2022 OCR AS Level Chemistry B (Salters) - H033 模擬試題連答案詳解

1. Calculate energy transferred to water: \(q = m c \Delta T\) where \(m = 150.0\text{ g}\), \(c = 4.18\text{ J g}^{-1}\ \text{K}^{-1}\), and \(\Delta T = 55.5 - 19.5 = 36.0\text{ K}\).
\(q = 150.0 \times 4.18 \times 36.0 = 22572\text{ J} = 22.572\text{ kJ}\).
2. Calculate moles of pentan-1-ol burned: \(n = \frac{1.10\text{ g}}{88.0\text{ g mol}^{-1}} = 0.0125\text{ mol}\).
3. Calculate enthalpy change of combustion: \(\Delta_c H = -\frac{q}{n} = -\frac{22.572}{0.0125} = -1805.76\text{ kJ mol}^{-1} \approx -1810\text{ kJ mol}^{-1}\).

[1 mark] award for correct calculation of heat transferred (q = 22.57 kJ), moles of pentan-1-ol (0.0125 mol), and combining these with the negative sign to yield -1810 kJ mol^-1.

- The broad band at \(2500\text{--}3300\text{ cm}^{-1}\) is characteristic of an \(\text{O-H}\) stretch of a carboxylic acid.
- The strong, sharp band at \(1715\text{ cm}^{-1}\) represents a \(\text{C=O}\) carbonyl stretch.
- Together, these bands indicate a carboxylic acid group, \(\text{-COOH}\).
- The fragment at \(m/z = 45\) corresponds to the species \(\text{[COOH]}^+\) (molecular mass = \(12 + 32 + 1 = 45\)), which is typical of carboxylic acids.
- Among the options, only butanoic acid is a carboxylic acid with the molecular formula \(\text{C}_4\text{H}_8\text{O}_2\).

[1 mark] award for identifying that the spectrum matches a carboxylic acid with four carbons, which is butanoic acid.

- As temperature increases, the average kinetic energy of the molecules increases, shifting the peak of the Maxwell-Boltzmann distribution curve to the right and making it flatter/lower (since the total number of particles/area remains constant).
- The activation energy, \(E_a\), is a constant characteristic of the reaction pathway and is unaffected by temperature.
- The increased temperature means a greater fraction of molecules have energy \(\ge E_a\), which increases the frequency of successful collisions and thus increases the rate of reaction.

[1 mark] award for choosing the option that accurately describes the physical shift in the Maxwell-Boltzmann distribution curve and the chemical reason for increased rate.

- Down Group 2, the number of electron shells increases, which means the outermost electron is further from the nucleus (increased atomic radius) and experiences more shielding from inner shells.
- Although the nuclear charge increases, the effects of distance and shielding dominate.
- Thus, the electrostatic attraction between the positive nucleus and the outer electron decreases, and less energy is required to remove the first electron.

[1 mark] award for identifying the correct trend (decrease) and the correct explanation (increased shielding and atomic radius outweighing increased nuclear charge).

- Write the expression for \(K_c\): \(K_c = \frac{[\text{C}]^2}{[\text{A}][\text{B}]^2}\).
- Substitute the equilibrium concentrations: \(K_c = \frac{(0.60)^2}{0.20 \times (0.30)^2} = \frac{0.36}{0.20 \times 0.09} = \frac{0.36}{0.018} = 20\text{ dm}^3\text{ mol}^{-1}\).
- The forward reaction is exothermic (\(\Delta H < 0\)).
- According to Le Chatelier's principle, increasing the temperature shifts the equilibrium in the endothermic (reverse) direction to absorb heat.
- This decreases the concentrations of products and increases those of reactants, thus decreasing the value of \(K_c\).

[1 mark] award for correctly calculating Kc = 20 and identifying that Kc decreases with increasing temperature due to the reaction being exothermic in the forward direction.

- Radical substitution mechanisms feature initiation, propagation, and termination steps.
- A termination step involves the combination of two radical species to form a single, non-radical, stable molecule.
- Option d shows an ethyl radical and a chlorine radical combining to form chloroethane. Since there are no radicals in the product side, this represents termination.

[1 mark] award for selecting the equation where two radicals combine to form a non-radical product.

- The reaction of \(\text{Cl}_2\) with cold, dilute \(\text{NaOH}\) is: \(\text{Cl}_2 + 2\text{NaOH} \rightarrow \text{NaCl} + \text{NaClO} + \text{H}_2\text{O}\).
- This is a disproportionation reaction.
- In \(\text{NaCl}\), the oxidation state of chlorine is \(-1\).
- In \(\text{NaClO}\), since sodium is \(+1\) and oxygen is \(-2\), the chlorine atom has an oxidation state of \(+1\).

[1 mark] award for identifying the products NaCl and NaClO and assigning the correct oxidation states of -1 and +1.

- The \(\text{HO-C}_6\text{H}_4\text{-}\) group represents a phenol group because the hydroxyl group is bonded directly to the aromatic benzene ring.
- The \(\text{-COO-}\) group is an ester group.
- The \(\text{-NH}_2\) group is a primary amine group.

[1 mark] award for identifying all three functional groups correctly: phenol, ester, and amine.

The equation for the formation of propan-1-ol is:
\(3C(s) + 4H_2(g) + \frac{1}{2}O_2(g) \rightarrow C_3H_7OH(l)\)

Using a Hess's cycle with combustion products:
\(\Delta_f H^\theta = \sum \Delta_c H^\theta(\text{reactants}) - \sum \Delta_c H^\theta(\text{products})\)
\(\Delta_f H^\theta = [3 \times (-394) + 4 \times (-286)] - (-2021)\)
\(\Delta_f H^\theta = [-1182 - 1144] + 2021 = -2326 + 2021 = -305 \text{ kJ mol}^{-1}\)

[1] Award 1 mark for the correct calculation leading to -305 kJ mol-1. Reject +305 kJ mol-1.

The very broad absorption band between \(2500 - 3300 \text{ cm}^{-1}\) is characteristic of an O-H stretch in a carboxylic acid. The sharp band at \(1715 \text{ cm}^{-1}\) corresponds to the C=O stretch of a carbonyl group. A molecular formula of \(C_3H_6O_2\) containing both these groups represents a carboxylic acid with three carbons: propanoic acid, \(CH_3CH_2COOH\).

[1] Correctly identifies propanoic acid based on functional group assignments from the infrared data.

The Maxwell-Boltzmann distribution curve is determined solely by temperature, so adding a catalyst does not change the shape or position of the curve. However, the catalyst provides an alternative reaction pathway with a lower activation energy, effectively establishing a new activation energy line, \(E_{a, \text{cat}}\), situated to the left of the original \(E_a\) on the energy axis.

[1] Correctly identifies that the curve is unchanged while the activation energy decreases.

Down Group 2, the cationic radius increases while the charge remains \(2+\). This leads to a lower charge density on the cation, meaning it has a weaker polarising effect on the carbonate ion's electron cloud. As polarization of the carbonate ion decreases, the C-O covalent bond within the carbonate ion is weakened less, requiring more thermal energy to decompose it. Thus, thermal stability increases down the group.

[1] Correctly identifies that increasing cation radius down the group leads to less polarization of the anion, causing thermal stability to increase.

Determine equilibrium amounts:
- \(NH_3\) equilibrium moles = \(0.20\) mol
- \(N_2\) equilibrium moles = \(0.40 - \frac{0.20}{2} = 0.30\) mol
- \(H_2\) equilibrium moles = \(1.20 - (3 \times 0.10) = 0.90\) mol

Calculate equilibrium concentrations in \(2.0 \text{ dm}^3\):
- \([NH_3] = \frac{0.20}{2.0} = 0.10 \text{ mol dm}^{-3}\)
- \([N_2] = \frac{0.30}{2.0} = 0.15 \text{ mol dm}^{-3}\)
- \([H_2] = \frac{0.90}{2.0} = 0.45 \text{ mol dm}^{-3}\)

Substitute into the \(K_c\) expression:
\(K_c = \frac{[NH_3]^2}{[N_2][H_2]^3} = \frac{(0.10)^2}{0.15 \times (0.45)^3} = \frac{0.01}{0.15 \times 0.091125} = \frac{0.01}{0.01366875} \approx 0.73 \text{ dm}^6 \text{ mol}^{-2}\)

[1] Calculates concentrations at equilibrium correctly by taking volume into account and calculates Kc to be 0.73.

1. Calculate moles of \(NaN_3\):
\(n(NaN_3) = \frac{\text{mass}}{M_r} = \frac{13.0}{65.0} = 0.20 \text{ mol}\)

2. Determine moles of \(N_2\) from stoichiometry:
\(n(N_2) = 0.20 \times \frac{3}{2} = 0.30 \text{ mol}\)

3. Calculate volume of \(N_2\):
\(V = 0.30 \times 24.0 \text{ dm}^3 \text{ mol}^{-1} = 7.20 \text{ dm}^3\)

[1] Deduces moles of nitrogen (0.30 mol) and correctly calculates volume as 7.20 dm3.

The carbonate ion, \(CO_3^{2-}\), consists of a central carbon atom bonded to three oxygen atoms with zero lone pairs on the carbon atom. The three regions of negative charge repel each other equally to minimize repulsion, resulting in a symmetric trigonal planar molecular geometry with bond angles of exactly \(120^\circ\). \(NH_3\) is trigonal pyramidal (\(\approx 107^\circ\)), \(H_2O\) is non-linear/bent (\(\approx 104.5^\circ\)), and \(SF_6\) is octahedral (\(90^\circ\)).

[1] Correctly identifies the carbonate ion as trigonal planar with bond angles of exactly 120 degrees.

In the electrophilic addition of \(HBr\) to propene, the hydrogen atom in \(H-Br\) is electron-deficient due to the highly electronegative bromine atom, forming a dipole \(H^{\delta+} - Br^{\delta-}\). Therefore, the \(H^{\delta+}\) acts as the electrophile. The addition of \(H^+\) to the double bond can form either a primary or a secondary carbocation. The secondary carbocation \(CH_3CH^+CH_3\) is more stable than the primary carbocation \(CH_3CH_2CH_2^+\) due to the electron-donating inductive effect of two methyl groups, meaning it is formed as the major intermediate.

[1] Correctly identifies Hδ+ as the electrophile and the secondary carbocation as the major intermediate.

First, calculate the heat energy absorbed by the water: \(q = m c \Delta T = 150.0\text{ g} \times 4.18\text{ J g}^{-1}\text{ K}^{-1} \times (41.5 - 18.5)\text{ K} = 150.0 \times 4.18 \times 23.0 = 14421\text{ J} = 14.421\text{ kJ}\). Next, find the amount in moles of ethanol burned: \(n = \frac{\text{mass}}{\text{molar mass}} = \frac{0.690\text{ g}}{46.0\text{ g mol}^{-1}} = 0.0150\text{ mol}\). Finally, calculate the enthalpy change of combustion: \(\Delta_c H = -\frac{q}{n} = -\frac{14.421\text{ kJ}}{0.0150\text{ mol}} = -961.4\text{ kJ mol}^{-1}\), which rounds to \(-961\text{ kJ mol}^{-1}\).

Award 1 mark for the correct answer A.

The broad peak at \(2500\text{--}3300\text{ cm}^{-1}\) is characteristic of an \(\text{O--H}\) stretch in carboxylic acids. The strong, sharp peak at \(1715\text{ cm}^{-1}\) is due to a \(\text{C=O}\) stretch. Together, these peaks confirm the presence of a carboxylic acid functional group. Since X has the molecular formula \(\text{C}_4\text{H}_8\text{O}_2\), it must be a 4-carbon carboxylic acid, which is butanoic acid.

Award 1 mark for the correct answer B.

A catalyst provides an alternative pathway with a lower activation energy, which shifts the activation energy (\(E_a\)) to a lower value (to the left on the Maxwell-Boltzmann distribution graph). Because the temperature is constant, the Maxwell-Boltzmann distribution curve of molecular energies remains completely unchanged.

Award 1 mark for the correct answer C.

The equilibrium constant, \(K_c\), is only affected by changes in temperature. For an exothermic forward reaction (\(\Delta H < 0\)), decreasing the temperature shifts the equilibrium position to the right (favouring the exothermic direction to release heat). This increases the concentration of the products and decreases the concentration of the reactants at equilibrium, thereby increasing \(K_c\).

Award 1 mark for the correct answer B.

(i) Heat absorbed by water, \(q = m c \Delta T = 150.0 \times 4.18 \times 32.5 = 20377.5\text{ J} = 20.3775\text{ kJ}\). Moles of pentan-1-ol burned = \(1.10 / 88.0 = 0.0125\text{ mol}\). \(\Delta_c H = -q / n = -20.3775 / 0.0125 = -1630.2\text{ kJ mol}^{-1}\) (or \(-1630\text{ kJ mol}^{-1}\)). (ii) Heat loss to surroundings, incomplete combustion, or evaporation of the fuel. (iii) Bonds broken: \(4 \times (C-C) + 11 \times (C-H) + 1 \times (C-O) + 1 \times (O-H) + 7.5 \times (O=O) = 4(347) + 11(413) + 358 + 463 + 7.5(498) = 1388 + 4543 + 358 + 463 + 3735 = 10487\text{ kJ mol}^{-1}\). Bonds formed: \(10 \times (C=O) + 12 \times (O-H) = 10(805) + 12(463) = 8050 + 5556 = 13606\text{ kJ mol}^{-1}\). \(\Delta_c H = \text{Bonds broken} - \text{Bonds formed} = 10487 - 13606 = -3119\text{ kJ mol}^{-1}\).

(i) 4 marks: \(q = 20.38\text{ kJ}\) (1 mark), moles = \(0.0125\) (1 mark), dividing energy by moles (1 mark), final value of \(-1630\) with correct negative sign (1 mark). (ii) 3 marks: any two valid reasons such as heat loss to surroundings (1.5 marks) and incomplete combustion (1.5 marks). (iii) 5.5 marks: list of bonds broken and total sum of \(10487\) (2 marks), list of bonds formed and total sum of \(13606\) (2 marks), final subtraction with correct sign to give \(-3119\text{ kJ mol}^{-1}\) (1.5 marks).

(i) Ratio of atoms: \(C = 60.0 / 12.0 = 5.0\); \(H = 8.0 / 1.0 = 8.0\); \(O = 32.0 / 16.0 = 2.0\). Thus, the empirical formula ratio is \(C:H:O = 5:8:2\), giving the empirical formula \(C_5H_8O_2\). (ii) The empirical formula mass is \(5(12.0) + 8(1.0) + 2(16.0) = 100\). Since the molecular ion peak is at \(m/z = 100\), the molecular formula is also \(C_5H_8O_2\). The fragment at \(m/z = 43\) is due to the acylium ion, \(CH_3CO^+\). Equation: \([C_5H_8O_2]^{+ \bullet} \rightarrow CH_3CO^+ + C_3H_5O^\bullet\). (iii) The absorption at \(3300\text{ cm}^{-1}\) indicates an alcohol \(-OH\) group, and the absorption at \(1715\text{ cm}^{-1}\) indicates a carbonyl \(C=O\) group. Since the molecular formula contains only two oxygen atoms, the compound must be a hydroxy-ketone. A consistent structure is 5-hydroxypentan-2-one: \(CH_3COCH_2CH_2CH_2OH\).

(i) 4 marks: division of percentages by relative atomic masses (1.5 marks), obtaining ratio \(5:8:2\) (1.5 marks), writing final empirical formula (1 mark). (ii) 4.5 marks: molecular formula \(C_5H_8O_2\) linked to \(m/z = 100\) (1.5 marks), identifying \(CH_3CO^+\) with charge (1.5 marks), correct fragmentation equation with positive charge and radical (1.5 marks). (iii) 4 marks: identification of \(-OH\) group (1 mark), identification of \(C=O\) group (1 mark), drawing/writing a valid structural formula consistent with all data (2 marks).

(i) The reaction proceeds via an \(S_N1\) mechanism. The first step, which is the slow rate-determining step, involves the heterolytic fission of the \(C-Br\) bond to form a carbocation intermediate and a bromide ion. This step only involves the halogenoalkane. The hydroxide ion attacks the carbocation in a subsequent rapid step, meaning its concentration does not affect the rate. (ii) From \(0.100\text{ mol dm}^{-3}\) to \(0.050\text{ mol dm}^{-3}\) takes \(100\text{ s}\). From \(0.050\text{ mol dm}^{-3}\) to \(0.025\text{ mol dm}^{-3}\) takes another \(100\text{ s}\). Since successive half-lives are constant at \(100\text{ s}\), the reaction must be first-order. The rate constant \(k = \ln(2) / t_{1/2} = 0.693 / 100 = 6.93 \times 10^{-3}\text{ s}^{-1}\). If the temperature increases, the rate constant \(k\) increases. (iii) When temperature increases, the Maxwell-Boltzmann distribution curve shifts to the right and has a lower peak. The total area under the curve remains constant, but a significantly larger fraction of molecules possess energy equal to or greater than the activation energy \(E_a\). Therefore, there are more frequent successful collisions per unit time, increasing the reaction rate.

(i) 4 marks: reference to \(S_N1\) mechanism (1 mark), stating that the slow step is rate-determining (1 mark), stating that the slow step involves only the halogenoalkane breaking down to a carbocation (1 mark), stating that hydroxide reacts in a subsequent fast step (1 mark). (ii) 4.5 marks: identifying constant half-life of \(100\text{ s}\) to prove first order (1.5 marks), calculation of \(k = 6.93 \times 10^{-3}\) with unit \(\text{s}^{-1}\) (1.5 marks), stating that \(k\) increases with temperature (1.5 marks). (iii) 4 marks: describing shift of distribution curve to the right and lower peak (1.5 marks), identifying that more molecules have energy \(\ge E_a\) (1.5 marks), relating this to more frequent successful collisions (1 mark).

(i) Thermal stability of Group 2 carbonates increases down the group. As you go down the group, the ionic radius of the Group 2 cation increases while the charge remains \(2+\). Therefore, the charge density of the cation decreases, reducing its polarising power. The cation is less able to polarise/distort the carbonate ion's electron cloud, which weakens the \(C-O\) bond less, making the carbonate more stable to heat. (ii) \(n(CO_2) = 240 / 24000 = 0.010\text{ mol}\). Since \(MCO_3 \rightarrow MO + CO_2\), \(n(MCO_3) = 0.010\text{ mol}\). \(M_r(MCO_3) = 1.97 / 0.010 = 197\text{ g mol}^{-1}\). \(A_r(M) = 197 - 12.0 - 3(16.0) = 137\text{ g mol}^{-1}\). The metal is Barium (Ba). (iii) The solubility of Group 2 hydroxides increases down the group. Ionic equation: \(Mg^{2+}(aq) + 2OH^-(aq) \rightarrow Mg(OH)_2(s)\).

(i) 4 marks: trend of thermal stability increasing (1 mark), cation ionic radius increases down the group / charge density decreases (1 mark), decreased polarising power of cation (1 mark), less distortion of carbonate anion/weakening of \(C-O\) bond (1 mark). (ii) 4.5 marks: \(n(CO_2) = 0.010\text{ mol}\) (1.5 marks), calculating \(M_r(MCO_3) = 197\text{ g mol}^{-1}\) (1.5 marks), relative atomic mass of \(M = 137\) and identifying Barium (1.5 marks). (iii) 4 marks: solubility of hydroxides increases down the group (1.5 marks), correct formula of species in ionic equation (1.5 marks), correct state symbols (1 mark).

(a) Equation:
\(\text{C}_5\text{H}_{11}\text{OH}(\text{l}) + 7.5\text{O}_2(\text{g}) \rightarrow 5\text{CO}_2(\text{g}) + 6\text{H}_2\text{O}(\text{l})\)
Alternatively, using whole numbers:
\(2\text{C}_5\text{H}_{11}\text{OH}(\text{l}) + 15\text{O}_2(\text{g}) \rightarrow 10\text{CO}_2(\text{g}) + 12\text{H}_2\text{O}(\text{l})\)

(b)(i) \(q = m c \Delta T\)
\(m = 150.0\text{ g}\)
\(\Delta T = 45.5 - 21.0 = 24.5\,^{\circ}\text{C}\)
\(q = 150.0 \times 4.18 \times 24.5 = 15361.5\text{ J} = 15.36\text{ kJ}\)

(ii) Mass of pentan-1-ol burned = \(184.25 - 183.40 = 0.850\text{ g}\)
Molar mass of pentan-1-ol (\(\text{C}_5\text{H}_{11}\text{OH}\)):
\(M_r = (5 \times 12.0) + (12 \times 1.0) + 16.0 = 88.0\text{ g mol}^{-1}\)
\(n = \frac{0.850}{88.0} = 0.009659\text{ mol}\)

(iii) \(\Delta_c H = -\frac{q}{n} = -\frac{15.3615}{0.009659} = -1590.38\text{ kJ mol}^{-1}\)
Rounding to 3 significant figures: \(-1590\text{ kJ mol}^{-1}\).

(c) Reasons for discrepancy:
1. Heat loss to the surroundings (air, copper can, draft shield).
2. Incomplete combustion of pentan-1-ol (indicated by soot formation on the bottom of the can).
3. Evaporation of pentan-1-ol from the wick after extinguishing but before weighing.
4. Non-standard conditions.

(d) Bonds broken:
- 4 \(\text{C-C}\) bonds = \(4 \times 347 = 1388\text{ kJ}\)
- 11 \(\text{C-H}\) bonds = \(11 \times 413 = 4543\text{ kJ}\)
- 1 \(\text{C-O}\) bond = \(1 \times 358 = 358\text{ kJ}\)
- 1 \(\text{O-H}\) bond = \(1 \times 463 = 463\text{ kJ}\)
- 7.5 \(\text{O=O}\) bonds = \(7.5 \times 495 = 3712.5\text{ kJ}\)
Total Energy In = \(1388 + 4543 + 358 + 463 + 3712.5 = 10464.5\text{ kJ mol}^{-1}\)

Bonds formed:
- 10 \(\text{C=O}\) bonds (in \(5\text{CO}_2\)) = \(10 \times 805 = 8050\text{ kJ}\)
- 12 \(\text{O-H}\) bonds (in \(6\text{H}_2\text{O}\)) = \(12 \times 463 = 5556\text{ kJ}\)
Total Energy Out = \(8050 + 5556 = 13606\text{ kJ mol}^{-1}\)

\(\Delta_c H = \text{Total Energy In} - \text{Total Energy Out} = 10464.5 - 13606 = -3141.5\text{ kJ mol}^{-1}\)

(a) Correct formula of pentan-1-ol, oxygen, carbon dioxide and water [1]
Correct balancing and state symbols [1]

(b)(i) Correct temperature change calculation (\(24.5\,^{\circ}\text{C}\)) [0.5]
Correct calculation of \(q\) with matching units (\(15.4\text{ kJ}\) or \(15.36\text{ kJ}\)) [1.5]

(b)(ii) Correct mass of fuel burned (\(0.850\text{ g}\)) and calculation of \(M_r\) of pentan-1-ol (\(88.0\)) [1]
Correct calculation of moles (\(0.00966\text{ mol}\)) [0.5]

(b)(iii) Correct division of \(q\) by \(n\) [1]
Negative sign included [1]
Final answer to 3 significant figures (\(-1590\text{ kJ mol}^{-1}\)) [1]

(c) Any three distinct points: [1 mark each, max 3]
- Heat loss to surrounding air / calorimeter [1]
- Incomplete combustion of fuel [1]
- Loss of fuel through evaporation from wick [1]

(d) Calculation of bonds broken in pentan-1-ol and oxygen:
- \(4 \times 347 + 11 \times 413 + 358 + 463\) = \(6752\text{ kJ}\) [1.5]
- \(7.5 \times 495\) = \(3712.5\text{ kJ}\) (Total broken = \(10464.5\text{ kJ}\)) [1]
Calculation of bonds formed in products:
- \(10 \times 805 = 8050\text{ kJ}\) [1]
- \(12 \times 463 = 5556\text{ kJ}\) (Total formed = \(13606\text{ kJ}\)) [1]
Final subtraction (broken - formed): \(10464.5 - 13606 = -3141.5\text{ kJ mol}^{-1}\) [1.5]

(a) Empirical Formula Calculation:
- Carbon: \(\frac{58.8}{12.0} = 4.90\)
- Hydrogen: \(\frac{9.8}{1.0} = 9.80\)
- Oxygen: \(\frac{31.4}{16.0} = 1.96\)

Divide by the smallest value (1.96):
- C: \(\frac{4.90}{1.96} = 2.5\)
- H: \(\frac{9.80}{1.96} = 5.0\)
- O: \(\frac{1.96}{1.96} = 1.0\)

Multiply by 2 to get whole numbers:
Empirical Formula = \(\text{C}_5\text{H}_{10}\text{O}_2\)

(b) Empirical formula mass of \(\text{C}_5\text{H}_{10}\text{O}_2 = (5 \times 12.0) + (10 \times 1.0) + (2 \times 16.0) = 60.0 + 10.0 + 32.0 = 102.0\text{ g mol}^{-1}\).
Since the molecular ion peak in the mass spectrum is at \(m/z = 102\), the molecular mass matches the empirical formula mass.
Therefore, the molecular formula is also \(\text{C}_5\text{H}_{10}\text{O}_2\).

(c) IR Analysis:
- Broad absorption band in the region \(2500 - 3300\text{ cm}^{-1}\) indicates the presence of an \(\text{O-H}\) bond in a carboxylic acid.
- Strong, sharp absorption band at \(1710\text{ cm}^{-1}\) indicates the presence of a \(\text{C=O}\) bond (carbonyl group).
- Together, these show the presence of a carboxylic acid functional group.

(d) Structure and Fragmentation:
Structure: Pentanoic acid, \(\text{CH}_3\text{CH}_2\text{CH}_2\text{CH}_2\text{COOH}\) (accept branched isomers like 3-methylbutanoic acid).

Fragmentation pathways:
- Fragment at \(m/z = 45\):
Equation: \([\text{C}_4\text{H}_9\text{COOH}]^{+\bullet} \rightarrow \text{C}_4\text{H}_9^{\bullet} + \text{COOH}^+\) (or equivalent)
Fragment ion structural formula: \([\text{C}(=\text{O})\text{OH}]^+\)

- Fragment at \(m/z = 57\):
Equation: \([\text{C}_4\text{H}_9\text{COOH}]^{+\bullet} \rightarrow \text{COOH}^{\bullet} + \text{C}_4\text{H}_9^+\) (or equivalent)
Fragment ion structural formula: \([\text{CH}_3\text{CH}_2\text{CH}_2\text{CH}_2]^+\) (butyl cation)

(e) Reagent: Aqueous sodium carbonate (\(\text{Na}_2\text{CO}_3\)) or sodium hydrogencarbonate (\(\text{NaHCO}_3\)).
Observation: Effervescence / bubbling (due to carbon dioxide release).

(a) Divide by relative atomic masses [1]
Calculate ratio (2.5 : 5.0 : 1.0) [1]
State correct empirical formula: \(\text{C}_5\text{H}_{10}\text{O}_2\) [1]

(b) Calculate empirical mass as 102 [1]
State that molecular mass = empirical mass, hence molecular formula is \(\text{C}_5\text{H}_{10}\text{O}_2\) [1]

(c) Correct assignment of \(2500 - 3300\text{ cm}^{-1}\) as O-H in carboxylic acid [1.5]
Correct assignment of \(1710\text{ cm}^{-1}\) as C=O carbonyl [1.5]
Identify the functional group as a carboxylic acid [1.5]

(d) Correct structure drawn/written of pentanoic acid or a structural isomer [2]
Equation and positive ion formula for \(m/z = 45\): \([\text{COOH}]^+\) [2]
Equation and positive ion formula for \(m/z = 57\): \([\text{C}_4\text{H}_9]^+\) [2]
*(Deduct 1 mark overall if positive charges are missing from fragment ions)*

(e) Correct reagent (soluble carbonate/hydrogencarbonate or reactive metal like Mg) [1]
Correct observation (fizzing/effervescence) [1]

(a) Over time, the concentration of hydrogen peroxide molecules decreases because they are being consumed in the reaction. This reduces the number of reactant particles per unit volume, which decreases the frequency of collisions between reactant particles. Consequently, the frequency of successful collisions decreases, leading to a decreased reaction rate.

(b) The Maxwell-Boltzmann distribution diagram should show:
- y-axis labeled as 'Number of molecules' or 'Fraction of molecules' and x-axis labeled as 'Kinetic energy' or 'Energy'.
- Curve \(T_1\) starting at the origin, rising to a peak, and then tailing off towards the x-axis but never touching it.
- \(E_{mp}\) labeled at the peak of the \(T_1\) curve.
- Curve \(T_2\) starting at the origin, with its peak shifted to the right (higher energy) and lower height than curve \(T_1\), crossing the first curve and remaining higher at higher energies.
- Activation energy \(E_a\) marked as a vertical line towards the high-energy end of the x-axis.
- Catalyzed activation energy \(E_c\) marked as a vertical line to the left of \(E_a\).

(c)(i) When the temperature increases to \(T_2\), the average kinetic energy of the molecules increases, shifting the distribution curve to the right. This significantly increases the proportion of molecules with energy greater than or equal to the activation energy (\(E \ge E_a\)), represented by the larger area under the curve to the right of the \(E_a\) line. Consequently, there is a greater frequency of successful collisions. (Additionally, molecules move faster so collide more frequently, though the increase in successful fraction is the dominant factor).

(ii) The manganese(IV) oxide catalyst provides an alternative reaction pathway with a lower activation energy, \(E_c\). As \(E_c < E_a\), a much larger fraction of the molecules now possess sufficient energy to react upon collision (represented by the area under the curve between \(E_c\) and \(E_a\) in addition to the area past \(E_a\)). This increases the rate of successful collisions, thus increasing the rate of reaction.

(d) The student could monitor the reaction by placing the reaction flask on a digital mass balance and measuring the continuous loss of mass over time. As the oxygen gas is produced, it escapes from the open flask (plugged loosely with cotton wool), causing the total mass of the flask to decrease.

(a) Concentration of reactant/hydrogen peroxide decreases [1]
Fewer particles per unit volume leading to lower frequency of collisions [1]
Lower frequency of successful/effective collisions [1]

(b) Fully labeled axes (y: fraction/number of molecules, x: kinetic energy/energy) [1]
Curve \(T_1\) starts at origin, rises, and does not touch x-axis at high energy [1]
Peak of \(T_2\) is lower and shifted to the right of \(T_1\) [1]
\(E_{mp}\) correctly labeled at the maximum of curve \(T_1\) [1]
Correct positions of \(E_a\) and \(E_c\) (with \(E_c\) to the left of \(E_a\)) [2]

(c)(i) Peak shifts to the right / average energy increases [1]
Greater fraction of molecules have energy \(\ge E_a\) [1]
More frequent successful collisions [1]
Reference to shading or increased area under curve to the right of \(E_a\) [0.5]

(c)(ii) Catalyst provides alternative route with lower activation energy (\(E_c\)) [1]
More molecules have energy \(\ge E_c\) [1]
Greater frequency of successful collisions [1]

(d) Monitor mass loss using a digital balance [1]
Mass decreases as oxygen gas escapes [1]
(Accept monitoring color change/absorbance if a colored indicator/reagent is specified, but reject temperature unless related directly to rate quantification).

(a) Trend: The thermal stability of Group 2 carbonates increases down the group.
Explanation:
- Down the group, the ionic radius of the Group 2 cation (\(\text{M}^{2+}\)) increases (while the charge remains \(2+\)).
- Therefore, the charge density of the cation decreases down the group.
- A smaller cation with higher charge density (like \(\text{Mg}^{2+}\)) has a stronger polarizing effect on the large, electron-rich carbonate ion (\(\text{CO}_3^{2-}\)).
- This polarization distorts the electron cloud of the carbonate ion, weakening the \(\text{C-O}\) bond within the carbonate.
- Down the group, the weaker polarizing ability of the larger cations means the \(\text{C-O}\) bonds are less weakened, requiring more thermal energy to break them and decompose the carbonate.

(b) \(\text{CaCO}_3(\text{s}) \xrightarrow{\Delta} \text{CaO}(\text{s}) + \text{CO}_2(\text{g})\)

(c) Solubility of Group 2 hydroxides increases down the group. Because the solubility increases, more hydroxide ions (\(\text{OH}^-\)) dissolve and go into solution down the group. This causes the concentration of hydroxide ions to increase, resulting in an increase in the pH of the saturated solution down the group (solutions become more alkaline).

(d)(i) Distinguishing test:
- Filter the barium sulfate suspension to obtain a clear filtrate (water/extremely dilute barium solution).
- Add aqueous sodium hydroxide (\(\text{NaOH}(\text{aq})\)) or sodium carbonate to samples of both solutions.
- With the magnesium sulfate solution, addition of \(\text{NaOH}(\text{aq})\) will produce a thick white precipitate of magnesium hydroxide, \(\text{Mg(OH)}_2\).
- With the filtered barium suspension filtrate, no precipitate (or a very faint trace) will form because barium hydroxide is highly soluble.
- Alternatively: Conduct a flame test on the solid residue/solutions. Magnesium gives no flame color, whereas barium gives an apple-green flame.

(d)(ii) Precipitation ionic equation:
\(\text{Mg}^{2+}(\text{aq}) + 2\text{OH}^-(\text{aq}) \rightarrow \text{Mg(OH)}_2(\text{s})\)
(or alternative depending on test used, e.g. sulfate test using \(\text{Ba}^{2+}\)).

(a) State that thermal stability increases down the group [1]
Identify that the cationic size increases / ionic radius increases [1]
Identify that the charge density of the cation decreases [1]
Explain that the smaller cation / higher charge density has a greater polarizing effect [1]
Explain that polarization distorts/weakens the C-O bond in the carbonate ion [1]
Conclude that less energy is required to decompose carbonates at the top of the group / more energy at the bottom [1]

(b) Correct chemical formula for reactants and products [1]
Correct state symbols: \(\text{CaCO}_3(\text{s})\), \(\text{CaO}(\text{s})\), \(\text{CO}_2(\text{g})\) [1]

(c) State that solubility of hydroxides increases down the group [1]
State that hydroxide ion concentration increases [1]
Conclude that pH increases down the group [1.5]

(d)(i) Describe filtration of the suspension [1]
Add aqueous sodium hydroxide (or flame test protocol) [1]
Observation for magnesium: white precipitate (or no flame color) [1]
Observation for barium: no precipitate/remains clear (or apple-green flame) [1]

(d)(ii) Correct reactants and products: \(\text{Mg}^{2+}\) and \(\text{OH}^-\) forming \(\text{Mg(OH)}_2\) [1]
Correct state symbols included: \((\text{aq})\) for reactants, \((\text{s})\) for product [1]