \(7.40 \times 10^{-3}\text{ mol}\). Molar mass of hydrated salt = \(\frac{1.624\text{ g}}{7.40 \times 10^{-3}\text{ mol}} = 219.46\text{ g mol}^{-1}\). Molar mass of anhydrous \(\text{Cu}(C_3H_2O_4)\) = \(63.5 + 3(12.0) + 2(1.0) + 4(16.0) = 165.5\text{ g mol}^{-1}\). Mass of water of crystallisation per mole = \(219.46 - 165.5 = 53.96\text{ g mol}^{-1}\). Value of \(x = \frac{53.96}{18.0} = 2.998 \approx 3\).
Marking scheme
(a) Blue-black / dark blue to off-white precipitate (1 mark). (b) Starch forms an insoluble/irreversible complex with high concentrations of iodine (1 mark); adding it late ensures all iodine can be titrated, producing a sharp end-point (1 mark). (c) Calculates moles of thiosulfate correctly: \(7.40 \times 10^{-4}\text{ mol}\) (1 mark); states moles of copper is the same and calculates concentration as \(0.0296\text{ mol dm}^{-3}\) (1 mark). (d) Moles of salt in \(250\text{ cm}^3\) is \(7.40 \times 10^{-3}\text{ mol}\) (1 mark); calculates experimental molar mass as \(219.5\text{ g mol}^{-1}\) (1 mark); calculates anhydrous molar mass as \(165.5\text{ g mol}^{-1}\) (1 mark); calculates \(x = 3\) (1 mark).
Question 17 · structured
9 marks
A student synthesises cyclohexene by the dehydration of cyclohexanol using concentrated phosphoric acid. The procedure is: 1. Place \(10.0\text{ cm}^3\) of cyclohexanol (density = \(0.962\text{ g cm}^{-3}\)) and \(4\text{ cm}^3\) of concentrated phosphoric(V) acid into a round-bottomed flask. Add anti-bumping granules. 2. Heat the mixture and distil off the liquid boiling below \(100^\circ\text{C}\). 3. Transfer the distillate to a separating funnel, wash with saturated sodium chloride solution, and discard the aqueous layer. 4. Transfer the organic layer to a conical flask, add an anhydrous inorganic salt to dry it, then decant the liquid. 5. Distil the dry liquid and collect the fraction boiling between \(81\text{--}83^\circ\text{C}\) to obtain \(4.12\text{ g}\) of pure cyclohexene. (a) Explain the purpose of the anti-bumping granules. (b) Explain why sodium hydrogencarbonate solution is preferred over sodium chloride solution by some students if traces of acid carried over, and write an ionic equation for this reaction. (c) State the name of a suitable anhydrous inorganic salt that could be used as a drying agent in step 4, and describe how the student would know the cyclohexene is dry. (d) Calculate the percentage yield of cyclohexene achieved in this synthesis. [Molar masses: Cyclohexanol, \(C_6H_{11}OH = 100.0\text{ g mol}^{-1}\); Cyclohexene, \(C_6H_{10} = 82.0\text{ g mol}^{-1}\)]
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Worked solution
(a) Anti-bumping granules provide nucleation sites to promote smooth and even boiling, preventing large bubbles and violent bumping. (b) Sodium hydrogencarbonate is basic and neutralises any remaining phosphoric/acidic impurities carried over in the distillation. Equation: \(\text{HCO}_3^-(aq) + \text{H}^+(aq) \rightarrow \text{CO}_2(g) + \text{H}_2\text{O}(l)\). (c) Suitable drying agent: Anhydrous calcium chloride, \(\text{CaCl}_2\), or anhydrous magnesium sulfate, \(\text{MgSO}_4\). The student would know the liquid is dry when the cloudy suspension becomes completely clear / translucent. (d) Mass of cyclohexanol = \(10.0\text{ cm}^3 \times 0.962\text{ g cm}^{-3} = 9.62\text{ g}\). Moles of cyclohexanol = \(\frac{9.62}{100.0} = 0.0962\text{ mol}\). Theoretical mass of cyclohexene = \(0.0962\text{ mol} \times 82.0\text{ g mol}^{-1} = 7.8884\text{ g}\). Percentage yield = \(\frac{4.12}{7.8884} \times 100 = 52.23\% \approx 52.2\%\).
Marking scheme
(a) Promotes smooth boiling/prevents violent bumping by providing nucleation sites (1 mark). (b) Neutralises acidic impurities (1 mark); correct ionic equation: \(\text{HCO}_3^- + \text{H}^+ \rightarrow \text{H}_2\text{O} + \text{CO}_2\) (1 mark). (c) Correct drying agent e.g. anhydrous \(\text{MgSO}_4\) / \(\text{CaCl}_2\) / \(\text{Na}_2\text{SO}_4\) (1 mark); the liquid changes from cloudy to clear (1 mark). (d) Calculates mass of cyclohexanol (\(9.62\text{ g}\)) (1 mark); calculates moles of cyclohexanol (\(0.0962\text{ mol}\)) (1 mark); calculates theoretical mass of cyclohexene (\(7.89\text{ g}\)) (1 mark); calculates yield as \(52.2\%\) (accept \(52\%\)) (1 mark).
Question 18 · structured
9 marks
A student wants to prepare a buffer solution with a pH of 4.85. They have access to \(0.200\text{ mol dm}^{-3}\) propanoic acid (\(CH_3CH_2COOH\), \(K_a = 1.35 \times 10^{-5}\text{ mol dm}^{-3}\)) and solid sodium propanoate (\(CH_3CH_2COONa\), \(M_r = 96.0\)). (a) Define the term 'buffer solution'. (b) Calculate the mass of sodium propanoate that must be dissolved in \(500\text{ cm}^3\) of the \(0.200\text{ mol dm}^{-3}\) propanoic acid to obtain a buffer solution of pH 4.85. Assume that the addition of solid sodium propanoate does not change the volume of the solution. (c) Write an ionic equation to show how this buffer solution minimises pH changes when a small amount of aqueous nitric acid is added.
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Worked solution
(a) A buffer solution is a solution that resists changes in pH when small amounts of an acid or an alkali are added. (b) For a pH of 4.85, the hydrogen ion concentration is: \([\text{H}^+] = 10^{-4.85} = 1.4125 \times 10^{-5}\text{ mol dm}^{-3}\). Using the Henderson-Hasselbalch equation or the \(K_a\) expression: \(K_a = \frac{[\text{H}^+][\text{A}^-]}{[\text{HA}]}\). Rearranging to solve for the salt concentration: \([\text{A}^-] = \frac{K_a \times [\text{HA}]}{[\text{H}^+]} = \frac{1.35 \times 10^{-5} \times 0.200}{1.4125 \times 10^{-5}} = 0.19115\text{ mol dm}^{-3}\). In \(500\text{ cm}^3\) (\(0.500\text{ dm}^3\)), moles of salt needed: \(n = 0.19115\text{ mol dm}^{-3} \times 0.500\text{ dm}^3 = 0.09558\text{ mol}\). Mass of sodium propanoate = \(0.09558\text{ mol} \times 96.0\text{ g mol}^{-1} = 9.18\text{ g}\) (to 3 significant figures). (c) When nitric acid is added, the propanoate ions react with the added hydrogen ions: \(\text{CH}_3\text{CH}_2\text{COO}^-(aq) + \text{H}^+(aq) \rightarrow \text{CH}_3\text{CH}_2\text{COOH}(aq)\).
Marking scheme
(a) A solution that resists changes in pH (1 mark) when small amounts of acid or base are added (1 mark). (b) Calculates \([\text{H}^+] = 1.41 \times 10^{-5}\text{ mol dm}^{-3}\) (1 mark); states/rearranges \(K_a\) expression (1 mark); calculates \([\text{A}^-] = 0.191\text{ mol dm}^{-3}\) (1 mark); calculates moles in \(500\text{ cm}^3 = 0.0956\text{ mol}\) (1 mark); calculates mass = \(9.18\text{ g}\) (1 mark). (c) Correct ionic equation: \(\text{CH}_3\text{CH}_2\text{COO}^- + \text{H}^+ \rightarrow \text{CH}_3\text{CH}_2\text{COOH}\) (2 marks; 1 mark for correct reactants and products, 1 mark for state symbols).
Question 19 · structured
9 marks
A student investigates the kinetics of the reaction between peroxodisulfate(VIII) ions, \(S_2O_8^{2-}\), and iodide ions, \(I^-\), by a clock reaction. They determine the rate constant, \(k\), at several different absolute temperatures, \(T\). The student plots a graph of \(\ln(k)\) against \(\frac{1}{T}\). The gradient of the line of best fit is determined to be \(-6450\text{ K}\). (a) Write the Arrhenius equation in logarithmic form, explaining the meaning of all terms. (b) Calculate the activation energy, \(E_a\), for this reaction in \(\text{kJ mol}^{-1}\) to 3 significant figures. (Gas constant \(R = 8.31\text{ J K}^{-1}\text{ mol}^{-1}\)) (c) The rate-determining step for this reaction is: \(S_2O_8^{2-}(aq) + I^-(aq) \rightarrow \text{intermediate products}\). Explain why this step is slow, referring to collision theory and the charges of the reacting ions.
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Worked solution
(a) The Arrhenius equation in logarithmic form is: \(\ln(k) = -\frac{E_a}{RT} + \ln(A)\), where \(k\) is the rate constant, \(E_a\) is the activation energy, \(R\) is the gas constant, \(T\) is the temperature in Kelvin, and \(A\) is the pre-exponential / Arrhenius factor. (b) From the equation, the gradient of the graph is equal to \(-\frac{E_a}{R}\). Therefore: \(-\frac{E_a}{R} = -6450\), so \(E_a = 6450 \times R = 6450 \times 8.31 = 53600\text{ J mol}^{-1} = 53.6\text{ kJ mol}^{-1}\). (c) Both reacting species (\(S_2O_8^{2-}\) and \(I^-\)) are negatively charged ions. There is a strong electrostatic repulsion between like-charged ions, which creates a high energy barrier (activation energy) for collision. Therefore, fewer collisions have energy greater than or equal to \(E_a\), leading to a slow rate of reaction.
Marking scheme
(a) Identifies correct equation: \(\ln(k) = -\frac{E_a}{RT} + \ln(A)\) (1 mark); defines \(k\) and \(T\) (1 mark); defines \(E_a\) and \(A\) (1 mark). (b) Relates gradient to \(-\frac{E_a}{R}\) (1 mark); calculates value as \(53599.5\text{ J mol}^{-1}\) (1 mark); rounds correctly to \(53.6\text{ kJ mol}^{-1}\) (1 mark). (c) Identifies both ions are negatively charged (1 mark); notes that they repel each other / high repulsion (1 mark); links repulsion to high activation energy / low frequency of successful collisions (1 mark).
Question 20 · structured
9 marks
An organic compound X has the molecular formula \(C_4H_8O_2\). A student is given the following spectroscopic data: - The IR spectrum of X shows a strong, sharp absorption at \(1740\text{ cm}^{-1}\), but no broad absorption in the region \(2500\text{--}3300\text{ cm}^{-1}\) or \(3200\text{--}3600\text{ cm}^{-1}\). - The \(^1\text{H}\) NMR spectrum of X displays three environments: a triplet at \(\delta = 1.25\text{ ppm}\) (relative area 3), a singlet at \(\delta = 2.05\text{ ppm}\) (relative area 3), and a quartet at \(\delta = 4.12\text{ ppm}\) (relative area 2). - In the mass spectrum, there is a prominent peak at \(m/z = 43\). (a) Identify the functional group present in X using the IR data. State which bonds are responsible for this and the one that is clearly absent. (b) Explain the splitting patterns and relative areas of the three peaks in the \(^1\text{H}\) NMR spectrum, including the structural fragments they represent. (c) Deduce the structural formula of X and identify the fragment responsible for the peak at \(m/z = 43\).
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Worked solution
(a) The strong, sharp absorption at \(1740\text{ cm}^{-1}\) corresponds to the \(\text{C}=\text{O}\) (carbonyl) group. The absence of broad bands at \(2500\text{--}3300\text{ cm}^{-1}\) and \(3200\text{--}3600\text{ cm}^{-1}\) indicates that there is no \(\text{O-H}\) group (acid or alcohol). Therefore, X contains an ester functional group. (b) - Singlet at \(\delta = 2.05\text{ ppm}\) with area 3 represents a \(\text{CH}_3\) group adjacent to zero hydrogens (typically adjacent to carbonyl, \(\text{CH}_3\text{CO-}\)). - Triplet at \(\delta = 1.25\text{ ppm}\) with area 3 represents a \(\text{CH}_3\) group adjacent to 2 hydrogens (part of an ethyl group, \(\text{-CH}_2\text{CH}_3\)). - Quartet at \(\delta = 4.12\text{ ppm}\) with area 2 represents a \(\text{CH}_2\) group adjacent to 3 hydrogens (the other part of the ethyl group, \(\text{-OCH}_2\text{CH}_3\)), downfield due to being attached directly to the oxygen. (c) Combining the fragments gives ethyl ethanoate, \(\text{CH}_3\text{COOCH}_2\text{CH}_3\). The peak at \(m/z = 43\) is due to the acylium ion fragment, \([\text{CH}_3\text{CO}]^+\).
Marking scheme
(a) Identifies ester / carbonyl (1 mark); notes absence of \(\text{O-H}\) group from lacking IR absorbances in the specified regions (1 mark). (b) Explains singlet at \(2.05\text{ ppm}\) as \(\text{CH}_3\) adjacent to zero protons / adjacent to \(\text{C}=\text{O}\) (1 mark); explains triplet at \(1.25\text{ ppm}\) as \(\text{CH}_3\) coupled to 2 protons / adjacent to \(\text{CH}_2\) (1 mark); explains quartet at \(4.12\text{ ppm}\) as \(\text{CH}_2\) coupled to 3 protons / adjacent to \(\text{CH}_3\) (1 mark); links chemical shift of quartet to attachment to electronegative oxygen (1 mark). (c) Deduces structural formula as \(\text{CH}_3\text{COOCH}_2\text{CH}_3\) (2 marks; 1 mark if name is given but not structure); identifies the \(m/z = 43\) fragment as \([\text{CH}_3\text{CO}]^+\) (including positive charge) (1 mark).
Question 21 · structured
9 marks
A student uses a colorimeter to determine the concentration of copper(II) ions in an unknown solution. To make the colour more intense, they add excess ammonia to form the deep-blue complex \([\text{Cu}(\text{H}_2\text{O})_2(\text{NH}_3)_4]^{2+}\). (a) Write the balanced equation for the reaction of \([\text{Cu}(\text{H}_2\text{O})_6]^{2+}\) with excess aqueous ammonia, and state the type of reaction. (b) Describe how a student would use a colorimeter and standard solutions of copper(II) ions to find the concentration of an unknown copper(II) solution. (c) Suggest, with a reason, a suitable colour of filter that should be used in the colorimeter for this analysis. (d) State why colorimetry is a more suitable method for determining low concentrations of copper(II) ions than gravimetric analysis.
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Worked solution
(a) Equation: \([\text{Cu}(\text{H}_2\text{O})_6]^{2+}(aq) + 4\text{NH}_3(aq) \rightarrow [\text{Cu}(\text{H}_2\text{O})_2(\text{NH}_3)_4]^{2+}(aq) + 4\text{H}_2\text{O}(l)\). Reaction type: Ligand substitution. (b) 1. Prepare a series of standard solutions of copper(II) ions of known concentration. 2. Add the same excess volume of ammonia to each standard and the unknown. 3. Measure the absorbance of each standard solution in the colorimeter. 4. Plot a calibration curve of absorbance against concentration. 5. Measure the absorbance of the unknown solution and read the concentration from the calibration curve. (c) A yellow or orange filter. The solution is deep blue because it absorbs yellow/orange light (complementary colour) most strongly. (d) Colorimetry is much more sensitive than gravimetric analysis, where the mass of precipitate obtained would be too small to weigh accurately.
Marking scheme
(a) Correct balanced equation (1 mark); reaction type: ligand substitution (1 mark). (b) Prepare standard solutions of known concentrations (1 mark); measure absorbance for each and plot calibration curve of absorbance against concentration (1 mark); measure unknown absorbance and read concentration from curve (1 mark); mentions adding same excess ammonia to all (1 mark). (c) Yellow or orange filter (1 mark); complementary to blue/wavelength is strongly absorbed by the complex (1 mark). (d) Sensitvity: mass of precipitate would be too low to weigh accurately (1 mark).
Question 22 · structured
9 marks
A student prepares methyl 3-nitrobenzoate by nitrating methyl benzoate using a mixture of concentrated nitric acid and concentrated sulfuric acid. (a) Write the equations showing how the active electrophile, \(\text{NO}_2^+\), is generated in the nitrating mixture. (b) Explain why the temperature must be kept below \(15^\circ\text{C}\) during the addition of the nitrating mixture. (c) Describe the practical steps required to carry out the recrystallisation of the crude methyl 3-nitrobenzoate.
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Worked solution
(a) The formation of the nitronium ion (\(\text{NO}_2^+\)) is given by: \(\text{HNO}_3 + 2\text{H}_2\text{SO}_4 \rightarrow \text{NO}_2^+ + 2\text{HSO}_4^- + \text{H}_3\text{O}^+\) (or in two steps: \(\text{HNO}_3 + \text{H}_2\text{SO}_4 \rightleftharpoons \text{H}_2\text{NO}_3^+ + \text{HSO}_4^-\), then \(\text{H}_2\text{NO}_3^+ \rightarrow \text{NO}_2^+ + \text{H}_2\text{O}\)). (b) The temperature must be kept below \(15^\circ\text{C}\) to prevent further nitration of the ring (e.g. dinitration to form methyl 3,5-dinitrobenzoate) and to prevent other side reactions/decomposition. (c) Recrystallisation steps: 1. Dissolve the crude solid in the minimum volume of hot ethanol / solvent. 2. Filter the hot solution (to remove insoluble impurities). 3. Allow the filtrate to cool slowly (to room temperature and then in ice) to allow crystals to reform. 4. Filter the crystals under reduced pressure using a Buchner flask and funnel. 5. Wash the crystals with a small amount of ice-cold solvent, and dry them.
Marking scheme
(a) First step protonation of nitric acid (1 mark); second step dehydration to produce nitronium ion (1 mark). (Accept overall equation for 2 marks). (b) Prevents further nitration / dinitration (1 mark); minimises side reactions / decomposition / safety reasons (1 mark). (c) Dissolve in minimum volume of hot solvent (1 mark); hot filtration to remove insoluble impurities (1 mark); cool in ice to reform crystals (1 mark); filter under reduced pressure / Buchner funnel (1 mark); wash with ice-cold solvent and dry (1 mark).
Question 23 · structured
9 marks
A student is asked to determine the experimental lattice energy of calcium chloride, \(\text{CaCl}_2\), using a Born-Haber cycle and the following thermodynamic data: Enthalpy of atomisation of calcium, \(\Delta H_{\text{at}}^\ominus[\text{Ca}(s)] = +178\text{ kJ mol}^{-1}\); First ionisation energy of calcium, \(\text{IE}_1[\text{Ca}] = +590\text{ kJ mol}^{-1}\); Second ionisation energy of calcium, \(\text{IE}_2[\text{Ca}] = +1145\text{ kJ mol}^{-1}\); Enthalpy of atomisation of chlorine, \(\Delta H_{\text{at}}^\ominus[\text{Cl}_2(g)] = +122\text{ kJ mol}^{-1}\); Electron affinity of chlorine, \(\text{EA}[\text{Cl}] = -349\text{ kJ mol}^{-1}\); Enthalpy of formation of calcium chloride, \(\Delta H_f^\ominus[\text{CaCl}_2(s)] = -796\text{ kJ mol}^{-1}\). (a) Define the term 'lattice energy'. (b) Calculate the experimental lattice energy, \(\Delta H_{\text{latt}}^\ominus\), of \(\text{CaCl}_2(s)\) (defined as the formation of the solid lattice from gaseous ions). (c) The theoretical lattice energy of \(\text{CaCl}_2\) calculated using an ionic model is \(-2223\text{ kJ mol}^{-1}\). Explain why the experimental value differs from this theoretical value, and what this tells us about the bonding in calcium chloride.
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Worked solution
(a) Lattice energy is the enthalpy change when 1 mole of an ionic compound is formed from its constituent gaseous ions under standard conditions. (b) Using Hess's law around the Born-Haber cycle: \(\Delta H_f^\ominus = \Delta H_{\text{at}}^\ominus[\text{Ca}] + \text{IE}_1[\text{Ca}] + \text{IE}_2[\text{Ca}] + 2 \times \Delta H_{\text{at}}^\ominus[\text{Cl}_2] + 2 \times \text{EA}[\text{Cl}] + \Delta H_{\text{latt}}^\ominus\). Substituting the values: \(-796 = +178 + 590 + 1145 + 2(122) + 2(-349) + \Delta H_{\text{latt}}^\ominus\). \(-796 = 178 + 590 + 1145 + 244 - 698 + \Delta H_{\text{latt}}^\ominus\). \(-796 = 1459 + \Delta H_{\text{latt}}^\ominus\). \(\Delta H_{\text{latt}}^\ominus = -796 - 1459 = -2255\text{ kJ mol}^{-1}\). (c) The experimental lattice energy (\(-2255\text{ kJ mol}^{-1}\)) is more exothermic than the theoretical value (\(-2223\text{ kJ mol}^{-1}\)). This indicates that the bonding has some covalent character. The small, highly charged \(\text{Ca}^{2+}\) ion polarises the electron cloud of the \(\text{Cl}^-\) ions, leading to some sharing of electron density, which strengthens the lattice beyond the predictions of the pure ionic model.
Marking scheme
(a) Enthalpy change when one mole of solid ionic lattice is formed (1 mark) from gaseous ions (1 mark). (b) Correct multipliers for Cl atomisation (\(2 \times 122\)) and EA (\(2 \times -349\)) (1 mark); correct algebraic expression of cycle (1 mark); correct calculation of sum as \(+1459\) (1 mark); final answer \(-2255\text{ kJ mol}^{-1}\) (including unit and sign) (1 mark). (c) Notes experimental is more exothermic/negative than theoretical (1 mark); identifies covalent character (1 mark); explains polarisation of chloride ion by calcium ion (1 mark).
Question 24 · Structured Practical
9 marks
A student carried out an experiment to determine the percentage by mass of copper in a sample of brass.
The student used the following procedure: 1. Weigh out exactly \(2.00\text{ g}\) of brass into a beaker. 2. Dissolve the brass sample in a minimum volume of concentrated nitric acid. 3. Transfer the resulting solution and washings to a volumetric flask and make up to the mark with deionised water so that the total volume is \(250.0\text{ cm}^3\). 4. Pipette \(25.0\text{ cm}^3\) of this solution into a conical flask and neutralize excess acid by adding sodium carbonate solution dropwise until a faint precipitate forms, then dissolve it with a few drops of ethanoic acid. 5. Add an excess of potassium iodide solution to the flask. 6. Titrate the liberated iodine with \(0.100\text{ mol dm}^{-3}\) sodium thiosulfate solution.
(a) State **one** safety precaution, other than wearing safety goggles and a lab coat, that must be taken in Step 2, and explain why it is necessary. (2 marks)
(b) Write the ionic equation for the reaction occurring in Step 5. State symbols are not required. (1 mark)
(c) Describe how the student should use starch indicator in the titration in Step 6 to determine the end-point accurately. Include the stage at which the indicator is added and the colour change observed at the end-point. (3 marks)
(d) The student repeated the titration and obtained a mean titre of \(22.00\text{ cm}^3\) of \(0.100\text{ mol dm}^{-3}\) sodium thiosulfate solution. Calculate the percentage by mass of copper in the brass sample. Give your answer to three significant figures. The relative atomic mass of copper, \(A_r(Cu) = 63.5\). (3 marks)
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Worked solution
(a) Precaution: Perform the dissolving step in a fume cupboard. Explanation: The reaction of brass with concentrated nitric acid produces nitrogen dioxide (\(NO_2\)) gas, which is toxic and highly irritating to the respiratory system.
(c) The starch indicator should be added only when the reaction mixture is pale yellow (straw-coloured) rather than at the start. At this stage, adding starch turns the mixture blue-black. The titration is then continued dropwise until the blue-black colour completely disappears, leaving an off-white precipitate (or a virtually colourless solution with white precipitate) as the end-point.
(d) Moles of \(S_2O_3^{2-}\) used in the titration: \(n(S_2O_3^{2-}) = 0.100\text{ mol dm}^{-3} \times \frac{22.00}{1000}\text{ dm}^3 = 2.20 \times 10^{-3}\text{ mol}\)
From the equations: \(2Cu^{2+} + 4I^- \rightarrow 2CuI + I_2\) \(2S_2O_3^{2-} + I_2 \rightarrow S_4O_6^{2-} + 2I^-\) This gives a 1:1 stoichiometry between copper(II) ions and thiosulfate ions: \(n(Cu^{2+}) = n(S_2O_3^{2-})\).
Moles of \(Cu^{2+}\) in the \(25.0\text{ cm}^3\) aliquot = \(2.20 \times 10^{-3}\text{ mol}\) Moles of \(Cu^{2+}\) in the original \(250.0\text{ cm}^3\) solution = \(2.20 \times 10^{-3}\text{ mol} \times 10 = 2.20 \times 10^{-2}\text{ mol}\)
Mass of copper in the sample: \(\text{mass} = 2.20 \times 10^{-2}\text{ mol} \times 63.5\text{ g mol}^{-1} = 1.397\text{ g}\)
Percentage by mass of copper in the brass: \(\%Cu = \frac{1.397\text{ g}}{2.00\text{ g}} \times 100\% = 69.85\% \approx 69.9\%\)
Marking scheme
Part (a): - 1 mark for stating that the reaction must be performed in a fume cupboard / fume hood. - 1 mark for explaining that toxic/hazardous nitrogen dioxide gas is produced.
Part (b): - 1 mark for a balanced ionic equation: \(2Cu^{2+} + 4I^- \rightarrow 2CuI + I_2\) (accept multiples or \(Cu_2I_2\)).
Part (c): - 1 mark for adding starch when the mixture is pale yellow/straw-coloured (not at the start). - 1 mark for stating the starting color with starch is blue-black. - 1 mark for stating the end-point color change is to off-white / white precipitate / colorless solution.
Part (d): - 1 mark for calculating moles of thiosulfate: \(2.20 \times 10^{-3}\text{ mol}\). - 1 mark for scaling up to find moles of copper in \(250\text{ cm}^3\): \(2.20 \times 10^{-2}\text{ mol}\) (by using 1:1 ratio with thiosulfate and multiplying by 10). - 1 mark for calculating percentage by mass to 3 SF: \(69.9\%\). (Allow TE for incorrect moles, provided correct steps and 3 SF are used. Correct answer with no working scores 3 marks).
Question 25 · Structured Practical
9 marks
The reaction between propanone and iodine in acidic conditions is represented by the equation: \(CH_3COCH_3(aq) + I_2(aq) \xrightarrow{H^+(aq)} CH_3COCH_2I(aq) + H^+(aq) + I^-(aq)\)
A student investigated the kinetics of this reaction.
(a) (i) Explain why the samples withdrawn at regular intervals must be 'quenched' immediately. (1 mark) (ii) Suggest a reagent that could be added to quench the reaction. Explain how this reagent stops the reaction, writing an ionic equation for the process. (3 marks)
(b) In an alternative method, the student used a colorimeter to monitor the concentration of iodine continuously. Explain how a colorimeter can be used to monitor the concentration of iodine over time, including how the absorbance data is converted into concentration. (2 marks)
(c) The initial concentration of acid was held constant at \([H^+] = 0.0500\text{ mol dm}^{-3}\). The following initial rates were obtained:
Using these data: - Deduce the order of reaction with respect to propanone and with respect to iodine. - Given that the reaction is first-order with respect to \(H^+\), write the overall rate equation. - Calculate the rate constant, \(k\), using the data from Experiment 1, and state its units. (3 marks)
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Worked solution
(a) (i) Quenching stops the reaction so that the concentrations of reactants do not change while the sample is being analysed / titrated. (a) (ii) Reagent: Sodium hydrogencarbonate (\(NaHCO_3\)) or sodium carbonate (\(Na_2CO_3\)). Explanation: The reaction is catalysed by acid (\(H^+\)). The added base neutralises the acid catalyst, stopping the reaction. Ionic equation: \(HCO_3^- + H^+ \rightarrow CO_2 + H_2O\) (or \(CO_3^{2-} + 2H^+ \rightarrow CO_2 + H_2O\)).
(b) Since iodine is coloured (brown/yellow/orange) and the other species are colourless, the absorbance of the solution is directly proportional to the iodine concentration. The student should first plot a calibration curve of absorbance against known concentrations of iodine. Then, they measure absorbance over time and read the iodine concentration from the calibration curve.
(c) - Order wrt propanone: comparing Exp 1 and Exp 2, doubling \([CH_3COCH_3]\) whilst holding \([I_2]\) constant doubles the initial rate, so the order is 1. - Order wrt iodine: comparing Exp 2 and Exp 3, doubling \([I_2]\) whilst holding \([CH_3COCH_3]\) constant has no effect on the initial rate, so the order is 0. - Rate equation: \(\text{Rate} = k[CH_3COCH_3][H^+]\). - Calculation of \(k\): Using Exp 1: \(1.20 \times 10^{-5} = k \times 0.400 \times 0.0500\) \(k = \frac{1.20 \times 10^{-5}}{0.0200} = 6.00 \times 10^{-4}\) - Units: \(\text{dm}^3\text{ mol}^{-1}\text{ s}^{-1}\).
Marking scheme
Part (a)(i): - 1 mark: To stop the reaction / freeze concentrations at that specific time.
Part (a)(ii): - 1 mark: Suggests sodium hydrogencarbonate / sodium carbonate (accept any suitable weak base; reject strong bases like NaOH as they can cause alkaline hydrolysis / side reactions). - 1 mark: Neutralises the acid catalyst. - 1 mark: Correct ionic equation: \(HCO_3^- + H^+ \rightarrow CO_2 + H_2O\) (allow state symbols, allow full balanced equations if ionic species are correct).
Part (b): - 1 mark: Iodine is coloured / absorbs light, so absorbance decreases as iodine is used up. - 1 mark: Convert absorbance to concentration using a calibration curve (of known iodine concentrations).
Part (c): - 1 mark: Deduces correct orders (propanone = 1, iodine = 0) with brief reasoning (or showing calculations). - 1 mark: Correct rate equation: \(\text{Rate} = k[CH_3COCH_3][H^+]\) (allow ecf from incorrect orders, but must include \(H^+\)). - 1 mark: Correct value of \(k = 6.00 \times 10^{-4}\) and units of \(\text{dm}^3\text{ mol}^{-1}\text{ s}^{-1}\) (allow ecf from rate equation).
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