2023 OCR GCSE Twenty First Century Science - Chemistry B - J258 Practice Paper with Answers

First, calculate the relative formula masses (\(M_r\)) of the reactant and product:
\(M_r(\text{TiO}_2) = 47.9 + (2 \times 16.0) = 79.9\text{ g/mol}\)
\(M_r(\text{TiCl}_4) = 47.9 + (4 \times 35.5) = 189.9\text{ g/mol}\)

Next, find the moles of the reactant, \(\text{TiO}_2\):
\(\text{moles of TiO}_2 = \frac{12.0\text{ g}}{79.9\text{ g/mol}} \approx 0.1502\text{ mol}\)

From the 1:1 balanced equation stoichiometry, the theoretical moles of \(\text{TiCl}_4\) produced is also \(0.1502\text{ mol}\).

Calculate the theoretical yield mass of \(\text{TiCl}_4\):
\(\text{theoretical mass} = 0.1502\text{ mol} \times 189.9\text{ g/mol} \approx 28.52\text{ g}\)

Finally, calculate the percentage yield:
\(\text{Percentage Yield} = \frac{\text{Actual Yield}}{\text{Theoretical Yield}} \times 100 = \frac{21.4\text{ g}}{28.52\text{ g}} \times 100 \approx 75.03\% \approx 75\%\)

Award marks as follows (total 9 marks):
- 1 mark: Correct relative formula mass for \(\text{TiO}_2\) (\(79.9\)).
- 1 mark: Correct relative formula mass for \(\text{TiCl}_4\) (\(189.9\)).
- 2 marks: Moles of \(\text{TiO}_2\) calculated correctly as \(0.150\text{ mol}\).
- 2 marks: Theoretical mass of \(\text{TiCl}_4\) calculated correctly as \(28.5\text{ g}\).
- 2 marks: Percentage yield calculation set up and calculated correctly as \(75\%\).
- 1 mark: Correct multiple-choice option B selected.

To find the rate of change of \(\text{CO}_2\) concentration:
\(\text{Total Change in CO}_2 = 415\text{ ppm} - 325\text{ ppm} = 90\text{ ppm}\)
\(\text{Time Period} = 50\text{ years}\)
\(\text{Rate of Change} = \frac{90\text{ ppm}}{50\text{ years}} = 1.80\text{ ppm/year}\)

Observing the variables over time, as \(\text{CO}_2\) concentration increases from \(325\text{ ppm}\) to \(415\text{ ppm}\), the temperature anomaly increases from \(-0.05\text{ }^\circ\text{C}\) to \(+0.85\text{ }^\circ\text{C}\). This demonstrates a strong positive correlation.

Award marks as follows (total 9 marks):
- 2 marks: Correct calculation of the total change in carbon dioxide (\(90\text{ ppm}\)).
- 2 marks: Correct calculation of the rate per year (\(1.80\text{ ppm/year}\)).
- 2 marks: Accurate identification of a positive correlation between \(\text{CO}_2\) concentration and temperature anomaly.
- 2 marks: Explanation that correlation does not automatically equal causation, but the correlation matches the physical mechanism of the greenhouse effect.
- 1 mark: Correct multiple-choice option A selected.

During the electrolysis of concentrated aqueous sodium chloride (brine):
1. At the anode (positive electrode): Chloride ions (\(\text{Cl}^-\)) are discharged in preference to hydroxide ions (\(\text{OH}^-\)) because of their high concentration, producing chlorine gas (\(\text{Cl}_2\)).
2. At the cathode (negative electrode): Hydrogen ions (\(\text{H}^+\)) from water are discharged in preference to sodium ions (\(\text{Na}^+\)) because sodium is more reactive than hydrogen. This produces hydrogen gas (\(\text{H}_2\)).
3. Cathode half-equation: \(2\text{H}^+ + 2\text{e}^- \rightarrow \text{H}_2\).
4. Effect on solution: Hydroxide ions (\(\text{OH}^-\)) and sodium ions (\(\text{Na}^+\)) remain in solution, forming sodium hydroxide (\(\text{NaOH}\)), which is alkaline. Therefore, the pH of the remaining solution increases.

Award marks as follows (total 9 marks):
- 2 marks: Correctly identifying that chlorine gas is produced at the anode because halide ions are discharged.
- 2 marks: Correctly identifying that hydrogen gas is produced at the cathode because hydrogen is less reactive than sodium.
- 2 marks: Providing the correct, balanced ionic half-equation at the cathode: \(2\text{H}^+ + 2\text{e}^- \rightarrow \text{H}_2\).
- 2 marks: Explaining that hydroxide ions left behind cause the solution to become alkaline, resulting in a pH increase.
- 1 mark: Correct multiple-choice option B selected.

Calculate energy required to break bonds (reactants):
- \(4 \times \text{C-H} = 4 \times 413 = 1652\text{ kJ}\)
- \(2 \times \text{O=O} = 2 \times 498 = 996\text{ kJ}\)
- \(\text{Total bond breaking energy} = 1652 + 996 = 2648\text{ kJ/mol}\)

Calculate energy released when making bonds (products):
- \(2 \times \text{C=O} = 2 \times 805 = 1610\text{ kJ}\)
- \(4 \times \text{O-H} = 4 \times 464 = 1856\text{ kJ}\)
- \(\text{Total bond making energy} = 1610 + 1856 = 3466\text{ kJ/mol}\)

Overall energy change (\(\Delta H\)):
\(\Delta H = 2648 - 3466 = -818\text{ kJ/mol}\)

Since the value is negative, the reaction is exothermic. More energy is released when the new bonds are made in the products than is absorbed to break the chemical bonds in the reactants.

Award marks as follows (total 9 marks):
- 2 marks: Correctly calculating reactant bond energy total as \(2648\text{ kJ/mol}\).
- 2 marks: Correctly calculating product bond energy total as \(3466\text{ kJ/mol}\).
- 2 marks: Accurate subtraction of product energy from reactant energy to get \(-818\text{ kJ/mol}\).
- 2 marks: Describing the reaction as exothermic because bond making releases more energy than bond breaking requires.
- 1 mark: Correct multiple-choice option B selected.

First, convert the volumes from \(\text{cm}^3\) to \(\text{dm}^3\):
- Volume of \(\text{NaOH} = 25.0 / 1000 = 0.0250\text{ dm}^3\)
- Volume of \(\text{H}_2\text{SO}_4 = 18.75 / 1000 = 0.01875\text{ dm}^3\)

Calculate the number of moles of \(\text{NaOH}\) used:
\(\text{moles of NaOH} = \text{concentration} \times \text{volume} = 0.150\text{ mol/dm}^3 \times 0.0250\text{ dm}^3 = 0.00375\text{ mol}\)

Determine the moles of \(\text{H}_2\text{SO}_4\) using the reaction stoichiometry (1 mole of acid reacts with 2 moles of alkali):
\(\text{moles of H}_2\text{SO}_4 = \frac{0.00375\text{ mol}}{2} = 0.001875\text{ mol}\)

Calculate the concentration of the sulfuric acid:
\(\text{concentration} = \frac{\text{moles}}{\text{volume}} = \frac{0.001875\text{ mol}}{0.01875\text{ dm}^3} = 0.100\text{ mol/dm}^3\)

Award marks as follows (total 9 marks):
- 2 marks: Correctly converting both volumes to \(\text{dm}^3\) (\(0.0250\text{ dm}^3\) and \(0.01875\text{ dm}^3\)).
- 2 marks: Correctly calculating the moles of sodium hydroxide (\(0.00375\text{ mol}\)).
- 2 marks: Correct application of the 1:2 molar ratio to find moles of sulfuric acid (\(0.001875\text{ mol}\)).
- 2 marks: Correct calculation of sulfuric acid concentration (\(0.100\text{ mol/dm}^3\)).
- 1 mark: Correct multiple-choice option B selected.

Find the relative formula mass (\(M_r\)) of the reactant, glucose:
\(M_r(\text{C}_6\text{H}_{12}\text{O}_6) = (6 \times 12.0) + (12 \times 1.0) + (6 \times 16.0) = 72.0 + 12.0 + 96.0 = 180.0\text{ g/mol}\)

Find the relative formula mass of the desired product, ethanol:
\(M_r(\text{C}_2\text{H}_5\text{OH}) = (2 \times 12.0) + (6 \times 1.0) + (1 \times 16.0) = 24.0 + 6.0 + 16.0 = 46.0\text{ g/mol}\)

Calculate the total mass of desired products produced in the stoichiometric equation:
\(\text{Total mass of desired product} = 2 \times 46.0 = 92.0\text{ g/mol}\)

Calculate the atom economy:
\(\text{Atom Economy} = \frac{\text{Total } M_r \text{ of desired products}}{\text{Total } M_r \text{ of all reactants}} \times 100\)

\(\text{Atom Economy} = \frac{92.0}{180.0} \times 100 \approx 51.11\% \approx 51.1\%\)

Award marks as follows (total 9 marks):
- 2 marks: Correct calculation of \(M_r\) of glucose as \(180.0\).
- 2 marks: Correct calculation of \(M_r\) of ethanol as \(46.0\).
- 2 marks: Multiplying ethanol's mass by 2 to account for the stoichiometric coefficient (\(92.0\)).
- 2 marks: Correct calculation of percentage atom economy (\(51.1\%\)).
- 1 mark: Correct multiple-choice option B selected.

Cracking decane (\(\text{C}_{10}\text{H}_{22}\)) into octane (\(\text{C}_8\text{H}_{18}\)) results in another hydrocarbon. To balance the atoms:
\(\text{C}_{10}\text{H}_{22} \rightarrow \text{C}_8\text{H}_{18} + \text{C}_2\text{H}_4\)

The molecule \(\text{C}_2\text{H}_4\) is an alkene named ethene. Alkenes contain a unsaturated carbon-carbon double bond, which can be chemically tested by adding orange bromine water. The bromine adds across the double bond, causing the solution to turn from orange to colourless.

Award marks as follows (total 9 marks):
- 2 marks: Deducing the molecular formula of the other product as \(\text{C}_2\text{H}_4\) by balancing the equation.
- 2 marks: Identifying \(\text{C}_2\text{H}_4\) as the alkene named ethene.
- 2 marks: Stating that bromine water is used to test for unsaturation/alkenes.
- 2 marks: Describing the positive result of the test (colour change from orange to colourless).
- 1 mark: Correct multiple-choice option A selected.

For a cube with side length \(L\):
\(\text{Surface Area} = 6 \times L^2\)
\(\text{Volume} = L^3\)
\(\text{SA:V ratio} = \frac{6 \times L^2}{L^3} = \frac{6}{L}\)

For the nanoparticle (\(L = 10\text{ nm}\)):
\(\text{SA:V} = \frac{6}{10\text{ nm}} = 0.6\text{ nm}^{-1}\)

For the microparticle (\(L = 1000\text{ nm}\)):
\(\text{SA:V} = \frac{6}{1000\text{ nm}} = 0.006\text{ nm}^{-1}\)

Comparison of the ratios:
\(\frac{\text{Nanoparticle Ratio}}{\text{Microparticle Ratio}} = \frac{0.6}{0.006} = 100\)

Therefore, the nanoparticle's surface area-to-volume ratio is 100 times greater than that of the microparticle.

Award marks as follows (total 9 marks):
- 2 marks: Correctly calculating the surface area-to-volume ratio of the nanoparticle as \(0.6\text{ nm}^{-1}\).
- 2 marks: Correctly calculating the surface area-to-volume ratio of the microparticle as \(0.006\text{ nm}^{-1}\).
- 2 marks: Determining that the nanoparticle's ratio is \(100\) times larger by dividing the two ratios.
- 2 marks: Explaining that a larger surface area-to-volume ratio increases the proportion of atoms exposed at the surface, which explains why nanoparticles can be highly reactive compared to bulk material.
- 1 mark: Correct multiple-choice option A selected.

(a) To prepare and use the burette accurately:
1. Rinse the burette with the hydrochloric acid solution before filling it.
2. Fill the burette and run a small amount of acid through the tap to ensure the jet space is filled and there are no air bubbles.
3. Record the initial and final volumes with your eye level aligned with the bottom of the meniscus to avoid parallax error.
4. Swirl the conical flask constantly during the titration and add the acid dropwise near the end-point until the indicator shows a permanent colour change.

(b) Mean titre calculation:
\(\text{Mean titre} = \frac{22.30 + 22.40 + 22.35}{3} = 22.35\text{ cm}^3\)

(c) Sodium hydroxide concentration calculation:
- Step 1: Calculate the amount of \(\text{HCl}\) in moles:
\(\text{Moles of HCl} = \text{concentration} \times \text{volume (in dm}^3\text{)}\)
\(\text{Moles of HCl} = 0.100\text{ mol/dm}^3 \times \frac{22.35}{1000}\text{ dm}^3 = 2.235 \times 10^{-3}\text{ mol}\)

- Step 2: Use the stoichiometry of the reaction:
Since \(1\text{ mole}\) of \(\text{HCl}\) reacts with \(1\text{ mole}\) of \(\text{NaOH}\), then:
\(\text{Moles of NaOH} = 2.235 \times 10^{-3}\text{ mol}\)

- Step 3: Calculate the concentration of the \(\text{NaOH}\) solution:
\(\text{Concentration of NaOH} = \frac{\text{moles}}{\text{volume (in dm}^3\text{)}}\)
\(\text{Volume of NaOH} = 25.0\text{ cm}^3 = 0.0250\text{ dm}^3\)
\(\text{Concentration} = \frac{2.235 \times 10^{-3}\text{ mol}}{0.0250\text{ dm}^3} = 0.0894\text{ mol/dm}^3\) (to 3 s.f.)

(a) [Total: 4 marks]
- 1 mark: Rinse the burette with the acid solution.
- 1 mark: Fill the jet space / ensure no air bubbles are trapped in the jet.
- 1 mark: Read the meniscus at eye level / avoid parallax error.
- 1 mark: Add acid dropwise near the end-point / swirl the flask.

(b) [Total: 1 mark]
- 1 mark: For \(22.35\text{ cm}^3\).

(c) [Total: 4 marks]
- 1 mark: Calculate moles of acid: \(2.235 \times 10^{-3}\text{ mol}\) (allow ecf from part b).
- 1 mark: State the mole ratio is 1:1, hence moles of \(\text{NaOH} = 2.235 \times 10^{-3}\text{ mol}\).
- 1 mark: Divide moles by volume in \(\text{dm}^3\) (\(0.0250\text{ dm}^3\)).
- 1 mark: Correct final concentration of \(0.0894\text{ mol/dm}^3\) rounded to 3 significant figures.

(a) At the negative electrode (cathode), copper metal is produced because copper ions are less reactive than hydrogen ions in the solution and are preferentially discharged.
Half-equation: \(\text{Cu}^{2+}\text{(aq)} + 2\text{e}^- \rightarrow \text{Cu(s)}\)

(b) The colourless gas produced at the positive electrode (anode) is oxygen, \(\text{O}_2\), which forms from the discharge of hydroxide ions.
Test: Place a glowing splint into a tube containing the collected gas.
Result: The glowing splint relights.

(c) The blue colour of the solution is caused by hydrated copper(II) ions, \(\text{Cu}^{2+}\text{(aq)}\). As electrolysis proceeds, these copper(II) ions are discharged at the cathode to form solid copper metal, decreasing their concentration in solution, which makes the blue colour fade.
The remaining ions in solution are hydrogen ions, \(\text{H}^+\text{(aq)}\), and sulfate ions, \(\text{SO}_4^{2-}\text{(aq)}\). Together, these form sulfuric acid, \(\text{H}_2\text{SO}_4\text{(aq)}\).

(a) [Total: 3 marks]
- 1 mark: Identify product as copper / \(\text{Cu}\).
- 1 mark: Correctly balanced half-equation: \(\text{Cu}^{2+} + 2\text{e}^- \rightarrow \text{Cu}\).
- 1 mark: Correct state symbols: \(\text{Cu}^{2+}\text{(aq)}\) and \(\text{Cu(s)}\).

(b) [Total: 3 marks]
- 1 mark: Identify gas as oxygen / \(\text{O}_2\).
- 1 mark: Describe test using a glowing splint.
- 1 mark: Describe result that the splint relights (reject burns with a squeaky pop / burns with a blue flame).

(c) [Total: 3 marks]
- 1 mark: State that the blue colour is due to copper(II) ions / \(\text{Cu}^{2+}\).
- 1 mark: Explain that these ions are discharged / reduced to copper atoms, so their concentration decreases.
- 1 mark: Identify the remaining acid as sulfuric acid (accept formula \(\text{H}_2\text{SO}_4\)).

Part (a):
1. Electromagnetic radiation from the Sun (mainly short-wavelength ultraviolet and visible light) passes through the atmosphere and reaches the Earth's surface.
2. The Earth absorbs this radiation and warms up, re-emitting it as longer-wavelength infrared (IR) radiation.
3. Greenhouse gases (like \(\text{CO}_2\) and \(\text{CH}_4\)) absorb this outgoing long-wave infrared radiation.
4. The greenhouse gas molecules then re-emit the infrared radiation in all directions, including back towards the Earth's surface, trapping thermal energy and keeping the planet warmer than it would otherwise be.

Part (b):
- Evidence for natural cycles: Historically, Earth's climate has warmed and cooled due to orbital changes (Milankovitch cycles) and solar activity. Solar cycles do influence temperature, but only on a minor scale (typically around 11-year cycles).
- Evidence against natural cycles / for human activity: Since the Industrial Revolution (around 1750), the concentration of atmospheric \(\text{CO}_2\) has risen exponentially, correlating strongly with global average temperature increases. Isotopic evidence shows that the rising \(\text{CO}_2\) comes primarily from burning fossil fuels (human activity), not natural sources like volcanoes. Additionally, satellite measurements show solar energy output has been stable or slightly declining since the 1970s while global temperatures have risen sharply.
- Conclusion: There is an overwhelming scientific consensus that human greenhouse gas emissions, not solar cycles, are the primary driver of modern global warming.

Part (a) [4 marks]:
- Mark 1: Short-wavelength radiation from the Sun passes through the atmosphere. (1)
- Mark 2: Earth's surface absorbs this radiation and re-emits it as long-wavelength / infrared radiation. (1)
- Mark 3: Greenhouse gases in the atmosphere absorb the outgoing infrared radiation. (1)
- Mark 4: The gases re-emit this radiation in all directions (including back to Earth), trapping heat. (1)

Part (b) [5 marks]:
- Mark 1: Identifies that natural solar cycles or orbital variations exist and can cause small periodic temperature changes. (1)
- Mark 2: States that there is a strong correlation between rising global temperatures and the exponential increase in atmospheric carbon dioxide levels since the Industrial Revolution. (1)
- Mark 3: Explains that solar output has remained relatively constant/declining in recent decades, while global temperatures have risen (evidence against solar cycles being the sole cause). (1)
- Mark 4: Mentions physical evidence linking greenhouse gases to human activity (e.g., carbon isotope ratios in atmospheric \(\text{CO}_2\) indicating fossil fuel origin). (1)
- Mark 5: Draws a clear, scientifically supported conclusion that human activity is the primary driver of current climate change, supported by overwhelming scientific consensus. (1)

Part (a):
With inert graphite electrodes, hydroxide ions (or water molecules) are discharged at the anode in preference to sulfate ions. This produces oxygen gas, so bubbling/effervescence is observed.
Equation: \(2\text{H}_2\text{O(l)} \rightarrow \text{O}_2\text{(g)} + 4\text{H}^+\text{(aq)} + 4\text{e}^-\) or \(4\text{OH}^-\text{(aq)} \rightarrow \text{O}_2\text{(g)} + 2\text{H}_2\text{O(l)} + 4\text{e}^-\).

Part (b):
With active copper electrodes, the anode itself reacts. Copper atoms lose electrons and go into solution as copper(II) ions, so the anode gradually decreases in mass / dissolves.
Equation: \(\text{Cu(s)} \rightarrow \text{Cu}^{2+}\text{(aq)} + 2\text{e}^-\).
This reaction is oxidation because copper atoms lose electrons.

Part (c):
In the industrial purification of copper:
1. The anode is made of impure copper, and the cathode is made of a thin sheet of pure copper.
2. The electrolyte is copper(II) sulfate solution.
3. Copper atoms from the impure anode dissolve into the solution as copper(II) ions, while at the cathode, copper(II) ions from the solution are reduced to pure copper metal and deposit onto the cathode. Impurities fall to the bottom as anode slime.

Part (a) [2 marks]:
- Mark 1: Observations of effervescence / bubbles of colourless gas at the anode. (1)
- Mark 2: Correct balanced half-equation: \(4\text{OH}^- \rightarrow \text{O}_2 + 2\text{H}_2\text{O} + 4\text{e}^-\) (accept equivalent using \(\text{H}_2\text{O}\)). (1)

Part (b) [4 marks]:
- Mark 1: Observation that the anode dissolves / loses mass / gets smaller. (1)
- Mark 2: Explanation that copper atoms are oxidized / lose electrons to become copper(II) ions. (1)
- Mark 3: Correct half-equation: \(\text{Cu(s)} \rightarrow \text{Cu}^{2+}\text{(aq)} + 2\text{e}^-\). (1)
- Mark 4: State that this process is oxidation. (1)

Part (c) [3 marks]:
- Mark 1: The anode is impure copper and the cathode is pure copper. (1)
- Mark 2: Copper ions travel through the electrolyte from the anode to the cathode, where they deposit as pure copper. (1)
- Mark 3: Impurities do not deposit and fall to the bottom as 'anode sludge' / 'slime', leaving pure copper on the cathode. (1)

Part (a):
1. Calculate moles of NaOH used:
\(\text{Moles of NaOH} = \text{concentration} \times \text{volume in dm}^3\)
\(\text{Volume of NaOH} = 22.40 \text{ cm}^3 = 0.02240 \text{ dm}^3\)
\(\text{Moles} = 0.150 \text{ mol/dm}^3 \times 0.02240 \text{ dm}^3 = 0.00336 \text{ moles}\)

2. Use the balanced equation to find moles of \(\text{H}_2\text{SO}_4\):
The mole ratio of \(\text{H}_2\text{SO}_4\) to \(\text{NaOH}\) is 1:2.
\(\text{Moles of H}_2\text{SO}_4 = 0.00336 / 2 = 0.00168 \text{ moles}\)

3. Calculate the concentration of \(\text{H}_2\text{SO}_4\):
\(\text{Volume of acid} = 25.0 \text{ cm}^3 = 0.0250 \text{ dm}^3\)
\(\text{Concentration} = \text{moles} / \text{volume} = 0.00168 / 0.0250 = 0.0672 \text{ mol/dm}^3\)

Part (b):
1. To convert mol/dm³ to g/dm³:
\(\text{Concentration in g/dm}^3 = \text{concentration in mol/dm}^3 \times M_r\)
\(\text{Concentration} = 0.0672 \text{ mol/dm}^3 \times 98.1 = 6.59232 \text{ g/dm}^3\)
Rounding to 3 significant figures gives \(6.59 \text{ g/dm}^3\).

Part (c):
- The volumetric pipette is designed to measure out exactly one fixed volume (in this case, 25.0 cm³) with very high precision, which is required for the sample to be analyzed.
- The burette is designed to measure variable volumes, allowing the user to add the reactant dropwise and read off the exact volume used (the titre) when the indicator changes colour at the neutralisation point.

Part (a) [4 marks]:
- Mark 1: Calculates the moles of NaOH: \(0.150 \times 0.02240 = 0.00336 \text{ mol}\). (1)
- Mark 2: Uses the stoichiometric ratio (1:2) to find the moles of \(\text{H}_2\text{SO}_4\): \(0.00336 / 2 = 0.00168 \text{ mol}\). (1)
- Mark 3: Converts the volume of sulfuric acid to dm³: \(25.0 / 1000 = 0.0250 \text{ dm}^3\). (1)
- Mark 4: Calculates final concentration: \(0.00168 / 0.0250 = 0.0672 \text{ mol/dm}^3\). (1)

Part (b) [2 marks]:
- Mark 1: Multiplies their answer from (a) by 98.1. (1)
- Mark 2: Correctly calculated value, e.g., \(6.59 \text{ g/dm}^3\) (accept transfer of error from part a). (1)

Part (c) [3 marks]:
- Mark 1: States that a pipette is used to deliver a single, fixed, highly accurate volume of solution. (1)
- Mark 2: States that a burette allows for the delivery of a variable volume drop-by-drop. (1)
- Mark 3: Connects this to finding the precise end-point of the reaction where neutralisation occurs. (1)

Part (a):
1. Identify the bonds broken (reactants):
- 4 \(\text{C}-\text{H}\) bonds: \(4 \times 413 = 1652 \text{ kJ}\)
- 2 \(\text{O}=\text{O}\) bonds: \(2 \times 498 = 996 \text{ kJ}\)
Total energy input to break bonds = \(1652 + 996 = 2648 \text{ kJ/mol}\).

2. Identify the bonds made (products):
- 2 \(\text{C}=\text{O}\) bonds: \(2 \times 805 = 1610 \text{ kJ}\)
- 4 \(\text{O}-\text{H}\) bonds: \(4 \times 464 = 1856 \text{ kJ}\)
Total energy released from making bonds = \(1610 + 1856 = 3466 \text{ kJ/mol}\).

3. Calculate the overall energy change:
\(\Delta H = \text{Energy of bonds broken} - \text{Energy of bonds made}\)
\(\Delta H = 2648 - 3466 = -818 \text{ kJ/mol}\).

Part (b):
- Bond breaking is an endothermic process that requires an input of energy from the surroundings.
- Bond making is an exothermic process that releases energy to the surroundings.
- In this reaction, the energy released when the new bonds (in the carbon dioxide and water molecules) are formed (3466 kJ/mol) is greater than the energy taken in to break the original bonds (in the methane and oxygen molecules) (2648 kJ/mol).
- Therefore, there is a net release of heat energy to the surroundings, making the reaction exothermic.

Part (a) [5 marks]:
- Mark 1: Correctly calculates energy needed to break C-H bonds: \(4 \times 413 = 1652 \text{ kJ}\). (1)
- Mark 2: Correctly calculates energy needed to break O=O bonds: \(2 \times 498 = 996 \text{ kJ}\), leading to total bond breaking energy of \(2648 \text{ kJ/mol}\). (1)
- Mark 3: Correctly calculates energy released in making C=O bonds: \(2 \times 805 = 1610 \text{ kJ}\). (1)
- Mark 4: Correctly calculates energy released in making O-H bonds: \(4 \times 464 = 1856 \text{ kJ}\), leading to total bond making energy of \(3466 \text{ kJ/mol}\). (1)
- Mark 5: Correctly calculates final energy change: \(2648 - 3466 = -818 \text{ kJ/mol}\) (value and negative sign required for mark). (1)

Part (b) [4 marks]:
- Mark 1: States that breaking bonds is endothermic / requires energy. (1)
- Mark 2: States that making bonds is exothermic / releases energy. (1)
- Mark 3: Explicitly compares the energy quantities (more energy is released during bond making than is absorbed during bond breaking). (1)
- Mark 4: Concludes that this leads to a net release of energy / decrease in enthalpy, which makes the reaction exothermic. (1)

Part (a):
(i) Increasing the temperature: The forward reaction is exothermic (\(\Delta H = -92 \text{ kJ/mol}\)). According to Le Chatelier's principle, if temperature is increased, the system will shift in the endothermic direction (to the left) to absorb the excess thermal energy. This decreases the yield of ammonia.
(ii) Increasing the pressure: There are 4 moles of gas on the reactant side (\(\text{N}_2 + 3\text{H}_2\)) and only 2 moles of gas on the product side (\(2\text{NH}_3\)). According to Le Chatelier's principle, increasing the pressure shifts the equilibrium to the side with fewer gas molecules to decrease the pressure. This shifts the equilibrium to the right, increasing the yield of ammonia.

Part (b):
- Temperature compromise: A low temperature would give a very high yield of ammonia at equilibrium, but the rate of reaction would be too slow to be economically viable. A high temperature would make the reaction run quickly, but would result in an extremely low yield of ammonia. Therefore, 450 °C is chosen as a compromise temperature that provides a reasonable yield at an acceptable rate.
- Pressure compromise: A very high pressure would give both a high yield of ammonia and a high rate of reaction. However, building and maintaining equipment that can safely withstand extremely high pressures is very expensive and poses significant safety risks. 200 atmospheres is chosen as a compromise between maximizing yield/rate and minimizing capital cost and safety hazards.
- Catalyst role: The iron catalyst allows the reaction to reach equilibrium much faster at any given temperature, which makes it possible to use the compromise temperature of 450 °C without the reaction taking too long.

Part (a) [4 marks]:
- Mark 1: (Temperature) Yield decreases because the forward reaction is exothermic / reverse reaction is endothermic. (1)
- Mark 2: (Temperature) The system shifts to the left to oppose the temperature increase. (1)
- Mark 3: (Pressure) Yield increases because there are fewer moles of gas on the right-hand side (2 moles) than the left-hand side (4 moles). (1)
- Mark 4: (Pressure) The system shifts to the right to decrease the pressure. (1)

Part (b) [5 marks]:
- Mark 1: Explains that lower temperatures increase yield but decrease the rate of reaction, making the process too slow. (1)
- Mark 2: Explains that 450 °C is a compromise to achieve a reasonable yield in a short period of time. (1)
- Mark 3: Explains that higher pressures increase both yield and rate of reaction. (1)
- Mark 4: Explains that 200 atm is a compromise because higher pressures are too expensive (due to energy/equipment costs) and present safety risks/hazards. (1)
- Mark 5: Explains that the iron catalyst increases the rate of both forward and backward reactions, enabling the use of a lower, safer compromise temperature. (1)

Part (a):
- The Experiment: Positively charged alpha particles were fired at an extremely thin sheet of gold foil in a vacuum. A detector was placed around the foil to observe the path of the alpha particles.
- Observations:
1. Most alpha particles passed straight through the gold foil without any deflection.
2. A small number of alpha particles were deflected through small angles.
3. A very small fraction (about 1 in 8000) were deflected through extremely large angles, or bounced back towards the source.
- Interpretations and Rejection of Plum Pudding:
- In the plum pudding model, positive charge and mass were thought to be spread out evenly throughout the atom, so alpha particles should have passed straight through with minimal deflection.
- The fact that most passed straight through showed that the atom consists mostly of empty space.
- The large deflections showed that there must be a concentrated region of positive charge and mass in the centre of the atom (the nucleus) that repelled the positive alpha particles.
- Because only a few particles were deflected back, this nucleus must be extremely small compared to the overall size of the atom. This led to the creation of the nuclear model.

Part (b):
- Rutherford's model had electrons orbiting the nucleus like planets, but classical physics predicted these electrons would spiral into the nucleus.
- Niels Bohr suggested that electrons can only orbit the nucleus in fixed shells (or energy levels) at specific distances.
- The evidence supporting Bohr's model came from the emission spectra of atoms. When atoms are heated or excited, they emit light at specific, discrete wavelengths (spectral lines). This corresponds to energy emitted as electrons drop from a higher energy level to a lower one.

Part (a) [6 marks]:
This is a Level of Response question. Use the following criteria:
- Level 3 (5-6 marks): A detailed and coherent description of the experiment, all three main observations, and their corresponding logical deductions that directly explain why the plum pudding model was rejected and the nuclear model accepted. Excellent use of scientific terminology.
- Level 2 (3-4 marks): A clear description of the experiment, at least two observations, and some attempt to relate these to the structure of the nuclear model. Some logical gaps or minor errors.
- Level 1 (1-2 marks): Simple statements about firing alpha particles at gold foil, or basic knowledge of the plum pudding/nuclear models. Highly descriptive with little linking of evidence to conclusions.
- 0 marks: No relevant content.

Points to look for:
- Alpha particles (positive helium nuclei) fired at gold foil.
- Observation 1: Most went straight through -> deduction: atom is mostly empty space.
- Observation 2: Some deflected -> deduction: nucleus has a positive charge.
- Observation 3: Very few bounced back -> deduction: nucleus contains almost all the mass and is extremely small.
- Comparison: Explains that the plum pudding model (homogenous positive charge) predicted no large-angle deflections, so it was disproven.

Part (b) [3 marks]:
- Mark 1: States Bohr proposed electrons orbit in shells / fixed energy levels at specific distances from the nucleus. (1)
- Mark 2: Mentions that this prevents electrons from spiraling into the nucleus / collapsing. (1)
- Mark 3: Explains that this model was supported by experimental evidence from emission spectra / discrete spectral lines of elements. (1)

Part (a):
1. Crude oil is heated until it vaporises and enters the bottom of a fractionating column.
2. The fractionating column has a temperature gradient: it is very hot at the bottom and becomes progressively cooler towards the top.
3. The hydrocarbon vapours rise up the column. When they reach a region where the temperature is below their specific boiling point, they condense back into a liquid and are tapped off.
4. Molecular size effects: Small hydrocarbon molecules have weak intermolecular forces, so they have low boiling points and remain as gases until they reach the cool top of the column. They have low viscosity.
5. Large hydrocarbon molecules have strong intermolecular forces, so they have high boiling points and condense quickly near the hot bottom of the column. They have high viscosity.

Part (b):
(i) Equation:
\(\text{C}_{10}\text{H}_{22} \rightarrow \text{C}_8\text{H}_{18} + \text{C}_2\text{H}_4\)

(ii) The other product is ethene (\(\text{C}_2\text{H}_4\)), which is an alkene.
To distinguish ethene from octane:
- Add orange-brown bromine water to both samples.
- Octane is an alkane (saturated), so there is no immediate reaction; the bromine water remains orange-brown.
- Ethene is an alkene (unsaturated), so it undergoes an addition reaction with bromine; the bromine water is rapidly decolourised (turns from orange-brown to colourless).

Part (a) [5 marks]:
- Mark 1: Crude oil is vaporized before entering the column. (1)
- Mark 2: Column is hotter at the bottom and cooler at the top (temperature gradient). (1)
- Mark 3: Vapours rise and condense when they reach their boiling point. (1)
- Mark 4: Smaller molecules have lower boiling points and condense at the top, while larger molecules have higher boiling points and condense at the bottom. (1)
- Mark 5: Links larger molecules to stronger intermolecular forces and higher viscosity / less ease of flow. (1)

Part (b) [4 marks]:
- Mark 1: (i) Correct equation: \(\text{C}_{10}\text{H}_{22} \rightarrow \text{C}_8\text{H}_{18} + \text{C}_2\text{H}_4\). (1)
- Mark 2: (ii) Identifies the other product as ethene. (1)
- Mark 3: Describes the test: Add bromine water. (1)
- Mark 4: States correct results: Ethene decolourises bromine water / turns it colourless, whereas octane leaves it orange-brown. (1)

Part (a):
- Increasing the concentration of hydrochloric acid means there are more acid particles (hydrogen ions) per unit volume.
- This decreases the average distance between reactant particles, so they collide more frequently.
- This results in a higher frequency of successful collisions (more successful collisions per second), which increases the overall rate of reaction.

Part (b):
- Increasing the temperature increases the average kinetic energy of the reacting particles, making them move faster.
- This increases the frequency of collisions. However, this effect is relatively small.
- The major reason for the large increase in rate is that a small increase in temperature greatly increases the proportion of particles with energy equal to or greater than the activation energy (\(E_a\)).
- Therefore, a much larger fraction of collisions are successful because they have enough energy to break bonds, leading to a much higher frequency of successful collisions.

Part (c):
(i) A catalyst works by providing an alternative reaction pathway that has a lower activation energy.
(ii) On a reaction profile diagram, a catalyst decreases the height of the energy barrier / activation energy peak, while leaving the energy levels of the reactants and products unchanged.

Part (a) [3 marks]:
- Mark 1: Higher concentration means more reactant particles per unit volume. (1)
- Mark 2: This leads to a higher frequency of collisions (more collisions per unit time). (1)
- Mark 3: This results in a higher frequency of successful collisions / faster reaction rate. (1)

Part (b) [4 marks]:
- Mark 1: Higher temperature increases the kinetic energy / speed of the particles. (1)
- Mark 2: This causes particles to collide more frequently. (1)
- Mark 3: Mentions that a much larger proportion of colliding particles now have energy greater than or equal to the activation energy. (1)
- Mark 4: Explains that this leads to a highly increased frequency of successful collisions. (1)

Part (c) [2 marks]:
- Mark 1: (i) States that the catalyst provides an alternative pathway with a lower activation energy. (1)
- Mark 2: (ii) States that it lowers the activation energy peak / curve on a reaction profile diagram. (1)

Part (a):
To perform a precise titration:
1. Use a volumetric pipette and pipette filler to measure exactly \(25.0\text{ cm}^3\) of the sodium hydroxide solution into a clean conical flask.
2. Add a few drops of a suitable indicator (such as phenolphthalein) to the flask.
3. Rinse a burette with the sulfuric acid, and then fill it with the acid, ensuring the jet space below the tap is filled with no air bubbles. Record the initial volume reading at eye level from the bottom of the meniscus.
4. Place the conical flask on a white tile under the burette to clearly see the color change.
5. Add the acid from the burette to the flask while continuously swirling the flask until the indicator permanently changes color (e.g., from pink to colourless for phenolphthalein). This is a rough titration.
6. Repeat the titration carefully, adding the acid dropwise near the end-point. Repeat until at least two concordant results (titres within \(0.10\text{ cm}^3\) of each other) are achieved.

Part (b):
1. Calculate moles of \(\text{H}_2\text{SO}_4\) reacted:
\(n(\text{H}_2\text{SO}_4) = \text{concentration} \times \text{volume in dm}^3\)
\(n(\text{H}_2\text{SO}_4) = 0.0500\text{ mol/dm}^3 \times \frac{18.80}{1000}\text{ dm}^3 = 9.40 \times 10^{-4}\text{ mol}\)

2. Determine moles of \(\text{NaOH}\) in the volumetric flask using the reaction stoichiometry (\(1:2\)):
\(n(\text{NaOH}) = 2 \times 9.40 \times 10^{-4}\text{ mol} = 1.88 \times 10^{-3}\text{ mol}\)

3. Calculate concentration of \(\text{NaOH}\) in \(\text{mol/dm}^3\):
\(C(\text{NaOH}) = \frac{1.88 \times 10^{-3}\text{ mol}}{0.0250\text{ dm}^3} = 0.0752\text{ mol/dm}^3\)

4. Convert concentration to \(\text{g/dm}^3\):
\(\text{Concentration in g/dm}^3 = 0.0752\text{ mol/dm}^3 \times 40.0\text{ g/mol} = 3.008\text{ g/dm}^3\)
Rounding to 3 significant figures gives \(3.01\text{ g/dm}^3\).

Part (a) Level of Response (6 marks):
- Level 3 (5-6 marks): Detailed, logical method described. Correct apparatus specified (pipette, burette, conical flask, indicator). Multiple accurate techniques explained (rinsing with solutions, reading meniscus at eye level, white tile, dropwise addition, swirling). Clear explanation of obtaining concordant results (within \(0.10\text{ cm}^3\)).
- Level 2 (3-4 marks): Clear method described. Mentions major apparatus (pipette and burette) and at least one technique to ensure accuracy. Mentions repeating to get close values.
- Level 1 (1-2 marks): Simple description of adding acid to alkali with indicator. No explanation of techniques to ensure precision or accuracy.

Part (b) Calculation (3 marks):
- 1 mark: Calculation of moles of acid = \(9.40 \times 10^{-4}\text{ mol}\).
- 1 mark: Mole ratio used correctly to find moles of alkali = \(1.88 \times 10^{-3}\text{ mol}\) (or concentration = \(0.0752\text{ mol/dm}^3\)).
- 1 mark: Final concentration correctly converted to \(\text{g/dm}^3\) and rounded to 3 s.f. = \(3.01\text{ g/dm}^3\) (accept \(3.008\text{ g/dm}^3\)).

Part (a):
1. Calculate the energy required to break bonds in the reactants (for 1 mole of \(\text{CH}_3\text{OH}\) and 1.5 moles of \(\text{O}_2\)):
- Bonds broken in 1 mole of methanol (\(\text{CH}_3\text{OH}\)):
- 3 \(\text{C}-\text{H}\) bonds: \(3 \times 413 = 1239\text{ kJ}\)
- 1 \(\text{C}-\text{O}\) bond: \(1 \times 358 = 358\text{ kJ}\)
- 1 \(\text{O}-\text{H}\) bond: \(1 \times 463 = 463\text{ kJ}\)
- Bonds broken in 1.5 moles of oxygen (\(\text{O}_2\)):
- 1.5 \(\text{O}=\text{O}\) bonds: \(1.5 \times 495 = 742.5\text{ kJ}\)
- Total energy input (breaking bonds) = \(1239 + 358 + 463 + 742.5 = 2802.5\text{ kJ}\).

2. Calculate the energy released when bonds are formed in the products (for 1 mole of \(\text{CO}_2\) and 2 moles of \(\text{H}_2\text{O}\)):
- Bonds formed in 1 mole of carbon dioxide (\(\text{CO}_2\)):
- 2 \(\text{C}=\text{O}\) bonds: \(2 \times 799 = 1598\text{ kJ}\)
- Bonds formed in 2 moles of water (\(\text{H}_2\text{O}\)):
- 4 \(\text{O}-\text{H}\) bonds: \(4 \times 463 = 1852\text{ kJ}\)
- Total energy output (forming bonds) = \(1598 + 1852 = 3450\text{ kJ}\).

3. Calculate the overall enthalpy change:
\(\Delta H = \text{energy input} - \text{energy output}\)
\(\Delta H = 2802.5 - 3450 = -647.5\text{ kJ/mol}\).

Part (b):
Reasons for difference:
1. Heat loss to the surrounding environment (air, container holding the water) instead of being fully transferred to the water.
2. Incomplete combustion of methanol due to restricted air supply, producing carbon monoxide or soot (carbon) and releasing less energy.

Improvements:
1. Use draft shields to reduce heat loss to the air, or a copper calorimeter with a lid to improve thermal conduction and reduce heat escaping.
2. Use a bomb calorimeter with a pure oxygen supply to ensure complete combustion and prevent heat loss.

Part (a) Calculation (5 marks):
- 1 mark: Calculating bonds broken in methanol = \(2060\text{ kJ}\) (or 3 \(\text{C}-\text{H}\) + 1 \(\text{C}-\text{O}\) + 1 \(\text{O}-\text{H}\)).
- 1 mark: Calculating bonds broken in oxygen = \(742.5\text{ kJ}\) (for \(1.5\text{ O}_2\)) or \(1485\text{ kJ}\) (for \(3\text{ O}_2\)).
- 1 mark: Calculating bonds formed in products = \(3450\text{ kJ}\) (for \(1\text{ CO}_2 + 2\text{ H}_2\text{O}\)) or \(6900\text{ kJ}\) (for \(2\text{ CO}_2 + 4\text{ H}_2\text{O}\)).
- 1 mark: Correct subtraction of \(\text{energy output}\) from \(\text{energy input}\) (e.g., \(2802.5 - 3450\) or \(5605 - 6900\)).
- 1 mark: Final value with negative sign and correct units = \(-647.5\text{ kJ/mol}\) (or \(-1295\text{ kJ}\) divided by 2).

Part (b) Explanation (4 marks):
- 1 mark: Heat loss to surroundings / beaker.
- 1 mark: Incomplete combustion of fuel.
- 2 marks: Two suitable improvements linked to the causes (e.g., use a lid/draft shields/copper can, use a bomb calorimeter/supply pure oxygen).