2024 IB DP Chemistry Practice Paper with Answers

\(\text{NO}_2^-\) has 3 electron domains around the central nitrogen atom (one lone pair and two bonding domains), giving it a trigonal planar electron domain geometry. Its molecular geometry is bent, with a bond angle slightly less than \(120^\circ\) (approximately \(115^\circ\)). The other species have tetrahedral electron domain geometries with bond angles of around \(104.5^\circ\) (\(\text{H}_2\text{O}\)), \(107^\circ\) (\(\text{H}_3\text{O}^+\)), or \(109.5^\circ\) (\(\text{NH}_4^+\)).

Award [1] for the correct option A. No partial marks.

The structural formula is \(\text{H}_2\text{C}=\text{CH}-\text{C}\equiv\text{N}\). There are 3 \(\text{C}-\text{H}\) single bonds (3 \(\sigma\)), 1 \(\text{C}=\text{C}\) double bond (1 \(\sigma\) and 1 \(\pi\)), 1 \(\text{C}-\text{C}\) single bond (1 \(\sigma\)), and 1 \(\text{C}\equiv\text{N}\) triple bond (1 \(\sigma\) and 2 \(\pi\)). Total \(\sigma\) bonds = 3 + 1 + 1 + 1 = 6. Total \(\pi\) bonds = 1 + 2 = 3.

Award [1] for the correct option C. No partial marks.

Assume 100 g of the compound. Mass of \(\text{M}\) = 70.0 g, mass of \(\text{O}\) = 30.0 g. Moles of \(\text{M}\) = 70.0 / 56.0 = 1.25 mol. Moles of \(\text{O}\) = 30.0 / 16.0 = 1.875 mol. Dividing by the smaller value: ratio of \(\text{M} : \text{O}\) is 1 : 1.5. Multiplying by 2 gives the whole-number empirical ratio of 2 : 3, corresponding to \(\text{M}_2\text{O}_3\).

Award [1] for the correct option C. No partial marks.

Determine the total moles of atoms in each sample:
A: 4.0 g of \(\text{H}_2\) contains (4.0 / 2.0) = 2.0 mol of molecules. Total atoms = 2.0 * 2 = 4.0 mol of atoms.
B: 16.0 g of \(\text{CH}_4\) contains (16.0 / 16.0) = 1.0 mol of molecules. Total atoms = 1.0 * 5 = 5.0 mol of atoms.
C: 28.0 g of \(\text{CO}\) contains (28.0 / 28.0) = 1.0 mol of molecules. Total atoms = 1.0 * 2 = 2.0 mol of atoms.
D: 44.0 g of \(\text{CO}_2\) contains (44.0 / 44.0) = 1.0 mol of molecules. Total atoms = 1.0 * 3 = 3.0 mol of atoms.
Thus, the 16.0 g sample of methane contains the greatest number of atoms.

Award [1] for the correct option B. No partial marks.

Using the dilution formula \(C_1 V_1 = C_2 V_2\): 0.50 * 100 = 0.10 * \(V_2\), which gives \(V_2\) = 500 \(\text{cm}^3\). The volume of distilled water that must be added is the final volume minus the initial volume: 500 - 100 = 400 \(\text{cm}^3\).

Award [1] for the correct option B. No partial marks.

Using powdered calcium carbonate instead of chips increases the surface area of the reactant, increasing the rate of collision and therefore the initial rate of reaction. Since the calcium carbonate is in excess, the limiting reactant is hydrochloric acid. Because the concentration and volume of hydrochloric acid remain unchanged, the number of moles of acid is unchanged, resulting in the same total yield of gas.

Award [1] for the correct option A. No partial marks.

A catalyst lowers the activation energy of the reaction, which shifts the activation energy barrier to the left on the energy axis. It does not alter the actual distribution of molecular kinetic energies, so the Maxwell-Boltzmann distribution curve itself remains unchanged.

Award [1] for the correct option C. No partial marks.

The instantaneous rate of a reaction is defined as the change in concentration per unit time at one specific instant. Graphically, this is found by constructing a tangent to the concentration-time curve at that specific time and calculating its gradient (slope).

Award [1] for the correct option A. No partial marks.

Nitrogen in \( \text{NF}_3 \) has 5 valence electrons. It forms three single covalent bonds with fluorine atoms, leaving one lone pair on the nitrogen atom. This gives a total of four electron domains, leading to a tetrahedral electron domain geometry and a trigonal pyramidal molecular geometry. Because of its asymmetrical shape and highly electronegative fluorine atoms, the molecular dipoles do not cancel, making \( \text{NF}_3 \) polar. \( \text{BF}_3 \) and \( \text{SO}_3 \) are trigonal planar (non-polar), while \( \text{CF}_4 \) is tetrahedral (non-polar).

[1 mark] for selecting option C. No partial credit.

In carbon dioxide (\( \text{CO}_2 \)), there are two double bonds, giving a carbon-oxygen bond order of 2. In the carbonate ion (\( \text{CO}_3^{2-} \)), the negative charge is delocalized over three resonance structures, resulting in a bond order of \( 4/3 \approx 1.33 \). Since a lower bond order means weaker and longer bonds, the C–O bonds in \( \text{CO}_3^{2-} \) are longer and weaker than those in \( \text{CO}_2 \).

[1 mark] for selecting option B. No partial credit.

The structure of acrylonitrile is \( \text{H}_2\text{C}=\text{CH}-\text{C}\equiv\text{N} \). Counting the single and multiple bonds:
- Three C–H single bonds: 3 \( \sigma \)
- One C=C double bond: 1 \( \sigma \) and 1 \( \pi \)
- One C–C single bond: 1 \( \sigma \)
- One C\( \equiv \)N triple bond: 1 \( \sigma \) and 2 \( \pi \)
Total \( \sigma \) bonds = \( 3 + 1 + 1 + 1 = 6 \).
Total \( \pi \) bonds = \( 1 + 2 = 3 \).

[1 mark] for selecting option C. No partial credit.

To find the empirical formula, divide the percentage of each element by its molar mass:
- \( n(\text{C}) = \frac{40.0}{12.0} \approx 3.33 \text{ mol} \)
- \( n(\text{H}) = \frac{6.7}{1.0} \approx 6.7 \text{ mol} \)
- \( n(\text{O}) = \frac{53.3}{16.0} \approx 3.33 \text{ mol} \)
Divide by the smallest value (3.33):
- C: \( \frac{3.33}{3.33} = 1 \)
- H: \( \frac{6.7}{3.33} \approx 2 \)
- O: \( \frac{3.33}{3.33} = 1 \)
The empirical formula is \( \text{CH}_2\text{O} \).

[1 mark] for selecting option B. No partial credit.

Calculate the total moles of nitrogen atoms in each sample:
- Option A: \( 1.0\text{ mol} \times 1 = 1.0\text{ mol of N} \) atoms
- Option B: \( 0.6\text{ mol} \times 2 = 1.2\text{ mol of N} \) atoms
- Option C: \( 0.5\text{ mol} \times 2 = 1.0\text{ mol of N} \) atoms
- Option D: \( 0.3\text{ mol} \times 2 = 0.6\text{ mol of N} \) atoms
Option B contains the highest amount of nitrogen atoms (\( 1.2\text{ mol} \)).

[1 mark] for selecting option B. No partial credit.

According to collision theory and the Maxwell-Boltzmann distribution, increasing the temperature increases the average kinetic energy of the particles. Consequently, a significantly larger fraction of reactant particles have kinetic energy equal to or greater than the activation energy (\( E_a \)), leading to more frequent successful collisions. The activation energy itself remains unchanged (only a catalyst can lower \( E_a \)).

[1 mark] for selecting option B. No partial credit.

Comparing Experiment 1 and Experiment 2: when the concentration of \( \text{X} \) is doubled and the concentration of \( \text{Y} \) is held constant, the rate increases by a factor of 4. This indicates that the reaction is second-order with respect to \( \text{X} \).

Comparing Experiment 2 and Experiment 3: when the concentration of \( \text{Y} \) is doubled and the concentration of \( \text{X} \) is held constant, the rate doubles. This indicates that the reaction is first-order with respect to \( \text{Y} \).

Therefore, the rate equation is \( \text{Rate} = k[\text{X}]^2[\text{Y}] \).

[1 mark] for selecting option B. No partial credit.

For an endothermic reaction, the energy of the products is higher than the energy of the reactants, so \( \Delta H \) is positive. The activation energy (\( E_a \)) is the energy difference between the transition state (the peak of the potential energy curve) and the reactants, which must also be positive. Because the transition state is always at a higher energy level than both the reactants and the products, \( E_a \) must be greater than the overall enthalpy change (\( E_a > \Delta H \)).

[1 mark] for selecting option A. No partial credit.

First, calculate the number of moles of water molecules:

\( n(\text{H}_2\text{O}) = \frac{\text{mass}}{\text{molar mass}} = \frac{1.80\text{ g}}{18.02\text{ g mol}^{-1}} \approx 0.100\text{ mol} \)

Next, calculate the number of water molecules:

\( N(\text{H}_2\text{O}) = 0.100\text{ mol} \times 6.02 \times 10^{23}\text{ mol}^{-1} = 6.02 \times 10^{22} \text{ molecules} \)

Each molecule of water, \( \text{H}_2\text{O} \), contains 3 atoms (2 hydrogen atoms and 1 oxygen atom).

Total number of atoms:

\( 3 \times 6.02 \times 10^{22} = 1.81 \times 10^{23} \text{ atoms} \)

Award 1 mark for the correct option C.
- Award 0 marks for incorrect calculations leading to A, B, or D.

Determine the amount (in mol) of hydrated copper(II) sulfate:

\( n(\text{CuSO}_4 \cdot 5\text{H}_2\text{O}) = \frac{4.99\text{ g}}{249.7\text{ g mol}^{-1}} = 0.0200\text{ mol} \)

Since 1 mole of hydrated copper(II) sulfate yields 1 mole of anhydrous copper(II) sulfate:

\( n(\text{CuSO}_4) = 0.0200\text{ mol} \)

Calculate the mass of anhydrous copper(II) sulfate:

\( m(\text{CuSO}_4) = 0.0200\text{ mol} \times 159.6\text{ g mol}^{-1} = 3.19\text{ g} \)

Award 1 mark for the correct answer B.
- Option A represents an incorrect calculation or misinterpretation.
- Option C is obtained if water mass is incorrectly subtracted without stoichiometric steps.
- Option D is close to the starting mass with incorrect water loss subtraction.

- \( \text{BF}_3 \) has 3 bonding electron domains and 0 lone pairs on the central B atom, leading to a trigonal planar geometry.
- \( \text{PCl}_3 \) has 3 bonding electron domains and 1 lone pair on the central P atom, giving it a trigonal pyramidal geometry.
- \( \text{CO}_3^{2-} \) has 3 bonding domains around the central C atom and 0 lone pairs, resulting in a trigonal planar geometry.
- \( \text{ClF}_3 \) has 3 bonding domains and 2 lone pairs on the central Cl atom, giving it a T-shaped molecular geometry.

Award 1 mark for selecting B.
- All other options represent geometries other than trigonal pyramidal.

The Lewis structure of ozone can be represented as resonance structures: \( \text{O}=\ddot{\text{O}}^+\text{--}\ddot{\text{O}}:^- \).

To find the formal charge (FC) of the central oxygen atom:
- Valence electrons of oxygen (V) = 6
- Non-bonding valence electrons on central oxygen (N) = 2 (1 lone pair)
- Bonding electrons shared by central oxygen (B) = 6 (one double bond and one single bond)

\( \text{FC} = V - N - \frac{B}{2} = 6 - 2 - \frac{6}{2} = +1 \)

Award 1 mark for the correct formal charge C.
- Option A corresponds to the terminal single-bonded oxygen atom.
- Option B corresponds to the terminal double-bonded oxygen atom in an uncharged state.
- Option D is an incorrect calculation.

When the temperature increases:
- The average kinetic energy of the molecules increases, shifting the peak of the Maxwell-Boltzmann distribution curve to the right (and flattening it).
- The activation energy, \( E_a \), remains constant because it is a characteristic property of the reaction path.
- A larger fraction of the reactant molecules possess kinetic energy greater than or equal to the activation energy (\( E \ge E_a \)), leading to a higher frequency of successful collisions and an increased reaction rate.

Award 1 mark for B.
- Options A and C are incorrect because the peak shifts to the right at higher temperatures.
- Options C and D are incorrect because temperature changes do not alter the activation energy (\( E_a \)) itself.

1. Compare Experiment 1 and Experiment 2:
- \( [\text{B}] \) is constant.
- \( [\text{A}] \) doubles (from 0.10 to 0.20).
- The rate doubles (from \( 2.0 \times 10^{-4} \) to \( 4.0 \times 10^{-4} \)).
- Therefore, the reaction is first-order with respect to A: order = 1.

2. Compare Experiment 1 and Experiment 3:
- \( [\text{A}] \) is constant.
- \( [\text{B}] \) doubles (from 0.10 to 0.20).
- The rate quadruples (from \( 2.0 \times 10^{-4} \) to \( 8.0 \times 10^{-4} \)).
- Therefore, the reaction is second-order with respect to B: order = 2.

3. Calculate the overall order of the reaction:
- \( \text{Overall order} = 1 + 2 = 3 \).

Award 1 mark for the correct overall order C.
- Option A assumes first-order overall.
- Option B incorrectly deduces first-order for both reactants.
- Option D incorrectly deduces second-order for both reactants.

Lattice enthalpy depends on the ionic charges and ionic radii:

\( \Delta H_{\text{lat}} \propto \frac{q_+ q_-}{r_+ + r_-} \)

1. Compare the charges of the constituent ions:
- \( \text{NaCl} \): \( +1 \) and \( -1 \)
- \( \text{MgO} \): \( +2 \) and \( -2 \)
- \( \text{CaO} \): \( +2 \) and \( -2 \)
- \( \text{MgF}_2 \): contains divalent \( \text{Mg}^{2+} \) and monovalent \( \text{F}^- \) ions.

Since \( \text{MgO} \) and \( \text{CaO} \) both have \( +2 \) and \( -2 \) ions, they will have significantly more exothermic lattice enthalpies than \( \text{NaCl} \) and \( \text{MgF}_2 \).

2. Compare ionic radii between \( \text{MgO} \) and \( \text{CaO} \):
- The ionic radius of \( \text{Mg}^{2+} \) is smaller than that of \( \text{Ca}^{2+} \).
- Thus, the interionic distance in \( \text{MgO} \) is smaller than in \( \text{CaO} \), resulting in a stronger electrostatic attraction and the most exothermic lattice enthalpy.

Award 1 mark for selecting B.
- Option A has low charge product.
- Option C has larger cation size than B.
- Option D has lower anion charge than B.

According to the Brønsted–Lowry theory, a base is a proton (\( \text{H}^+ \)) acceptor.

- In the forward reaction, \( \text{HCO}_3^- \) accepts a proton from \( \text{H}_2\text{PO}_4^- \) to form \( \text{H}_2\text{CO}_3 \). Thus, \( \text{HCO}_3^- \) is the base.
- In the reverse reaction, \( \text{HPO}_4^{2-} \) accepts a proton from \( \text{H}_2\text{CO}_3 \) to form \( \text{H}_2\text{PO}_4^- \). Thus, \( \text{HPO}_4^{2-} \) is the base.

Award 1 mark for A.
- Option B lists the Brønsted-Lowry acids.
- Options C and D mix Brønsted-Lowry acids and bases.

To determine the bond angle, we look at the Lewis structure and valence shell electron pair repulsion (VSEPR) theory. In \(\text{H}_3\text{O}^+\), the central oxygen atom has 3 bonding pairs and 1 lone pair. This gives it a tetrahedral electron domain geometry and a trigonal pyramidal molecular geometry. Due to lone-pair-bonding-pair repulsion, the bond angle is reduced from the ideal \(109.5^\circ\) to approximately \(107^\circ\). In contrast, \(\text{NH}_4^+\) is tetrahedral with an angle of \(109.5^\circ\), \(\text{BF}_3\) is trigonal planar with \(120^\circ\), and \(\text{CO}_2\) is linear with \(180^\circ\).

Award 1 mark for selecting the correct option A. No partial marks are awarded for multiple-choice questions.

One formula unit of iron(III) sulfate, \(\text{Fe}_2(\text{SO}_4)_3\), dissociates to produce 5 ions in total: 2 \(\text{Fe}^{3+}\) ions and 3 \(\text{SO}_4^{2-}\) ions. Therefore, \(0.10\text{ mol}\) of \(\text{Fe}_2(\text{SO}_4)_3\) contains \(0.10 \times 5 = 0.50\text{ mol}\) of ions. The total number of ions is calculated by multiplying the moles of ions by Avogadro's constant: \(0.50\text{ mol} \times 6.0 \times 10^{23}\text{ mol}^{-1} = 3.0 \times 10^{23}\).

Award 1 mark for selecting the correct option C. No partial marks are awarded for multiple-choice questions.

A catalyst increases the rate of a chemical reaction by providing an alternative reaction pathway that has a lower activation energy (\(E_a\)). This allows a greater proportion of colliding reactant particles to have kinetic energy equal to or greater than the activation energy, increasing the frequency of successful collisions. Catalysts do not increase the average kinetic energy of the particles (which depends only on temperature), nor do they change the enthalpy change (\(\Delta H\)) of the reaction.

Award 1 mark for selecting the correct option C. No partial marks are awarded for multiple-choice questions.

Let us count the bonds in \(\text{CH}_2=\text{CHCOOCH}_3\). There are 6 \(\text{C}-\text{H}\) single bonds (all are \(\sigma\) bonds). There is 1 \(\text{C}-\text{C}\) single bond (1 \(\sigma\) bond). There is 1 \(\text{C}=\text{C}\) double bond (1 \(\sigma\) and 1 \(\pi\) bond). There is 1 \(\text{C}=\text{O}\) double bond (1 \(\sigma\) and 1 \(\pi\) bond). There are 2 \(\text{C}-\text{O}\) single bonds (2 \(\sigma\) bonds). Summing these up gives: Total \(\sigma\) bonds = \(6 + 1 + 1 + 1 + 2 = 11\). Total \(\pi\) bonds = \(1 + 1 = 2\). Therefore, the molecule has 11 \(\sigma\) and 2 \(\pi\) bonds.

Award 1 mark for selecting the correct option B. No partial marks are awarded for multiple-choice questions.

First, find the percentage of hydrogen by mass: \(100.0\% - 85.7\% = 14.3\%\). Next, calculate the relative molar amounts by dividing each percentage by its molar mass: Moles of C = \(85.7 / 12.01 \approx 7.14\text{ mol}\), and Moles of H = \(14.3 / 1.01 \approx 14.16\text{ mol}\). Finally, find the simplest whole-number ratio by dividing both values by the smallest number of moles (7.14): Carbon ratio = \(7.14 / 7.14 = 1\), and Hydrogen ratio = \(14.16 / 7.14 \approx 2\). This gives an empirical formula of \(\text{CH}_2\).

Award 1 mark for selecting the correct option B. No partial marks are awarded for multiple-choice questions.

The units of reaction rate are \(\text{mol dm}^{-3}\text{ s}^{-1}\). Rearranging the rate expression for \(k\) gives: \(k = \text{Rate} / ([\text{A}][\text{B}]^2)\). Substituting the units into this expression: \(\text{Units of } k = (\text{mol dm}^{-3}\text{ s}^{-1}) / ((\text{mol dm}^{-3}) \times (\text{mol dm}^{-3})^2) = (\text{mol dm}^{-3}\text{ s}^{-1}) / (\text{mol}^3\text{ dm}^{-9}) = \text{mol}^{-2}\text{ dm}^6\text{ s}^{-1}\), which can be written as \(\text{dm}^6\text{ mol}^{-2}\text{ s}^{-1}\).

Award 1 mark for selecting the correct option C. No partial marks are awarded for multiple-choice questions.

(a) To draw the Lewis structure of \( \text{XeF}_4 \):
- Total valence electrons = \( 8\text{ (from Xe)} + 4 \times 7\text{ (from F)} = 36 \) electrons (18 pairs).
- Xe forms four single bonds with the four F atoms (using 8 electrons).
- Each F atom is completed with three lone pairs (using 24 electrons).
- This leaves 4 electrons (2 pairs), which are placed on the central Xe atom as lone pairs.
- The central Xe atom has an expanded octet with 12 valence electrons surrounding it.

(b) The central Xe atom has 6 electron domains (4 bonding pairs and 2 lone pairs). Under VSEPR theory, this corresponds to an octahedral electron domain geometry. To minimize repulsion, the two lone pairs occupy opposite positions, resulting in a square planar molecular geometry. The F–Xe–F bond angles in the plane are exactly 90°.

(c) Although fluorine is much more electronegative than xenon, making each Xe–F bond highly polar, the molecular geometry is square planar. This highly symmetrical arrangement results in the individual bond dipole moments cancelling each other out completely, giving a net dipole moment of zero.

(a) [2 marks]
- 1 mark for showing four single covalent bonds between Xe and F and complete octets on all four F atoms.
- 1 mark for showing two lone pairs on the central Xe atom.

(b) [2 marks]
- 1 mark for stating "square planar".
- 1 mark for stating "90°" (accept "90° and 180°").

(c) [3 marks]
- 1 mark for mentioning the symmetrical shape/geometry of the molecule.
- 1 mark for stating that bond dipoles are equal and opposite / point in opposite directions.
- 1 mark for stating that the dipoles cancel each other out (or net dipole moment is zero).

(a) Calculate masses:
- Mass of C in \( 2.64\text{ g } \text{CO}_2 \):
\( m(\text{C}) = 2.64\text{ g} \times \frac{12.01\text{ g mol}^{-1}}{44.01\text{ g mol}^{-1}} = 0.72\text{ g} \)
- Mass of H in \( 1.44\text{ g } \text{H}_2\text{O} \):
\( m(\text{H}) = 1.44\text{ g} \times \frac{2 \times 1.01\text{ g mol}^{-1}}{18.02\text{ g mol}^{-1}} = 0.16\text{ g} \)
- Mass of O by subtraction:
\( m(\text{O}) = 1.20\text{ g} - 0.72\text{ g} - 0.16\text{ g} = 0.32\text{ g} \)

(b) Determine empirical formula:
- Moles of C: \( \frac{0.72\text{ g}}{12.01\text{ g mol}^{-1}} = 0.060\text{ mol} \)
- Moles of H: \( \frac{0.16\text{ g}}{1.01\text{ g mol}^{-1}} = 0.158\text{ mol} \approx 0.160\text{ mol} \)
- Moles of O: \( \frac{0.32\text{ g}}{16.00\text{ g mol}^{-1}} = 0.020\text{ mol} \)
- Divide by the smallest value (0.020):
\( \text{C} = \frac{0.060}{0.020} = 3 \)
\( \text{H} = \frac{0.160}{0.020} = 8 \)
\( \text{O} = \frac{0.020}{0.020} = 1 \)
- The empirical formula is \( \text{C}_3\text{H}_8\text{O} \).

(c) Determine molecular formula:
- Empirical formula mass of \( \text{C}_3\text{H}_8\text{O} = (3 \times 12.01) + (8 \times 1.01) + 16.00 = 60.11\text{ g mol}^{-1} \).
- Ratio \( N = \frac{\text{Molar Mass}}{\text{Empirical Mass}} = \frac{120.20}{60.11} \approx 2 \).
- Molecular formula = \( 2 \times (\text{C}_3\text{H}_8\text{O}) = \text{C}_6\text{H}_{16}\text{O}_2 \).

(a) [3 marks]
- 1 mark for correct calculation of mass of C (0.72 g).
- 1 mark for correct calculation of mass of H (0.16 g).
- 1 mark for correct calculation of mass of O by subtraction (0.32 g). Award full marks if correct final masses are obtained through alternative correct pathways.

(b) [2 marks]
- 1 mark for converting masses to moles of C (0.060), H (0.160), and O (0.020).
- 1 mark for the correct simplest whole-number ratio and final empirical formula (\( \text{C}_3\text{H}_8\text{O} \)).

(c) [2 marks]
- 1 mark for calculating the empirical formula mass (60.11 g/mol).
- 1 mark for the correct molecular formula (\( \text{C}_6\text{H}_{16}\text{O}_2 \)). Award 2 marks if molecular formula is derived directly and correctly.

(a) Determination of reaction orders:
- **For \( \text{NO} \)**: Compare Experiments 1 and 2. \( [\text{Cl}_2] \) is held constant at \( 0.10\text{ mol dm}^{-3} \), while \( [\text{NO}] \) is doubled (from \( 0.10 \) to \( 0.20\text{ mol dm}^{-3} \)). The rate increases by a factor of \( \frac{4.8 \times 10^{-3}}{1.2 \times 10^{-3}} = 4 \). Since \( 2^2 = 4 \), the reaction is second order with respect to \( \text{NO} \).
- **For \( \text{Cl}_2 \)**: Compare Experiments 1 and 3. \( [\text{NO}] \) is held constant at \( 0.10\text{ mol dm}^{-3} \), while \( [\text{Cl}_2] \) is doubled (from \( 0.10 \) to \( 0.20\text{ mol dm}^{-3} \)). The rate increases by a factor of \( \frac{2.4 \times 10^{-3}}{1.2 \times 10^{-3}} = 2 \). Since \( 2^1 = 2 \), the reaction is first order with respect to \( \text{Cl}_2 \).

(b) The rate expression combines these findings:
\( \text{Rate} = k[\text{NO}]^2[\text{Cl}_2] \)

(c) Calculation of \( k \) and units:
- Substitute data from Experiment 1:
\( 1.2 \times 10^{-3}\text{ mol dm}^{-3}\text{ s}^{-1} = k (0.10\text{ mol dm}^{-3})^2 (0.10\text{ mol dm}^{-3}) \)
\( 1.2 \times 10^{-3} = k (0.010) (0.10) \)
\( 1.2 \times 10^{-3} = k (1.0 \times 10^{-3}) \)
\( k = 1.2 \)
- Determination of units:
\( \text{Units of } k = \frac{\text{mol dm}^{-3}\text{ s}^{-1}}{(\text{mol dm}^{-3})^3} = \text{dm}^6\text{ mol}^{-2}\text{ s}^{-1} \)

(a) [4 marks]
- 1 mark for identifying that comparing Exp 1 and 2 isolates \( [\text{NO}] \).
- 1 mark for deducing second order with respect to \( \text{NO} \).
- 1 mark for identifying that comparing Exp 1 and 3 isolates \( [\text{Cl}_2] \).
- 1 mark for deducing first order with respect to \( \text{Cl}_2 \).

(b) [1 mark]
- 1 mark for the correct rate expression: \( \text{Rate} = k[\text{NO}]^2[\text{Cl}_2] \) (Accept "rate" or "R" for Rate).

(c) [2 marks]
- 1 mark for calculating the correct numerical value of \( k = 1.2 \).
- 1 mark for the correct units: \( \text{dm}^6\text{ mol}^{-2}\text{ s}^{-1} \) (accept \( \text{dm}^6\text{/mol}^2\text{s} \)).

(a) Salicylic acid (2-hydroxybenzoic acid) contains:
1. A carboxylic acid group (\( -\text{COOH} \))
2. A phenol / hydroxyl group (\( -\text{OH} \) bonded to a benzene ring)
During acetylation, the phenol / hydroxyl group reacts with ethanoic anhydride to form the ester link in aspirin.

(b) Atom economy is calculated using the formula:
\( \text{Atom Economy} = \frac{\text{Molar mass of desired product}}{\text{Total molar mass of reactants}} \times 100\% \)
- Desired product = Aspirin (\( 180.17\text{ g mol}^{-1} \))
- Total mass of reactants = \( M(\text{salicylic acid}) + M(\text{ethanoic anhydride}) = 138.13 + 102.10 = 240.23\text{ g mol}^{-1} \)
- \( \text{Atom Economy} = \frac{180.17}{240.23} \times 100\% = 75.00\% \)

(c) Melting point determination:
- Heat a small sample of the dry product in a melting point apparatus.
- Record the temperature range over which it melts.
- Pure aspirin melts sharply at a specific temperature (around 136 °C).
- If impurities are present, the melting point will be lowered and the melting range will be wider (broad).

(a) [3 marks]
- 1 mark for "carboxylic acid".
- 1 mark for "phenol" (accept "hydroxyl" / "hydroxyl group on a benzene ring").
- 1 mark for stating that the "phenol / hydroxyl" group reacts.

(b) [2 marks]
- 1 mark for calculating the total mass of reactants (240.23 g/mol).
- 1 mark for the correct atom economy calculation and final percentage (75.00% or 75.0%).

(c) [2 marks]
- 1 mark for stating that impurities lower the melting point and broaden the melting range.
- 1 mark for stating that the experimental melting point is compared to the literature value / pure standard to check for deviations.

(a) Hybridization determination:
- In \( \text{CO}_3^{2-} \), the central carbon atom has 3 electron domains (3 single-bonding pathways, no lone pairs in the resonance average). Thus, the hybridization of the carbon atom is \( \text{sp}^2 \).
- In \( \text{CO}_2 \), the central carbon atom has 2 electron domains (2 double bonds, no lone pairs). Thus, the hybridization of the carbon atom is \( \text{sp} \).

(b) Structuring the comparison and explanation:
- **Comparison**: The carbon-to-oxygen bonds in the carbonate ion are longer and weaker than the carbon-to-oxygen bonds in carbon dioxide.
- **Explanation**:
1. In \( \text{CO}_2 \), the carbon-oxygen bonds are localized double bonds (bond order = 2.0).
2. In \( \text{CO}_3^{2-} \), there are three resonance structures. The pi-bonding electrons are delocalized over all three carbon-oxygen positions.
3. This results in an intermediate bond order of \( 1.33 \) (or \( 4/3 \)) for all three C–O bonds. Since a higher bond order correlates with shorter and stronger bonds, the bonds in \( \text{CO}_3^{2-} \) (bond order 1.33) are longer and weaker than those in \( \text{CO}_2 \) (bond order 2).

(a) [2 marks]
- 1 mark for identifying the hybridization of carbon in \( \text{CO}_3^{2-} \) as \( \text{sp}^2 \).
- 1 mark for identifying the hybridization of carbon in \( \text{CO}_2 \) as \( \text{sp} \).

(b) [5 marks]
- 1 mark for stating that C–O bonds in \( \text{CO}_3^{2-} \) are longer (than in \( \text{CO}_2 \)).
- 1 mark for stating that C–O bonds in \( \text{CO}_3^{2-} \) are weaker (than in \( \text{CO}_2 \)).
- 1 mark for stating that \( \text{CO}_2 \) has double bonds / bond order of 2.
- 1 mark for mentioning that \( \text{CO}_3^{2-} \) exhibits resonance / delocalization of pi electrons.
- 1 mark for stating that the bond order in \( \text{CO}_3^{2-} \) is intermediate / equal to 1.33 / between a single and a double bond.

(a) Dehydration equation:
\( \text{CaSO}_4 \cdot x\text{H}_2\text{O(s)} \rightarrow \text{CaSO}_4\text{(s)} + x\text{H}_2\text{O(g)} \)
(Accepting the version with the calculated value of \( x = 2 \): \( \text{CaSO}_4 \cdot 2\text{H}_2\text{O(s)} \rightarrow \text{CaSO}_4\text{(s)} + 2\text{H}_2\text{O(g)} \)).

(b) Calculation:
- Mass of hydrated salt = \( 21.94\text{ g} - 18.50\text{ g} = 3.44\text{ g} \)
- Mass of anhydrous salt (\( \text{CaSO}_4 \)) = \( 21.22\text{ g} - 18.50\text{ g} = 2.72\text{ g} \)
- Mass of water lost = \( 21.94\text{ g} - 21.22\text{ g} = 0.72\text{ g} \)
- Moles of \( \text{CaSO}_4 = \frac{2.72\text{ g}}{136.15\text{ g mol}^{-1}} = 0.0200\text{ mol} \)
- Moles of \( \text{H}_2\text{O} = \frac{0.72\text{ g}}{18.02\text{ g mol}^{-1}} = 0.03996\text{ mol} \approx 0.0400\text{ mol} \)
- Mole ratio \( \text{H}_2\text{O} : \text{CaSO}_4 = \frac{0.0400}{0.0200} = 2 \)
- Therefore, \( x = 2 \).

(c) Heating to constant mass involves heating the sample, allowing it to cool (ideally in a desiccator), and weighing it. This process is repeated until there is no further change in mass between consecutive weighings. This is essential to guarantee that all the water of crystallization has been completely evaporated. If the heating is insufficient, some water remains, leading to an underestimated mass of water lost and a lower calculated value of \( x \).

(a) [1 mark]
- 1 mark for correct chemical equation including state symbols (accept balanced equation with either \( x \) or 2).

(b) [3 marks]
- 1 mark for calculating correct masses of anhydrous salt (2.72 g) and water lost (0.72 g).
- 1 mark for converting both masses into moles (0.020 mol of \( \text{CaSO}_4 \) and 0.040 mol of \( \text{H}_2\text{O} \)).
- 1 mark for the correct mole ratio leading to \( x = 2 \).

(c) [3 marks]
- 1 mark for describing the process (heat, cool, weigh, repeat).
- 1 mark for stating that repetition stops when consecutive mass measurements are the same / constant.
- 1 mark for explaining that this ensures all water is removed / reaction has gone to completion.

(a & b) Visualizing the Maxwell-Boltzmann Distribution:
- **Axes**: The vertical axis (y-axis) must be labeled "Number of particles" or "Fraction of particles with energy \( E \)". The horizontal axis (x-axis) must be labeled "Kinetic energy" (or "Energy").
- **Curves**:
- The curve for \( T_1 \) starts at the origin, rises to a peak, and slopes down asymmetrically, never quite touching the x-axis.
- The curve for \( T_2 \) (higher temperature) must have its peak shifted to the right, have a lower peak height, and be flatter (broader) than the \( T_1 \) curve, crossing the \( T_1 \) curve once.
- **Activation energies**:
- \( E_a \) (uncatalyzed) is a vertical line on the right-hand tail of the curves.
- \( E_{a,\text{cat}} \) (catalyzed) is a vertical line positioned to the left of \( E_a \).

(c) Explanation using collision theory:
- **Catalyst**: A catalyst provides an alternative reaction pathway with a lower activation energy (represented by \( E_{a,\text{cat}} \) being to the left of \( E_a \)). This means that at any given temperature, a significantly larger fraction of molecules possess kinetic energy greater than or equal to this lower activation energy (the area under the curve to the right of \( E_{a,\text{cat}} \) is larger). This results in a higher frequency of successful collisions, thus increasing the rate of reaction.
- **Temperature Increase**: Increasing the temperature from \( T_1 \) to \( T_2 \) increases the average kinetic energy of the particles. The distribution curve flattens and shifts to the right, meaning a greater fraction of the particles have kinetic energy greater than or equal to the activation energy, \( E_a \) (the area under the \( T_2 \) curve to the right of \( E_a \) is larger than that for the \( T_1 \) curve). This increases the frequency of successful collisions, and hence the reaction rate.

(a) [2 marks]
- 1 mark for correctly labeled axes (y-axis: Fraction / Number of molecules/particles; x-axis: Kinetic Energy / Energy).
- 1 mark for drawing a correct asymmetrical curve starting at the origin and marking \( E_a \).

(b) [2 marks]
- 1 mark for drawing the \( T_2 \) curve correctly (peak lower, shifted to the right, and broader than the \( T_1 \) curve).
- 1 mark for marking \( E_{a,\text{cat}} \) to the left of \( E_a \).

(c) [3 marks]
- 1 mark for stating that a catalyst lowers the activation energy / provides an alternative pathway, increasing the fraction of particles with enough energy to react.
- 1 mark for stating that increasing temperature increases the average kinetic energy of the particles, so a larger fraction of particles have energy \( \ge E_a \).
- 1 mark for connecting both effects to an increase in the frequency of successful / fruitful collisions.

(a) Doubling the concentration of \(\text{H}_2\text{O}_2\) (e.g., from \(0.100\) to \(0.200\text{ mol dm}^{-3}\)) doubles the initial rate (from \(1.60 \times 10^{-4}\) to \(3.20 \times 10^{-4}\text{ mol dm}^{-3}\text{ s}^{-1}\)). Therefore, the reaction is first order with respect to \(\text{H}_2\text{O}_2\).

(b) \(\text{Rate} = k[\text{H}_2\text{O}_2]\). Using Run 1:
\(1.60 \times 10^{-4}\text{ mol dm}^{-3}\text{ s}^{-1} = k \times 0.100\text{ mol dm}^{-3}\)
\(k = 1.60 \times 10^{-3}\text{ s}^{-1}\).

(c) Percentage uncertainty = \(\frac{0.5}{25.0} \times 100\% = 2.0\%\).

(d) The increase in temperature increases the rate of reaction. Since the calculated rate constant is determined from \(k = \frac{\text{Rate}}{[\text{H}_2\text{O}_2]}\), the calculated value of \(k\) will also be larger. At higher temperatures, particles have more kinetic energy, leading to a higher frequency of successful collisions as a greater fraction of molecules have energy \(E \ge E_a\).

(a) First order [1 mark]; because doubling the concentration of reactant doubles the rate of reaction [1 mark].
(b) \(k = 1.60 \times 10^{-3}\) [1 mark]; units of \(\text{s}^{-1}\) [1 mark].
(c) \(2.0\%\) (accept \(2\%\)) [1 mark].
(d) Rate and rate constant both increase [1 mark]; because there is a higher frequency of successful collisions / more particles have energy greater than or equal to the activation energy [1 mark].

(a) The student heated the residue three times to ensure "constant mass" was reached, verifying that all of the water of crystallization has been completely evaporated.

(b) Mass of anhydrous \(\text{MgSO}_4\) = \(19.653\text{ g} - 18.450\text{ g} = 1.203\text{ g}\).
Mass of water lost = \(20.915\text{ g} - 19.653\text{ g} = 1.262\text{ g}\).

(c) Moles of \(\text{MgSO}_4\) = \(\frac{1.203\text{ g}}{120.38\text{ g mol}^{-1}} = 0.009993\text{ mol}\).
Moles of \(\text{H}_2\text{O}\) = \(\frac{1.262\text{ g}}{18.02\text{ g mol}^{-1}} = 0.07003\text{ mol}\).
Ratio \(x = \frac{0.07003}{0.009993} = 7.008 \approx 7\).
Therefore, \(x = 7\).

(d) If cooled without a lid, the anhydrous \(\text{MgSO}_4\) residue would absorb water vapor/moisture from the air. This increases the measured final mass of the residue, reducing the calculated mass of water lost and resulting in an underestimated value of \(x\).

(a) To reach constant mass / to ensure all water of crystallization is driven off [1 mark].
(b) Mass of anhydrous salt = \(1.203\text{ g}\) [1 mark]; Mass of water lost = \(1.262\text{ g}\) [1 mark].
(c) Moles of anhydrous salt = \(0.0100\text{ mol}\) AND moles of water lost = \(0.0700\text{ mol}\) [1 mark]; molar ratio calculation [1 mark]; \(x = 7\) [1 mark].
(d) Anhydrous residue absorbs water from the air [1 mark]; this leads to a higher residue mass, representing a lower calculated mass of water lost, which underestimates \(x\) [1 mark].

To recrystallize crude aspirin: 1. Dissolve the crude product in the minimum volume of hot solvent (such as ethanol or water). 2. Filter the hot solution to remove any insoluble impurities. 3. Allow the filtrate to cool slowly to room temperature (and then in an ice bath) to recrystallize the aspirin while keeping soluble impurities dissolved in the solvent. 4. Filter the crystals using a Buchner funnel under reduced pressure, wash with a small portion of ice-cold solvent, and dry. To verify purity: Measure the melting point of the dry product using a melting point apparatus. Compare the experimental melting point range with the literature value of pure aspirin (approximately 136 °C). Pure substances have a sharp melting point close to the literature value, whereas impurities depress (lower) the melting point and broaden the melting range.

Award 1 mark for each of the following up to 4 marks: [1] Dissolve crude aspirin in minimum volume of hot solvent. [1] Cool the solution to recrystallize AND filter the crystals (e.g., using a Buchner funnel), washing with cold solvent. [1] Measure the melting point range of the dry crystals. [1] State that pure aspirin has a sharp melting point close to the literature value (136 °C) OR that impurities lower the melting point and widen the range.

Morphine has two polar hydroxyl (-OH) groups attached to its ring structure. In diamorphine, these two hydroxyl groups are converted into less polar ester (-OCOCH3) groups via an esterification reaction with ethanoic anhydride. Because of these ester groups, diamorphine is significantly less polar and more lipid-soluble than morphine. The blood-brain barrier is composed of a lipid membrane, which easily allows non-polar, lipid-soluble molecules to diffuse through. Therefore, diamorphine crosses the blood-brain barrier much more rapidly and in greater concentrations than morphine. Once inside the brain, diamorphine is metabolized back into morphine to bind to opioid receptors, making diamorphine a much more potent and faster-acting analgesic.

Award 1 mark for each of the following up to 4 marks: [1] Morphine contains two hydroxyl groups, whereas diamorphine contains two ester/acetyl groups. [1] Diamorphine is less polar / more lipid-soluble than morphine. [1] Diamorphine crosses the blood-brain barrier more easily/rapidly. [1] Consequently, diamorphine is more potent / faster-acting.

Step 1: Write the balanced chemical equation: \text{Mg(OH)}_2\text{(s)} + 2\text{HCl(aq)} \rightarrow \text{MgCl}_2\text{(aq)} + 2\text{H}_2\text{O(l)}. Step 2: Calculate the moles of \text{HCl} present in the stomach acid: \(n(\text{HCl}) = \text{concentration} \times \text{volume} = 2.0 \times 10^{-2}\text{ mol dm}^{-3} \times 0.150\text{ dm}^3 = 3.0 \times 10^{-3}\text{ mol}\). Step 3: Use the stoichiometric ratio from the balanced equation (1 mol of \text{Mg(OH)}_2 reacts with 2 mol of \text{HCl}) to find the moles of \text{Mg(OH)}_2 required: \(n(\text{Mg(OH)}_2) = \frac{3.0 \times 10^{-3}\text{ mol}}{2} = 1.5 \times 10^{-3}\text{ mol}\). Step 4: Calculate the volume of \text{Mg(OH)}_2 suspension needed: \(V = \frac{n}{C} = \frac{1.5 \times 10^{-3}\text{ mol}}{0.10\text{ mol dm}^{-3}} = 0.015\text{ dm}^3 = 15\text{ cm}^3\).

Award 1 mark for each of the following: [1] Correctly balanced chemical equation: \text{Mg(OH)}_2 + 2\text{HCl} \rightarrow \text{MgCl}_2 + 2\text{H}_2\text{O} (state symbols not required). [1] Calculation of moles of \text{HCl}: \(3.0 \times 10^{-3}\text{ mol}\). [1] Calculation of moles of \text{Mg(OH)}_2: \(1.5 \times 10^{-3}\text{ mol}\) (using 1:2 ratio). [1] Correct final volume of magnesium hydroxide: \(15\text{ cm}^3\) (or \(0.015\text{ dm}^3\)).

Neuraminidase is an enzyme on the surface of the influenza virus that cleaves sialic acid receptors on the host cell membrane, allowing newly replicated viral particles to be released (bud off) and infect neighboring cells. Antiviral drugs like oseltamivir and zanamivir act as neuraminidase inhibitors. They have structures similar to sialic acid and bind to the active site of neuraminidase, preventing the enzyme from functioning. Consequently, the viruses remain trapped inside or attached to the host cell and cannot spread. Comparing structures: Zanamivir contains polar hydroxyl groups (-OH), a carboxylic acid group (-COOH), and a guanidino group. Oseltamivir lacks these (it contains an ester group instead of carboxylic acid, and ether/alkoxy groups instead of hydroxyl groups, and does not have a guanidino group).

Award 1 mark for each of the following up to 4 marks: [1] They bind to the active site of the neuraminidase enzyme. [1] This prevents the release of newly formed viral particles from the host cell. [1] This limits the spread of the virus to other cells. [1] Identify any one functional group present in zanamivir but absent in oseltamivir: hydroxyl (alcohol) OR carboxylic acid OR guanidino group (accept formula or name).

High-level waste (HLW) is characterized by high levels of radioactivity and isotopes with very long half-lives. Low-level waste (LLW) is characterized by low levels of radioactivity and isotopes with short half-lives. Due to their differences, disposal methods differ significantly. LLW is typically stored in secure, shielded containers on-site or in specialized facilities until the radioactivity has decayed to background levels (due to the short half-lives). Once safe, it can be disposed of in conventional landfills or municipal waste systems. In contrast, HLW cannot simply be allowed to decay in the short term; it must be vitrified (mixed with molten glass) or encased in concrete, placed in heavily shielded casks, and buried deep underground in stable geological formations for thousands of years.

Award 1 mark for each of the following up to 4 marks: [1] HLW has high activity and long half-lives, while LLW has low activity and short half-lives. [1] LLW disposal: Stored in shielded containers/on-site until radioactivity decays. [1] LLW disposal: Disposed of as conventional waste / in landfills once safe. [1] HLW comparison: HLW requires vitrification/shielding and deep geological burial (whereas LLW does not).