2022 OCR GCSE Gateway Science - Chemistry A - J248 Practice Paper with Answers

At the anode (positive electrode), oxidation of negative ions occurs. Since chloride ions are present, they are discharged in preference to hydroxide ions, forming chlorine gas: \(2\text{Cl}^-\text{(aq)} \rightarrow \text{Cl}_2\text{(g)} + 2\text{e}^-\).

1 mark for the correct answer A.

Using a fine powder increases the surface area, which increases the rate of reaction. Since the mass of the limiting reactant (calcium carbonate) remains constant, the total volume of gas produced does not change.

1 mark for the correct answer B.

Silicon dioxide has a giant covalent structure with many strong covalent bonds that require large amounts of energy to break, leading to a high melting point. It does not conduct electricity because it contains no free-moving ions or delocalised electrons.

1 mark for the correct answer A.

Relative atomic mass is calculated by: \(A_{\text{r}} = \frac{(75.0 \times 35) + (25.0 \times 37)}{100} = \frac{2625 + 925}{100} = 35.5\).

1 mark for the correct answer B.

The retardation factor (\(R_{\text{f}}\)) is calculated using the formula: \(R_{\text{f}} = \frac{\text{distance moved by spot}}{\text{distance moved by solvent}} = \frac{6.0\text{ cm}}{8.0\text{ cm}} = 0.75\).

1 mark for the correct answer B.

Alkenes are unsaturated hydrocarbons containing at least one carbon-carbon double bond with the general formula \(\text{C}_n\text{H}_{2n}\). Propene (\(\text{C}_3\text{H}_6\)) is an alkene, and butene (\(\text{C}_4\text{H}_8\)) is also an alkene in the same series. Ethane and butane are alkanes.

1 mark for the correct answer B.

Greenhouse gases allow short-wavelength electromagnetic radiation (like visible light) from the sun to pass through the atmosphere. The Earth's surface absorbs this radiation and re-emits it as longer-wavelength infrared radiation. Greenhouse gases absorb this infrared radiation and re-radiate it, trapping heat in the atmosphere.

1 mark for the correct answer B.

To increase the yield of ammonia, we want to shift the equilibrium to the right. Because the forward reaction is exothermic, decreasing the temperature will favour the forward direction. Because there are more moles of gas on the reactant side (4 moles) than the product side (2 moles), increasing the pressure will shift the equilibrium to the side with fewer gas molecules (the right).

1 mark for the correct answer B.

A liquid at room temperature that does not conduct electricity as a liquid or in solution indicates a simple molecular structure. Covalent bonds exist within the molecules, but the weak intermolecular forces between molecules allow it to have a low boiling point, meaning it is a liquid at room temperature.

1 mark for selecting option C.

At the anode (positive electrode), negative ions migrate. In aqueous sodium sulfate, the anions present are sulfate (\(\text{SO}_4^{2-}\)) and hydroxide (\(\text{OH}^-\)). Hydroxide ions are more readily discharged than sulfate ions, producing oxygen gas and water: \(4\text{OH}^- \rightarrow 2\text{H}_2\text{O} + \text{O}_2 + 4\text{e}^-\).

1 mark for selecting option C.

The formula for calculating the \(R_{\text{f}}\) value is: \(R_{\text{f}} = \frac{\text{Distance travelled by substance}}{\text{Distance travelled by solvent}}\). Rearranging this to find the distance travelled by the dye: \text{Distance} = R_{\text{f}} \times \text{Distance of solvent} = 0.35 \times 8.0\text{ cm} = 2.8\text{ cm}.

1 mark for selecting option A.

In a solid, particles are in fixed positions and vibrate. When heated, they gain kinetic energy and vibrate more rapidly and with greater amplitude. This causes the average distance between the particles to increase slightly, resulting in expansion. The particles themselves do not change size, and they do not overcome their bonds or slide past one another until melting occurs.

1 mark for selecting option C.

The number of protons determines the atomic number of the element. The element with atomic number 19 is Potassium (\(\text{K}\)). Since the number of electrons is 18 (one fewer than the 19 protons), it has a \(1+\) charge. Therefore, the species is \(\text{K}^+\).

1 mark for selecting option A.

1. Calculate moles of \(\text{Mg}\): \(\text{Moles of Mg} = 4.8 / 24 = 0.2\text{ mol}\). 2. According to the balanced equation, \(1\text{ mol of Mg}\) produces \(1\text{ mol of H}_2\). Thus, \(0.2\text{ mol of Mg}\) produces \(0.2\text{ mol of H}_2\). 3. Calculate the volume of \(\text{H}_2\) gas: \(\text{Volume} = 0.2\text{ mol} \times 24\text{ dm}^3/\text{mol} = 4.8\text{ dm}^3\).

1 mark for selecting option B.

1. Energy required to break reactant bonds: \(\text{H-H} + \text{Cl-Cl} = 436 + 243 = 679\text{ kJ/mol}\). 2. Energy released making product bonds: \(2 \times \text{H-Cl} = 2 \times 432 = 864\text{ kJ/mol}\). 3. Energy change = \(\text{Energy to break} - \text{Energy released} = 679 - 864 = -185\text{ kJ/mol}\).

1 mark for selecting option A.

During the electrolysis of aqueous sodium chloride, both sodium ions (\(\text{Na}^+\)) and hydrogen ions (\(\text{H}^+\)) are attracted to the negative cathode. Because sodium is more reactive than hydrogen in the reactivity series, hydrogen ions are more easily reduced (gain electrons) than sodium ions. Therefore, hydrogen ions discharge to form hydrogen gas, leaving the sodium ions in the solution. The half-equation for this process is: \(2\text{H}^+ + 2\text{e}^- \rightarrow \text{H}_2\).

1 mark: Identify that sodium is more reactive than hydrogen (or hydrogen ions are preferentially reduced/discharged). 1 mark: State that hydrogen ions gain electrons at the cathode. 1 mark: Correctly balanced half-equation: \(2\text{H}^+ + 2\text{e}^- \rightarrow \text{H}_2\) (accept state symbols, accept fractions: \(\text{H}^+ + \text{e}^- \rightarrow \frac{1}{2}\text{H}_2\)).

When the concentration of hydrochloric acid is increased, there are more reactant particles (hydrogen ions) present in the same volume of solution. This means the particles are closer together, which increases the frequency of collisions between the acid particles and the magnesium ribbon. Consequently, there are more successful collisions per unit time, which increases the overall rate of reaction.

1 mark: State that there are more particles in a given volume / unit volume (do not accept just 'more particles'). 1 mark: State that collisions become more frequent / higher frequency of collisions (do not accept 'more collisions' without time reference). 1 mark: State that this leads to a higher rate of successful collisions (collisions with energy greater than the activation energy).

Step 1: Calculate the moles of each element in a 100g sample. Moles of \(\text{C} = \frac{40.0}{12.0} = 3.33\text{ mol}\). Moles of \(\text{H} = \frac{6.7}{1.0} = 6.70\text{ mol}\). Moles of \(\text{O} = \frac{53.3}{16.0} = 3.33\text{ mol}\). Step 2: Divide each mole value by the smallest calculated mole value (3.33). Ratio of \(\text{C} = \frac{3.33}{3.33} = 1\). Ratio of \(\text{H} = \frac{6.70}{3.33} = 2\). Ratio of \(\text{O} = \frac{3.33}{3.33} = 1\). The empirical formula is \(\text{CH}_2\text{O}\).

1 mark: Correct calculation of moles for all three elements (\(\text{C} = 3.33\), \(\text{H} = 6.7\), \(\text{O} = 3.33\)). 1 mark: Correct division by the smallest value to obtain the simplest whole number ratio (\(1 : 2 : 1\)). 1 mark: Correct final empirical formula: \(\text{CH}_2\text{O}\).

In addition polymerisation, the monomers used must be unsaturated molecules containing a carbon-to-carbon double bond (\(\text{C}=\text{C}\)), and only one product (the polymer) is formed. In condensation polymerisation, the monomers used contain two different functional groups (or there are two different types of monomers, each with two of the same functional groups). This process produces the polymer and also releases a small molecule, such as water (\(\text{H}_2\text{O}\)) or hydrogen chloride (\(\text{HCl}\)), as a byproduct.

1 mark: Identify that addition polymerisation requires monomers with a \(\text{C}=\text{C}\) double bond, whereas condensation polymerisation requires monomers with two functional groups. 1 mark: Identify that addition polymerisation results in only one product (the polymer). 1 mark: Identify that condensation polymerisation yields the polymer and a small molecule (such as water or hydrogen chloride).

Silicon dioxide has a giant covalent macromolecular structure. Every silicon atom is covalently bonded to four oxygen atoms in a massive tetrahedral lattice. Breaking these strong covalent bonds requires a large amount of thermal energy, leading to a high melting point. In contrast, carbon dioxide exists as simple covalent molecules. While the double bonds within each \(\text{CO}_2\) molecule are strong, the intermolecular forces between individual molecules are very weak. Minimal energy is needed to overcome these weak forces, explaining why carbon dioxide is a gas at room temperature.

1 mark: Describe silicon dioxide as having a giant covalent structure / lattice containing many strong covalent bonds. 1 mark: Describe carbon dioxide as consisting of simple molecular structures with weak intermolecular forces (forces between molecules). 1 mark: Explain that breaking the covalent bonds in silicon dioxide requires a large amount of energy, whereas overcoming the weak intermolecular forces in carbon dioxide requires very little energy.

According to Le Chatelier's Principle, if the temperature of a system at equilibrium is increased, the system will oppose this change by favoring the endothermic reaction to absorb heat. Since the forward reaction is exothermic, the backward reaction is endothermic. Therefore, increasing the temperature shifts the position of equilibrium to the left (towards the reactants). This results in a decreased yield of sulfur trioxide (\(\text{SO}_3\)).

1 mark: State that increasing the temperature shifts the position of equilibrium to the left (or in the reverse/backwards direction). 1 mark: Explain that the equilibrium shifts to favor the endothermic direction to oppose the temperature increase. 1 mark: State that the yield of sulfur trioxide (\(\text{SO}_3\)) decreases.

Short-wavelength electromagnetic radiation (such as ultraviolet and visible light) emitted by the Sun passes through the Earth's atmosphere and is absorbed by the Earth's surface, warming it. The Earth then re-radiates this energy as longer-wavelength infrared (IR) radiation. Greenhouse gases like carbon dioxide in the atmosphere absorb this outgoing infrared radiation instead of letting it escape into space. The gas molecules then re-emit this thermal energy in all directions, including back towards the Earth, trapping heat and warming the atmosphere.

1 mark: State that short-wavelength radiation from the Sun passes through the atmosphere and warms the Earth's surface. 1 mark: State that the Earth's surface re-emits longer-wavelength infrared radiation. 1 mark: Explain that greenhouse gases absorb this infrared radiation and re-emit / re-radiate it in all directions, trapping heat in the atmosphere.

First, calculate the retention factor (\(R_f\)) value using the formula: \(R_f = \frac{\text{distance moved by substance}}{\text{distance moved by solvent front}}\) which gives \(R_f = \frac{5.2}{8.0} = 0.65\). To identify the substance, the student can compare this calculated \(R_f\) value with known reference \(R_f\) values from a database. Crucially, the comparison must be done under identical conditions, using the exact same solvent (mobile phase) and chromatography paper (stationary phase).

1 mark: Correct calculation of the \(R_f\) value as \(0.65\) (accept no units; reject if greater than 1). 1 mark: State that the calculated value must be compared to known reference \(R_f\) values of substances. 1 mark: Specify that the comparison must be conducted using the same solvent and chromatography conditions (stationary phase).

1. Identify that both hydroxide (\(\text{OH}^-\)) and chloride (\(\text{Cl}^-\)) ions migrate to the anode. 2. Explain that in a concentrated halide solution, halide ions (\(\text{Cl}^-\)) are preferentially discharged over hydroxide ions. 3. Write the balanced oxidation half-equation: \(2\text{Cl}^- \rightarrow \text{Cl}_2 + 2\text{e}^-\) (or \(2\text{Cl}^- - 2\text{e}^- \rightarrow \text{Cl}_2\)).

1 mark: State that both chloride and hydroxide ions migrate to the anode, but chloride ions are discharged because they are in a high concentration (halide ions discharge preferentially in concentrated solutions). 1 mark: Identify chlorine gas is formed. 1 mark: Correct balanced half-equation: \(2\text{Cl}^- \rightarrow \text{Cl}_2 + 2\text{e}^-\) (allow state symbols, but they are not required).

1. Describe silicon dioxide: it has a giant covalent lattice structure. To melt it, many strong covalent bonds between atoms must be broken, which requires a massive amount of thermal energy. 2. Describe carbon dioxide: it consists of simple molecules. To melt/boil it, only weak intermolecular forces between the molecules need to be overcome, which requires very little energy. 3. Conclude that this explains the vast difference in melting points.

1 mark: Identify that silicon dioxide is a giant covalent structure with many strong covalent bonds that require a large amount of energy to break. 1 mark: Identify that carbon dioxide has a simple molecular structure with weak intermolecular forces between molecules. 1 mark: State that very little energy is needed to overcome these weak intermolecular forces compared to breaking strong covalent bonds.

1. Calculate the number of moles of magnesium: \(\text{moles} = \frac{\text{mass}}{\text{Ar}} = \frac{4.80\text{ g}}{24.0\text{ g/mol}} = 0.20\text{ mol}\). 2. Determine moles of hydrogen gas produced: the stoichiometric ratio of \(\text{Mg} : \text{H}_2\) is 1:1, so \(0.20\text{ mol}\) of \(\text{Mg}\) produces \(0.20\text{ mol}\) of \(\text{H}_2\). 3. Calculate gas volume: \(\text{volume} = \text{moles} \times 24.0\text{ dm}^3/\text{mol} = 0.20 \times 24.0 = 4.80\text{ dm}^3\).

1 mark: Calculate the correct number of moles of magnesium (\(0.20\text{ mol}\)). 1 mark: Identify the 1:1 molar ratio, meaning \(0.20\text{ mol}\) of \(\text{H}_2\) is produced. 1 mark: Correct calculation of volume of hydrogen gas as \(4.8\text{ dm}^3\) (accept \(4.80\text{ dm}^3\)). Award 3 marks for correct final answer without working shown.

1. Use the \(R_f\) formula: \(\text{distance travelled by dye} = R_f \times \text{distance travelled by solvent}\). 2. Substitute the values: \(\text{distance} = 0.65 \times 8.0\text{ cm} = 5.2\text{ cm}\). 3. Explain that ink is soluble in the chromatography solvent and would run or separate, interfering with the results, whereas pencil (graphite) is insoluble and will stay on the baseline.

1 mark: Calculate the distance travelled by the dye correctly as \(5.2\text{ cm}\). 1 mark: State that pencil (graphite) is insoluble in the solvent. 1 mark: Explain that ink is soluble and would dissolve, run, and contaminate or interfere with the chromatogram.

Step 1: Calculate the relative formula mass (\(M_r\)) of \(\text{TiCl}_4\): \(M_r = 48.0 + (4 \times 35.5) = 190.0\).
Step 2: Calculate the number of moles of \(\text{TiCl}_4\) used: \(\text{moles} = \frac{9.50}{190.0} = 0.050\text{ mol}\).
Step 3: Use the balanced equation to find the moles of titanium produced. The ratio of \(\text{TiCl}_4 : \text{Ti}\) is 1:1, so 0.050 mol of \(\text{Ti}\) is produced.
Step 4: Calculate the mass of titanium produced: \(\text{mass} = 0.050 \times 48.0 = 2.40\text{ g}\).

1. Correctly calculates the relative formula mass of \(\text{TiCl}_4\) as 190 [1 mark].
2. Calculates the moles of \(\text{TiCl}_4\) as 0.050 mol [1 mark].
3. Uses the 1:1 stoichiometry to deduce that 0.050 mol of \(\text{Ti}\) is produced [1 mark].
4. Calculates the final mass of titanium as 2.40 g [1 mark]. (Accept 2.4 g. Award full marks for correct final answer with no working.)

Step 1: Calculate the moles of sulfuric acid reacted: \(\text{moles} = \text{concentration} \times \text{volume in dm}^3 = 0.120 \times \frac{18.50}{1000} = 0.00222\text{ mol}\).
Step 2: Use the stoichiometry of the equation to find the moles of sodium hydroxide. The ratio of \(\text{H}_2\text{SO}_4\) to \(\text{NaOH}\) is 1:2. Moles of \(\text{NaOH} = 0.00222 \times 2 = 0.00444\text{ mol}\).
Step 3: Convert the volume of sodium hydroxide to \(\text{dm}^3\): \(\frac{25.0}{1000} = 0.0250\text{ dm}^3\).
Step 4: Calculate the concentration of sodium hydroxide: \(\text{concentration} = \frac{\text{moles}}{\text{volume}} = \frac{0.00444}{0.0250} = 0.1776\text{ mol/dm}^3\). Rounded to 3 significant figures, this is 0.178 \(\text{mol/dm}^3\).

1. Calculates moles of \(\text{H}_2\text{SO}_4\) as 0.00222 mol [1 mark].
2. Multiplies moles by 2 to find moles of \(\text{NaOH}\) as 0.00444 mol [1 mark].
3. Divides by volume of \(\text{NaOH}\) in \(\text{dm}^3\) (0.0250) [1 mark].
4. Obtains final answer of 0.178 (or 0.1776) \(\text{mol/dm}^3\) [1 mark]. (Award full marks for correct final answer with no working. Accept rounding to 0.18 if correct intermediate steps shown.)

Step 1: Calculate the relative formula mass (\(M_r\)) of the reactant, glucose: \(M_r(\text{C}_6\text{H}_{12}\text{O}_6) = (6 \times 12.0) + (12 \times 1.0) + (6 \times 16.0) = 180.0\).
Step 2: Calculate the relative formula mass of the desired product, ethanol: \(M_r(\text{C}_2\text{H}_5\text{OH}) = (2 \times 12.0) + (6 \times 1.0) + 16.0 = 46.0\).
Step 3: Calculate the total mass of the desired product: \(2 \times 46.0 = 92.0\).
Step 4: Calculate the atom economy: \(\text{Atom Economy} = \frac{\text{mass of desired product}}{\text{total mass of reactants}} \times 100 = \frac{92.0}{180.0} \times 100 = 51.11...\%\). Rounded to 1 decimal place, this is 51.1%.

1. Calculates the \(M_r\) of glucose as 180 and the \(M_r\) of ethanol as 46 [1 mark].
2. Determines the total mass of desired product as 92 [1 mark].
3. Correctly sets up the equation: \(\frac{92}{180} \times 100\) [1 mark].
4. Calculates the final atom economy as 51.1% (or 51%) [1 mark]. (Award full marks for correct final answer. If total mass of products was used as the denominator, allow method mark.)

Step 1: Calculate the energy required to break bonds in the reactants. Bonds broken: 4 \(\text{C-H}\) bonds and 1 \(\text{Cl-Cl}\) bond. \(\text{Energy in} = (4 \times 413) + 243 = 1652 + 243 = 1895\text{ kJ/mol}\). (Alternatively, considering only the bonds that change: 1 \(\text{C-H}\) and 1 \(\text{Cl-Cl}\) broken = \(413 + 243 = 656\text{ kJ/mol}\)).
Step 2: Calculate the energy released when new bonds are formed in the products. Bonds formed: 3 \(\text{C-H}\) bonds, 1 \(\text{C-Cl}\) bond, and 1 \(\text{H-Cl}\) bond. \(\text{Energy out} = (3 \times 413) + 327 + 432 = 1239 + 327 + 432 = 1998\text{ kJ/mol}\). (Alternatively, considering only the bonds that change: 1 \(\text{C-Cl}\) and 1 \(\text{H-Cl}\) formed = \(327 + 432 = 759\text{ kJ/mol}\)).
Step 3: Calculate the overall energy change: \(\text{Energy change} = \text{Energy in} - \text{Energy out} = 1895 - 1998 = -103\text{ kJ/mol}\) (or \(656 - 759 = -103\text{ kJ/mol}\)).

1. Calculates total energy needed to break reactant bonds as 1895 kJ/mol (or 656 kJ/mol) [1 mark].
2. Calculates total energy released from making product bonds as 1998 kJ/mol (or 759 kJ/mol) [1 mark].
3. Correct subtraction method: \(\text{energy to break} - \text{energy to make}\) [1 mark].
4. Final answer of -103 kJ/mol [1 mark]. (Award 3 marks for 103 without the minus sign. Award full marks for correct final answer with sign and working.)

Step 1: Calculate the relative formula mass (\(M_r\)) of \(\text{CaCO}_3\): \(M_r = 40.0 + 12.0 + (3 \times 16.0) = 100.0\).
Step 2: Calculate the number of moles of \(\text{CaCO}_3\) reacted: \(\text{moles} = \frac{5.00}{100.0} = 0.050\text{ mol}\).
Step 3: Deduce the moles of carbon dioxide gas produced. The mole ratio of \(\text{CaCO}_3 : \text{CO}_2\) is 1:1, so 0.050 mol of \(\text{CO}_2\) is produced.
Step 4: Calculate the volume of \(\text{CO}_2\) at RTP: \(\text{volume} = 0.050 \times 24.0 = 1.20\text{ dm}^3\).

1. Correctly calculates the relative formula mass of \(\text{CaCO}_3\) as 100 [1 mark].
2. Calculates moles of \(\text{CaCO}_3\) as 0.050 mol [1 mark].
3. Recognises 1:1 mole ratio of reactant to gas product [1 mark].
4. Multiplies by 24.0 to obtain 1.20 \(\text{dm}^3\) (accept 1.2 or 1.20) [1 mark]. (Award full marks for correct final answer with no working.)

### Indicative Content

#### 1. Temperature
* **Effect on Yield**: The forward reaction is exothermic. According to Le Chatelier's principle, lowering the temperature shifts the position of equilibrium to the right to produce more heat, increasing the yield of ammonia. Thus, a high temperature decreases the equilibrium yield.
* **Effect on Rate**: Lowering the temperature decreases the kinetic energy of the reacting particles. This results in fewer successful collisions per second, slowing down the rate of reaction.
* **Compromise**: \(450\ ^\circ\text{C}\) is a compromise temperature. It is low enough to ensure a reasonable yield of ammonia, but high enough to maintain an acceptable rate of reaction.

#### 2. Pressure
* **Effect on Yield**: There are 4 moles of gas on the reactant side and only 2 moles of gas on the product side. Increasing the pressure shifts the equilibrium to the side with fewer gas molecules (the right), increasing the yield of ammonia.
* **Effect on Rate**: Increasing the pressure brings the gas particles closer together, increasing the frequency of collisions, which increases the rate of reaction.
* **Compromise**: While higher pressure increases both rate and yield, extremely high pressures are incredibly expensive to maintain (requiring strong pipes, compressors, and lots of energy) and present significant safety hazards (risk of explosion). Therefore, \(200\text{ atm}\) is chosen as a safe and economically viable compromise.

#### 3. Iron Catalyst
* **Role**: The catalyst increases the rate of both the forward and reverse reactions equally by providing an alternative reaction pathway with a lower activation energy.
* **Effect on Yield**: It does NOT affect the position of equilibrium or the yield of ammonia.
* **Benefit**: It allows the reaction to reach dynamic equilibrium much faster, meaning a lower compromise temperature can be utilized, saving energy costs.

### Level of Response Criteria

* **Level 3 (5–6 marks)**:
* Explains the effects of both temperature and pressure on rate AND equilibrium yield clearly, using correct chemistry terminology (e.g., collision theory, exothermic forward reaction, gas mole ratio, Le Chatelier's principle).
* Explains the compromise for both temperature and pressure in terms of yield, rate, and cost/safety.
* Explains the role and economic benefit of the iron catalyst.
* *Line of reasoning is well-developed, clear, and logically structured.*

* **Level 2 (3–4 marks)**:
* Explains the effect of temperature and/or pressure on rate and yield, with some reference to compromise conditions.
* Mentions the role of the catalyst (e.g., speeds up reaction).
* *There is a line of reasoning with some structure. Supporting points are mostly relevant but may not cover all parts of the question in equal depth.*

* **Level 1 (1–2 marks)**:
* Identifies basic relationships (e.g., high temperature makes reaction faster; high pressure increases yield).
* Mentions that a catalyst is used to speed up the reaction.
* *Information is basic and lacks detailed chemical explanations or discussion of compromises.*

* **0 marks**: No response or no creditworthy content.

The forward reaction is exothermic (\(\Delta H = -91\text{ kJ/mol}\)), so a low temperature shifts the position of equilibrium to the right to oppose the change, increasing the yield of methanol. There are 3 moles of gas on the reactant side and 1 mole of gas on the product side. Therefore, a high pressure shifts the position of equilibrium to the right (the side with fewer moles of gas) to decrease the pressure, also increasing the yield of methanol.

[1 mark] C - Low temperature and high pressure
- Reject other combinations.

In aqueous solution, sodium ions (\(\text{Na}^+\)) and hydrogen ions (\(\text{H}^+\)) migrate to the cathode. Since sodium is more reactive than hydrogen, hydrogen gas is produced at the cathode. Sulfate ions (\(\text{SO}_4^{2-}\)) and hydroxide ions (\(\text{OH}^-\)) migrate to the anode. Hydroxide ions are preferentially discharged over sulfate ions to form oxygen gas and water at the anode.

[1 mark] A - Anode: Oxygen; Cathode: Hydrogen
- Reject other combinations.

Increasing the concentration increases the number of acid particles per unit volume. This increases the frequency of collisions between the reactant particles, which in turn increases the rate of reaction. It does not affect the activation energy, the average kinetic energy of the particles, or the proportion of particles with energy greater than the activation energy (which is influenced by temperature and catalysts).

[1 mark] B - It increases the frequency of collisions between reactant particles
- Reject options suggesting changes in energy or activation energy.

Condensation polymerisation to form a polyester requires two monomers that each have two functional groups at their ends (bifunctional). One must be a dicarboxylic acid (containing two \(\text{-COOH}\) groups) and the other must be a diol (containing two \(\text{-OH}\) groups). Option A represents propanedioic acid (\(\text{HOOC-CH}_2\text{-COOH}\)) and ethane-1,2-diol (\(\text{HO-CH}_2\text{CH}_2\text{-OH}\)), which will form ester links at both ends to create a polymer chain.

[1 mark] A - \(\text{HOOC-CH}_2\text{-COOH}\) and \(\text{HO-CH}_2\text{CH}_2\text{-OH}\)
- Reject monomer pairs that do not contain two of each functional group.

Portion 1: The reaction with acid producing carbon dioxide gas (which turns limewater cloudy) indicates carbonate ions (\(\text{CO}_3^{2-}\)) are present. Portion 2: No white precipitate with barium chloride means sulfate ions (\(\text{SO}_4^{2-}\)) are absent. Portion 3: A cream precipitate with silver nitrate indicates bromide ions (\(\text{Br}^-\)) are present. Therefore, the salt solution contains carbonate and bromide ions.

[1 mark] C - Carbonate and bromide
- Reject other combinations.

Atom Economy is calculated using the formula:
\(\text{Atom Economy} = \frac{\text{Total } M_r \text{ of desired product}}{\text{Total } M_r \text{ of all reactants}} \times 100\)

Desired product is \(3\text{H}_2\), mass = \(3 \times 2.0 = 6.0\).
Total mass of reactants = \(\text{CH}_4 + \text{H}_2\text{O} = 16.0 + 18.0 = 34.0\).

\(\text{Atom Economy} = \frac{6.0}{34.0} \times 100 = 17.65\% \approx 17.6\%\).

[1 mark] B - \(17.6\%\)
- Allow 18% if rounded.
- Reject 82.4% (atom economy of byproduct CO) or 5.9% (using 1 mole of H2 instead of 3).

Ionic substances (giant ionic lattices) have high melting points due to strong electrostatic attractions between oppositely charged ions. They cannot conduct electricity when solid because the ions are fixed in a giant lattice, but they do conduct when molten or dissolved in water because the ions are free to move and carry charge.

[1 mark] B - Giant ionic lattice with ionic bonding
- Reject other structures.

Step 1: Calculate moles of \(\text{NaOH}\):
\(\text{moles} = 0.200\text{ mol/dm}^3 \times 0.0250\text{ dm}^3 = 0.0050\text{ mol}\).

Step 2: Use the stoichiometry of the equation:
\(2\text{ moles of NaOH}\) react with \(1\text{ mole of H}_2\text{SO}_4\).
So, \(\text{moles of H}_2\text{SO}_4 = \frac{0.0050\text{ mol}}{2} = 0.0025\text{ mol}\).

Step 3: Calculate the concentration of \(\text{H}_2\text{SO}_4\):
\(\text{concentration} = \frac{\text{moles}}{\text{volume(dm}^3\text{)}} = \frac{0.0025\text{ mol}}{0.0125\text{ dm}^3} = 0.200\text{ mol/dm}^3\).

[1 mark] B - \(0.200\text{ mol/dm}^3\)
- Reject other numerical values.

At the positive anode, negative anions are attracted. In aqueous copper(II) sulfate, these are \(OH^{-}\) and \(SO_{4}^{2-}\). Since \(OH^{-}\) is more easily oxidized than \(SO_{4}^{2-}\), oxygen gas is produced at the anode according to the half-equation: \(4OH^{-} \rightarrow O_{2} + 2H_{2}O + 4e^{-}\).

1 mark for selecting B.

The forward reaction is endothermic (\(\Delta H > 0\)), so increasing the temperature shifts the equilibrium position to the right to favor the formation of \(NO_{2}\). There is 1 mole of gas on the left and 2 moles of gas on the right, so decreasing the pressure shifts the equilibrium position to the right (the direction with more moles of gas) to produce more brown \(NO_{2}\) gas.

1 mark for selecting B.

Condensation polymerization to form a polyamide requires monomers with two amine functional groups (diamines) and monomers with two carboxylic acid functional groups (dicarboxylic acids), which react to form amide linkages with the loss of water molecules.

1 mark for selecting B.

Atom economy is calculated as \(\frac{\text{total } M_r \text{ of desired product}}{\text{total } M_r \text{ of all reactants}} \times 100\). The desired product is \(2Fe\), with mass \(2 \times 56 = 112\). The total mass of reactants is \(Fe_{2}O_{3} + 3CO = (2 \times 56 + 3 \times 16) + 3 \times (12 + 16) = 160 + 84 = 244\). Atom economy = \(\frac{112}{244} \times 100 = 45.9\%\).

1 mark for selecting B.

Increasing the temperature increases the kinetic energy of the particles. This results in more frequent collisions because the particles move faster, but more importantly, a much larger proportion of the colliding particles possess energy equal to or greater than the activation energy, leading to a higher rate of successful collisions.

1 mark for selecting B.

Silicon dioxide is a giant covalent structure where all atoms are held together by strong covalent bonds. It has no free-moving charged particles (ions or delocalized electrons) in either solid or liquid states, so it cannot conduct electricity. Sodium chloride is a giant ionic lattice; in the solid state, its ions are fixed, but when molten, the electrostatic forces are overcome, and the ions are free to move and conduct electricity.

1 mark for selecting C.

Relative atomic mass \(A_r = \frac{(63 \times 69.2) + (65 \times 30.8)}{100} = \frac{4359.6 + 2002.0}{100} = \frac{6361.6}{100} = 63.616\). Rounded to 1 decimal place, this is 63.6.

1 mark for selecting B.

Le Chatelier's principle states that increasing temperature shifts the position of equilibrium in the endothermic direction to absorb the added thermal energy. Since the forward reaction is exothermic, the reverse reaction is endothermic. Therefore, increasing the temperature will cause the equilibrium to shift to the left, which decreases the yield of the product \( \text{SO}_3 \).

1 mark: State that the yield of \( \text{SO}_3 \) decreases (or equilibrium shifts to the left). 1 mark: Identify that the forward reaction is exothermic or the reverse reaction is endothermic. 1 mark: Explain that increasing the temperature shifts the equilibrium in the endothermic direction to absorb heat.

In aqueous solution, both hydrogen ions (\( \text{H}^+ \)) and sodium ions (\( \text{Na}^+ \)) migrate to the negative electrode (cathode). Since sodium is more reactive than hydrogen, sodium ions are more stable and remain in solution, while hydrogen ions are preferentially reduced (discharged) by gaining electrons to form diatomic hydrogen gas molecules: \( 2\text{H}^+ + 2\text{e}^- \rightarrow \text{H}_2 \).

1 mark: State the correct half-equation: \( 2\text{H}^+ + 2\text{e}^- \rightarrow \text{H}_2 \) (allow state symbols to be omitted). 1 mark: Explain that both hydrogen ions and sodium ions are attracted to the negative electrode. 1 mark: Explain that hydrogen is discharged because sodium is more reactive than hydrogen (or hydrogen ions are more easily reduced/discharged).

The rate of reaction is equal to the gradient of the tangent line drawn at that specific time. The gradient is calculated using the formula: \( \text{Gradient} = \frac{\text{change in } y}{\text{change in } x} \). Using the coordinates provided: \( \text{Gradient} = \frac{48 - 12}{50 - 0} = \frac{36}{50} = 0.72 \). Since the vertical axis measures volume in cubic centimeters and the horizontal axis measures time in seconds, the unit is cubic centimeters per second (\( \text{cm}^3/\text{s} \)).

1 mark: Show substitution into the gradient formula: \( \frac{48 - 12}{50 - 0} \) or equivalent working. 1 mark: Correct calculation of the value: 0.72. 1 mark: State the correct unit: \( \text{cm}^3/\text{s} \) or \( \text{cm}^3\text{ s}^{-1} \).

1. Calculate the number of moles of magnesium: \( \text{moles of Mg} = \frac{\text{mass}}{\text{A}_r} = \frac{4.80}{24.3} \approx 0.198\text{ mol} \). 2. Calculate the number of moles of hydrochloric acid: \( \text{moles of HCl} = \text{concentration} \times \text{volume in dm}^3 = 1.50 \times 0.100 = 0.150\text{ mol} \). 3. Look at the stoichiometry of the reaction: \( 1 \text{ mole of Mg} \) reacts with \( 2 \text{ moles of HCl} \). Therefore, \( 0.198\text{ mol} \) of Mg requires \( 2 \times 0.198 = 0.396\text{ mol} \) of HCl. Since only \( 0.150\text{ mol} \) of HCl is present, HCl is in short supply and is the limiting reactant.

1 mark: Calculate the moles of magnesium as 0.198 mol (accept 0.20 mol). 1 mark: Calculate the moles of HCl as 0.150 mol. 1 mark: Show comparison using the 1:2 reaction ratio to correctly conclude that HCl is the limiting reactant.

Polyesters are formed via condensation polymerization. The reaction requires two types of monomers: a dicarboxylic acid, which has two carboxylic acid functional groups (\( -\text{COOH} \)), and a diol, which has two alcohol functional groups (\( -\text{OH} \)). The carboxylic acid group of one monomer reacts with the alcohol group of another, forming an ester link (\( -\text{COO}- \)) that binds the monomers together into a long chain. Each time an ester link is formed, a small molecule of water (\( \text{H}_2\text{O} \)) is eliminated.

1 mark: Identify the monomers as dicarboxylic acids (containing two \( -\text{COOH} \) groups) and diols (containing two \( -\text{OH} \) groups). 1 mark: Explain that these groups react together to form ester links that connect the monomers into a chain. 1 mark: State that a small molecule of water (\( \text{H}_2\text{O} \)) is released/eliminated for each ester link formed.

Because the forward reaction is exothermic, a lower temperature shifts the equilibrium to the right, increasing the yield of ammonia. However, at low temperatures, the reactant particles have less kinetic energy, leading to fewer successful collisions per second and a rate of reaction that is too slow to be economically viable. A high temperature increases the rate but severely lowers the yield of ammonia. Therefore, a compromise temperature of \( 450^\circ\text{C} \) is chosen to achieve a fast enough rate of reaction while still obtaining an acceptable yield of ammonia.

1 mark: Explain that a lower temperature would increase the equilibrium yield of ammonia because the forward reaction is exothermic. 1 mark: Explain that a higher temperature is needed to increase the rate of reaction (particles have more energy and collide more frequently). 1 mark: State that \( 450^\circ\text{C} \) is a compromise to obtain a reasonable yield at an acceptable rate.

Silicon dioxide forms a giant macromolecular covalent lattice. To melt it, many strong covalent bonds between silicon and oxygen atoms must be broken, which requires a massive amount of thermal energy. Carbon dioxide exists as simple covalent molecules. The strong covalent bonds within the molecule are not broken during sublimation; instead, only the weak intermolecular forces between individual \( \text{CO}_2 \) molecules must be overcome, which requires very little energy.

1 mark: State that silicon dioxide has a giant covalent lattice with many strong covalent bonds that must be broken. 1 mark: State that carbon dioxide has a simple molecular structure with weak intermolecular forces between the molecules. 1 mark: Link the structural differences to the energy needed to melt/sublime them (strong covalent bonds require a lot of energy to break, whereas weak intermolecular forces require minimal energy).

The percentage atom economy is calculated using the formula: \( \text{Atom Economy} = \frac{\text{Total } M_r \text{ of desired product(s)}}{\text{Total } M_r \text{ of all reactants}} \times 100 \). 1. Calculate the total relative mass of the reactants: \( \text{Total reactant mass} = M_r(\text{CH}_4) + M_r(\text{H}_2\text{O}) = 16.0 + 18.0 = 34.0 \). 2. Calculate the mass of the desired product (hydrogen): \( 3 \times M_r(\text{H}_2) = 3 \times 2.0 = 6.0 \). 3. Calculate percentage atom economy: \( \frac{6.0}{34.0} \times 100 = 17.647...\% \). Rounding to 3 significant figures gives 17.6%.

1 mark: Calculate the total relative mass of reactants as 34.0 (or identify the total mass of products as 34.0). 1 mark: Calculate the relative mass of the desired product (3 moles of \( \text{H}_2 \)) as 6.0. 1 mark: Calculate the final percentage atom economy to 3 significant figures: 17.6% (accept 17.65%).

In an aqueous solution of sodium chloride, both sodium ions (\(\text{Na}^+\)) and hydrogen ions (\(\text{H}^+\)) are present. Since hydrogen is lower in the reactivity series than sodium, hydrogen ions are more easily reduced. Therefore, hydrogen ions gain electrons to form hydrogen gas, whilst sodium ions remain in solution. The balanced half-equation at the cathode is: \(2\text{H}^+ + 2\text{e}^- \rightarrow \text{H}_2\) (or \(2\text{H}_2\text{O} + 2\text{e}^- \rightarrow \text{H}_2 + 2\text{OH}^-\)).

Mark 1: Identifies that both hydrogen ions and sodium ions are present, and hydrogen is less reactive than sodium (or sodium is more reactive).
Mark 2: States that hydrogen ions are preferentially discharged/reduced (gain electrons).
Mark 3: Correct, balanced half-equation: \(2\text{H}^+ + 2\text{e}^- \rightarrow \text{H}_2\) (accept state symbols, ignore charge on electron if written as \(\text{e}\)).

Hydration of ethene is a continuous process that is very fast and yields highly pure ethanol without the need for fractional distillation. However, it relies on ethene which is obtained from non-renewable crude oil and requires a high energy input. The reaction requires a temperature of approximately 300 \(^{\circ}\text{C}\), a pressure of 60-70 atm, and a concentrated phosphoric acid catalyst.

Mark 1 (Advantage): Any one from: Fast rate of reaction / continuous process / high purity ethanol (no distillation needed).
Mark 2 (Disadvantage): Any one from: Uses non-renewable resources (crude oil) / high temperature and pressure requirement makes it expensive/high energy demand.
Mark 3 (Condition): Any one from: High temperature (accept 250 to 350 \(^{\circ}\text{C}\)) / high pressure (accept 60 to 70 atm) / phosphoric acid (\(\text{H}_3\text{PO}_4\)) catalyst.

According to Le Chatelier's principle, increasing the pressure shifts the position of equilibrium to the side with fewer moles of gas. There are 4 moles of gas on the reactant side and only 2 moles on the product side, so the yield of ammonia increases.
For temperature, since the forward reaction is exothermic, lowering the temperature would shift the equilibrium to the right, increasing the yield. However, at low temperatures, the reactant particles have less kinetic energy, resulting in a rate of reaction that is too slow for commercial viability. Therefore, 450 \(^{\circ}\text{C}\) is a compromise temperature that balances a reasonable yield with a sufficiently fast rate of reaction.

Mark 1 (Pressure): States that increasing pressure shifts equilibrium to the right (increasing yield of ammonia) because there are fewer moles/molecules of gas on the product side (2 vs 4).
Mark 2 (Temperature - Yield): States that a lower temperature would favour the exothermic forward reaction, giving a higher equilibrium yield.
Mark 3 (Temperature - Rate): States that a very low temperature decreases the rate of reaction too much (fewer successful collisions), so 450 \(^{\circ}\text{C}\) is a compromise to achieve a fast rate.

First, calculate the total formula mass of all the reactants (or all products):
\(M_r(\text{TiCl}_4) = 47.9 + (4 \times 35.5) = 189.9\)
Mass of \(4\text{Na} = 4 \times 23.0 = 92.0\)
Total mass of reactants = \(189.9 + 92.0 = 281.9\)
(Alternatively, using products: \(M_r(\text{Ti}) + 4 \times M_r(\text{NaCl}) = 47.9 + 4 \times (23.0 + 35.5) = 47.9 + 234.0 = 281.9\))

The mass of the desired product (Titanium) is 47.9.

Now, calculate the percentage atom economy:
\(\text{Atom economy} = \frac{\text{Mass of desired product}}{\text{Total mass of all reactants}} \times 100\)
\(\text{Atom economy} = \frac{47.9}{281.9} \times 100 \approx 16.9918\%\)
To 3 significant figures, this is \(17.0\%\).

Mark 1: Correctly calculates total formula mass of reactants (or products) as 281.9 (or shows the correct working: \(189.9 + 92.0\)).
Mark 2: Correctly sets up the fraction for atom economy: \(\frac{47.9}{281.9} \times 100\).
Mark 3: Final answer of 17.0% (accept 17% or 16.99%).

1. Calculate the relative formula mass of \(\text{TiCl}_4\):
\(M_r(\text{TiCl}_4) = 48.0 + (4 \times 35.5) = 190.0\)

2. Calculate the theoretical yield of titanium:
\(\text{Moles of TiCl}_4 = \frac{9.50}{190.0} = 0.0500\text{ mol}\)
From the equation, 1 mole of \(\text{TiCl}_4\) produces 1 mole of \(\text{Ti}\).
\(\text{Theoretical mass of Ti} = 0.0500\text{ mol} \times 48.0\text{ g/mol} = 2.40\text{ g}\)

3. Calculate the percentage yield:
\(\text{Percentage yield} = \frac{\text{Actual yield}}{\text{Theoretical yield}} \times 100\)
\(\text{Percentage yield} = \frac{1.80}{2.40} \times 100 = 75.0\%\)

M1: Correct calculation of \(M_r\) of \(\text{TiCl}_4 = 190\) (1)
M2: Correct determination of moles of \(\text{TiCl}_4\) or \(\text{Ti} = 0.0500\text{ mol}\) (or theoretical mass of titanium = \(2.40\text{ g}\)) (1)
M3: Correct percentage yield calculation method: \(\frac{1.80}{\text{their theoretical mass}} \times 100\) (1)
M4: Correct evaluation to 3 significant figures: \(75.0\%\) (1)
[Accept alternative pathways using ratios. Allow error carried forward (ECF) from M1 or M2.]

1. Calculate moles of sulfuric acid reacted:
\(\text{Moles} = \text{concentration} \times \text{volume}\)
\(\text{Moles of H}_2\text{SO}_4 = 0.120\text{ mol/dm}^3 \times \frac{18.50}{1000}\text{ dm}^3 = 0.00222\text{ mol}\)

2. Determine moles of NaOH using the stoichiometric ratio (2:1):
\(\text{Moles of NaOH} = 2 \times 0.00222 = 0.00444\text{ mol}\)

3. Calculate concentration of NaOH in \(\text{mol/dm}^3\):
\(\text{Concentration} = \frac{\text{moles}}{\text{volume in dm}^3} = \frac{0.00444}{0.0250} = 0.1776\text{ mol/dm}^3\)

4. Convert concentration to \(\text{g/dm}^3\):
\(\text{Concentration in g/dm}^3 = 0.1776\text{ mol/dm}^3 \times 40.0\text{ g/mol} = 7.104 \approx 7.10\text{ g/dm}^3\)

M1: Correct calculation of moles of \(\text{H}_2\text{SO}_4 = 0.00222\text{ mol}\) (1)
M2: Correct moles of \(\text{NaOH} = 0.00444\text{ mol}\) (ECF 2 x M1) (1)
M3: Correct calculation of concentration of \(\text{NaOH}\) in \(\text{mol/dm}^3 = 0.1776\text{ mol/dm}^3\) (1)
M4: Correct conversion to \(\text{g/dm}^3\) to 3 sig figures: \(7.10\text{ g/dm}^3\) (1)
[Accept alternative titration formula methods if clearly laid out. Allow ECF throughout.]

1. Calculate energy required to break bonds (reactants):
Four \(\text{C}-\text{H}\) bonds and two \(\text{O}=\text{O}\) bonds are broken.
\(\text{Energy in} = (4 \times 413) + (2 \times 498) = 1652 + 996 = 2648\text{ kJ/mol}\)

2. Calculate energy released when making bonds (products):
Two \(\text{C}=\text{O}\) bonds and four \(\text{O}-\text{H}\) bonds are formed.
\(\text{Energy out} = (2 \times 805) + (4 \times 463) = 1610 + 1852 = 3462\text{ kJ/mol}\)

3. Calculate the overall energy change:
\(\text{Energy change} = \text{Energy in} - \text{Energy out}\)
\(\text{Energy change} = 2648 - 3462 = -814\text{ kJ/mol}\)

M1: Correct calculation of total bond-breaking energy = \(2648\text{ kJ/mol}\) (1)
M2: Correct calculation of total bond-making energy = \(3462\text{ kJ/mol}\) (1)
M3: Subtraction of bond-making energy from bond-breaking energy (1)
M4: Final answer of \(-814\text{ kJ/mol}\) (including negative sign) (1)
[Allow ECF if correct subtraction logic applied to incorrect intermediate values.]

1. Calculate the number of moles of calcium carbonate:
\(\text{Moles of CaCO}_3 = \frac{\text{mass}}{M_r} = \frac{4.00}{100} = 0.0400\text{ mol}\)

2. Determine the moles of carbon dioxide produced:
From the balanced equation, the ratio of \(\text{CaCO}_3 : \text{CO}_2\) is 1:1.
Therefore, \(0.0400\text{ mol}\) of \(\text{CO}_2\) is produced.

3. Calculate the volume of \(\text{CO}_2\) in \(\text{dm}^3\):
\(\text{Volume in dm}^3 = 0.0400\text{ mol} \times 24.0\text{ dm}^3\text{/mol} = 0.960\text{ dm}^3\)

4. Convert the volume to \(\text{cm}^3\):
\(\text{Volume in cm}^3 = 0.960 \times 1000 = 960\text{ cm}^3\)

M1: Moles of \(\text{CaCO}_3 = 0.0400\text{ mol}\) (1)
M2: Correct molar ratio (1:1) leading to \(0.0400\text{ mol}\) of \(\text{CO}_2\) (1)
M3: Volume in \(\text{dm}^3 = 0.960\text{ dm}^3\) (1)
M4: Correct conversion to \(\text{cm}^3 = 960\text{ cm}^3\) (1)
[Allow ECF throughout. Award 4 marks for correct final answer of 960 with or without units.]

1. Find the mass of oxygen in the sample:
\(\text{Mass of O} = 5.80\text{ g} - 4.20\text{ g} = 1.60\text{ g}\)

2. Calculate the moles of each element:
\(\text{Moles of Fe} = \frac{4.20}{56.0} = 0.0750\text{ mol}\)
\(\text{Moles of O} = \frac{1.60}{16.0} = 0.100\text{ mol}\)

3. Determine the simplest whole-number ratio of \(\text{Fe} : \text{O}\):
Divide both by the smaller value (0.0750):
\(\text{Fe} = \frac{0.0750}{0.0750} = 1.00\)
\(\text{O} = \frac{0.100}{0.0750} = 1.33\)

Multiply by 3 to achieve whole numbers:
\(\text{Fe} = 1.00 \times 3 = 3\)
\(\text{O} = 1.33 \times 3 = 4\)

Therefore, the empirical formula is \(\text{Fe}_3\text{O}_4\).

M1: Mass of oxygen = \(1.60\text{ g}\) (1)
M2: Correct moles of both elements: \(\text{Fe} = 0.075\text{ mol}\) AND \(\text{O} = 0.100\text{ mol}\) (1)
M3: Correct ratio calculation (e.g. 1 : 1.33 or 3 : 4) (1)
M4: Final empirical formula \(\text{Fe}_3\text{O}_4\) (1)
[Allow ECF if subtraction or division error occurs but logic remains correct. Accept other clear layouts.]

To explain the compromise conditions of the Haber process, we must look at three main factors: Temperature, Pressure, and the Catalyst.

1. **Temperature (\(450\ ^\circ\text{C}\))**:
* **Equilibrium (Yield)**: The forward reaction is exothermic (\(\Delta H = -92 \text{ kJ/mol}\)). According to Le Chatelier's principle, lowering the temperature shifts the equilibrium position to the right (the exothermic direction) to release heat, thereby increasing the yield of ammonia. Therefore, a low temperature is preferred for high yield.
* **Rate of Reaction**: At low temperatures, the reactant particles have less kinetic energy, meaning they collide less frequently and with less energy, resulting in fewer successful collisions per second. The rate of reaction would be too slow to be commercially viable.
* **Compromise**: A temperature of \(450\ ^\circ\text{C}\) is a compromise. It is high enough to ensure a fast, economically viable rate of reaction, but low enough to still produce a reasonable yield of ammonia (about 15% per pass, with unreacted gases recycled).

2. **Pressure (\(200\text{ atm}\))**:
* **Equilibrium (Yield)**: There are 4 moles of gaseous reactants on the left (\(1\text{ N}_2 + 3\text{H}_2\)) and 2 moles of gaseous product on the right (\(2\text{NH}_3\)). According to Le Chatelier's principle, increasing the pressure shifts the equilibrium to the side with fewer gas moles (the right) to reduce pressure, increasing the yield of ammonia.
* **Rate of Reaction**: Higher pressure increases the concentration of gas particles, resulting in more frequent collisions and a faster rate of reaction.
* **Costs and Safety**: Operating at extremely high pressures requires highly reinforced, expensive equipment and high electricity/energy costs to compress the gases. It also poses significant safety risks of explosion.
* **Compromise**: \(200\text{ atm}\) is a compromise pressure that provides a high rate and high yield without the prohibitive construction, maintenance, and safety costs of running at even higher pressures.

3. **Iron Catalyst**:
* **Effect**: The iron catalyst speeds up both the forward and reverse reactions equally by providing an alternative reaction pathway with a lower activation energy.
* **Significance**: It does not affect the yield of ammonia, but it allows the reaction to reach dynamic equilibrium much faster. This enables the use of the lower compromise temperature (\(450\ ^\circ\text{C}\)) while still maintaining an acceptable rate, saving significant energy costs.

**Level 3 (5–6 marks)**
- Explains the effect of BOTH temperature and pressure on BOTH rate and equilibrium yield.
- Relates these factors clearly to Le Chatelier's principle and collision theory.
- Discusses industrial compromises including energy costs, equipment costs, and safety.
- Explains the role of the iron catalyst in lowering operating costs/temperatures.

**Level 2 (3–4 marks)**
- Explains the effect of temperature and/or pressure on both rate and yield, but may lack detail or fail to connect one of the variables fully.
- Mentions the compromise nature of the conditions but does not fully integrate cost/safety considerations.
- Mentions the catalyst increases rate.

**Level 1 (1–2 marks)**
- Identifies basic trends (e.g., higher temperature increases rate, higher pressure increases yield).
- Demonstrates isolated knowledge of the Haber process conditions without deep explanation of the compromises or principles involved.

**0 marks**
- No response or no relevant chemistry content.