2024 OCR GCSE Twenty First Century Science - Chemistry B - J258 模擬試題連答案詳解

The forward reaction is exothermic (\(\Delta H = -91\text{ kJ/mol}\)), so a lower temperature shifts the position of equilibrium to the right to increase the yield. There are 3 moles of gas on the reactant (left) side and 1 mole of gas on the product (right) side. A higher pressure shifts the position of equilibrium to the side with fewer moles of gas, which is the right side. Therefore, a combination of low temperature and high pressure maximizes the equilibrium yield of methanol.

1.5 marks: 0.5 marks for identifying that low temperature favors the exothermic forward reaction; 0.5 marks for identifying that high pressure shifts the equilibrium to the side with fewer gas moles; 0.5 marks for selecting the correct option (C).

Step 1: Calculate the number of moles of calcium carbonate: Moles = Mass / Mr = 5.0 g / 100 g/mol = 0.05 mol. Step 2: From the reaction stoichiometry, 1 mole of \(CaCO_3\) produces 1 mole of \(CO_2\). Therefore, moles of \(CO_2\) produced = 0.05 mol. Step 3: Calculate the volume of \(CO_2\): Volume = Moles * 24.0 dm3/mol = 0.05 * 24.0 = 1.2 dm3.

1 mark for calculating the correct moles of calcium carbonate (0.05 mol). 1 mark for multiplying the moles of carbon dioxide by 24.0 to calculate the volume (1.2). 0.5 mark for providing the correct unit (dm3).

Step 1: Calculate the total Mr of all reactants. Mr of \(CH_4\) = 12 + (4 * 1) = 16. Mr of \(H_2O\) = (2 * 1) + 16 = 18. Total mass of reactants = 16 + 18 = 34. Step 2: Calculate the total Mr of the desired product (3 moles of \(H_2\)). Mass of desired product = 3 * (2 * 1) = 6. Step 3: Calculate the atom economy: (6 / 34) * 100% = 17.647...% which rounds to 17.6%.

1 mark for calculating the correct total mass of reactants (34) or total products (34). 1 mark for the correct substitution of values into the atom economy formula: (6 / 34) * 100. 0.5 mark for rounding to 3 significant figures (17.6%).

During the electrolysis of brine, chloride ions (\(Cl^-\)) are attracted to the positive anode where they lose electrons (oxidation) to form chlorine gas (\(Cl_2\)). The half-equation is \(2Cl^- \rightarrow Cl_2 + 2e^-\). Because chlorine is a gas, bubbles are observed, and chlorine is a pale green-yellow gas with a sharp, pungent smell.

1 mark for the correct reactants and products in the half-equation: 2Cl- and Cl2. 1 mark for correct balancing with 2e- on the product side (or - 2e- on the reactant side). 0.5 mark for a valid observation (e.g., bubbles, effervescence, or a pale green/green-yellow gas).

Silicon dioxide exists as a giant covalent macromolecular lattice where each silicon atom is strongly bonded to oxygen atoms by many covalent bonds. Breaking these strong bonds requires a huge amount of thermal energy, giving it a very high melting point. In contrast, carbon dioxide exists as simple molecules. Although the covalent bonds within the molecules are strong, there are only weak intermolecular forces holding the molecules together, which require very little energy to overcome at room temperature.

1 mark for identifying that silicon dioxide has a giant covalent structure with many strong covalent bonds that need a lot of energy to break. 1 mark for identifying that carbon dioxide has a simple molecular structure with weak intermolecular forces. 0.5 mark for explaining that breaking covalent bonds in silicon dioxide requires significantly more energy than overcoming intermolecular forces in carbon dioxide.

Step 1: Calculate energy needed to break reactant bonds: (1 * H-H) + (1 * Cl-Cl) = 436 + 243 = 679 kJ/mol. Step 2: Calculate energy released when product bonds form: 2 * (H-Cl) = 2 * 432 = 864 kJ/mol. Step 3: Calculate overall energy change: Energy change = Energy in - Energy out = 679 - 864 = -185 kJ/mol.

1 mark for calculating total energy to break bonds as 679 kJ/mol. 1 mark for calculating total energy released as 864 kJ/mol. 0.5 mark for the correct final value of -185 kJ/mol (must include the negative sign).

A catalyst increases the reaction rate by providing an alternative reaction pathway that has a lower activation energy. Consequently, a greater proportion of reactant particles have the minimum energy required to react when they collide, leading to more successful collisions per second. On a reaction profile diagram, this lower activation energy is represented by a lower peak connecting the reactants to the products compared to the uncatalyzed pathway.

1 mark for explaining that the catalyst provides an alternative pathway with a lower activation energy. 1 mark for connecting lower activation energy to a higher rate of successful collisions. 0.5 mark for explaining that it is shown as a lower peak/curve on the reaction profile diagram.

The formula for the Rf value is: Rf = distance moved by substance / distance moved by solvent front. Substituting the given values: Rf = 3.6 cm / 8.0 cm = 0.45. Rf values are ratios and therefore do not have units.

1 mark for stating the correct Rf formula (distance moved by substance / distance moved by solvent). 1 mark for substituting the correct values: 3.6 / 8.0. 0.5 mark for the correct final answer of 0.45 (do not award the final 0.5 mark if units are included).

Extraction by heating with carbon is a displacement reaction where carbon acts as a reducing agent. Carbon is more reactive than iron, so it can displace iron from iron oxide, reducing the iron ions to iron metal. However, aluminium is more reactive than carbon, meaning carbon cannot displace or reduce aluminium from its oxide. Therefore, aluminium must be extracted using electrolysis.

1 mark for explaining that iron is less reactive than carbon, so carbon can reduce/displace iron from iron oxide. 1 mark for explaining that aluminium is more reactive than carbon, so carbon cannot reduce/displace aluminium from aluminium oxide. 0.5 mark for stating that electrolysis is required instead for aluminium.

1. Increasing the pressure shifts the position of equilibrium to the side with fewer gas molecules. In this reaction, there are 4 moles of gas on the left-hand side and 2 moles on the right-hand side, so the equilibrium shifts to the right.
2. This shift increases the percentage yield of ammonia.
3. High pressure is used in industry because it also increases the rate of reaction, achieving a compromise between maximizing yield, maintaining a fast reaction rate, and managing the high safety and building costs associated with high-pressure equipment.

Award [1 mark] for identifying that increasing pressure shifts equilibrium to the right because there are fewer molecules of gas on the right (2 compared to 4 on the left).
Award [1 mark] for stating this increases the percentage yield of ammonia.
Award [0.5 marks] for explaining that a high pressure increases the rate of reaction or makes the process more economically viable despite the high cost of equipment.

1. Calculate the relative formula mass (\(M_r\)) of \(\text{CaCO}_3\):
\(M_r = 40.0 + 12.0 + (3 \times 16.0) = 100.0\)

2. Calculate the relative formula mass (\(M_r\)) of \(\text{CaO}\):
\(M_r = 40.0 + 16.0 = 56.0\)

3. Calculate the moles of \(\text{CaCO}_3\) used:
\(\text{moles} = \frac{2.50\text{ g}}{100.0\text{ g/mol}} = 0.025\text{ mol}\)

4. From the balanced equation, the mole ratio of \(\text{CaCO}_3 : \text{CaO}\) is \(1 : 1\).
Therefore, \(0.025\text{ mol}\) of \(\text{CaO}\) is produced.

5. Calculate the mass of \(\text{CaO}\):
\(\text{mass} = 0.025\text{ mol} \times 56.0\text{ g/mol} = 1.40\text{ g}\)

Award [0.5 marks] for correct calculation of both formula masses (\(M_r\) of \(\text{CaCO}_3 = 100.0\) and \(\text{CaO} = 56.0\)).
Award [1 mark] for calculating the moles of reactant (\(0.025\text{ mol}\)).
Award [1 mark] for the correct final theoretical mass of \(1.40\text{ g}\) (accept \(1.4\text{ g}\)).

1. The gas produced at the anode (positive electrode) is chlorine.
2. The half-equation for the oxidation of chloride ions is: \(2\text{Cl}^- \rightarrow \text{Cl}_2 + 2\text{e}^-\) (or \(2\text{Cl}^- - 2\text{e}^- \rightarrow \text{Cl}_2\)).
3. In aqueous solution, water dissociates into \(\text{H}^+\) and \(\text{OH}^-\) ions. At the cathode (negative electrode), \(\text{H}^+\) ions are discharged in preference to \(\text{Na}^+\) ions because hydrogen is less reactive than sodium, making \(\text{H}^+\) ions easier to reduce.

Award [0.5 marks] for identifying the anode product as chlorine.
Award [1 mark] for the correct balanced half-equation (including state symbols if given, but not required).
Award [1 mark] for explaining that hydrogen ions are discharged preferentially because sodium is more reactive than hydrogen (or hydrogen ions are more easily reduced than sodium ions).

1. A catalyst speeds up a reaction by providing an alternative reaction pathway.
2. This alternative pathway has a lower activation energy, meaning more reactant particles have enough energy to react when they collide.
3. On a reaction profile diagram, the reaction pathway is shown as a curve going from reactants to products. The presence of a catalyst is represented by a second, lower curve with a lower peak, indicating a decreased activation energy barrier.

Award [1 mark] for stating that the catalyst provides an alternative reaction pathway.
Award [1 mark] for explaining that this pathway has a lower activation energy.
Award [0.5 marks] for describing that on a reaction profile, this is represented by a lower curve or lower peak.

1. Calculate the total volume of the reacting mixture:
\(\text{Volume} = 50.0\text{ cm}^3 + 50.0\text{ cm}^3 = 100.0\text{ cm}^3\)

2. Convert the volume to mass using the density (\(1.0\text{ g/cm}^3\)):
\(\text{Mass } (m) = 100.0\text{ g}\)

3. Use the energy equation \(Q = mc\Delta T\):
\(Q = 100.0\text{ g} \times 4.2\text{ J/g/}^\circ\text{C} \times 6.5\text{ }^\circ\text{C}\)
\(Q = 2730\text{ J}\)

Award [0.5 marks] for determining the correct total mass of the mixture (\(100.0\text{ g}\)).
Award [1 mark] for correctly substituting values into the \(Q = mc\Delta T\) formula.
Award [1 mark] for the correct calculation of \(2730\text{ J}\) (accept \(2.73\text{ kJ}\)).

1. Calculate the retention factor (\(R_f\)) using the formula:
\(R_f = \frac{\text{Distance travelled by substance}}{\text{Distance travelled by solvent front}} = \frac{6.4\text{ cm}}{8.0\text{ cm}} = 0.80\)

2. The baseline must be drawn in pencil because graphite (pencil) is insoluble in the chromatography solvents used.
3. If ink were used, it would dissolve in the solvent and run up the paper alongside the spots, contaminating the chromatogram and rendering the results invalid.

Award [1 mark] for the correct \(R_f\) calculation of \(0.80\) (or \(0.8\)).
Award [1 mark] for stating that pencil is insoluble in the solvent.
Award [0.5 marks] for explaining that ink is soluble and would dissolve, run, or contaminate the chromatography results.

1. Write and balance the reaction equation:
\(\text{Fe}_2\text{O}_3(\text{s}) + 3\text{CO}(\text{g}) \rightarrow 2\text{Fe}(\text{l}) + 3\text{CO}_2(\text{g})\)

2. This is a redox reaction because both oxidation and reduction occur simultaneously:
- Reduction is the loss of oxygen: \(\text{Fe}_2\text{O}_3\) loses oxygen to become iron (\(\text{Fe}\)).
- Oxidation is the gain of oxygen: \(\text{CO}\) gains oxygen to become carbon dioxide (\(\text{CO}_2\)).

Award [1 mark] for the correctly balanced equation: \(\text{Fe}_2\text{O}_3 + 3\text{CO} \rightarrow 2\text{Fe} + 3\text{CO}_2\).
Award [0.75 marks] for explaining that iron oxide is reduced because it loses oxygen.
Award [0.75 marks] for explaining that carbon monoxide is oxidized because it gains oxygen.

1. Both diamond and graphite are allotropes of carbon with giant covalent structures.
2. In diamond, each carbon atom is covalently bonded to four other carbon atoms in a tetrahedral arrangement. All valence electrons are locked in localized covalent bonds, meaning there are no free-moving charged particles to conduct electricity.
3. In graphite, each carbon atom is covalently bonded to only three other carbon atoms in hexagonal layers. This leaves one delocalized valence electron per carbon atom. These delocalized electrons are free to move between the layers and can flow to carry an electric current.

Award [0.5 marks] for stating that both have giant covalent structures made of carbon.
Award [1 mark] for explaining that in diamond, each carbon atom forms 4 bonds, leaving no free or delocalized electrons.
Award [1 mark] for explaining that in graphite, each carbon atom forms 3 bonds, leaving one delocalized electron per carbon atom that is free to move and carry charge.

1. Calculate the relative formula masses: \(M_r(\text{CaCO}_3) = 40.0 + 12.0 + (3 \times 16.0) = 100.0\) and \(M_r(\text{CaO}) = 40.0 + 16.0 = 56.0\). 2. Find the moles of \(\text{CaCO}_3\): \(\text{moles} = \frac{25.0}{100.0} = 0.250\text{ mol}\). 3. Since the reacting ratio is 1:1, moles of \(\text{CaO}\) produced = \(0.250\text{ mol}\). Mass of \(\text{CaO} = 0.250 \times 56.0 = 14.0\text{ g}\).

First mark: Calculate the relative formula masses of \(\text{CaCO}_3\) (100.0) and \(\text{CaO}\) (56.0). Second mark: Calculate the moles of \(\text{CaCO}_3\) (0.250 mol). Third mark: Calculate the correct mass of \(\text{CaO}\) (14.0 g). Accept 14 g.

1. Calculate moles of reactant: \(\text{moles of N}_2 = \frac{14.0}{28.0} = 0.50\text{ mol}\). 2. Calculate theoretical yield of \(\text{NH}_3\): 1 mole of \(\text{N}_2\) yields 2 moles of \(\text{NH}_3\), so theoretical moles of \(\text{NH}_3 = 2 \times 0.50 = 1.00\text{ mol}\). Theoretical mass of \(\text{NH}_3 = 1.00 \times 17.0 = 17.0\text{ g}\). 3. Calculate percentage yield: \(\text{yield} = \frac{4.25}{17.0} \times 100 = 25.0\\%\).

First mark: Correct calculation of nitrogen moles (0.50 mol) or theoretical moles of ammonia (1.00 mol). Second mark: Correct calculation of theoretical mass of ammonia (17.0 g). Third mark: Correct percentage yield (25.0%). Accept 25%.

1. Determine total mass of reactants = \(180.0\). 2. Determine mass of desired product (ethanol): \(2 \times 46.0 = 92.0\). 3. Calculate percentage atom economy: \(\frac{92.0}{180.0} \times 100 = 51.1\\%\).

First mark: Calculate the total mass of desired product (2 x 46.0 = 92.0). Second mark: Identify total mass of reactants/products as 180.0. Third mark: Correct calculation of atom economy to 3 significant figures (51.1%).

1. Determine the relative formula mass of \(\text{NaOH}\): \(M_r = 23.0 + 16.0 + 1.0 = 40.0\). 2. Calculate the amount of \(\text{NaOH}\) in moles: \(\text{moles} = \frac{4.00}{40.0} = 0.100\text{ mol}\). 3. Convert the volume to \(\text{dm}^3\) and calculate concentration: \(\text{volume} = 0.250\text{ dm}^3\); \(\text{concentration} = \frac{0.100}{0.250} = 0.400\text{ mol/dm}^3\).

First mark: Correct calculation of relative formula mass of NaOH (40.0). Second mark: Correct calculation of moles of NaOH (0.100 mol) and conversion of volume to 0.250 dm³. Third mark: Correct concentration value (0.400 mol/dm³). Accept 0.4 mol/dm³ or 0.4.

1. Calculate the change in volume of gas: \(48\text{ cm}^3 - 12\text{ cm}^3 = 36\text{ cm}^3\). 2. Calculate the time elapsed: \(50\text{ s} - 20\text{ s} = 30\text{ s}\). 3. Calculate the mean rate of reaction: \(\text{rate} = \frac{36}{30} = 1.2\text{ cm}^3/\text{s}\).

First mark: Correct calculation of volume change (36 cm³). Second mark: Correct calculation of time difference (30 s). Third mark: Correct calculation of mean rate (1.2 cm³/s).

1. Energy to break reactant bonds: \(4 \times (\text{C}-\text{H}) + 2 \times (\text{O}=\text{O}) = (4 \times 413) + (2 \times 498) = 1652 + 996 = 2648\text{ kJ/mol}\). 2. Energy released by forming product bonds: \(2 \times (\text{C}=\text{O}) + 4 \times (\text{O}-\text{H}) = (2 \times 805) + (4 \times 464) = 1610 + 1856 = 3466\text{ kJ/mol}\). 3. Overall energy change: \(2648 - 3466 = -818\text{ kJ/mol}\).

First mark: Correct total energy for breaking bonds (2648 kJ/mol). Second mark: Correct total energy for forming bonds (3466 kJ/mol). Third mark: Correct overall energy change with negative sign (-818 kJ/mol).

1. Find the moles of \(\text{HCl}\): \(\text{moles} = 0.100\text{ mol/dm}^3 \times 0.0250\text{ dm}^3 = 0.00250\text{ mol}\). 2. Since the reacting ratio is 1:1, moles of \(\text{NaOH}\) = \(0.00250\text{ mol}\). 3. Calculate concentration of \(\text{NaOH}\): \(\text{concentration} = \frac{0.00250\text{ mol}}{0.0200\text{ dm}^3} = 0.125\text{ mol/dm}^3\).

First mark: Calculate the moles of HCl (0.00250 mol). Second mark: State or use that the moles of NaOH is equal to the moles of HCl (0.00250 mol). Third mark: Correct calculation of the concentration of NaOH (0.125 mol/dm³).

1. Find the mass of \(\text{Fe}_2\text{O}_3\) in the ore: \(500\text{ tonnes} \times 0.700 = 350\text{ tonnes}\) . 2. Calculate the fraction of iron in \(\text{Fe}_2\text{O}_3\): \(M_r(\text{Fe}_2\text{O}_3) = (2 \times 56.0) + (3 \times 16.0) = 160.0\). Fraction of iron = \(\frac{112.0}{160.0} = 0.700\) (or \(70.0\\%\)). 3. Calculate the mass of iron extracted: \(350\text{ tonnes} \times 0.700 = 245\text{ tonnes}\).

First mark: Correct calculation of the mass of iron(III) oxide in the ore (350 tonnes). Second mark: Correct calculation of the percentage or fraction of iron in iron(III) oxide (70.0% or 0.7). Third mark: Correct final mass of iron (245 tonnes). Accept 245.

An increase in temperature always increases the rate of reaction because particles have more kinetic energy and collide more frequently with energy greater than the activation energy. Since the forward reaction is exothermic, increasing the temperature shifts the equilibrium position to the left (the endothermic direction) to oppose the change, which decreases the equilibrium yield of sulfur trioxide.

1.66 marks: B is the correct option. A, C, and D represent incorrect combinations of rate and equilibrium principles.

First, find the moles of each reactant: Moles of Fe2O3 = 16.0 / 160.0 = 0.10 mol. Moles of CO = 5.60 / 28.0 = 0.20 mol. According to the balanced equation, 1 mole of Fe2O3 requires 3 moles of CO. Therefore, 0.10 mol of Fe2O3 would require 0.30 mol of CO. Since we only have 0.20 mol of CO, CO is the limiting reactant. Using the stoichiometry, 3 moles of CO produce 2 moles of Fe. So, 0.20 mol of CO will produce 0.20 * (2/3) = 0.133 mol of Fe. Mass of Fe = 0.133 * 56.0 = 7.47 g.

1.66 marks: B is the correct answer. Give 1 mark for identifying CO as the limiting reactant and 0.66 marks for calculating the correct mass of iron.

The Rf value is calculated by dividing the distance travelled by the substance by the distance travelled by the solvent front, both measured from the baseline. Distance travelled by the solvent front = 9.5 cm - 1.5 cm = 8.0 cm. Distance travelled by the yellow dye = 7.9 cm - 1.5 cm = 6.4 cm. Rf = 6.4 / 8.0 = 0.80.

1.66 marks: B is correct. 1 mark for calculating correct distances from the baseline (6.4 cm and 8.0 cm). 0.66 marks for dividing 6.4 by 8.0 to get 0.80.

In aqueous solution, the ions present are Na+, SO4(2-), H+, and OH-. At the negative cathode, both Na+ and H+ are attracted. Since hydrogen is less reactive than sodium, hydrogen ions are preferentially reduced to form hydrogen gas: 2H+ + 2e- -> H2. At the positive anode, both SO4(2-) and OH- are attracted. Hydroxide ions are more easily oxidised than sulfate ions, forming oxygen gas and water: 4OH- -> O2 + 2H2O + 4e-. Therefore, oxygen is produced at the anode and hydrogen at the cathode.

1.66 marks: A is correct. 1 mark for identifying the correct reactions/products at both electrodes based on reactivity and ease of discharge. 0.66 marks for selecting the correct option.

Silicon dioxide forms a giant covalent structure (similar to diamond) where every silicon atom is covalently bonded to four oxygen atoms, requiring significant energy to break these strong covalent bonds. Carbon dioxide exists as simple molecules, which have strong covalent bonds within the molecules but only very weak intermolecular forces holding the molecules together in the bulk substance, requiring very little energy to overcome.

1.66 marks: A is correct. 1 mark for referencing giant covalent lattice vs simple molecular structures, 0.66 marks for correctly identifying that covalent bonds are broken in silicon dioxide while intermolecular forces are overcome in carbon dioxide.

Energy required to break bonds (reactants) = 1 * (H-H) + 1 * (Cl-Cl) = 436 + 243 = 679 kJ/mol. Energy released when making bonds (products) = 2 * (H-Cl) = 2 * 432 = 864 kJ/mol. Enthalpy change = Energy in - Energy out = 679 - 864 = -185 kJ/mol.

1.66 marks: B is correct. 1 mark for showing correct calculation of reactant bonds broken (679 kJ/mol) and product bonds formed (864 kJ/mol). 0.66 marks for obtaining -185 kJ/mol.

1. Moles of hydrogen gas = mass / \(M_r\) = \(12.0 / 2.0 = 6.0\) mol. 2. From the balanced equation, 3 moles of \(\text{H}_2\) produce 2 moles of \(\text{NH}_3\). Thus, 6.0 mol of \(\text{H}_2\) theoretically yields 4.0 mol of \(\text{NH}_3\). 3. Theoretical mass of \(\text{NH}_3\) = \(4.0\) mol * 17.0 g/mol = 68.0 g. 4. Percentage yield = (actual mass / theoretical mass) * 100 = \((17.0 / 68.0) \times 100 = 25.0\)%.

1 mark for calculating moles of hydrogen or theoretical moles of ammonia. 1 mark for calculating theoretical mass of ammonia (68.0 g). 0.5 marks for correct final percentage yield (25.0%).

1. Moles of \(\text{CuO}\) = mass / \(M_r\) = \(4.00 / (63.5 + 16.0) = 4.00 / 79.5 = 0.0503\) mol. 2. Moles of \(\text{HCl}\) = concentration * volume = \(1.00\) \(\text{mol/dm}^3\) * 0.0500 \(\text{dm}^3\) = 0.0500 mol. 3. The balanced equation shows 1 mole of \(\text{CuO}\) reacts with 2 moles of \(\text{HCl}\). Thus, 0.0503 mol of \(\text{CuO}\) would require 0.1006 mol of \(\text{HCl}\). Since only 0.0500 mol of \(\text{HCl}\) is available, \(\text{HCl}\) is the limiting reactant.

1 mark for calculating moles of \(\text{CuO}\) (0.0503 mol). 1 mark for calculating moles of \(\text{HCl}\) (0.0500 mol). 0.5 marks for identifying hydrochloric acid as the limiting reactant with a valid comparison.

At the positive electrode (anode), oxidation occurs. Hydroxide ions (\(\text{OH}^-\)) from the water are discharged in preference to sulfate ions (\(\text{SO}_4^{2-}\)) because hydroxide ions lose electrons more readily (are more easily oxidized). The balanced half-equation is: \(4\text{OH}^-(\text{aq}) \rightarrow 2\text{H}_2\text{O(l)} + \text{O}_2(\text{g}) + 4\text{e}^-\).

1 mark for correct anode half-equation. 1 mark for identifying that hydroxide ions (or water) are discharged in preference to sulfate. 0.5 marks for stating that hydroxide ions are more easily oxidized.

1. A catalyst provides an alternative reaction pathway with a lower activation energy. 2. This means a higher proportion of reacting particles have energy greater than or equal to the activation energy, leading to more frequent successful collisions. 3. The catalyst has no effect on the overall enthalpy change (\(\Delta H\)) of the reaction.

1 mark for explaining lower activation energy / alternative pathway. 1 mark for linking this to a higher frequency of successful collisions. 0.5 marks for stating no change in enthalpy change.

1. Energy required to break bonds (reactants) = \((\text{H}-\text{H}) + (\text{Cl}-\text{Cl}) = 436 + 243 = 679\) kJ/mol. 2. Energy released when making bonds (products) = \(2 \times (\text{H}-\text{Cl}) = 2 \times 432 = 864\) kJ/mol. 3. Overall energy change = energy in - energy out = \(679 - 864 = -185\) kJ/mol. Since the energy change is negative, the reaction is exothermic.

1 mark for energy absorbed to break bonds (679 kJ/mol). 1 mark for energy released to form bonds (864 kJ/mol). 0.5 marks for correct calculation of -185 kJ/mol and stating that it is exothermic.

1. Bioleaching uses bacteria to break down low-grade copper ores, producing an acidic solution called a leachate containing copper(II) ions (\(\text{Cu}^{2+}\)). 2. Copper metal is then recovered from the leachate by displacement using scrap iron. 3. The ionic half-equation for the reduction of copper(II) ions is \(\text{Cu}^{2+}(\text{aq}) + 2\text{e}^- \rightarrow \text{Cu(s)}\).

1 mark for describing that bacteria are used to produce a leachate containing copper ions. 1 mark for the correct ionic half-equation with state symbols. 0.5 marks for describing displacement using scrap iron.

In graphite, each carbon atom is covalently bonded to three others in layers, leaving one delocalized valence electron per carbon atom. These delocalized electrons are free to move throughout the structure and carry electrical charge. In diamond, each carbon atom is covalently bonded to four others in a rigid tetrahedral structure. All valence electrons are localized in covalent bonds, so there are no free electrons to carry electrical charge.

1 mark for explaining graphite's structure and delocalized electrons. 1 mark for explaining diamond's structure and lack of free electrons. 0.5 marks for linking delocalized electrons to electrical conductivity.

1. Rf value = distance moved by substance / distance moved by solvent front = \(5.2\text{ cm} / 8.0\text{ cm} = 0.65\). 2. A pure substance consists of only one compound and will produce a single spot on the chromatogram in any given solvent. A mixture consists of multiple compounds and will separate into two or more spots.

1 mark for correct calculation of Rf value (0.65). 1 mark for explaining that a pure substance produces only one spot. 0.5 marks for explaining that a mixture produces multiple spots.

Increasing the pressure shifts the position of equilibrium to the side with fewer moles of gas. Since the reactants have 3 moles of gas and the products have 2, the system shifts to the right, increasing the yield of sulfur trioxide. A compromise pressure of 1-2 atm is used because building and maintaining equipment to withstand extremely high pressures is highly expensive and poses severe safety hazards.

1 mark: Correctly identifies that increasing pressure shifts equilibrium to the right because there are fewer moles of gas on the right side. 1 mark: States this increases the yield of sulfur trioxide. 0.5 marks: Explains that extremely high pressures are avoided because of safety concerns or high economic/energy costs of building high-pressure equipment.

Step 1: Calculate the moles of magnesium carbonate. Moles = mass / molar mass = \(4.86\text{ g} / 84.3\text{ g/mol} = 0.05765\text{ mol}\). Step 2: Use the stoichiometry of the equation. 1 mole of \(\text{MgCO}_3\) produces 1 mole of \(\text{CO}_2\). Therefore, moles of \(\text{CO}_2 = 0.05765\text{ mol}\). Step 3: Calculate the volume. Volume = moles \(\times\) molar volume = \(0.05765\text{ mol} \times 24.0\text{ dm}^3/\text{mol} = 1.3836\text{ dm}^3\). To 3 significant figures, this is 1.38 \(\text{dm}^3\).

1 mark: Calculation of moles of magnesium carbonate (\(4.86 / 84.3 = 0.0577\)). 1 mark: Calculation of volume of gas (\(0.05765 \times 24.0 = 1.38\)). 0.5 marks: Stating the final answer to 3 significant figures as 1.38 (accept 1.38 to 1.40 if rounding occurred early).

At the cathode, water dissociates slightly to produce hydrogen ions (\(\text{H}^+\)) alongside sodium ions (\(\text{Na}^+\)). The half-equation for hydrogen formation is \(2\text{H}^+ + 2\text{e}^- \rightarrow \text{H}_2\). Sodium is higher in the reactivity series (more reactive) than hydrogen. This means hydrogen ions accept electrons more easily (are more easily reduced) than sodium ions, so hydrogen gas is discharged while sodium ions remain in solution.

1 mark: Correct balanced ionic half-equation showing reduction of hydrogen ions. 1 mark: Explaining that both sodium ions and hydrogen ions are present in aqueous solution. 0.5 marks: Explaining that sodium is more reactive than hydrogen, so hydrogen ions are preferentially discharged.

To find the rate at 20 seconds, the student must draw a tangent line to the curve at the 20-second point. They then calculate the gradient of this tangent line by dividing the change in volume (y-axis) by the change in time (x-axis). As the reaction proceeds, the reactant particles are consumed, which reduces their concentration. This lower concentration means there are fewer reactant particles per unit volume, leading to a lower frequency of successful collisions between particles per second, thus decreasing the rate.

1 mark: Describes drawing a tangent at 20 seconds and calculating its gradient. 1 mark: Explains that concentration of reactants decreases as they are used up. 0.5 marks: Relates this to a lower frequency of successful collisions per second.

Step 1: Energy required to break bonds (reactants) = \(\text{H}-\text{H} + \text{Cl}-\text{Cl} = 436 + 243 = 679\text{ kJ/mol}\). Step 2: Energy released when making bonds (products) = \(2 \times \text{H}-\text{Cl} = 2 \times 432 = 864\text{ kJ/mol}\). Step 3: Overall energy change = Energy in - Energy out = \(679 - 864 = -185\text{ kJ/mol}\). Since the overall energy change is negative (energy released is greater than energy absorbed), the reaction is exothermic.

1 mark: Correct calculation of total bond-breaking energy (679) and total bond-making energy (864). 1 mark: Calculation of overall energy change as -185 kJ/mol. 0.5 marks: Identifying the reaction as exothermic due to the negative energy change.

In phytoextraction, plants are grown on soil containing low concentrations of copper compounds. The plants absorb the metal ions through their root systems and accumulate them in their leaves and shoots. The plants are then harvested and burned to produce ash, which contains a high concentration of copper compounds. Copper can then be extracted from the ash, for example by electrolysis or displacement with scrap iron. This method is preferred over traditional open-cast mining because it causes far less noise and visual pollution, preserves high-grade ore reserves, and does not produce huge heaps of waste rock.

1 mark: Describes plants absorbing copper ions from soil and accumulating them. 1 mark: Describes harvesting and burning the plants to produce ash from which copper is extracted. 0.5 marks: Identifies an advantage such as less habitat destruction, less visual/noise pollution, or ability to utilize low-grade sources.

Step 1: Calculate the \(R_f\) value: \(R_f = \text{distance moved by spot} / \text{distance moved by solvent front} = 5.4\text{ cm} / 8.2\text{ cm} = 0.6585...\). Rounded to 2 decimal places, this is 0.66. Step 2: In TLC, the components in the mixture partition between the mobile phase (the moving solvent) and the stationary phase (the silica plate). Components that have a higher solubility in the solvent (higher affinity for the mobile phase) and lower adsorption to the plate (lower affinity for the stationary phase) will move further up the plate, causing separation.

1 mark: Correct calculation of \(R_f\) as 0.66 (must be to 2 decimal places). 1 mark: Explains that components partition between the mobile phase (solvent) and stationary phase (plate). 0.5 marks: Explains that substances with higher affinity for the mobile phase travel further.

Early photosynthetic organisms, such as cyanobacteria and early plants, used carbon dioxide and water to produce glucose and oxygen via photosynthesis. This process significantly reduced carbon dioxide levels and increased atmospheric oxygen levels over millions of years. The geological evidence for this rise in oxygen is found in banded iron formations (red beds). Dissolved iron in the early oceans reacted with the newly produced oxygen to form insoluble iron oxides, which settled and formed distinct red-coloured sedimentary rock layers. These layers are only found in rocks younger than about 2.4 billion years, marking when oxygen levels became significant.

1 mark: Explains that photosynthesis took in carbon dioxide and released oxygen. 1 mark: Identifies banded iron formations (red beds/iron oxides in rocks) as the geological evidence. 0.5 marks: Explains that these formed when dissolved iron in early oceans reacted with newly produced oxygen.

1. Calculate the moles of each element in 100 g of the compound:
- Moles of titanium (\(\text{Ti}\)) = \(\frac{60.0}{48.0} = 1.25\text{ mol}\)
- Moles of oxygen (\(\text{O}\)) = \(\frac{40.0}{16.0} = 2.50\text{ mol}\)

2. Determine the simplest ratio by dividing by the smallest value (1.25):
- \(\text{Ti} = \frac{1.25}{1.25} = 1\)
- \(\text{O} = \frac{2.50}{1.25} = 2\)

3. Write the empirical formula:
- The empirical formula is \(\text{TiO}_2\).

- **1 Mark**: Correct calculation of moles for both elements (\(1.25\text{ mol}\) of \(\text{Ti}\) and \(2.50\text{ mol}\) of \(\text{O}\)).
- **1 Mark**: Finding the simplest whole-number ratio (\(1 : 2\)).
- **1 Mark**: Correct final empirical formula written as \(\text{TiO}_2\).

1. Calculate the relative formula mass of nitrogen gas (\(\text{N}_2\)):
\(M_r(\text{N}_2) = 14.0 \times 2 = 28.0\text{ g/mol}\)

2. Calculate the number of moles of nitrogen gas reacting:
\(\text{Moles of } \text{N}_2 = \frac{5.60\text{ g}}{28.0\text{ g/mol}} = 0.20\text{ mol}\)

3. Use the stoichiometric ratio from the balanced equation to find the moles of hydrogen gas needed:
From the equation, \(1\text{ mol}\) of \(\text{N}_2\) reacts with \(3\text{ mol}\) of \(\text{H}_2\).
\(\text{Moles of } \text{H}_2 = 0.20\text{ mol} \times 3 = 0.60\text{ mol}\)

4. Calculate the volume of hydrogen gas:
\(\text{Volume of } \text{H}_2 = 0.60\text{ mol} \times 24.0\text{ dm}^3/\text{mol} = 14.4\text{ dm}^3\).

- **1 Mark**: Correct calculation of moles of \(\text{N}_2\) (\(0.20\text{ mol}\)).
- **1 Mark**: Correct calculation of moles of \(\text{H}_2\) based on stoichiometry (\(0.60\text{ mol}\)).
- **1 Mark**: Correct calculation of gas volume (\(14.4\text{ dm}^3\)).

1. Calculate the moles of ethanoic acid used:
\(\text{Moles of ethanoic acid} = \frac{12.0\text{ g}}{60.0\text{ g/mol}} = 0.20\text{ mol}\)

2. Find the theoretical yield of ethyl ethanoate:
Since the molar ratio is 1:1, the maximum expected moles of ethyl ethanoate is \(0.20\text{ mol}\).
\(\text{Theoretical mass of ethyl ethanoate} = 0.20\text{ mol} \times 88.0\text{ g/mol} = 17.6\text{ g}\)

3. Calculate the percentage yield:
\(\text{Percentage yield} = \left( \frac{\text{Actual Yield}}{\text{Theoretical Yield}} \right) \times 100 = \left( \frac{13.2\text{ g}}{17.6\text{ g}} \right) \times 100 = 75\%\).

- **1 Mark**: Correct calculation of the theoretical moles of product (\(0.20\text{ mol}\)).
- **1 Mark**: Correct calculation of the theoretical mass of product (\(17.6\text{ g}\)).
- **1 Mark**: Correct percentage yield calculation leading to \(75\%\).

1. Energy required to break bonds (reactants):
- \(1 \times \text{H-H} = 436\text{ kJ/mol}\)
- \(1 \times \text{Cl-Cl} = 243\text{ kJ/mol}\)
- Total energy input = \(436 + 243 = 679\text{ kJ/mol}\)

2. Energy released in forming bonds (products):
- \(2 \times \text{H-Cl} = 2 \times 431 = 862\text{ kJ/mol}\)

3. Overall energy change:
- \(\text{Energy change} = \text{Energy input} - \text{Energy output} = 679 - 862 = -183\text{ kJ/mol}\).

- **1 Mark**: Correctly calculated total energy input for bond breaking (\(679\text{ kJ/mol}\)).
- **1 Mark**: Correctly calculated total energy output for bond making (\(862\text{ kJ/mol}\)).
- **1 Mark**: Correctly calculated overall energy change, including the negative sign (\(-183\text{ kJ/mol}\)).

1. Calculate the number of moles of sulfuric acid reacted:
\(\text{Moles of } \text{H}_2\text{SO}_4 = \text{concentration} \times \text{volume in } \text{dm}^3\)
\(\text{Moles of } \text{H}_2\text{SO}_4 = 0.050\text{ mol/dm}^3 \times 0.0200\text{ dm}^3 = 0.0010\text{ mol}\)

2. Determine the moles of sodium hydroxide reacted using the balanced chemical equation (1:2 ratio):
\(\text{Moles of } \text{NaOH} = 2 \times \text{moles of } \text{H}_2\text{SO}_4\)
\(\text{Moles of } \text{NaOH} = 2 \times 0.0010\text{ mol} = 0.0020\text{ mol}\)

3. Calculate the concentration of the sodium hydroxide solution:
\(\text{Concentration of } \text{NaOH} = \frac{\text{moles}}{\text{volume in } \text{dm}^3} = \frac{0.0020\text{ mol}}{0.0250\text{ dm}^3} = 0.080\text{ mol/dm}^3\).

- **1 Mark**: Correct calculation of moles of sulfuric acid (\(0.0010\text{ mol}\)).
- **1 Mark**: Correct deduction of moles of sodium hydroxide (\(0.0020\text{ mol}\)).
- **1 Mark**: Correct concentration of NaOH (\(0.080\text{ mol/dm}^3\) or \(0.08\text{ mol/dm}^3\)).

1. Calculate the total mass of reactants:
\(\text{Mass of reactants} = 2 \times M_r(\text{CuO}) + A_r(\text{C})\)
\(= 2 \times (63.5 + 16.0) + 12.0 = 2 \times 79.5 + 12.0 = 159.0 + 12.0 = 171.0\text{ g/mol}\)

2. Calculate the mass of the desired product (copper):
\(\text{Mass of desired product} = 2 \times A_r(\text{Cu}) = 2 \times 63.5 = 127.0\text{ g/mol}\)

3. Calculate percentage atom economy:
\(\text{Atom economy} = \left( \frac{\text{Mass of desired product}}{\text{Total mass of reactants}} \right) \times 100\)
\(\text{Atom economy} = \left( \frac{127.0}{171.0} \right) \times 100 \approx 74.269\%\)

Rounding to 3 significant figures gives \(74.3\%\).

- **1 Mark**: Correct calculation of total formula mass of reactants (\(171.0\)).
- **1 Mark**: Correct calculation of total formula mass of desired product (\(127.0\)).
- **1 Mark**: Correct calculation of percentage atom economy to 3 significant figures (\(74.3\%\)).

1. Identify the formula for calculating the gradient (rate) from a tangent line:
\(\text{Rate} = \frac{\text{Change in } y}{\text{Change in } x} = \frac{\Delta y}{\Delta x}\)

2. Use the given coordinates on the tangent line:
- \(\Delta y = 45\text{ cm}^3 - 15\text{ cm}^3 = 30\text{ cm}^3\)
- \(\Delta x = 50\text{ s} - 10\text{ s} = 40\text{ s}\)

3. Calculate the gradient:
\(\text{Rate} = \frac{30\text{ cm}^3}{40\text{ s}} = 0.75\text{ cm}^3/\text{s}\).

- **1 Mark**: Correct calculation of the vertical change (\(\Delta y = 30\)).
- **1 Mark**: Correct calculation of the horizontal change (\(\Delta x = 40\)).
- **1 Mark**: Correct final calculated rate value (\(0.75\)) with the correct unit (\(\text{cm}^3/\text{s}\) or \(\text{cm}^3\text{ s}^{-1}\)).

For molten sodium chloride: Ions present are \(\text{Na}^+\) and \(\text{Cl}^-\). At the cathode, sodium ions gain electrons to form liquid sodium: \(\text{Na}^+\text{(l)} + \text{e}^- \rightarrow \text{Na(l)}\). At the anode, chloride ions lose electrons to form chlorine gas: \(2\text{Cl}^-\text{(l)} \rightarrow \text{Cl}_2\text{(g)} + 2\text{e}^-\). For aqueous sodium chloride (brine): Ions present are \(\text{Na}^+\), \(\text{Cl}^-\), \(\text{H}^+\), and \(\text{OH}^-\). At the cathode, hydrogen gas is produced because hydrogen ions are discharged in preference to sodium ions as hydrogen is less reactive than sodium: \(2\text{H}^+\text{(aq)} + 2\text{e}^- \rightarrow \text{H}_2\text{(g)}\). At the anode, chlorine gas is produced because halide ions are discharged in preference to hydroxide ions when concentrated: \(2\text{Cl}^-\text{(aq)} \rightarrow \text{Cl}_2\text{(g)} + 2\text{e}^-\). The presence of water introduces \(\text{H}^+\) and \(\text{OH}^-\) ions, which changes the cathode product to hydrogen gas due to competitive discharge.

Level 3 (5-6 marks): A detailed and coherent response that correctly identifies all four products. Writes accurate balanced half-equations with state symbols for both systems. Explains why hydrogen is discharged instead of sodium in the aqueous system by referring to the relative reactivity of sodium and hydrogen. Level 2 (3-4 marks): Correctly identifies most products. Writes some correct half-equations (may omit state symbols or have minor balancing errors). Mentions that hydrogen ions from water are discharged but the reactivity comparison may be incomplete. Level 1 (1-2 marks): Identifies some products for at least one process. Shows a basic understanding that water contains hydrogen and hydroxide ions, or writes one correct electrode reaction. 0 marks: No response worthy of credit.

Temperature: The forward reaction is exothermic, so a lower temperature shifts the equilibrium to the right, increasing the yield of ammonia. However, a lower temperature slows down the rate of reaction. A temperature of \(450^\circ\text{C}\) is a compromise that yields a reasonable amount of ammonia at a sufficiently fast rate. Pressure: There are fewer moles of gas on the product side (2 moles) than on the reactant side (4 moles). A high pressure shifts the equilibrium to the right, increasing yield, and also increases the rate of reaction by increasing collision frequency. However, extremely high pressures require expensive equipment to maintain and present significant safety hazards (explosions). A pressure of \(200\text{ atm}\) is a compromise between high yield, fast rate, and manageable safety/capital costs. Catalyst: An iron catalyst increases the rate of both the forward and reverse reactions equally by providing an alternative pathway with a lower activation energy. It does not affect the position of equilibrium or yield, but it allows the compromise temperature of \(450^\circ\text{C}\) to be viable, saving energy costs.

Level 3 (5-6 marks): Detailed and structured explanation of all three factors (temperature, pressure, and catalyst). Explains temperature in terms of the exothermic forward reaction and rate of collision. Explains pressure in terms of molecular count (moles of gas) and safety/cost factors. Explains how the catalyst speeds up the reaction rate without affecting the equilibrium yield. Level 2 (3-4 marks): Explains two factors well, or all three with some omissions (such as omitting the molecular count explanation for pressure or omitting the exothermic nature of the reaction). Level 1 (1-2 marks): Simple descriptions of how temperature, pressure, or a catalyst affect rate or yield, without linking them to equilibrium shifts or industrial compromise. 0 marks: No response worthy of credit.