2025 AQA IAS-Level Chemistry (9620) Practice Paper with Answers

A ground-state chromium atom has the electronic configuration \(1s^2 2s^2 2p^6 3s^2 3p^6 3d^5 4s^1\). When forming a chromium(III) ion, three electrons are removed. The single electron in the \(4s\) subshell is lost first, followed by two electrons from the \(3d\) subshell. This yields the final electronic configuration of \(1s^2 2s^2 2p^6 3s^2 3p^6 3d^3\).

1 mark: Correct electronic configuration (\(1s^2 2s^2 2p^6 3s^2 3p^6 3d^3\)). 1 mark: Correctly identifying the \(4s\) subshell as losing electrons first.

First, convert all units to standard SI units: \(P = 100,000\text{ Pa}\), \(V = 2.48 \times 10^{-4}\text{ m}^3\), and \(T = 298\text{ K}\). Next, calculate the moles of gas using the ideal gas equation: \(n = \frac{PV}{RT} = \frac{100,000 \times 2.48 \times 10^{-4}}{8.31 \times 298} = 0.010014\text{ mol}\). Finally, determine the relative molecular mass: \(M_r = \frac{\text{mass}}{n} = \frac{0.581}{0.010014} = 58.0\text{ g mol}^{-1}\) (to 3 significant figures).

1 mark: Correct calculation of gas amount, showing \(0.0100\text{ mol}\) (or unrounded intermediate). 1 mark: Correct calculation of \(M_r\) as 58.0 (must be to 3 significant figures).

The central chlorine atom in \(\text{ClF}_2^+\) has 6 valence electrons after accounting for the positive charge. It forms 2 single covalent bonds with fluorine atoms, leaving 4 non-bonding electrons (which make up 2 lone pairs). The 4 electron pairs adopt a tetrahedral distribution, resulting in a bent (or V-shaped) molecular geometry. Strong repulsion from the two lone pairs reduces the bond angle from the standard tetrahedral angle of \(109.5^\circ\) to approximately \(104.5^\circ\) (accept values in the range of \(104^\circ\) to \(105^\circ\)).

1 mark: Stating the shape is bent (or V-shaped). 1 mark: Stating a bond angle in the range \(104^\circ\) to \(105^\circ\).

The equation for the complete combustion of propan-1-ol is \(\text{C}_3\text{H}_7\text{OH(l)} + 4.5\text{O}_2\text{(g)} \rightarrow 3\text{CO}_2\text{(g)} + 4\text{H}_2\text{O(l)}\). Using Hess's Law: \(\Delta H_c^{\theta} = \sum \Delta H_f^{\theta}\text{(products)} - \sum \Delta H_f^{\theta}\text{(reactants)}\). Therefore: \(\Delta H_c^{\theta} = [3(-394) + 4(-286)] - [-303 + 0] = [-1182 - 1144] + 303 = -2023\text{ kJ mol}^{-1}\).

1 mark: Correctly setting up the Hess's cycle or expression, e.g. \([3(-394) + 4(-286)] - [-303]\). 1 mark: Correct final value of \(-2023\text{ kJ mol}^{-1}\) (negative sign is required).

Acidified barium chloride contains barium ions, which react with aqueous sulfate ions to form an insoluble white precipitate of barium sulfate. The ionic equation including state symbols is \(\text{Ba}^{2+}(\text{aq}) + \text{SO}_4^{2-}(\text{aq}) \rightarrow \text{BaSO}_4(\text{s})\).

1 mark: Correct ionic equation with state symbols: \(\text{Ba}^{2+}(\text{aq}) + \text{SO}_4^{2-}(\text{aq}) \rightarrow \text{BaSO}_4(\text{s})\). 1 mark: Stating the correct observation of a white precipitate (reject incorrect colors).

Iodide ions are powerful reducing agents. They reduce the sulfur atom in concentrated sulfuric acid from an oxidation state of +6 to -2, producing hydrogen sulfide gas (\(\text{H}_2\text{S}\)), which has the characteristic smell of bad eggs.

1 mark: Identifying the gas as hydrogen sulfide or \(\text{H}_2\text{S}\). 1 mark: Identifying the role of the iodide ions as a reducing agent or reductant.

A catalyst provides an alternative reaction pathway with a lower activation energy (\(E_a\)). On a Maxwell-Boltzmann distribution curve, this shifts the activation energy threshold to the left. As a result, a larger fraction (or proportion) of the reactant molecules possess energy equal to or greater than this new, lower activation energy, leading to a higher frequency of successful collisions.

1 mark: Stating that a catalyst provides an alternative pathway with a lower activation energy. 1 mark: Explaining that a greater fraction or proportion of molecules now have energy equal to or greater than the new activation energy.

The reduction half-equation is: \(\text{Cr}_2\text{O}_7^{2-} + 14\text{H}^+ + 6\text{e}^- \rightarrow 2\text{Cr}^{3+} + 7\text{H}_2\text{O}\). The oxidation half-equation is: \(\text{Fe}^{2+} \rightarrow \text{Fe}^{3+} + \text{e}^-\). Multiplying the oxidation half-equation by 6 to balance the transferred electrons and combining the equations gives: \(\text{Cr}_2\text{O}_7^{2-} + 6\text{Fe}^{2+} + 14\text{H}^+ \rightarrow 2\text{Cr}^{3+} + 6\text{Fe}^{3+} + 7\text{H}_2\text{O}\).

1 mark: Identifying the correct species in the equation on correct sides (even if unbalanced). 1 mark: Correctly balanced overall ionic equation: \(\text{Cr}_2\text{O}_7^{2-} + 6\text{Fe}^{2+} + 14\text{H}^+ \rightarrow 2\text{Cr}^{3+} + 6\text{Fe}^{3+} + 7\text{H}_2\text{O}\).

The third ionization energy of magnesium corresponds to the removal of one mole of electrons from one mole of gaseous magnesium 2+ ions to form one mole of gaseous magnesium 3+ ions. The equation is: \(\text{Mg}^{2+}(g) \rightarrow \text{Mg}^{3+}(g) + e^-\). The electron configuration of magnesium is \(1s^2 2s^2 2p^6 3s^2\). Since the first two electrons are removed from the \(3s\) sub-shell, the third electron is removed from the \(2p\) sub-shell.

1 mark: Correct equation with state symbols (accept subtraction of electron on the left: \(\text{Mg}^{2+}(g) - e^- \rightarrow \text{Mg}^{3+}(g)\)). 1 mark: Correctly identifies the 2p sub-shell.

First, calculate the molar mass of sodium carbonate: \(M_r(\text{Na}_2\text{CO}_3) = 2(23.0) + 12.0 + 3(16.0) = 106.0\text{ g mol}^{-1}\). Next, calculate the amount in moles of \(\text{Na}_2\text{CO}_3\): \(n = \frac{5.30}{106.0} = 0.0500\text{ mol}\). The concentration of the sodium carbonate solution is \(c = \frac{0.0500}{0.250} = 0.200\text{ mol dm}^{-3}\). Since each formula unit of \(\text{Na}_2\text{CO}_3\) dissociates to yield two \(\text{Na}^+\) ions, the concentration of \(\text{Na}^+\) is \(2 \times 0.200 = 0.400\text{ mol dm}^{-3}\).

1 mark: Correctly calculates the concentration of the sodium carbonate solution as \(0.200\text{ mol dm}^{-3}\) (or finds \(0.0500\text{ mol}\) of \(\text{Na}_2\text{CO}_3\)). 1 mark: Multiplies by 2 to obtain the final concentration of \(0.400\text{ mol dm}^{-3}\) (or \(0.4\)).

The central chlorine atom has 7 valence electrons. The positive charge means one electron has been lost, leaving 6 valence electrons. Two of these are shared in covalent bonds with the two fluorine atoms, leaving 4 non-bonding electrons (2 lone pairs). The shape is determined by 4 electron pairs (2 bonding pairs and 2 lone pairs) which arrange themselves tetrahedrally to minimize repulsion. The presence of two lone pairs compresses the bond angle from the standard tetrahedral angle of \(109.5^\circ\) to approximately \(104.5^\circ\), resulting in a bent (or V-shaped) molecule.

1 mark: Identifies the shape as bent / V-shaped / non-linear. 1 mark: States a bond angle in the range \(104^\circ\) to \(105^\circ\) (inclusive).

The equation for the formation of two moles of ammonia is: \(\text{N}_2(g) + 3\text{H}_2(g) \rightarrow 2\text{NH}_3(g)\), for which the enthalpy change is \(\Delta H = 2 \times (-46) = -92\text{ kJ mol}^{-1}\). Using the formula \(\Delta H = \sum(\text{bonds broken}) - \sum(\text{bonds formed})\): \(-92 = [944 + 3(436)] - [6 \times E(\text{N}-\text{H})]\). This simplifies to \(-92 = 2252 - 6 E(\text{N}-\text{H})\). Rearranging gives \(6 E(\text{N}-\text{H}) = 2344\), so \(E(\text{N}-\text{H}) = +390.7\text{ kJ mol}^{-1}\) (or \(+391\text{ kJ mol}^{-1}\)).

1 mark: Correct substitution into the bond enthalpy equation, e.g., \(-92 = 2252 - 6x\) or \(-46 = 1126 - 3x\). 1 mark: Correct final calculation of \(+391\text{ kJ mol}^{-1}\) (accept \(390.7\) or \(391\); ignore sign, but must be positive value).

The reaction of bromide ions with concentrated sulfuric acid is a redox reaction where bromide is oxidized to bromine gas, and sulfuric acid is reduced to sulfur dioxide. The bromine gas produced appears as orange-brown fumes. The balanced molecular equation is: \(2\text{NaBr} + 2\text{H}_2\text{SO}_4 \rightarrow \text{Na}_2\text{SO}_4 + \text{SO}_2 + \text{Br}_2 + 2\text{H}_2\text{O}\). Alternatively, the balanced ionic equation is: \(2\text{Br}^- + 2\text{H}^+ + \text{H}_2\text{SO}_4 \rightarrow \text{Br}_2 + \text{SO}_2 + 2\text{H}_2\text{O}\).

1 mark: Observation of orange/brown fumes or vapor (accept brown liquid/solution). 1 mark: Correctly balanced molecular or ionic equation.

When the temperature of a gas sample increases, the average kinetic energy of the molecules increases. This shifts the distribution of molecular energies. Consequently, the Maxwell-Boltzmann curve changes in two ways: 1. The peak of the curve (the most probable energy) shifts to the right (to a higher energy value). 2. The peak of the curve is lower (flatter) because the total area under the curve, which represents the total number of molecules, must remain constant.

1 mark: Peak is lower / flatter / maximum height is reduced. 1 mark: Peak shifts to the right / to a higher energy.

The hydrochloric acid is added to react with and remove any dissolved carbonate (\(\text{CO}_3^{2-}\)) or sulfite (\(\text{SO}_3^{2-}\)) ions present in the test solution. If these are not removed, they will react with barium ions to form barium carbonate or barium sulfite, both of which are white, insoluble precipitates that would give a false positive result for sulfate. A positive result for the sulfate test is the formation of a white precipitate of insoluble barium sulfate.

1 mark: Explains that the acid reacts with / removes carbonate or sulfite ions (accept 'prevents the formation of false precipitates of barium carbonate / sulfite'). 1 mark: Identifies the observation as a white precipitate.

1. **Calculate the moles of \(\text{HCl}\) used in the titration:**
\(n(\text{HCl}) = C \times V = 0.100\text{ mol dm}^{-3} \times 0.02600\text{ dm}^3 = 0.00260\text{ mol}\)

2. **Deduce the moles of \(\text{Na}_2\text{CO}_3\) in the \(25.0\text{ cm}^3\) aliquot:**
The balanced equation for the reaction is:
\(\text{Na}_2\text{CO}_3\text{(aq)} + 2\text{HCl}\text{(aq)} \rightarrow 2\text{NaCl}\text{(aq)} + \text{H}_2\text{O}\text{(l)} + \text{CO}_2\text{(g)}\)

From the 1:2 reacting ratio:
\(n(\text{Na}_2\text{CO}_3) = \frac{1}{2} \times n(\text{HCl}) = \frac{1}{2} \times 0.00260 = 0.00130\text{ mol}\)

3. **Scale up to find moles of \(\text{Na}_2\text{CO}_3\) in the original \(250.0\text{ cm}^3\) solution:**
\(n_{\text{total}} = 0.00130\text{ mol} \times \frac{250.0}{25.0} = 0.0130\text{ mol}\)

4. **Calculate the mass of anhydrous \(\text{Na}_2\text{CO}_3\) in the sample:**
\(M_r(\text{Na}_2\text{CO}_3) = (2 \times 23.0) + 12.0 + (3 \times 16.0) = 106.0\text{ g mol}^{-1}\)
\(m(\text{Na}_2\text{CO}_3) = n \times M_r = 0.0130\text{ mol} \times 106.0\text{ g mol}^{-1} = 1.378\text{ g}\)

5. **Calculate the mass and moles of water of crystallisation:**
\(m(\text{H}_2\text{O}) = 3.72\text{ g} - 1.378\text{ g} = 2.342\text{ g}\)
\(n(\text{H}_2\text{O}) = \frac{2.342\text{ g}}{18.0\text{ g mol}^{-1}} = 0.1301\text{ mol}\)

6. **Determine the value of \(x\):**
\(x = \frac{n(\text{H}_2\text{O})}{n(\text{Na}_2\text{CO}_3)} = \frac{0.1301}{0.0130} = 10.0\)

Therefore, \(x = 10\).

- **M1:** Calculates moles of \(\text{HCl}\) correctly as \(0.00260\text{ mol}\). (1 mark)
- **M2:** Uses correct stoichiometry (1:2 ratio) to find moles of \(\text{Na}_2\text{CO}_3\) in \(25.0\text{ cm}^3\) as \(0.00130\text{ mol}\). (1 mark)
- **M3:** Scales up to the total volume (\(250.0\text{ cm}^3\)) to find total moles of \(\text{Na}_2\text{CO}_3\) as \(0.0130\text{ mol}\). (1 mark)
- **M4:** Calculates mass of anhydrous \(\text{Na}_2\text{CO}_3\) as \(1.378\text{ g}\) (or \(1.38\text{ g}\)). (1 mark)
- **M5:** Subtracts to find mass of water (\(2.342\text{ g}\)) and calculates moles of water as \(0.1301\text{ mol}\). (1 mark)
- **M6:** Divides moles of water by moles of anhydrous salt to find \(x = 10\). (1 mark)

1. **Write the chemical equation for the standard enthalpy of formation of gaseous pent-1-ene:**
\(5\text{C(s, graphite)} + 5\text{H}_2\text{(g)} \rightarrow \text{C}_5\text{H}_{10}\text{(g)}\)

2. **Construct a Hess's Law cycle or algebraic equation using enthalpies of combustion:**
By definition, the enthalpy of formation of liquid pent-1-ene is given by:
\(\Delta_f H^\theta [\text{C}_5\text{H}_{10}\text{(l)}] = 5 \times \Delta_c H^\theta [\text{C(s)}] + 5 \times \Delta_c H^\theta [\text{H}_2\text{(g)}] - \Delta_c H^\theta [\text{C}_5\text{H}_{10}\text{(l)}]\)

3. **Calculate \(\Delta_f H^\theta\) of liquid pent-1-ene:**
\(\sum \Delta_c H^\theta(\text{reactants}) = 5(-393.5) + 5(-285.8) = -1967.5 - 1429.0 = -3396.5\text{ kJ mol}^{-1}\)
\(\Delta_f H^\theta [\text{C}_5\text{H}_{10}\text{(l)}] = -3396.5 - (-3368.0) = -28.5\text{ kJ mol}^{-1}\)

4. **Incorporate the enthalpy of vaporisation to find \(\Delta_f H^\theta\) of gaseous pent-1-ene:**
The process is: \(5\text{C(s)} + 5\text{H}_2\text{(g)} \rightarrow \text{C}_5\text{H}_{10}\text{(l)} \rightarrow \text{C}_5\text{H}_{10}\text{(g)}\)
So:
\(\Delta_f H^\theta [\text{C}_5\text{H}_{10}\text{(g)}] = \Delta_f H^\theta [\text{C}_5\text{H}_{10}\text{(l)}] + \Delta_{vap} H^\theta\)
\(\Delta_f H^\theta [\text{C}_5\text{H}_{10}\text{(g)}] = -28.5 + 26.4 = -2.1\text{ kJ mol}^{-1}\)

- **M1:** Writes a balanced chemical equation for the formation of either gaseous or liquid pent-1-ene with correct state symbols. (1 mark)
- **M2:** Uses Hess's Law principle or sets up a correct algebraic equation for \(\Delta_f H\) from \(\Delta_c H\). (1 mark)
- **M3:** Correctly calculates the sum of the combustion enthalpies of the constituent elements: \(-3396.5\text{ kJ mol}^{-1}\). (1 mark)
- **M4:** Correctly calculates the enthalpy of formation of liquid pent-1-ene: \(-28.5\text{ kJ mol}^{-1}\). (1 mark)
- **M5:** Correctly links the formation of gaseous pent-1-ene by adding the vaporisation enthalpy to the liquid formation value: \(-28.5 + 26.4\). (1 mark)
- **M6:** Obtains the final value of \(-2.1\text{ kJ mol}^{-1}\) (with negative sign and correct units). (1 mark)

1. **Construct an ICE table (Initial, Change, Equilibrium) in moles:**
- **Initial moles:**
\(\text{SO}_2 = 1.50\text{ mol}\)
\(\text{O}_2 = 1.20\text{ mol}\)
\(\text{SO}_3 = 0.00\text{ mol}\)

- **Change in moles:**
Since equilibrium \(n(\text{SO}_3) = 0.80\text{ mol}\), the change for \(\text{SO}_3\) is \(+0.80\text{ mol}\).
Based on the stoichiometry of the reaction:
Change in \(n(\text{SO}_2) = -0.80\text{ mol}\)
Change in \(n(\text{O}_2) = -0.40\text{ mol}\)

- **Equilibrium moles:**
\(n(\text{SO}_2) = 1.50 - 0.80 = 0.70\text{ mol}\)
\(n(\text{O}_2) = 1.20 - 0.40 = 0.80\text{ mol}\)
\(n(\text{SO}_3) = 0.80\text{ mol}\)

2. **Calculate equilibrium concentrations (divide by \(V = 5.00\text{ dm}^3\)):**
\([\text{SO}_2] = \frac{0.70\text{ mol}}{5.00\text{ dm}^3} = 0.140\text{ mol dm}^{-3}\)
\([\text{O}_2] = \frac{0.80\text{ mol}}{5.00\text{ dm}^3} = 0.160\text{ mol dm}^{-3}\)
\([\text{SO}_3] = \frac{0.80\text{ mol}}{5.00\text{ dm}^3} = 0.160\text{ mol dm}^{-3}\)

3. **Write the expression for \(K_c\):**
\(K_c = \frac{[\text{SO}_3]^2}{[\text{SO}_2]^2 [\text{O}_2]}\)

4. **Calculate the value of \(K_c\):**
\(K_c = \frac{(0.160)^2}{(0.140)^2 \times 0.160} = \frac{0.160}{(0.140)^2} = \frac{0.160}{0.0196} = 8.163... \approx 8.16\)

5. **Determine the units of \(K_c\):**
\(\text{Units} = \frac{(\text{mol dm}^{-3})^2}{(\text{mol dm}^{-3})^2 \times (\text{mol dm}^{-3})} = \frac{1}{\text{mol dm}^{-3}} = \text{mol}^{-1}\text{ dm}^3\text{ or }\text{dm}^3\text{ mol}^{-1}\)

- **M1:** States the correct expression for \(K_c\): \(K_c = \frac{[\text{SO}_3]^2}{[\text{SO}_2]^2 [\text{O}_2]}\). (1 mark)
- **M2:** Calculates equilibrium moles of \(\text{SO}_2\) as \(0.70\text{ mol}\). (1 mark)
- **M3:** Calculates equilibrium moles of \(\text{O}_2\) as \(0.80\text{ mol}\). (1 mark)
- **M4:** Divides equilibrium moles by \(5.00\) to get equilibrium concentrations: \([\text{SO}_2] = 0.140\), \([\text{O}_2] = 0.160\), \([\text{SO}_3] = 0.160\). (1 mark)
- **M5:** Substitutes concentrations into the \(K_c\) expression and calculates value as \(8.16\) (accept \(8.16\) to \(8.17\)). (1 mark)
- **M6:** Correctly deduces the units as \(\text{dm}^3\text{ mol}^{-1}\) or \(\text{mol}^{-1}\text{ dm}^3\). (1 mark)

1. **Calculate the volume change of oxygen gas produced:**
\(\Delta V = 38.5\text{ cm}^3 - 12.0\text{ cm}^3 = 26.5\text{ cm}^3 = 26.5 \times 10^{-6}\text{ m}^3\)

2. **Calculate the moles of \(\text{O}_2\) produced using the ideal gas equation:**
\(P = 100\text{ kPa} = 100 \times 10^3\text{ Pa}\)
\(T = 298\text{ K}\)
\(n(\text{O}_2) = \frac{P \cdot \Delta V}{R \cdot T} = \frac{100 \times 10^3\text{ Pa} \times 26.5 \times 10^{-6}\text{ m}^3}{8.31\text{ J K}^{-1}\text{ mol}^{-1} \times 298\text{ K}} = \frac{2.65}{2476.38} = 1.0701 \times 10^{-3}\text{ mol}\)

3. **Use reaction stoichiometry to find moles of \(\text{H}_2\text{O}_2\) consumed:**
From the equation: \(2\text{H}_2\text{O}_2 \rightarrow 2\text{H}_2\text{O} + \text{O}_2\)
\(n(\text{H}_2\text{O}_2\text{ consumed}) = 2 \times n(\text{O}_2) = 2 \times 1.0701 \times 10^{-3} = 2.1402 \times 10^{-3}\text{ mol}\)

4. **Calculate change in concentration of \(\text{H}_2\text{O}_2\):**
\(\Delta [\text{H}_2\text{O}_2] = \frac{2.1402 \times 10^{-3}\text{ mol}}{0.0500\text{ dm}^3} = 0.042804\text{ mol dm}^{-3}\)

5. **Calculate the average rate of reaction (consumption of \(\text{H}_2\text{O}_2\)):**
\(\Delta t = 90.0\text{ s} - 30.0\text{ s} = 60.0\text{ s}\)
\(\text{Rate} = \frac{\Delta [\text{H}_2\text{O}_2]}{\Delta t} = \frac{0.042804\text{ mol dm}^{-3}}{60.0\text{ s}} = 7.134 \times 10^{-4}\text{ mol dm}^{-3}\text{ s}^{-1}\)

To 3 significant figures, the average rate of consumption of \(\text{H}_2\text{O}_2\) is \(7.13 \times 10^{-4}\text{ mol dm}^{-3}\text{ s}^{-1}\).

- **M1:** Calculates change in gas volume and converts to cubic metres: \(26.5 \times 10^{-6}\text{ m}^3\). (1 mark)
- **M2:** Uses ideal gas equation properly rearranged to find moles of oxygen gas: \(1.07 \times 10^{-3}\text{ mol}\). (1 mark)
- **M3:** Relates moles of oxygen to moles of hydrogen peroxide consumed (1:2 ratio): \(2.14 \times 10^{-3}\text{ mol}\). (1 mark)
- **M4:** Calculates concentration change of hydrogen peroxide in solution: \(0.0428\text{ mol dm}^{-3}\). (1 mark)
- **M5:** Calculates rate by dividing concentration change by time interval (\(60.0\text{ s}\)): \(7.13 \times 10^{-4}\). (1 mark)
- **M6:** Gives correct units: \(\text{mol dm}^{-3}\text{ s}^{-1}\). (1 mark)

**Method 1: Direct ratio method (simplest)**
Because kinetic energy (\(KE\)) and distance (\(d\)) are identical for both ions, the velocity \(v = \frac{d}{t}\) can be substituted into the kinetic energy equation:
\(KE = \frac{1}{2} m \left(\frac{d}{t}\right)^2 \Rightarrow t = d \sqrt{\frac{m}{2KE}}\)
This means time of flight is directly proportional to the square root of the mass of the ion:
\(t \propto \sqrt{m}\)

Therefore, we can establish the ratio:
\(\frac{t(^{113}\text{In}^+)}{t(^{115}\text{In}^+)} = \sqrt{\frac{113}{115}}\)

\(t(^{113}\text{In}^+) = 2.42 \times 10^{-5}\text{ s} \times \sqrt{\frac{113}{115}}\)
\(t(^{113}\text{In}^+) = 2.42 \times 10^{-5} \times 0.99127 = 2.3989 \times 10^{-5}\text{ s}\)
Rounding to 3 significant figures gives \(2.40 \times 10^{-5}\text{ s}\).

**Method 2: Multi-step calculation**
1. **Mass of \(^{115}\text{In}^+\):**
\(m(^{115}\text{In}^+) = \frac{115 \times 10^{-3}\text{ kg mol}^{-1}}{6.022 \times 10^{23}\text{ mol}^{-1}} = 1.9097 \times 10^{-25}\text{ kg}\)

2. **Velocity of \(^{115}\text{In}^+\):**
\(v = \frac{d}{t} = \frac{1.50}{2.42 \times 10^{-5}} = 61983\text{ m s}^{-1}\)

3. **Kinetic energy (\(KE\)):**
\(KE = \frac{1}{2} m v^2 = 0.5 \times (1.9097 \times 10^{-25}) \times (61983)^2 = 3.668 \times 10^{-16}\text{ J}\)

4. **Mass of \(^{113}\text{In}^+\):**
\(m(^{113}\text{In}^+) = \frac{113 \times 10^{-3}\text{ kg mol}^{-1}}{6.022 \times 10^{23}\text{ mol}^{-1}} = 1.8765 \times 10^{-25}\text{ kg}\)

5. **Velocity of \(^{113}\text{In}^+\):**
\(v(^{113}\text{In}^+) = \sqrt{\frac{2KE}{m}} = \sqrt{\frac{2 \times 3.668 \times 10^{-16}}{1.8765 \times 10^{-25}}} = 62530\text{ m s}^{-1}\)

6. **Time of flight of \(^{113}\text{In}^+\):**
\(t = \frac{d}{v} = \frac{1.50}{62530} = 2.3989 \times 10^{-5}\text{ s} \approx 2.40 \times 10^{-5}\text{ s}\)

**Using the multi-step route:**
- **M1:** Calculates mass of \(^{115}\text{In}^+\) ion in kg: \(1.91 \times 10^{-25}\text{ kg}\). (1 mark)
- **M2:** Calculates velocity of \(^{115}\text{In}^+\) ion: \(6.20 \times 10^4\text{ m s}^{-1}\). (1 mark)
- **M3:** Calculates constant kinetic energy: \(3.67 \times 10^{-16}\text{ J}\). (1 mark)
- **M4:** Calculates mass of \(^{113}\text{In}^+\) ion in kg: \(1.88 \times 10^{-25}\text{ kg}\). (1 mark)
- **M5:** Calculates velocity of \(^{113}\text{In}^+\) ion: \(6.25 \times 10^4\text{ m s}^{-1}\). (1 mark)
- **M6:** Correctly calculates time of flight to 3 sig figs: \(2.40 \times 10^{-5}\text{ s}\). (1 mark)

**Using the ratio route:**
- **M1:** Explains or shows that \(t \propto \sqrt{m}\) or \(t = k\sqrt{m}\). (1 mark)
- **M2 & M3:** Sets up ratio equation: \(t_2 = t_1 \times \sqrt{\frac{m_2}{m_1}}\). (2 marks)
- **M4:** Substitutes the mass numbers (113 and 115) into the expression. (1 mark)
- **M5:** Calculates ratio scale factor: \(\sqrt{\frac{113}{115}} = 0.9913\). (1 mark)
- **M6:** Correctly calculates final time of flight to 3 sig figs: \(2.40 \times 10^{-5}\text{ s}\). (1 mark)

1. Look for the largest increase (jump) between successive ionization energies. Here, the jump is between \(IE_3\) (2745 \(\text{kJ mol}^{-1}\)) and \(IE_4\) (11577 \(\text{kJ mol}^{-1}\)).
2. This indicates that the fourth electron is being removed from an inner core shell, meaning element \(X\) has 3 valence electrons and belongs to Group 13 (Aluminium in Period 3).
3. Therefore, \(X\) forms a \(3+\) ion (\(X^{3+}\)).
4. Oxygen forms a \(2-\) ion (\(\text{O}^{2-}\)).
5. The formula of the neutral oxide is \(X_2\text{O}_3\).

[1 mark] B - \(X_2\text{O}_3\)
Award 1 mark for the correct option choice.

1. Determine the Lewis structure and electron pair geometry of each species:
- \(\text{PCl}_3\): 3 bonding pairs, 1 lone pair on phosphorus. Trigonal pyramidal geometry with a bond angle of approximately \(107^\circ\).
- \(\text{CO}_3^{2-}\): 3 bonding areas (including double bonds), 0 lone pairs on carbon. Trigonal planar geometry with a bond angle of exactly \(120^\circ\) due to symmetrical resonance structures.
- \(\text{NH}_2^-\): 2 bonding pairs, 2 lone pairs on nitrogen. Bent geometry with a bond angle of approximately \(104.5^\circ\).
- \(\text{H}_3\text{O}^+\): 3 bonding pairs, 1 lone pair on oxygen. Trigonal pyramidal geometry with a bond angle of approximately \(107^\circ\).

[1 mark] B - \(\text{CO}_3^{2-}\)
Award 1 mark for the correct option choice.

1. Calculate the mass of sulfur in the compound:
\(\text{Mass of S} = 2.20\text{ g} - 1.24\text{ g} = 0.96\text{ g}\)

2. Calculate the moles of phosphorus and sulfur:
\(\text{Moles of P} = \frac{1.24}{31.0} = 0.040\text{ mol}\)
\(\text{Moles of S} = \frac{0.96}{32.1} \approx 0.030\text{ mol}\)

3. Determine the simplest whole-number ratio:
\(\text{Ratio P : S} = \frac{0.040}{0.030} : \frac{0.030}{0.030} = 1.33 : 1\)
Multiply by 3 to obtain integer values: \(4 : 3\).

Therefore, the empirical formula is \(\text{P}_4\text{S}_3\).

[1 mark] C - \(\text{P}_4\text{S}_3\)
Award 1 mark for the correct option choice.

1. Sulfuric acid (\(\text{H}_2\text{S}\text{O}_4\)) can be reduced by iodide ions (\(\text{I}^-\)) to various species because iodide is a very strong reducing agent.
2. The possible reduction products of \(\text{H}_2\text{S}\text{O}_4\) are:
- \(\text{SO}_2\) (sulfur dioxide): colorless choking gas
- \(\text{S}\) (sulfur): yellow solid
- \(\text{H}_2\text{S}\) (hydrogen sulfide): gas with a rotten egg smell
3. \(\text{I}_2\) is an oxidation product of iodide, not a reduction product of sulfuric acid.
4. Therefore, the yellow solid reduction product is elemental sulfur (\(\text{S}\)).

[1 mark] D - \(\text{S}\)
Award 1 mark for the correct option choice.

Stereoisomerism (specifically E/Z isomerism) occurs in hex-3-ene because: 1. There is restricted rotation about the carbon-carbon double bond (C=C) due to the presence of the pi bond. 2. Each of the carbon atoms in the C=C double bond is bonded to two different groups: a hydrogen atom and an ethyl group.

1 mark: Mention of restricted rotation about the C=C double bond. 1 mark: Mention that each carbon of the C=C double bond is attached to two different groups (H and ethyl / -CH2CH3).

When the temperature of a gas is increased: 1. The peak of the curve (representing the most probable energy) shifts to the right (higher energy) and is lower in height. 2. The curve flattens out, meaning it becomes broader, and the tail on the high-energy side rises higher, indicating that a larger proportion of molecules have high energies.

1 mark: Peak shifts to the right AND is lower. 1 mark: Curve is broader OR the tail at the high-energy end is higher.

The dehydration of butan-2-ol produces but-2-ene as the major product. Because but-2-ene has stereocenters, it exists as a pair of stereoisomers: (E)-but-2-ene (trans-but-2-ene) and (Z)-but-2-ene (cis-but-2-ene). (E)-but-2-ene is more stable because the bulky methyl groups are positioned on opposite sides of the double bond, which minimizes steric repulsion.

1 mark: Correctly identifies both (E)-but-2-ene and (Z)-but-2-ene (accept trans-but-2-ene and cis-but-2-ene). 1 mark: Correctly identifies (E)-but-2-ene (or trans-but-2-ene) as the more stable isomer.

Step 1: Calculate the heat energy transferred using q = m x c x \(\Delta T\). q = 100.0 g x 4.18 J g^-1 K^-1 x 22.5 K = 9405 J = 9.405 kJ. Step 2: Calculate the moles of methanol burned. n = 0.80 g / 32.0 g mol^-1 = 0.025 mol. Step 3: Calculate \(\Delta H_c\). \(\Delta H_c\) = -q / n = -9.405 kJ / 0.025 mol = -376.2 kJ mol^-1 (rounds to -376 kJ mol^-1).

1 mark: Correct calculation of heat energy (q = 9.405 kJ) and moles of methanol (n = 0.025 mol). 1 mark: Correct final calculation of \(\Delta H_c\) with negative sign (accept -376 to -376.2 kJ mol^-1). Reject positive values.

1-bromobutane reacts faster than 1-chlorobutane. The rate of nucleophilic substitution is determined by carbon-halogen bond strength (bond enthalpy), not bond polarity. Because the bromine atom has a larger atomic radius than chlorine, the C-Br bond is longer and weaker (has a lower bond enthalpy) than the C-Cl bond, meaning it requires less energy to break during the reaction.

1 mark: Correctly states that 1-bromobutane reacts faster (or has a higher rate). 1 mark: Explains that the C-Br bond is weaker / has a lower bond enthalpy than the C-Cl bond. Reject answers based on electronegativity or bond polarity.

The equilibrium constant expression is K_c = [CH3COOC2H5][H2O] / ([CH3COOH][C2H5OH]). Because there are equal moles of reactants and products (2 moles on each side), the concentration units cancel out completely, so K_c has no units.

1 mark: Correct expression for K_c. 1 mark: Correctly identifies that there are no units (or units cancel).

To distinguish between an aldehyde (propanal) and a ketone (propanone), Tollens' reagent can be used. Propanal is oxidized by Tollens' reagent, reducing silver ions to metallic silver, which forms a silver mirror on the inside of the tube. Propanone, being a ketone, is not oxidized, and the solution remains colourless. Alternatively, Fehling's solution can be used: propanal forms a brick-red precipitate of copper(I) oxide, while propanone remains blue.

1 mark: Identifies a suitable reagent (e.g., Tollens' reagent or Fehling's solution / Benedict's reagent or acidified potassium dichromate). 1 mark: Correctly states the observation for both compounds (e.g., silver mirror / red precipitate / orange to green color change for propanal, and no visible change / remains blue / remains orange for propanone).

In the propagation steps of free-radical chlorination of methane: 1. A chlorine radical reacts with methane to form a methyl radical and hydrogen chloride: .Cl + CH4 -> .CH3 + HCl. 2. The methyl radical then reacts with a chlorine molecule to produce chloromethane and regenerate the chlorine radical: .CH3 + Cl2 -> CH3Cl + .Cl.

1 mark: Correct equation for the first propagation step: .Cl + CH4 -> .CH3 + HCl (with radical dots on Cl and C). 1 mark: Correct equation for the second propagation step: .CH3 + Cl2 -> CH3Cl + .Cl (with radical dots on C and Cl).

In the first propagation step, a chlorine radical abstracts a hydrogen atom from ethane, producing an ethyl radical and hydrogen chloride: \(\text{CH}_3\text{CH}_3 + \text{Cl}^\bullet \rightarrow \text{CH}_3\text{CH}_2^\bullet + \text{HCl}\). In the second propagation step, the ethyl radical reacts with a chlorine molecule to produce chloroethane and regenerate the chlorine radical: \(\text{CH}_3\text{CH}_2^\bullet + \text{Cl}_2 \rightarrow \text{CH}_3\text{CH}_2\text{Cl} + \text{Cl}^\bullet\).

1 mark for propagation step 1: \(\text{CH}_3\text{CH}_3 + \text{Cl}^\bullet \rightarrow \text{CH}_3\text{CH}_2^\bullet + \text{HCl}\) (radical dot must be clearly on the carbon or indicating the radical species).
1 mark for propagation step 2: \(\text{CH}_3\text{CH}_2^\bullet + \text{Cl}_2 \rightarrow \text{CH}_3\text{CH}_2\text{Cl} + \text{Cl}^\bullet\).

The addition of \(\text{H}^+\) to 2-methylbut-2-ene can form either a secondary or a tertiary carbocation. The tertiary carbocation is more stable because of the electron-releasing inductive effect of the three surrounding alkyl groups, which helps disperse the positive charge. Therefore, the major product is formed by the subsequent nucleophilic attack of \(\text{Br}^-\text{\ pool}\) on this tertiary carbocation, giving 2-bromo-2-methylbutane.

1 mark for identifying the major product as 2-bromo-2-methylbutane (accept correct skeletal formula).
1 mark for explaining that it forms via the more stable tertiary carbocation (accept references to the greater inductive effect of three alkyl groups stabilizing the carbocation compared to two alkyl groups).

To promote elimination over nucleophilic substitution, the hydroxide ion must act as a base rather than a nucleophile. This is achieved by using an ethanolic solvent (dissolving potassium hydroxide or sodium hydroxide in ethanol) and heating the mixture under reflux.

1 mark for stating potassium hydroxide/sodium hydroxide in ethanol (allow 'ethanolic KOH' or 'ethanolic NaOH'; reject aqueous KOH).
1 mark for stating hot / high temperature / heat under reflux (reject 'warm' or 'room temperature').

Ethanol is a primary alcohol. To prevent further oxidation of the intermediate aldehyde (ethanal) to a carboxylic acid (ethanoic acid), ethanal must be removed from the reaction mixture immediately. This is achieved by heating ethanol and a limited amount of acidified potassium dichromate(VI) in a distillation setup; ethanal has a lower boiling point than ethanol and water, so it vaporizes immediately, is condensed, and is collected as the distillate.

1 mark for identifying acidified potassium dichromate(VI) as the oxidizing agent (accept dilute/limited oxidizing agent or formula \(\text{H}^+/\text{Cr}_2\text{O}_7^{2-}\)).
1 mark for stating that the product must be distilled off immediately / as it forms.

Infrared spectroscopy distinguishes molecules by detecting specific functional group bond vibrations. Cyclohexanone contains a carbonyl group which produces a sharp, strong peak at \(1680\text{–}1750\text{ cm}^{-1}\) and lacks an alcohol peak. Cyclohexanol contains an alcohol group which produces a wide, broad peak at \(3230\text{–}3550\text{ cm}^{-1}\) and lacks a carbonyl peak.

1 mark for identifying that cyclohexanone has a \(\text{C}=\text{O}\) peak at \(1680\text{–}1750\text{ cm}^{-1}\) (allow \(1630\text{–}1750\text{ cm}^{-1}\)).
1 mark for identifying that cyclohexanol has an alcohol \(\text{O}-\text{H}\) peak at \(3230\text{–}3550\text{ cm}^{-1}\) (allow \(3200\text{–}3600\text{ cm}^{-1}\)).

1. Calculate heat absorbed by water: \(q = m c \Delta T = 100.0 \times 4.18 \times (45.0 - 20.0) = 10450\text{ J} = 10.45\text{ kJ}\).
2. Calculate moles of methanol burned: \(n = \frac{\text{mass}}{M_r} = \frac{0.800}{32.0} = 0.0250\text{ mol}\).
3. Calculate enthalpy of combustion: \(\Delta H_c = -\frac{q}{n} = -\frac{10.45}{0.0250} = -418\text{ kJ mol}^{-1}\). The negative sign is required as combustion is exothermic.

1 mark for calculating correct heat energy transfer \(q = 10.45\text{ kJ}\) AND moles of methanol \(n = 0.0250\text{ mol}\).
1 mark for the correct final value of \(-418\text{ kJ mol}^{-1}\) (must have the negative sign; allow 2 or 3 significant figures, i.e., \(-418\) or \(-420\)).

When the temperature of a gas is increased, the average kinetic energy of the molecules increases. On the Maxwell-Boltzmann distribution, this causes the peak (the most probable energy) to shift to the right. Because the total number of molecules remains constant, the area under the curve must remain constant, meaning the peak becomes lower and the overall distribution becomes flatter and broader.

1 mark for stating that the peak shifts to the right / to higher energy.
1 mark for stating that the peak is lower / flatter (accept 'the curve becomes broader' or 'the distribution spreads out').

The forward reaction is exothermic. According to Le Chatelier's principle, if temperature is increased, the system will oppose the change by shifting in the endothermic direction (the reverse direction). This causes the equilibrium position to shift to the left, decreasing the concentration of \(\text{CH}_3\text{OH}\) and increasing the concentrations of \(\text{CO}\) and \(\text{H}_2\). Since \(K_c = \frac{[\text{CH}_3\text{OH}]}{[\text{CO}][\text{H}_2]^2}\), a decrease in products and increase in reactants decreases the value of \(K_c\).

1 mark for stating that the value of \(K_c\) decreases.
1 mark for explaining that increasing temperature shifts the equilibrium to the left / in the endothermic direction to oppose the temperature increase.

First, calculate the temperature change: \(dT = 58.7 - 20.2 = 38.5\text{ K}\). Next, calculate the heat energy transferred to the water using \(q = m \times c \times dT\): \(q = 150.0 \times 4.18 \times 38.5 = 24139.5\text{ J} = 24.1395\text{ kJ}\). Then, calculate the number of moles of propan-1-ol burned: \(n = 0.820 / 60.0 = 0.013667\text{ mol}\). Finally, calculate the enthalpy change per mole: \(dH_c = -q / n = -24.1395 / 0.013667 = -1766.3\text{ kJ mol}^{-1}\). Rounded to 3 significant figures, this is \(-1770\text{ kJ mol}^{-1}\).

M1: dT = 38.5 K (1 mark). M2: q = 24139.5 J or 24.14 kJ (1 mark). M3: Moles of propan-1-ol = 0.0137 mol (1 mark). M4: Division of q by moles (1 mark). M5: Correct final answer to 3 s.f. with negative sign: -1770 kJ mol-1 (1 mark). Accept range -1760 to -1770; deduct 1 mark if negative sign is missing.

First, find the equilibrium moles of reactants: moles of \(\text{CO} = 0.400 - 0.150 = 0.250\text{ mol}\); moles of \(\text{H}_2 = 0.800 - 2(0.150) = 0.500\text{ mol}\). Next, calculate equilibrium concentrations by dividing by the volume of \(5.00\text{ dm}^3\): \([\text{CO}] = 0.250 / 5.00 = 0.0500\text{ mol dm}^{-3}\), \([\text{H}_2] = 0.500 / 5.00 = 0.100\text{ mol dm}^{-3}\), \([\text{CH}_3\text{OH}] = 0.150 / 5.00 = 0.0300\text{ mol dm}^{-3}\). The expression for \(K_c\) is: \(K_c = [\text{CH}_3\text{OH}] / ([\text{CO}][\text{H}_2]^2)\). Substituting the values: \(K_c = 0.0300 / (0.0500 \times (0.100)^2) = 60.0\). The units of \(K_c\) are: \(\text{mol}^{-2}\text{ dm}^6\).

M1: Equilibrium moles of CO = 0.250 mol and H2 = 0.500 mol (1 mark). M2: Equilibrium concentrations of CO = 0.0500, H2 = 0.100, and methanol = 0.0300 mol dm-3 (1 mark). M3: Correct Kc expression (1 mark). M4: Correct calculation of numerical value = 60.0 (or 60) (1 mark). M5: Correct units: mol-2 dm6 (1 mark).

The enthalpy change of reaction is given by: \(\Delta H = \sum(\text{bonds broken}) - \sum(\text{bonds formed})\). Bonds broken: \(4 \times (\text{C-H}) + 1 \times (\text{Cl-Cl}) = 4(413) + 242 = 1894\text{ kJ mol}^{-1}\) (or net bonds broken: \(1 \times (\text{C-H}) + 1 \times (\text{Cl-Cl}) = 413 + 242 = 655\text{ kJ mol}^{-1}\)). Bonds formed: \(3 \times (\text{C-H}) + 1 \times (\text{C-Cl}) + 1 \times (\text{H-Cl}) = 3(413) + E(\text{C-Cl}) + 431 = 1670 + E(\text{C-Cl})\) (or net bonds formed: \(1 \times (\text{C-Cl}) + 1 \times (\text{H-Cl}) = E(\text{C-Cl}) + 431\)). Therefore: \(1894 - (1670 + E(\text{C-Cl})) = -99\), which simplifies to \(224 - E(\text{C-Cl}) = -99\). Solving for \(E(\text{C-Cl})\): \(E(\text{C-Cl}) = 224 + 99 = 323\text{ kJ mol}^{-1}\).

M1: Recall or use relation: Enthalpy change = bonds broken - bonds formed (1 mark). M2: Calculation of energy for bonds broken (1894 kJ mol-1 total or 655 kJ mol-1 net) (1 mark). M3: Algebraic expression for bonds formed (1670 + E(C-Cl) total or 431 + E(C-Cl) net) (1 mark). M4: Correct setup of equation: 224 - E(C-Cl) = -99 (or equivalent) (1 mark). M5: Correct final value with units: 323 kJ mol-1 (accept +323) (1 mark).

First, calculate the percentage yield. Moles of 2-methylpropan-2-ol starting material: \(n = 14.8 / 74.0 = 0.200\text{ mol}\). Since the stoichiometry is 1:1, the theoretical moles of 2-chloro-2-methylpropane expected is \(0.200\text{ mol}\). Theoretical mass of 2-chloro-2-methylpropane = \(0.200 \times 92.5 = 18.5\text{ g}\). Percentage yield = \((13.9 / 18.5) \times 100 = 75.135\%\), which is \(75.1\%\) to 3 significant figures. Next, calculate the percentage atom economy. The formula is: \(\text{Atom Economy} = (M_r(\text{desired product}) / \sum M_r(\text{all reactants})) \times 100\). Reactants are \((\text{CH}_3)_3\text{COH}\) (\(M_r = 74.0\)) and \(\text{HCl}\) (\(M_r = 36.5\)). Sum of reactant \(M_r\)'s = \(74.0 + 36.5 = 110.5\). Atom economy = \((92.5 / 110.5) \times 100 = 83.71\%\), which is \(83.7\%\) to 3 significant figures.

M1: Moles of 2-methylpropan-2-ol = 0.200 mol (1 mark). M2: Theoretical yield of 2-chloro-2-methylpropane = 18.5 g (or 0.200 mol) (1 mark). M3: Percentage yield = 75.1% (accept range 75.0% - 75.2%) (1 mark). M4: Sum of reactant Mr values = 110.5 (1 mark). M5: Percentage atom economy = 83.7% (accept range 83.6% - 83.8%) (1 mark).

For a molecule to exhibit \(E\)-\(-Z\) isomerism, it must have a \(C=C\) double bond where each carbon atom in the double bond is bonded to two different groups. In 2-methylbut-2-ene, \((CH_3)_2C=CHCH_3\), the left-hand carbon of the double bond is attached to two identical methyl groups. In pent-1-ene, \(CH_2=CHCH_2CH_2CH_3\), the terminal carbon is attached to two hydrogen atoms. In 3-methylpent-2-ene, \(CH_3CH=C(CH_3)CH_2CH_3\), the left carbon is bonded to \(-H\) and \(-CH_3\), and the right carbon is bonded to \(-CH_3\) and \(-CH_2CH_3\). Since both carbons have two different groups attached, this alkene exhibits \(E\)-\(Z\) isomerism. In 2,3-dimethylbut-2-ene, \((CH_3)_2C=C(CH_3)_2\), both carbons are bonded to identical methyl groups.

1 mark: Correctly identifying option C.

When temperature increases: The activation energy (\(E_a\)) remains constant because it is a fixed energy barrier for a given reaction pathway (option A is incorrect). The total area under the Maxwell-Boltzmann distribution curve remains unchanged because the total number of particles is constant (option B is incorrect). The peak of the curve flattens (becomes lower in height) and shifts to the right, which means the most probable energy of the molecules increases (option C is correct, option D is incorrect).

1 mark: Correctly identifying option C.

Alcohol \(X\) must be a secondary alcohol because secondary alcohols are oxidized to ketones (molecular formula \(C_nH_{2n}O\)), which do not react with Fehling's solution. Primary alcohols like butan-1-ol (option A) and 2-methylpropan-1-ol (option D) are oxidized under reflux with excess oxidizing agent to carboxylic acids, which have the molecular formula \(C_4H_8O_2\). Tertiary alcohols like 2-methylpropan-2-ol (option C) resist oxidation. Butan-2-ol (option B) is a secondary alcohol and is oxidized to butanone, \(C_4H_8O\), which is a ketone and does not react with Fehling's solution.

1 mark: Correctly identifying option B.

Sulfur trioxide is an acidic oxide. When it reacts with aqueous sodium hydroxide (a strong base), it forms sodium sulfate and water. The ionic equation is obtained by removing the spectator sodium ions: SO3(g) + 2OH-(aq) -> SO4^2-(aq) + H2O(l).

M1: Correct formulae of reactants and products (1 mark). M2: Correct balancing and charge balance (1 mark). State symbols are not required.

Addition of excess concentrated hydrochloric acid leads to ligand substitution. The hexaaquacopper(II) ion (blue) is converted into the tetrachlorocuprate(II) ion (yellow) due to replacement of water molecules by chloride ligands: [Cu(H2O)6]^2+ + 4Cl^- -> [CuCl4]^2- + 6H2O. The mixture often appears green due to the presence of both complex ions in equilibrium.

M1: Correct formula of the complex [CuCl4]^2- (including correct overall 2- charge) (1 mark). M2: Correct color change from blue to yellow, green-blue, or green (1 mark).

The rate equation is: Rate = k[A][B]^2. Substituting the values gives: 1.8 x 10^-4 = k * 0.15 * (0.20)^2. This simplifies to 1.8 x 10^-4 = k * 0.0060. Solving for k gives k = 0.030. The units of k are calculated as (mol dm^-3 s^-1) / ((mol dm^-3) * (mol dm^-3)^2) = dm^6 mol^-2 s^-1.

M1: Correct calculation of the value of k = 0.030 or 3.0 x 10^-2 (1 mark). M2: Correct units of dm^6 mol^-2 s^-1 (1 mark).

Using a Born-Haber cycle: Enthalpy of formation = Enthalpy of atomisation of Ca + First IE of Ca + Second IE of Ca + Bond dissociation of Cl2 + 2 * (Electron affinity of Cl) + Lattice enthalpy of formation. Substituting the values: -796 = 178 + 590 + 1145 + 242 + 2 * (-348) + Lattice Enthalpy. This gives -796 = 1459 + Lattice Enthalpy. Thus, Lattice Enthalpy of formation = -2255 kJ mol^-1.

M1: Correctly setting up the Born-Haber cycle calculation incorporating 2 * EA of Cl and 1 * Bond dissociation of Cl2 (1 mark). M2: Calculating -2255 kJ mol^-1 (1 mark). Accept +2255 kJ mol^-1 ONLY if clearly defined as lattice dissociation enthalpy.

For a weak acid, [H+] = square root of (Ka * [HA]). Substituting the values gives [H+] = square root of (1.50 x 10^-5 * 0.120) = square root of (1.80 x 10^-6) = 1.3416 x 10^-3 mol dm^-3. The pH = -log10[H+] = -log10(1.3416 x 10^-3) = 2.872, which rounds to 2.87.

M1: Correct calculation of [H+] = 1.34 x 10^-3 mol dm^-3 (1 mark). M2: Correct pH calculation to 2 decimal places = 2.87 (1 mark).

Copper(II) has a 3d^9 electronic configuration, which represents an incomplete d-subshell. This allows d-d electronic transitions to occur by absorbing specific wavelengths of visible light, making the complex colored. Zinc(II) has a 3d^10 configuration, which is a completely full d-subshell. Therefore, no d-d transitions can occur, and the complex is colorless.

M1: Copper(II) has an incomplete d-subshell or d-orbitals (3d^9) allowing d-d electronic transitions to occur by absorbing visible light (1 mark). M2: Zinc(II) has a complete d-subshell or d-orbitals (3d^10) preventing any d-d electronic transitions (1 mark).

The cell potential is E_cell = E_reduction - E_oxidation. The half-reaction with the more positive potential (+1.51 V) undergoes reduction, and the one with the less positive potential (+0.77 V) undergoes oxidation. Therefore, E_cell = +1.51 - (+0.77) = +0.74 V. Since the Fe^2+/Fe^3+ half-reaction is reversed (oxidised), Fe^2+ acts as the reducing agent.

M1: E_cell = +0.74 V (or 0.74 V) (1 mark). M2: Reducing agent: Fe^2+ (accept Fe^2+(aq) or iron(II) ions; reject Fe or Fe^3+) (1 mark).

The Kp expression is defined using partial pressures of the gas phase products divided by the reactants, raised to the power of their stoichiometric coefficients: Kp = p(CH3OH) / (p(CO) * p(H2)^2). The units are: kPa / (kPa * kPa^2) = kPa^-2.

M1: Correct Kp expression using partial pressure notation (1 mark). Square concentration brackets must not be used. M2: Correct units of kPa^-2 (1 mark).

At a higher temperature, the Maxwell-Boltzmann distribution curve shifts to the right and has a lower peak. This means a significantly larger fraction of molecules possess energy equal to or greater than the activation energy (\(E \ge E_a\)). Consequently, there is a much higher frequency of successful collisions (more successful collisions per unit time), which greatly increases the rate of reaction.

1 mark: For explaining that a much larger fraction of molecules have energy greater than or equal to the activation energy (\(E \ge E_a\)). 1 mark: For stating that this leads to a higher frequency of successful collisions (or more successful collisions per unit time).

The forward reaction is exothermic. According to Le Chatelier's principle, an increase in temperature shifts the position of equilibrium to the left (the endothermic direction) to absorb the added thermal energy. This decreases the concentration of products and increases the concentration of reactants, thereby decreasing the value of the equilibrium constant \(K_c\).

1 mark: For stating that the value of \(K_c\) decreases. 1 mark: For explaining that the forward reaction is exothermic, so increasing temperature shifts the equilibrium to the left (favouring the reverse reaction).

Phosphorus has the outer electron configuration \(3\text{p}^3\), meaning each of the three \(3\text{p}\) orbitals is singly occupied. Sulfur has the outer configuration \(3\text{p}^4\), meaning one of the \(3\text{p}\) orbitals contains a pair of electrons. The mutual repulsion between the two paired electrons in the same orbital of sulfur makes it easier to remove one of these electrons compared to removing an electron from a singly occupied orbital in phosphorus, despite sulfur having a larger nuclear charge.

1 mark: For identifying that sulfur has a paired electron in a \(3\text{p}\) orbital (or \(3\text{p}^4\)) while phosphorus has singly occupied orbitals (or \(3\text{p}^3\)). 1 mark: For explaining that mutual repulsion between these paired electrons in sulfur makes the outer electron easier to remove.

Chlorine is a stronger oxidising agent than bromine, so it oxidises bromide ions to bromine molecules, being itself reduced to chloride ions: \(\text{Cl}_2(\text{g}) + 2\text{Br}^-(\text{aq}) \rightarrow 2\text{Cl}^-(\text{aq}) + \text{Br}_2(\text{aq})\). The formation of aqueous bromine causes the solution to change from colorless to orange (or yellow/brown).

1 mark: For the correct balanced ionic equation including state symbols: \(\text{Cl}_2(\text{g}) + 2\text{Br}^-(\text{aq}) \rightarrow 2\text{Cl}^-(\text{aq}) + \text{Br}_2(\text{aq})\). 1 mark: For stating that the colorless solution turns orange/yellow/brown (reject red/brown precipitate).

The solubility of Group 2 sulfates decreases down the group. Although free barium ions are highly toxic, barium sulfate is extremely insoluble in water and digestive fluids. Consequently, it cannot dissolve or be absorbed into the bloodstream when ingested, passing safely through the gastrointestinal tract and acting as an effective contrast agent for X-rays.

1 mark: For stating that the solubility decreases down the group. 1 mark: For explaining that barium sulfate is highly insoluble (or virtually insoluble), so it cannot be absorbed into the body/bloodstream.

Sulfur is reduced from an oxidation state of +6 in \(\text{H}_2\text{SO}_4\) to -2 in \(\text{H}_2\text{S}\), which represents a gain of 8 electrons. To balance the atoms and charge: \(\text{H}_2\text{SO}_4 + 8\text{H}^+ + 8\text{e}^- \rightarrow \text{H}_2\text{S} + 4\text{H}_2\text{O}\). Alternatively, using the sulfate ion: \(\text{SO}_4^{2-} + 10\text{H}^+ + 8\text{e}^- \rightarrow \text{H}_2\text{S} + 4\text{H}_2\text{O}\).

1 mark: For writing correct reactants and products (\(\text{H}_2\text{SO}_4\) / \(\text{SO}_4^{2-}\) and \(\text{H}_2\text{S}\)). 1 mark: For a fully balanced equation with correct quantities of electrons, protons, and water molecules.

First, calculate the heat energy transferred to the water: \(q = m c \Delta T = 150 \times 4.18 \times (45.5 - 19.5) = 150 \times 4.18 \times 26.0 = 16302 \text{ J} = 16.302 \text{ kJ}\). Next, calculate the moles of methanol burned: \(n = \frac{\text{mass}}{M_r} = \frac{0.960}{32.0} = 0.0300 \text{ mol}\). Finally, calculate the molar enthalpy change: \(\Delta H_c = -\frac{q}{n} = -\frac{16.302}{0.0300} = -543.4 \text{ kJ mol}^{-1}\) (which rounds to \(-543 \text{ kJ mol}^{-1}\) to 3 significant figures).

1 mark: For correct calculation of heat energy \(q = 16.3\text{ kJ}\) (or \(16302\text{ J}\)) and moles of methanol \(n = 0.0300\text{ mol}\). 1 mark: For the final answer of \(-543\text{ kJ mol}^{-1}\) (or \(-543.4\text{ kJ mol}^{-1}\)). Must include the negative sign for credit; accept answers correct to 3 or 4 significant figures.

First, determine the mass of water lost: \(4.99\text{ g} - 3.19\text{ g} = 1.80\text{ g}\). Next, calculate the amount in moles of anhydrous \(\text{CuSO}_4\): \(n(\text{CuSO}_4) = \frac{3.19}{159.6} = 0.0200\text{ mol}\). Calculate the amount in moles of water lost: \(n(\text{H}_2\text{O}) = \frac{1.80}{18.0} = 0.100\text{ mol}\). Determine the molar ratio: \(x = \frac{n(\text{H}_2\text{O})}{n(\text{CuSO}_4)} = \frac{0.100}{0.0200} = 5.00\). Thus, \(x = 5\).

1 mark: For calculating correct moles of anhydrous salt (\(0.0200\text{ mol}\)) and water (\(0.100\text{ mol}\)). 1 mark: For the final value of \(x = 5\) with correct working shown.

When the temperature is increased, Le Chatelier's principle states that the system will oppose the change by shifting in the endothermic direction. Since the forward reaction is exothermic (\(\Delta H < 0\)), the reverse reaction is endothermic. Therefore, the equilibrium position shifts to the left. This decrease in the concentration of the product (\(\text{CH}_3\text{OH}\)) and increase in the concentrations of the reactants (\(\text{CO}\) and \(\text{H}_2\)) results in a smaller value of \(K_c = \frac{[\text{CH}_3\text{OH}]}{[\text{CO}][\text{H}_2]^2}\).

[1 mark] For stating that the value of \(K_c\) decreases.
[1 mark] For explaining that the equilibrium shifts to the left / in the endothermic direction to oppose the increase in temperature.

1. To construct the half-equation for the reduction of sulfuric acid to sulfur dioxide, we start with either sulfuric acid or sulfate ions:
\(\text{H}_2\text{SO}_4 + 2\text{H}^+ + 2\text{e}^- \rightarrow \text{SO}_2 + 2\text{H}_2\text{O}\)
Alternatively, using the sulfate ion:
\(\text{SO}_4^{2-} + 4\text{H}^+ + 2\text{e}^- \rightarrow \text{SO}_2 + 2\text{H}_2\text{O}\)

2. The bromide ions are oxidized to bromine gas (\(2\text{Br}^- \rightarrow \text{Br}_2 + 2\text{e}^-\)), donating electrons to reduce the sulfur atom from an oxidation state of +6 in \(\text{H}_2\text{SO}_4\) to +4 in \(\text{SO}_2\). Therefore, bromide ions act as a reducing agent.

[1 mark] For a correct, balanced ionic half-equation:
\(\text{H}_2\text{SO}_4 + 2\text{H}^+ + 2\text{e}^- \rightarrow \text{SO}_2 + 2\text{H}_2\text{O}\) OR \(\text{SO}_4^{2-} + 4\text{H}^+ + 2\text{e}^- \rightarrow \text{SO}_2 + 2\text{H}_2\text{O}\) (allow state symbols, but they are not required).
[1 mark] For stating that bromide ions act as a 'reducing agent' / 'reductant' / 'electron donor'.

1. Write down the Arrhenius relationship: \(\ln(k_2 / k_1) = \frac{E_a}{R} \left(\frac{1}{T_1} - \frac{1}{T_2}\right)\). 2. Calculate the ratio of the rate constants: \(k_2 / k_1 = (5.24 \times 10^{-3}) / (1.35 \times 10^{-4}) = 38.815\). 3. Calculate \(\ln(38.815) = 3.6588\). 4. Calculate the reciprocal temperature difference: \(\frac{1}{T_1} - \frac{1}{T_2} = \frac{1}{298.0} - \frac{1}{338.0} = 3.3557 \times 10^{-3} - 2.9586 \times 10^{-3} = 3.971 \times 10^{-4}\text{ K}^{-1}\). 5. Rearrange to solve for \(E_a\): \(E_a = \frac{3.6588 \times 8.314}{3.971 \times 10^{-4}} = 76598\text{ J mol}^{-1}\). 6. Convert to \(\text{kJ mol}^{-1}\): \(E_a = 76.6\text{ kJ mol}^{-1}\).

M1: Rearranging the Arrhenius equation to solve for \(E_a\) [1 mark]. M2: Calculating the ratio \(k_2/k_1\) as 38.8 (or ln of ratio as 3.66) [1 mark]. M3: Calculating the reciprocal temperature difference as \(3.97 \times 10^{-4}\text{ K}^{-1}\) [1 mark]. M4: Substituting the values correctly into the rearranged equation [1 mark]. M5: Correct calculation of \(E_a\) in \(\text{J mol}^{-1}\) as 76598 [1 mark]. M6: Converting to \(\text{kJ mol}^{-1}\) and rounding to 3 significant figures to give 76.6 [1 mark].

1. Calculate initial concentrations: \([A]_0 = 2.00 / 5.00 = 0.400\text{ mol dm}^{-3}\), \([B]_0 = 3.00 / 5.00 = 0.600\text{ mol dm}^{-3}\). 2. Determine equilibrium concentration of C: \([C]_{eq} = 0.320\text{ mol dm}^{-3}\). Since 2 moles of C are produced from 1 mole of A and 2 moles of B: Change in \([A] = -0.320 / 2 = -0.160\text{ mol dm}^{-3}\), Change in \([B] = -0.320\text{ mol dm}^{-3}\). 3. Calculate equilibrium concentrations: \([A]_{eq} = 0.400 - 0.160 = 0.240\text{ mol dm}^{-3}\), \([B]_{eq} = 0.600 - 0.320 = 0.280\text{ mol dm}^{-3}\). 4. Expression for \(K_c\): \(K_c = \frac{[C]^2}{[A][B]^2}\). 5. Calculate \(K_c\): \(K_c = \frac{(0.320)^2}{0.240 \times (0.280)^2} = \frac{0.1024}{0.018816} = 5.442\text{ dm}^3\text{ mol}^{-1}\). 6. Units: \(\text{dm}^3\text{ mol}^{-1}\).

M1: Calculating initial concentrations of A and B correctly [1 mark]. M2: Calculating the equilibrium concentration of A as \(0.240\text{ mol dm}^{-3}\) [1 mark]. M3: Calculating the equilibrium concentration of B as \(0.280\text{ mol dm}^{-3}\) [1 mark]. M4: Giving the correct expression for \(K_c\) [1 mark]. M5: Correct calculation of \(K_c\) value to 3 significant figures (5.44) [1 mark]. M6: Deducing the correct units as \(\text{dm}^3\text{ mol}^{-1}\) [1 mark].

1. Calculate initial moles of \(\text{HCl}\): \(n(\text{HCl})_{\text{initial}} = 0.0500\text{ dm}^3 \times 1.00\text{ mol dm}^{-3} = 0.0500\text{ mol}\). 2. Calculate moles of \(\text{NaOH}\) used: \(n(\text{NaOH}) = 0.03250\text{ dm}^3 \times 0.400\text{ mol dm}^{-3} = 0.0130\text{ mol}\). 3. Determine excess moles of \(\text{HCl}\): \(n(\text{HCl})_{\text{excess}} = n(\text{NaOH}) = 0.0130\text{ mol}\). 4. Calculate moles of \(\text{HCl}\) that reacted with calcium carbonate: \(n(\text{HCl})_{\text{reacted}} = 0.0500 - 0.0130 = 0.0370\text{ mol}\). 5. Use stoichiometry to find moles of \(\text{CaCO}_3\): \(\text{CaCO}_3 + 2\text{HCl} \rightarrow \text{CaCl}_2 + \text{CO}_2 + \text{H}_2\text{O}\), so \(n(\text{CaCO}_3) = 0.0370 / 2 = 0.0185\text{ mol}\). 6. Find mass of \(\text{CaCO}_3\): \(\text{mass} = 0.0185\text{ mol} \times 100.1\text{ g mol}^{-1} = 1.852\text{ g}\). 7. Calculate percentage by mass: \((1.852 / 2.25) \times 100 = 82.3\%\).

M1: Calculating initial moles of \(\text{HCl}\) as 0.0500 [1 mark]. M2: Calculating moles of \(\text{NaOH}\) (and hence excess \(\text{HCl}\)) as 0.0130 [1 mark]. M3: Calculating moles of reacted \(\text{HCl}\) as 0.0370 [1 mark]. M4: Deducing the 1:2 reacting ratio and calculating moles of \(\text{CaCO}_3\) as 0.0185 [1 mark]. M5: Calculating mass of \(\text{CaCO}_3\) as 1.852 g [1 mark]. M6: Calculating the percentage by mass to 3 significant figures as 82.3% [1 mark].

1. Moles of \(\text{FeSO}_4 \cdot 7\text{H}_2\text{O}\) dissolved: \(n = 6.95 / 277.9 = 0.0250\text{ mol}\). 2. Concentration of \(\text{Fe}^{2+}\) solution: \(C = 0.0250\text{ mol} / 0.250\text{ dm}^3 = 0.100\text{ mol dm}^{-3}\). 3. Moles of \(\text{Fe}^{2+}\) in \(25.0\text{ cm}^3\) sample: \(n(\text{Fe}^{2+}) = 0.0250\text{ dm}^3 \times 0.100\text{ mol dm}^{-3} = 2.50 \times 10^{-3}\text{ mol}\). 4. Write the ionic redox equation: \(\text{MnO}_4^- + 8\text{H}^+ + 5\text{Fe}^{2+} \rightarrow \text{Mn}^{2+} + 5\text{Fe}^{3+} + 4\text{H}_2\text{O}\). 5. Determine moles of \(\text{MnO}_4^-\): \(n(\text{MnO}_4^-) = 2.50 \times 10^{-3} / 5 = 5.00 \times 10^{-4}\text{ mol}\). 6. Calculate concentration of \(\text{KMnO}_4\): \(C = 5.00 \times 10^{-4}\text{ mol} / 0.01880\text{ dm}^3 = 0.0266\text{ mol dm}^{-3}\) (to 3 sig figs).

M1: Calculating the total moles of \(\text{FeSO}_4 \cdot 7\text{H}_2\text{O}\) as 0.0250 [1 mark]. M2: Calculating the concentration of \(\text{Fe}^{2+}\) as \(0.100\text{ mol dm}^{-3}\) or moles in \(25.0\text{ cm}^3\) directly as \(2.50 \times 10^{-3}\text{ mol}\) [1 mark]. M3: Stating or using the 1:5 ratio of \(\text{MnO}_4^-\) to \(\text{Fe}^{2+}\) [1 mark]. M4: Calculating moles of \(\text{MnO}_4^-\) as \(5.00 \times 10^{-4}\text{ mol}\) [1 mark]. M5: Dividing the moles of \(\text{MnO}_4^-\) by the titre volume in \(\text{dm}^3\) [1 mark]. M6: Calculating the final concentration as \(0.0266\text{ mol dm}^{-3}\) (accept \(0.02659\) to \(0.0266\)) [1 mark].

1. Set up the Born-Haber cycle equation: \(\Delta H_f^{\theta}(\text{CaCl}_2) = \Delta H_{at}^{\theta}(\text{Ca}) + 1^{st} \text{ IE}(\text{Ca}) + 2^{nd} \text{ IE}(\text{Ca}) + \Delta H_{diss}^{\theta}(\text{Cl}_2) + 2 \times EA(\text{Cl}) + \Delta H_{Latt}^{\theta}\). 2. Note that the bond dissociation enthalpy of \(\text{Cl}_2\) produces \(2\text{Cl}(g)\), so we use \(+242\text{ kJ mol}^{-1}\) directly without multiplying by 2. 3. Note that we must multiply the electron affinity of chlorine by 2: \(2 \times (-349) = -698\text{ kJ mol}^{-1}\). 4. Substitute values: \(-796 = +178 + 590 + 1145 + 242 + 2(-349) + \Delta H_{Latt}^{\theta}\). 5. Simplify the sum: \(-796 = 1457 + \Delta H_{Latt}^{\theta}\). 6. Solve for \(\Delta H_{Latt}^{\theta}\): \(\Delta H_{Latt}^{\theta} = -796 - 1457 = -2253\text{ kJ mol}^{-1}\).

M1: Showing a correct Born-Haber expression or cycle layout linking all enthalpy changes [1 mark]. M2: Using the correct value for the atomisation of Cl (using \(+242\) directly since it represents \(\text{Cl}_2(g) \rightarrow 2\text{Cl}(g)\)) [1 mark]. M3: Scaling the electron affinity of chlorine by a factor of 2 (\(-698\text{ kJ mol}^{-1}\)) [1 mark]. M4: Correctly substituting all values into the cycle equation [1 mark]. M5: Summing the intermediate values correctly to obtain \(+1457\text{ kJ mol}^{-1}\) [1 mark]. M6: Calculating the final lattice enthalpy of formation as \(-2253\text{ kJ mol}^{-1}\) (value and negative sign required) [1 mark].

1. Rearrange ideal gas equation: \(n = pV / RT\). Convert units: \(p = 100,000\text{ Pa}\), \(V = 1.24 \times 10^{-3}\text{ m}^3\), \(T = 298\text{ K}\). 2. Calculate \(n(\text{CO}_2) = (100000 \times 1.24 \times 10^{-3}) / (8.314 \times 298) = 124 / 2477.572 = 0.05005\text{ mol}\). 3. Moles of \(\text{MCO}_3 = 0.05005\text{ mol}\). 4. Relative formula mass of \(\text{MCO}_3 = \text{mass} / \text{moles} = 7.39 / 0.05005 = 147.7\text{ g mol}^{-1}\). 5. Relative atomic mass of \(\text{M}\): \(A_r(\text{M}) = 147.7 - 12.0 - 3(16.0) = 87.7\). 6. Comparing with Periodic Table: Strontium (\(\text{Sr}\)) has \(A_r \approx 87.6\). Therefore, \(\text{M}\) is \(\text{Sr}\).

M1: Converting pressure to Pa (\(1.00 \times 10^5\text{ Pa}\)) and volume to \(\text{m}^3\) (\(1.24 \times 10^{-3}\text{ m}^3\)) [1 mark]. M2: Calculating the number of moles of \(\text{CO}_2\) as 0.0500 mol (allow 0.0501) [1 mark]. M3: Deducing that the moles of \(\text{MCO}_3\) is equal to the moles of \(\text{CO}_2\) (1:1 ratio) [1 mark]. M4: Calculating the relative formula mass (\(M_r\)) of \(\text{MCO}_3\) as 147.7 (or within range 147-148) [1 mark]. M5: Subtracting the formula mass of the carbonate group (60.0) to find the relative atomic mass of \(\text{M}\) as 87.7 [1 mark]. M6: Correctly identifying the metal \(\text{M}\) as Strontium or \(\text{Sr}\) [1 mark].

At a higher temperature, the gaseous molecules have greater average kinetic energy. This causes the Maxwell-Boltzmann distribution curve to change in the following ways:
1. The peak of the curve (representing the most probable molecular energy) shifts to the right (higher energy).
2. The peak of the curve becomes lower to ensure the total area under the curve (which represents the total number of molecules) remains constant.

Additionally, the activation energy \(E_a\) of the reaction is a constant minimum energy requirement for a reaction to occur. It is independent of temperature and therefore remains unchanged unless a catalyst is added. Thus, option C is correct.

Score 1 mark for selecting C.
Reject all other options.
- Option A is incorrect because the peak must become lower, not higher, and the activation energy does not change.
- Option B is incorrect because the peak shifts to the right, not the left.
- Option D is incorrect because the peak shifts to the right and becomes lower.

To find the value of \(K_c\), follow these steps:

1. **Determine the equilibrium moles:**
- Initial moles: \(n(A) = 2.0\text{ mol}\), \(n(B) = 3.0\text{ mol}\), \(n(C) = 0.0\text{ mol}\).
- Since \(1.0\text{ mol}\) of \(A\) remains at equilibrium, the change in moles of \(A\) is \(-1.0\text{ mol}\).
- From the stoichiometry, the change in moles of \(B\) is \(-2 \times 1.0 = -2.0\text{ mol}\). Thus, equilibrium \(n(B) = 3.0 - 2.0 = 1.0\text{ mol}\).
- The change in moles of \(C\) is \(+2 \times 1.0 = +2.0\text{ mol}\). Thus, equilibrium \(n(C) = 2.0\text{ mol}\).

2. **Calculate equilibrium concentrations using the volume (\(V = 2.0\text{ dm}^3\)):**
- \([A] = \frac{1.0\text{ mol}}{2.0\text{ dm}^3} = 0.5\text{ mol dm}^{-3}\)
- \([B] = \frac{1.0\text{ mol}}{2.0\text{ dm}^3} = 0.5\text{ mol dm}^{-3}\)
- \([C] = \frac{2.0\text{ mol}}{2.0\text{ dm}^3} = 1.0\text{ mol dm}^{-3}\)

3. **Calculate the value of \(K_c\):**
- \(K_c = \frac{[C]^2}{[A][B]^2} = \frac{(1.0)^2}{(0.5) \times (0.5)^2} = \frac{1.0}{0.125} = 8.0\)

4. **Determine the units of \(K_c\):**
- \(\text{Units} = \frac{(\text{mol dm}^{-3})^2}{(\text{mol dm}^{-3}) \times (\text{mol dm}^{-3})^2} = \frac{1}{\text{mol dm}^{-3}} = \text{dm}^3\text{ mol}^{-1}\)

Thus, \(K_c = 8.0\text{ dm}^3\text{ mol}^{-1}\), which corresponds to option C.

Score 1 mark for selecting C.
Reject all other options:
- Option A is incorrect (corresponds to incorrect stoichiometry or a failure to square concentration terms).
- Option B is incorrect (corresponds to using moles directly instead of concentrations, which yields a value of \(4.0\)).
- Option D is incorrect (corresponds to mathematical or concentration conversion errors).