2024 Edexcel IGCSE Chemistry 模擬試題連答案詳解

The mass number is 31 and the atomic number (number of protons) is 15. The number of neutrons is calculated as \(31 - 15 = 16\). The negative charge of the phosphide ion only indicates three extra electrons and does not affect the number of neutrons in the nucleus.

1 mark: 16

Calculate the mass of each part of the formula: \(2 \times (\text{N} + 4 \times \text{H}) + \text{S} + 4 \times \text{O} = 2 \times (14 + 4 \times 1) + 32 + 4 \times 16 = 2 \times (18) + 32 + 64 = 36 + 32 + 64 = 132\).

1 mark: 132

An increase in temperature indicates that heat is being released to the surroundings, meaning the reaction is exothermic. For all exothermic reactions, the enthalpy change (\(\Delta H\)) is negative.

1 mark: negative (accept minus or -)

Clean, dry air is composed of approximately 78% nitrogen, 21% oxygen, 0.9% argon, and 0.04% carbon dioxide, along with trace amounts of other gases.

1 mark: 0.9% (accept any value from 0.9% to 1.0% or 0.9)

To form a neutral ionic compound, the total positive charge must equal the total negative charge. Two \(\text{Fe}^{3+}\) ions provide a total charge of \(+6\), which balances the total charge of \(-6\) provided by three \(\text{SO}_4^{2-}\) ions. Therefore, the formula is \(\text{Fe}_2(\text{SO}_4)_3\).

1 mark: \(\text{Fe}_2(\text{SO}_4)_3\) (accept Fe2(SO4)3, ignore case/subscript formatting errors as long as the ratio and elements are correct)

A catalyst increases the rate of reaction by providing an alternative reaction pathway that has a lower activation energy. This allows a greater proportion of reacting particles to have energy equal to or greater than the activation energy, leading to more frequent successful collisions.

1 mark: provides an alternative pathway with a lower activation energy (accept 'lowers the activation energy')

According to the Brønsted-Lowry theory, an acid is defined as a proton (\(\text{H}^+\)) donor, whereas a base is defined as a proton (\(\text{H}^+\)) acceptor.

1 mark: proton acceptor (accept \(\text{H}^+\) acceptor / hydrogen ion acceptor)

Alkanes follow the general homologous series formula \(\text{C}_n\text{H}_{2n+2}\). If \(n = 12\), the number of hydrogen atoms is \(2 \times 12 + 2 = 26\). The molecular formula is therefore \(\text{C}_{12}\text{H}_{26}\).

1 mark: \(\text{C}_{12}\text{H}_{26}\) (accept C12H26, ignore case/subscript formatting errors)

Fractional distillation is the technique used to separate miscible liquids based on their different boiling points, where the liquid with the lowest boiling point distils over first.

1 mark for fractional distillation. Reject 'distillation' or 'simple distillation'.

The number of protons in the iron atom is calculated by subtracting the number of neutrons from the mass number: \(56 - 30 = 26\). Since the atom is neutral, the number of electrons must equal the number of protons, which is 26.

1 mark for 26.

Using the formula: \(\text{amount in moles} = \frac{\text{number of particles}}{\text{Avogadro's constant}}\\text{amount in moles} = \frac{3.0 \times 10^{22}}{6.0 \times 10^{23}} = 0.05\text{ mol}\).

1 mark for 0.05 (or \(5.0 \times 10^{-2}\)).

To form a neutral compound, two aluminium ions with a total charge of \(+6\) (\(2 \times 3+\)) must combine with three oxide ions with a total charge of \(-6\) (\(3 \times 2-\)). Therefore, the empirical formula of the compound is \(\text{Al}_2\text{O}_3\).

1 mark for \(\text{Al}_2\text{O}_3\) (accept Al2O3, reject any incorrect casing such as AL2O3).

The activation energy is defined as the minimum energy that colliding reactant particles must have to overcome the energy barrier and undergo a successful chemical reaction.

1 mark for activation energy.

A decrease in temperature indicates that heat energy is being absorbed from the surroundings. This is an endothermic reaction, which has a positive enthalpy change (represented by the \(+\) sign).

1 mark for + or positive.

Methyl orange indicator is red in acidic solutions (pH < 3.1) and yellow in alkaline solutions. Since hydrochloric acid is a strong acid, the indicator turns red.

1 mark for red. Accept 'pink'.

Alkanes have the general molecular formula \(\text{C}_n\text{H}_{2n+2}\). If \(n = 15\), the number of hydrogen atoms is \(2 \times 15 + 2 = 32\).

1 mark for 32.

To calculate the relative formula mass, sum the relative atomic masses of all atoms in the formula: \(M_r = 63.5 + (2 \times 14) + (6 \times 16) + 3 \times (2 \times 1 + 16) = 63.5 + 28 + 96 + 54 = 241.5\).

Award 1 mark for the correct calculation and final value of 241.5.

Using a fine powder increases the surface area of the solid reactant, which leads to a greater frequency of successful collisions between the reactant particles, thereby increasing the rate of reaction.

Award 1 mark for stating that the rate increases / reaction is faster.

A chemical reaction that absorbs heat energy from the surroundings is called an endothermic reaction. This absorption of energy causes the temperature of the surroundings to decrease.

Award 1 mark for 'endothermic' (accept phonetic spelling, reject 'exothermic').

To form a neutral compound, the total positive charge must equal the total negative charge. Two aluminium ions have a total charge of \(2 \times (+3) = +6\), and three oxide ions have a total charge of \(3 \times (-2) = -6\). Therefore, the formula is \(\text{Al}_2\text{O}_3\).

Award 1 mark for \(\text{Al}_2\text{O}_3\) (accept Al2O3, reject AL2O3).

Clean, dry air is composed of approximately 78% nitrogen gas, 21% oxygen gas, 0.9% argon, and 0.04% carbon dioxide.

Award 1 mark for 78% (accept any percentage value in the range of 78% to 80%).

Sodium reacts vigorously with water to form sodium hydroxide (an alkaline solution) and hydrogen gas. The resulting alkaline solution has a high pH, which turns universal indicator purple (or blue).

Award 1 mark for 'purple' or 'blue' (accept violet).

The general formula for alkanes is \(\text{C}_n\text{H}_{2n+2}\). If \(n = 12\), the number of hydrogen atoms is \(2 \times 12 + 2 = 26\).

Award 1 mark for 26.

Fractional distillation is the technique used to separate miscible liquids based on their different boiling points using a fractionating column.

Award 1 mark for 'fractional distillation' (reject 'simple distillation' or 'distillation').

The formula for calculating the \(R_f\) value is:

\[R_f = \frac{\text{distance moved by the solute (spot)}}{\text{distance moved by the solvent front}}\]

Substituting the values:

\[R_f = \frac{2.8\text{ cm}}{8.0\text{ cm}} = 0.35\]

Award 1 mark for 0.35. Reject any other values.

1. Find the moles of each element:
- Moles of sulfur \(\text{S} = \frac{2.0\text{ g}}{32\text{ g/mol}} = 0.0625\text{ mol}\)
- Moles of oxygen \(\text{O} = \frac{3.0\text{ g}}{16\text{ g/mol}} = 0.1875\text{ mol}\)

2. Divide by the smallest mole value (0.0625) to get the ratio:
- Ratio of \(\text{S} = \frac{0.0625}{0.0625} = 1\)
- Ratio of \(\text{O} = \frac{0.1875}{0.0625} = 3\)

Thus, the empirical formula is \(\text{SO}_3\).

Award 1 mark for \(\text{SO}_3\) (also accept SO3).

A sodium atom (atomic number 11) has the electronic configuration of 2.8.1. When it forms a \(\text{Na}^+\) ion, it loses the single electron in its outermost shell. This leaves it with 10 electrons, giving an electronic configuration of 2.8.

Award 1 mark for 2.8 (also accept 2,8 or 2.8.0).

In the Haber process, nitrogen and hydrogen gases are reacted to produce ammonia. An iron catalyst is used to speed up the chemical reaction without being consumed itself.

Award 1 mark for 'iron' or 'Fe'. Reject 'iron oxide'.

According to the balanced equation, \(1\text{ mol}\) of \(\text{CH}_4\) reacts to produce \(1\text{ mol}\) of \(\text{CO}_2\).
Therefore, \(0.5\text{ mol}\) of \(\text{CH}_4\) will produce \(0.5\text{ mol}\) of \(\text{CO}_2\).

\[\text{Volume of } \text{CO}_2 = \text{moles} \times \text{molar volume} = 0.5\text{ mol} \times 24\text{ dm}^3/\text{mol} = 12\text{ dm}^3\]

Award 1 mark for 12 (or \(12\text{ dm}^3\)).

Ethene reacts with steam in an addition reaction (hydration) to form the alcohol ethanol:

\[\text{C}_2\text{H}_4 + \text{H}_2\text{O} \rightarrow \text{C}_2\text{H}_5\text{OH}\]

Award 1 mark for 'ethanol'. Reject 'alcohol' (too general).

Fractional distillation is used in oil refineries to separate the mixture of hydrocarbons in crude oil into useful fractions based on their different boiling point ranges.

Award 1 mark for 'fractional distillation'. Reject 'distillation' on its own.

Methyl orange indicator is red in acidic solutions, yellow in alkaline solutions, and orange in neutral solutions.

Award 1 mark for 'red'. Accept 'pink' or 'peach'. Reject 'orange' or 'yellow'.

First, find the relative formula mass of CO2: 12 + (16 * 2) = 44. Then, use mass = moles * Mr: 0.25 * 44 = 11.

1 mark for 11.

The minimum amount of energy required for colliding reactant particles to undergo a successful reaction is called the activation energy.

1 mark for activation energy.

A negative enthalpy change indicates that heat energy is released to the surroundings, which defines an exothermic reaction.

1 mark for exothermic.

To form a neutral ionic compound, the total positive charge must equal the total negative charge. Two aluminium ions (each with a 3+ charge) balance three oxide ions (each with a 2- charge), giving the empirical formula Al2O3.

1 mark for Al2O3.

A water molecule has two single covalent bonds (one between the oxygen atom and each of the two hydrogen atoms). Each single covalent bond consists of one shared pair of electrons, making a total of 2 shared pairs.

1 mark for 2.

Filtration is used to separate an insoluble solid from a liquid because the solid particles cannot pass through the filter paper, while the liquid can.

1 mark for filtration.

The total number of electrons in the atom is 2 + 8 + 6 = 16. In a neutral atom, the atomic number (number of protons) is equal to the number of electrons. The element with atomic number 16 is sulfur.

1 mark for sulfur (accept S).

Cracking (or catalytic cracking) is the thermal decomposition process used to break down larger, less useful hydrocarbon molecules into smaller, more useful ones.

1 mark for cracking (accept catalytic cracking).

Calculate the sum of relative atomic masses of all atoms in the formula: \(M_r = 2 \times [14 + (4 \times 1)] + 32 + (4 \times 16) = 2 \times 18 + 32 + 64 = 132\).

Award 1 mark for the correct answer of 132.

A catalyst increases the reaction rate by providing an alternative reaction pathway with a lower activation energy, which allows more collisions to be successful per second.

Award 1 mark for stating that it provides an alternative pathway with a lower activation energy.

An increase in temperature indicates an exothermic reaction, where heat is released to the surroundings. Exothermic reactions have a negative enthalpy change, \(\Delta H < 0\).

Award 1 mark for negative (accept - or < 0).

Aluminium ions have a \(3+\) charge and oxide ions have a \(2-\) charge. To form a neutral ionic compound, the ratio of ions must be 2 to 3, giving the formula \(\text{Al}_2\text{O}_3\).

Award 1 mark for \(\text{Al}_2\text{O}_3\) (accept Al2O3, ignore case if readable).

Kerosene is the fraction of crude oil that has a boiling point range suitable for use as jet fuel.

Award 1 mark for kerosene (accept paraffin).

During the electrolysis of concentrated sodium chloride solution (brine), chloride ions (\(\text{Cl}^-\)) migrate to the anode where they are oxidized to chlorine gas (\(\text{Cl}_2\)).

Award 1 mark for chlorine (accept \(\text{Cl}_2\)). Reject chloride.

Clean, dry air is composed of approximately 78% nitrogen gas, 21% oxygen gas, 0.9% argon, and 0.04% carbon dioxide.

Award 1 mark for 78% (accept 78 or any value from 78% to 79%).

Carbon dioxide gas reacts with limewater (aqueous calcium hydroxide) to form an insoluble precipitate of calcium carbonate, which turns the solution cloudy or milky.

Award 1 mark for mentioning both the test reagent (limewater) and the positive result (turns cloudy/milky/precipitate forms).

The formula of iron(III) sulfate is \(\text{Fe}_2(\text{SO}_4)_3\). It consists of 2 iron atoms, 3 sulfur atoms, and 12 oxygen atoms. Calculate the total mass: \(2 \times 56 + 3 \times 32 + 12 \times 16 = 112 + 96 + 192 = 400\).

[1] 400 (Award 1 mark for the correct calculation leading to 400).

Manganese(IV) oxide (or manganese dioxide) is the common catalyst used to speed up the decomposition of hydrogen peroxide into water and oxygen.

[1] manganese(IV) oxide / manganese dioxide / \(\text{MnO}_2\). Reject manganese.

A reaction that absorbs thermal energy from its surroundings, resulting in a temperature decrease of the mixture, is described as an endothermic reaction.

[1] endothermic

To form a neutral ionic compound, two aluminium ions with a charge of \(3+\) must combine with three oxide ions with a charge of \(2-\). This balances the charges as \(2 \times (+3) + 3 \times (-2) = 0\). The formula is \(\text{Al}_2\text{O}_3\).

[1] \(\text{Al}_2\text{O}_3\) (accept Al2O3). Reject Al^{2}O^{3} or other incorrect representations of subscripts.

A molecule of carbon dioxide has two double covalent bonds (\(\text{O}=\text{C}=\text{O}\)). Each double bond consists of 2 shared pairs of electrons, which is 4 shared electrons. Therefore, the total number of shared electrons is \(2 \times 4 = 8\).

[1] 8 (or eight)

During filtration, the insoluble solid that is left behind on the filter paper is called the residue.

[1] residue

The number of neutrons is found by subtracting the atomic number (number of protons) from the mass number: \(31 - 15 = 16\).

[1] 16

1. Powder has a larger surface area to volume ratio than large lumps. 2. Therefore, there are more exposed particles of calcium carbonate available to collide with acid particles. 3. This leads to a higher frequency of collisions (more collisions occurring per unit time). 4. Thus, more successful collisions occur per unit time, which increases the rate of reaction.

M1: Reference to larger surface area for powder [1 mark]. M2: More exposed particles / more particles available to react [1 mark]. M3: Greater frequency of collisions / more collisions per unit time [1 mark]. M4: More successful/effective collisions per unit time [0.5 marks].

1. Identify charges: \(\text{Mg}^{2+}\)/\(\text{O}^{2-}\) vs \(\text{Na}^+\)/\(\text{Cl}^-\). 2. Explain that higher charges result in stronger ionic bonding. 3. Electrostatic forces of attraction between oppositely charged ions are stronger in \(\text{MgO}\) than in \(\text{NaCl}\). 4. Therefore, much more thermal energy is needed to break these bonds in \(\text{MgO}\).

M1: Charges are \(\text{Mg}^{2+}\)/\(\text{O}^{2-}\) and \(\text{Na}^+\)/\(\text{Cl}^-\) [1 mark]. M2: Electrostatic attraction between oppositely charged ions is stronger in magnesium oxide [1 mark]. M3: Because the charges on the ions are larger/higher [1 mark]. M4: Hence, much more energy is needed to overcome these forces [0.5 marks].

1. Crude oil is heated and vaporised before entering the column. 2. The column has a temperature gradient (hot at the bottom, cool at the top). 3. Vapours rise up the column and condense when they reach a temperature below their boiling point. 4. Smaller hydrocarbons with lower boiling points condense near the top, while larger hydrocarbons with higher boiling points condense near the bottom.

M1: Crude oil is vaporised/heated [1 mark]. M2: The column has a temperature gradient / is hot at the bottom and cool at the top [1 mark]. M3: Vapours rise and condense at their boiling points [1 mark]. M4: Short chain/smaller molecules/lower boiling points at the top AND long chain/larger molecules/higher boiling points at the bottom [0.5 marks].

1. State that breaking bonds is an endothermic process that absorbs energy. 2. State that making bonds is an exothermic process that releases energy. 3. In this reaction, more energy is released during the formation of the new C-Cl and H-Cl bonds than is absorbed during the breaking of the C-H and Cl-Cl bonds. 4. This results in a net release of heat energy to the surroundings, making the reaction exothermic.

M1: Bond breaking requires energy / is endothermic [1 mark]. M2: Bond making releases energy / is exothermic [1 mark]. M3: The energy released in making bonds is greater than the energy taken in to break bonds [1 mark]. M4: Hence, net energy is released / exothermic reaction [0.5 marks].

1. Weigh the empty reduction tube/boat, and then weigh it with the copper(II) oxide. 2. Pass a stream of dry reducing gas (such as methane or hydrogen) through the tube before heating to flush out air (preventing explosions), and keep the gas flowing throughout the experiment. 3. Heat the tube strongly until the black copper(II) oxide has completely turned into pinkish-brown copper. 4. Let the tube cool with the gas still flowing, then record the final mass of the tube and the copper. Heat to constant mass to ensure completion.

M1: Measure mass of empty tube/boat, and mass of tube + copper(II) oxide [1 mark]. M2: Heat the oxide while passing a stream of reducing gas (hydrogen/methane/carbon monoxide) [1 mark]. M3: Continue passing the gas during cooling (to prevent the hot copper re-oxidising) [1 mark]. M4: Measure the final mass of tube + copper, and heat to constant mass to ensure reaction is complete [0.5 marks].

1. Aluminium is more reactive than carbon, so carbon is not a strong enough reducing agent to displace/reduce it. 2. Therefore, aluminium cannot be extracted by heating with carbon. 3. Cryolite is added to act as a solvent to dissolve the aluminium oxide. 4. This significantly lowers the melting temperature of the mixture from \(2000\ ^\circ\text{C}\) to around \(950\ ^\circ\text{C}\), reducing energy consumption and operating costs.

M1: Aluminium is more reactive than carbon [1 mark]. M2: Carbon cannot displace/reduce aluminium from aluminium oxide [1 mark]. M3: Cryolite dissolves aluminium oxide [1 mark]. M4: This lowers the operating temperature / melting point of the mixture, reducing energy costs [0.5 marks].

1. The reaction requires ultraviolet (UV) light. 2. During this substitution reaction, one of the hydrogen atoms in the methane molecule is replaced by a bromine atom. 3. This produces bromomethane (\(\text{CH}_3\text{Br}\)) and hydrogen bromide (\(\text{HBr}\)). 4. The main observation is that the reddish-brown/orange colour of the bromine fades and the mixture becomes colourless.

M1: Condition: ultraviolet (UV) light / radiation / sunlight [1 mark]. M2: Substitution explanation: an atom of hydrogen is replaced/substituted by an atom of bromine [1 mark]. M3: Products: bromomethane and hydrogen bromide (or correct chemical equation) [1 mark]. M4: Observation: The orange/red/brown colour of the bromine fades / decolourises [0.5 marks].

1. In diamond, each carbon atom is covalently bonded to four other carbon atoms in a giant tetrahedral structure. The strong covalent bonds in all directions make diamond extremely hard. 2. Diamond has no delocalised or free electrons because all four outer shell valence electrons are involved in covalent bonding. 3. In graphite, each carbon atom is covalently bonded to only three other carbon atoms in hexagonal layers. The layers are held together by weak intermolecular forces, allowing them to slide over each other easily, making graphite soft. 4. Each carbon atom in graphite has one delocalised electron that is free to move between the layers, carrying electrical charge.

M1: Diamond has 4 covalent bonds per carbon in a rigid giant covalent structure [1 mark]. M2: Graphite has 3 covalent bonds per carbon in layers with weak forces between layers, allowing layers to slide [1 mark]. M3: Graphite has delocalised/free electrons that can move and carry electrical charge [1 mark]. M4: Diamond has no free/delocalised electrons / all outer electrons are fixed in localized single covalent bonds [0.5 marks].

Silicon dioxide (\(\text{SiO}_2\)) has a giant covalent lattice structure. To melt it, many strong covalent bonds between silicon and oxygen atoms must be broken, which requires a very large amount of thermal energy. Carbon dioxide (\(\text{CO}_2\)) has a simple molecular structure. While there are strong covalent bonds within each molecule (intramolecular), there are only weak intermolecular forces of attraction between the individual molecules. Very little energy is needed to overcome these weak forces, which is why it is a gas at room temperature.

M1: Silicon dioxide has a giant covalent structure / giant lattice (1)
M2: Many strong covalent bonds must be broken, requiring a large amount of energy (1)
M3: Carbon dioxide has a simple molecular structure / is a simple covalent substance (1)
M4: Only weak intermolecular forces (forces between molecules) need to be overcome / require little energy to overcome (1)

Note: Reject any mention of covalent bonds breaking in \(\text{CO}_2\).

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

M1: More (acid / hydrogen) particles in a given volume / unit volume / space (1)
M2: More frequent collisions / higher frequency of collisions (between particles) (1)
M3: More successful / fruitful collisions per unit time / per second (1)

Reject: 'more collisions' without reference to time or frequency for M2.
Reject: 'particles move faster' or 'particles have more energy'.

1. Calculate the number of moles of each element in 100 g of the compound:
- Moles of \(\text{Fe} = \frac{70.0}{56} = 1.25\\ \text{mol}\)
- Mass of oxygen = \(100 - 70.0 = 30.0\\ \text{g}\)
- Moles of \(\text{O} = \frac{30.0}{16} = 1.875\\ \text{mol}\)

2. Determine the simplest ratio by dividing both values by the smaller number (1.25):
- \(\text{Fe} = \frac{1.25}{1.25} = 1\)
- \(\text{O} = \frac{1.875}{1.25} = 1.5\)

3. Multiply by 2 to obtain the simplest whole-number ratio:
- \(\text{Fe} : \text{O} = 2 : 3\)

Therefore, the empirical formula is \(\text{Fe}_2\text{O}_3\).

- **1 mark** for dividing percentages by relative atomic masses to find moles of Fe (1.25) and O (1.875).
- **1 mark** for finding the simplest whole-number ratio of 2:3.
- **0.5 mark** for writing the correct empirical formula (\(\text{Fe}_2\text{O}_3\)).

1. Calculate the temperature change:
\(\Delta T = 48.5 - 18.5 = 30.0\\ ^\circ\text{C}\)

2. Use the heat energy equation:
\(Q = m c \Delta T\)
\(Q = 100.0\\ \text{g} \times 4.18\\ \text{J/g/}^\circ\text{C} \times 30.0\\ ^\circ\text{C}\)
\(Q = 12540\\ \text{J}\)

3. Convert the energy from joules to kilojoules:
\(Q = \frac{12540}{1000} = 12.54\\ \text{kJ}\)

- **1 mark** for calculating the correct temperature difference of \(30.0\\ ^\circ\text{C}\).
- **1 mark** for calculating the heat energy in joules (\(12540\\ \text{J}\)).
- **0.5 mark** for converting the answer to kJ to get \(12.54\\ \text{kJ}\) (accept \(12.5\\ \text{kJ}\)).

1. The formula for percentage yield is:
\(\text{Percentage Yield} = \frac{\text{Actual Yield}}{\text{Theoretical Yield}} \times 100\)

2. Substitute the values into the formula:
\(\text{Percentage Yield} = \frac{9.50\\ \text{g}}{12.5\\ \text{g}} \times 100\)

3. Calculate the final percentage yield:
\(\text{Percentage Yield} = 76.0\\%\)

- **1.5 marks** for the correct substitution of values: \(\frac{9.50}{12.5} \times 100\).
- **1 mark** for the correct final percentage yield of \(76.0\\%\) (accept \(76\\%\)).

1. Calculate the number of moles of \(\text{HCl}\) used:
\(\text{Moles} = \text{Concentration} \times \text{Volume (in }\text{dm}^3\text{)}\)
\(\text{Moles of HCl} = 0.100\\ \text{mol/dm}^3 \times \frac{20.0}{1000}\\ \text{dm}^3 = 0.00200\\ \text{mol}\)

2. Determine the moles of \(\text{NaOH}\) using the 1:1 reacting ratio:
\(\text{Moles of NaOH} = 0.00200\\ \text{mol}\)

3. Calculate the concentration of the \(\text{NaOH}\) solution:
\(\text{Concentration} = \frac{\text{Moles}}{\text{Volume (in }\text{dm}^3\text{)}}\)
\(\text{Concentration of NaOH} = \frac{0.00200\\ \text{mol}}{\frac{25.0}{1000}\\ \text{dm}^3} = 0.0800\\ \text{mol/dm}^3\)

- **1 mark** for calculating the moles of hydrochloric acid: \(0.00200\\ \text{mol}\) (or \(2.00 \times 10^{-3}\\ \text{mol}\)).
- **1 mark** for setting up the concentration calculation: \(\frac{0.00200}{0.0250}\).
- **0.5 mark** for the correct final concentration of \(0.0800\\ \text{mol/dm}^3\) (accept \(0.08\\ \text{mol/dm}^3\)).

1. Use the weighted average formula for relative atomic mass:
\(A_r = \frac{\sum (\text{Isotopic Mass} \times \text{Abundance})}{\sum \text{Abundances}}\)

2. Substitute the given values:
\(A_r = \frac{(20 \times 90.5) + (22 \times 9.5)}{100}\)

3. Calculate the numerator:
\(20 \times 90.5 = 1810\)
\(22 \times 9.5 = 209\)

\(\text{Sum} = 1810 + 209 = 2019\)

4. Divide by 100:
\(A_r = 20.19\)

5. Round to 1 decimal place:
\(A_r \approx 20.2\)

- **1 mark** for setting up the weighted average expression correctly: \(\frac{(20 \times 90.5) + (22 \times 9.5)}{100}\).
- **1 mark** for calculating the raw value of \(20.19\).
- **0.5 mark** for rounding the final answer to one decimal place to get \(20.2\).

1. Calculate the number of moles of \(\text{CaCO}_3\) that reacted:
\(\text{Moles} = \frac{\text{Mass}}{M_r}\)
\(\text{Moles of CaCO}_3 = \frac{2.50\\ \text{g}}{100} = 0.0250\\ \text{mol}\)

2. Use the molar ratio to find the moles of \(\text{CO}_2\) produced:
Since the ratio of \(\text{CaCO}_3 : \text{CO}_2\) is 1:1, \(\text{Moles of CO}_2 = 0.0250\\ \text{mol}\).

3. Calculate the volume of \(\text{CO}_2\) gas produced at r.t.p.:
\(\text{Volume} = \text{Moles} \times \text{Molar Volume}\)
\(\text{Volume of CO}_2 = 0.0250\\ \text{mol} \times 24\\ \text{dm}^3/\text{mol} = 0.600\\ \text{dm}^3\)

- **1 mark** for calculating the moles of calcium carbonate: \(0.0250\\ \text{mol}\).
- **1 mark** for setting up the volume calculation: \(0.0250 \times 24\).
- **0.5 mark** for the correct final volume of \(0.600\\ \text{dm}^3\) (accept \(0.6\\ \text{dm}^3\)).

1. Calculate the number of moles of \(\text{Mg}\) reacted:
\(\text{Moles} = \frac{\text{Mass}}{A_r}\)
\(\text{Moles of Mg} = \frac{1.20\\ \text{g}}{24} = 0.0500\\ \text{mol}\)

2. Determine the moles of \(\text{MgO}\) produced:
The stoichiometry of the balanced equation is \(2\text{Mg} : 2\text{MgO}\), which simplifies to a 1:1 ratio.
Therefore, \(\text{Moles of MgO} = 0.0500\\ \text{mol}\).

3. Calculate the mass of \(\text{MgO}\) produced:
\(M_r\\ \text{of MgO} = 24 + 16 = 40\)
\(\text{Mass} = \text{Moles} \times M_r\)
\(\text{Mass of MgO} = 0.0500\\ \text{mol} \times 40 = 2.00\\ \text{g}\)

- **1 mark** for calculating the moles of magnesium: \(0.0500\\ \text{mol}\).
- **1 mark** for setting up the mass of MgO calculation: \(0.0500 \times 40\).
- **0.5 mark** for the correct final mass of \(2.00\\ \text{g}\) (accept \(2\\ \text{g}\)).

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

2. Calculate the energy released when making new bonds (products):
- Energy from making \(2 \times \text{H}-\text{Cl}\) bonds = \(2 \times 432 = 864\\ \text{kJ/mol}\)
- Total energy out = \(864\\ \text{kJ/mol}\)

3. Calculate the overall enthalpy change (\(\Delta H\)):
\(\Delta H = \text{Energy to break bonds} - \text{Energy from making bonds}\)
\(\Delta H = 679 - 864 = -185\\ \text{kJ/mol}\)

- **1 mark** for calculating total energy for bond breaking: \(679\\ \text{kJ/mol}\).
- **1 mark** for calculating total energy for bond making: \(864\\ \text{kJ/mol}\).
- **0.5 mark** for the correct final answer of \(-185\\ \text{kJ/mol}\) (the minus sign must be present).

An overall enthalpy change (\(\Delta H\)) that is negative indicates that the products have less chemical energy than the reactants, meaning energy is released to the surroundings. Therefore, the reaction is exothermic.

1 mark for: Exothermic

The minimum kinetic energy that colliding particles must have in order to react is defined as the activation energy (\(E_a\)).

1 mark for: Activation energy (accept \(E_a\); reject energy alone)

In concentrated aqueous sodium chloride, chloride ions (\(\text{Cl}^-\)) are selectively discharged at the positive electrode (anode) in preference to hydroxide ions, producing chlorine gas (\(\text{Cl}_2\)).

1 mark for: Chlorine (accept \(\text{Cl}_2\); reject chloride / chlorine ions)

Alkanes have the general molecular formula \(\text{C}_n\text{H}_{2n+2}\). If \(2n+2 = 14\), then \(2n = 12\), which gives \(n = 6\). Thus, the molecular formula of the alkane is \(\text{C}_6\text{H}_{14}\).

1 mark for: \(\text{C}_6\text{H}_{14}\) (accept C6H14; ignore case errors only if meaning is clear; numbers must represent subscripts)

First calculate the relative formula mass of \(\text{SO}_2\):
\(M_r(\text{SO}_2) = 32 + (2 \times 16) = 64\).
The mass of oxygen in one mole of \(\text{SO}_2\) is \(2 \times 16 = 32\).
Percentage of oxygen by mass = \(\frac{32}{64} \times 100 = 50.0\%\).

1 mark for: 50.0% (accept 50% or 50)

Clean, dry air is a mixture of gases of which nitrogen (\(\text{N}_2\)) makes up approximately 78% by volume.

1 mark for: 78% (accept any value in the range 78% to 79%, or 78, 79)

The reaction between ethanol and methanoic acid is an esterification reaction. The alcohol (ethanol) provides the 'ethyl' part of the name, and the carboxylic acid (methanoic acid) provides the 'methanoate' part. Therefore, the resulting ester is ethyl methanoate.

1 mark for: Ethyl methanoate (reject ethyl methanoate with spelling errors that change the functional group ending)

Crude oil is separated into fractions using fractional distillation. This process separates hydrocarbons based on their boiling points, which are related to their molecular sizes.

1 mark for: Fractional distillation (reject distillation alone)

The molecular formula of butane is \( \text{C}_4\text{H}_{10} \). To find the empirical formula, we find the simplest whole-number ratio of carbon atoms to hydrogen atoms. Dividing both subscripts by 2 gives the ratio \( 2 : 5 \), which corresponds to the empirical formula \( \text{C}_2\text{H}_5 \).

[1 mark] for \( \text{C}_2\text{H}_5 \) (accept C2H5, order of elements must be correct).

Dry, unpolluted air consists of approximately 78% nitrogen by volume, along with 21% oxygen, 0.9% argon, and 0.04% carbon dioxide.

[1 mark] for 78% (accept 78 or 79%).

A catalyst increases the rate of reaction by providing an alternative reaction pathway with a lower activation energy. This allows a greater proportion of colliding particles to have energy equal to or greater than the activation energy, increasing the rate of successful collisions.

[1 mark] for mentioning an alternative pathway/route with a lower activation energy.

Acidified potassium dichromate(VI) acts as an oxidizing agent. During the oxidation of ethanol to ethanoic acid, the orange dichromate(VI) ions are reduced to green chromium(III) ions.

[1 mark] for orange to green (both colors must be in the correct starting and ending order).

Methyl orange is an indicator that is red in highly acidic solutions (pH < 3.1), orange in the transition range, and yellow in alkaline solutions.

[1 mark] for red (reject pink or orange).

During the electrolysis of concentrated sodium chloride solution, both chloride ions (\( \text{Cl}^- \)) and hydroxide ions (\( \text{OH}^- \)) migrate to the positive anode. Because chloride ions are in high concentration, they are preferentially discharged to form chlorine gas (\( \text{Cl}_2 \)).

[1 mark] for chlorine (accept Cl2 or \( \text{Cl}_2 \), reject chloride).

An ester is named by taking the alkyl group from the alcohol (methanol gives 'methyl') followed by the carboxylate group from the carboxylic acid (propanoic acid gives 'propanoate'). Thus, the product is methyl propanoate.

[1 mark] for methyl propanoate (accept methylpropanoate).

In a pure metal, atoms are of uniform size and arranged in regular layers that can easily slide over one another when a force is applied. In an alloy, atoms of different sizes are introduced, which disrupts this regular structure and prevents the layers from sliding over each other easily.

[1 mark] for stating that different-sized atoms disrupt the regular layers/arrangement, preventing them from sliding.

To calculate the relative formula mass of \(\text{CoCl}_2\cdot6\text{H}_2\text{O}\): \(\text{Co} = 59\), \(2 \times \text{Cl} = 2 \times 35.5 = 71\), and \(6 \times \text{H}_2\text{O} = 6 \times (2 \times 1 + 16) = 6 \times 18 = 108\). Total \(M_r = 59 + 71 + 108 = 238\).

Award 1 mark for the correct calculation and final value of 238.

The activation energy is defined as the minimum energy that colliding particles must possess for a chemical reaction to occur.

Award 1 mark for 'activation energy'.

A negative value for enthalpy change (\(\Delta H\)) indicates that energy is released to the surroundings, meaning the reaction is exothermic.

Award 1 mark for 'exothermic' (or 'Exothermic').

The ammonium ion is a polyatomic cation with the formula \(\text{NH}_4^+\).

Award 1 mark for \(\text{NH}_4^+\) or NH4+.

Each covalent bond consists of a shared pair of electrons. Therefore, a triple covalent bond consists of three pairs of shared electrons, giving a total of 6 shared electrons.

Award 1 mark for 6.

Kerosene (also known as paraffin) is the fraction from crude oil used as fuel for jet engines and aircraft.

Award 1 mark for Kerosene (accept paraffin).

During the electrolysis of concentrated sodium chloride solution, chloride ions (\(\text{Cl}^-\)) are discharged at the anode to produce chlorine gas (\(\text{Cl}_2\)).

Award 1 mark for chlorine (accept Cl2).

Alkanes are saturated hydrocarbons with the general formula \(\text{C}_n\text{H}_{2n+2}\).

Award 1 mark for \(\text{C}_n\text{H}_{2n+2}\) (or CnH2n+2).

First, calculate the mass of oxygen reacted: 2.84 g - 1.24 g = 1.60 g. Next, calculate the moles of each element: Moles of P = 1.24 / 31 = 0.04 mol. Moles of O = 1.60 / 16 = 0.10 mol. Divide both values by the smaller number of moles (0.04): P = 0.04 / 0.04 = 1, O = 0.10 / 0.04 = 2.5. To convert this ratio to whole numbers, multiply both by 2, yielding P = 2 and O = 5. The empirical formula is P2O5.

1 mark for the correct empirical formula P2O5 (accept P_2O_5, ignore case as long as symbols are clear).

Enthalpy change (\(\Delta H\)) is calculated by dividing the heat energy change by the number of moles of reactant: \(\Delta H = -11.0 \text{ kJ} / 0.050 \text{ mol} = -220 \text{ kJ/mol}\). Because the reaction is exothermic (releases heat), the sign of the enthalpy change must be negative.

1 mark for the correct value with negative sign (-220). Reject +220 or 220 without a sign.

A catalyst increases the reaction rate by providing an alternative pathway that has a lower activation energy, allowing more reactant particles to have energy greater than or equal to the activation energy.

1 mark for stating that it provides an alternative pathway with a lower activation energy.

To form a neutral ionic compound, the total positive charge must equal the total negative charge. Two aluminium ions (2 x +3 = +6) balance three sulfide ions (3 x -2 = -6). The chemical formula is therefore Al2S3.

1 mark for the correct formula Al2S3. Reject AL2S3 or Al2s3.

In a molecule of carbon dioxide (O=C=O), the carbon atom forms a double covalent bond with each of the two oxygen atoms. Each double bond contains two shared pairs of electrons, giving a total of four shared pairs of electrons in the entire molecule.

1 mark for 4 (or four).

Dry, unpolluted air is a mixture of gases composed of approximately 78% nitrogen, 21% oxygen, 0.9% argon, and 0.04% carbon dioxide by volume.

1 mark for 78 (or 78%). Accept 78-79.

The general formula for alkanes is \(\text{C}_n\text{H}_{2n+2}\). Setting the number of hydrogen atoms to 14: \(2n + 2 = 14\), which simplifies to \(2n = 12\), and therefore \(n = 6\). The molecular formula of the alkane is C6H14.

1 mark for C6H14 (accept C_6H_{14}).

During the electrolysis of dilute sodium hydroxide, hydroxide ions (\(\text{OH}^-\)) are discharged at the anode to form oxygen gas and water, losing electrons in the process. The balanced ionic half-equation is: \(4\text{OH}^- \rightarrow \text{O}_2 + 2\text{H}_2\text{O} + 4\text{e}^-\).

1 mark for a fully correct balanced ionic half-equation. Accept \(4\text{OH}^- - 4\text{e}^- \rightarrow \text{O}_2 + 2\text{H}_2\text{O}\). Ignore state symbols.

The hydration of ethene to produce ethanol is carried out at a temperature of 300 degrees Celsius and a pressure of 60-70 atmospheres in the presence of a phosphoric acid catalyst.

1 mark for phosphoric acid (accept phosphoric(V) acid or H3PO4). Reject phosphate.

Argon makes up approximately 0.93% of the dry atmosphere by volume, making it the most abundant noble gas.

1 mark for argon (accept Ar).

First find the relative formula mass of CuO which is 64 + 16 = 80. Then calculate the percentage of oxygen: (16 / 80) * 100 = 20%.

1 mark for 20% (accept 20).

To explain the energetic changes: 1. Identify that energy is required to break covalent bonds in the reactants (C-H and Cl-Cl), meaning bond breaking is an endothermic process. 2. Identify that energy is released when new covalent bonds are formed in the products (C-Cl and H-Cl), meaning bond making is an exothermic process. 3. Compare the relative energies: the total energy released during bond making is greater than the total energy absorbed during bond breaking. Therefore, the overall reaction has a negative enthalpy change and is exothermic.

[1 mark] State that energy is taken in/required to break bonds (bond breaking is endothermic). [1 mark] State that energy is given out/released when bonds are formed (bond making is exothermic). [1 mark] State that more energy is released in making bonds than is taken in breaking bonds. [0.5 mark] Reference to the specific covalent bonds being broken (C-H / Cl-Cl) or formed (C-Cl / H-Cl).

1. Explain the role of the catalyst: it provides an alternative reaction pathway with a lower activation energy. 2. Connect this to particle energy: since the activation energy is lower, a greater fraction of the reacting particles possess energy equal to or exceeding this threshold energy. 3. Connect to collision frequency: this results in a higher frequency of successful collisions (more successful collisions per second), which increases the overall rate of reaction.

[1 mark] Provides an alternative reaction pathway. [1 mark] Lower activation energy. [1 mark] Higher frequency of successful collisions (or more successful collisions per unit time). [0.5 mark] Correctly links lower activation energy to a greater proportion of particles having sufficient energy to react.

1. Identify that both substances form giant ionic lattices. 2. Contrast the ionic charges: magnesium ions have a 2+ charge and oxide ions have a 2- charge, whereas sodium ions have a 1+ charge and chloride ions have a 1- charge. 3. Relate charge to force strength: the electrostatic forces of attraction between the doubly charged ions in MgO are significantly stronger than those between the singly charged ions in NaCl. 4. Relate force strength to energy: significantly more thermal energy is needed to overcome these stronger forces to melt MgO, resulting in a much higher melting point.

[1 mark] Identify that both magnesium oxide and sodium chloride have a giant ionic lattice structure. [1 mark] State that the charges on the ions in MgO are higher (\(\text{Mg}^{2+}\) and \(\text{O}^{2-}\)) than in NaCl (\(\text{Na}^+\) and \(\text{Cl}^-\)). [1 mark] State that the electrostatic forces of attraction between the ions in MgO are stronger. [0.5 mark] Conclude that significantly more heat/thermal energy is required to overcome these stronger forces in MgO.

1. Anaerobic conditions: In the presence of oxygen, yeast would respire aerobically to produce carbon dioxide and water instead of ethanol. Additionally, oxygen would oxidize the produced ethanol into ethanoic acid (vinegar). 2. Temperature: The reaction is catalyzed by enzymes in yeast. At temperatures above 40 °C, these enzymes are denatured because their active sites change shape, rendering them inactive. At low temperatures, the kinetic energy of the particles is low, leading to an extremely slow reaction rate. Therefore, 30 °C to 40 °C is the optimum compromise temperature.

[1 mark] Explain that anaerobic conditions prevent the oxidation of ethanol to ethanoic acid (or prevent aerobic respiration/production of carbon dioxide and water). [1 mark] State that yeast provides enzymes which catalyze the reaction. [1 mark] Explain that at high temperatures (above 40 °C) the enzymes are denatured, while at low temperatures the rate of reaction is too slow. [0.5 mark] Identify 30 °C to 40 °C as the optimum temperature for these yeast enzymes.

1. Identify the ions present in the solution: sodium ions (\(\text{Na}^+\)), chloride ions (\(\text{Cl}^-\)), hydrogen ions (\(\text{H}^+\)), and hydroxide ions (\(\text{OH}^-\)). 2. Explain cathode migration: both positive ions (\(\text{Na}^+\) and \(\text{H}^+\)) migrate to the negative electrode (cathode). 3. Apply the reactivity rule: hydrogen is less reactive than sodium on the reactivity series, meaning \(\text{H}^+\) ions gain electrons (are reduced) more readily than \(\text{Na}^+\) ions. 4. Write the half-equation: \(2\text{H}^+ + 2\text{e}^- \rightarrow \text{H}_2\).

[1 mark] State that both \(\text{Na}^+\) and \(\text{H}^+\) ions migrate to the cathode. [1 mark] Explain that hydrogen is lower in the reactivity series / less reactive than sodium, so \(\text{H}^+\) is preferentially discharged/reduced. [1 mark] Correct ionic half-equation: \(2\text{H}^+ + 2\text{e}^- \rightarrow \text{H}_2\) (accept \(2\text{H}_2\text{O} + 2\text{e}^- \rightarrow \text{H}_2 + 2\text{OH}^-\)). [0.5 mark] Correct state symbols or specifying that this is a reduction process (gain of electrons).

1. Solar radiation: Short-wavelength electromagnetic radiation (including ultraviolet and visible light) emitted by the Sun passes through the Earth's atmosphere. 2. Earth heating and emission: The Earth's surface absorbs this radiation, warms up, and re-emits longer-wavelength infrared (thermal) radiation back towards space. 3. Absorption and trapping: Greenhouse gas molecules in the atmosphere absorb this incoming longer-wavelength infrared radiation, preventing it from escaping into space. They then re-emit this thermal energy in all directions, including back towards the Earth's surface, trapping heat and raising global temperatures.

[1 mark] State that short-wavelength radiation from the Sun passes through the atmosphere. [1 mark] State that the Earth absorbs this and re-emits longer-wavelength/infrared radiation. [1 mark] Explain that greenhouse gases absorb this infrared radiation and re-emit it in all directions (including back to Earth). [0.5 mark] Conclude that this traps heat energy, resulting in global warming/temperature rise.

1. Find the mass of oxygen in the oxide:
\(3.48\text{ g} - 2.52\text{ g} = 0.96\text{ g}\)

2. Calculate the moles of iron and oxygen atoms:
\(\text{Moles of Fe} = \frac{2.52}{56} = 0.045\text{ mol}\)
\(\text{Moles of O} = \frac{0.96}{16} = 0.060\text{ mol}\)

3. Find the simplest whole-number ratio by dividing by the smallest value (0.045):
\(\text{Fe} = \frac{0.045}{0.045} = 1\)
\(\text{O} = \frac{0.060}{0.045} = 1.33\)

Multiply by 3 to obtain whole numbers:
\(\text{Fe} = 3\)
\(\text{O} = 4\)

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

M1: Calculate the mass of oxygen (0.96 g) AND the number of moles of Fe (0.045) and O (0.060) [1 mark]
M2: Show ratio conversion (divide by 0.045 and multiply by 3 to get 3:4) [1 mark]
M3: State correct empirical formula Fe3O4 [0.5 mark]

1. Calculate the moles of \(\text{NaHCO}_3\) reacted:
\(\text{Moles} = \frac{8.40}{84} = 0.100\text{ mol}\)

2. Use the stoichiometric ratio from the balanced equation:
The molar ratio of \(\text{NaHCO}_3 : \text{Na}_2\text{CO}_3\) is \(2 : 1\).
Therefore, moles of \(\text{Na}_2\text{CO}_3\) produced = \frac{0.100}{2} = 0.0500\text{ mol}\.

3. Calculate the mass of \(\text{Na}_2\text{CO}_3\) produced:
\(\text{Mass} = 0.0500 \times 106 = 5.30\text{ g}\).

M1: Moles of \(\text{NaHCO}_3\) = 0.100 mol [1 mark]
M2: Moles of \(\text{Na}_2\text{CO}_3\) = 0.0500 mol [1 mark]
M3: Correct final mass of 5.30 g (or 5.3 g) [0.5 mark]

1. Calculate the moles of \(\text{H}_2\text{SO}_4\) used:
\(\text{Moles} = 0.100 \times \frac{18.5}{1000} = 0.00185\text{ mol}\)

2. Use the equation to determine the moles of \(\text{NaOH}\):
The ratio of \(\text{NaOH} : \text{H}_2\text{SO}_4\) is \(2 : 1\).
\(\text{Moles of NaOH} = 2 \times 0.00185 = 0.00370\text{ mol}\)

3. Calculate the concentration of the \(\text{NaOH}\) solution:
\(\text{Concentration} = \frac{0.00370}{0.0250} = 0.148\text{ mol/dm}^3\).

M1: Moles of \(\text{H}_2\text{SO}_4\) = 0.00185 mol [1 mark]
M2: Moles of \(\text{NaOH}\) = 0.00370 mol [1 mark]
M3: Correct concentration calculation of 0.148 mol/dm^3 [0.5 mark]

1. Calculate the heat energy transferred to the water (\(Q\)):
\(Q = m \times c \times \Delta T\)
\(Q = 150 \times 4.18 \times 12.0 = 7524\text{ J} = 7.524\text{ kJ}\)

2. Calculate the moles of propan-1-ol burned:
\(\text{Moles} = \frac{0.60}{60} = 0.010\text{ mol}\)

3. Calculate the molar enthalpy change of combustion (\(\Delta H\)):
\(\Delta H = -\frac{Q}{\text{moles}} = -\frac{7.524}{0.010} = -752.4\text{ kJ/mol}\).

Since the temperature increased, the reaction is exothermic, so the sign must be negative.

M1: Calculate heat energy Q = 7.524 kJ (or 7524 J) [1 mark]
M2: Moles of propan-1-ol = 0.010 mol and divide Q by moles [1 mark]
M3: Correct value of -752.4 kJ/mol (or -752 kJ/mol) with negative sign [0.5 mark]

1. Calculate the moles of \(\text{CaCO}_3\) reacted:
\(\text{Moles} = \frac{5.00}{100} = 0.0500\text{ mol}\)

2. The mole ratio of \(\text{CaCO}_3 : \text{CO}_2\) is \(1 : 1\), so \(0.0500\text{ mol}\) of \(\text{CO}_2\) is produced.

3. Calculate the volume of \(\text{CO}_2\) gas at r.t.p.:
\(\text{Volume} = 0.0500 \times 24 = 1.20\text{ dm}^3\).

M1: Moles of \(\text{CaCO}_3\) = 0.0500 mol [1 mark]
M2: State that moles of \(\text{CO}_2\) is 0.0500 mol and multiply by 24 [1 mark]
M3: Correct final volume of 1.20 dm^3 (or 1.2 dm^3) [0.5 mark]

1. Calculate energy required to break bonds:
\(\text{Energy in} = (4 \times 413) + 243 = 1895\text{ kJ/mol}\)
(or considering only bonds broken: \(1 \times \text{C–H} + 1 \times \text{Cl–Cl} = 413 + 243 = 656\text{ kJ/mol}\))

2. Calculate energy released when bonds are formed:
\(\text{Energy out} = (3 \times 413) + 339 + 432 = 2010\text{ kJ/mol}\)
(or considering only bonds formed: \(1 \times \text{C–Cl} + 1 \times \text{H–Cl} = 339 + 432 = 771\text{ kJ/mol}\))

3. Calculate \(\Delta H\):
\(\Delta H = 1895 - 2010 = -115\text{ kJ/mol}\)
(or \(\Delta H = 656 - 771 = -115\text{ kJ/mol}\))

M1: Calculate total energy to break bonds (1895 or 656 kJ/mol) [1 mark]
M2: Calculate total energy released making bonds (2010 or 771 kJ/mol) [1 mark]
M3: Correct final answer of -115 kJ/mol (value and negative sign both required) [0.5 mark]