2023 Edexcel IGCSE Chemistry Practice Paper with Answers

(a) The iron reacts with the oxygen in the boiling tube, removing the oxygen gas from the air. This decreases the volume and pressure of the gas inside the tube, so water rises to replace the reacted oxygen.

(b) Oxygen (or air) and water.

(c)
- Decrease in volume of gas = \(50.0\text{ cm}^3 - 39.5\text{ cm}^3 = 10.5\text{ cm}^3\)
- Percentage of oxygen = \(\frac{10.5}{50.0} \times 100 = 21.0\%\)

(d) If there is not enough iron wool, the iron will be the limiting reactant and some oxygen will remain in the boiling tube at the end of the week. Therefore, the decrease in volume of gas will be less than expected, resulting in a lower calculated percentage of oxygen.

**(a)**
- An explanation that oxygen is used up / reacts with the iron (leaving less gas inside the tube) (1)

**(b)**
- Oxygen AND water (both required) (1)

**(c)**
- M1: Calculation of volume change: \(50.0 - 39.5 = 10.5\text{ cm}^3\) (1)
- M2: Percentage calculation setup: \(\frac{10.5}{50.0} \times 100\) (1)
- M3: Correct final answer: \(21.0\%\) (1)

**(d)**
- M1: Not all the oxygen reacts / some oxygen remains in the tube / iron is the limiting reactant (1)
- M2: The decrease in gas volume (or rise in water level) is smaller than it should be (1)

(a) (i) Heating a carbonate thermally decomposes it, producing carbon dioxide gas which turns limewater cloudy.

(ii) Copper(II) carbonate decomposes to form copper(II) oxide, which is a black solid. The chemical formula of copper(II) oxide is \(\text{CuO}\).

(iii) Since carbon dioxide gas was released, the anion present in the green solid (copper(II) carbonate) is the carbonate ion (\(\text{CO}_3^{2-}\)).

(b) (i) To test for transition metal cations such as copper(II), add sodium hydroxide solution (or ammonia solution) dropwise. A blue precipitate of copper(II) hydroxide, \(\text{Cu(OH)}_2\), will form.

(ii) When magnesium is added to the copper(II) nitrate solution:
- Magnesium is more reactive than copper and displaces it, forming a brown/pink copper solid deposit.
- As copper(II) ions are displaced and leave the solution, the blue colour fades and the solution eventually becomes colourless.
- The magnesium ribbon dissolves / gets smaller.
- Since the green solid was dissolved in dilute nitric acid, any excess acid reacts with the magnesium to produce bubbles of hydrogen gas.

(iii) Magnesium is higher in the reactivity series (more reactive) than copper, so it displaces copper from its salt in solution.

(a)(i)
- Carbon dioxide (1)
*Accept \(\text{CO}_2\)*

(a)(ii)
- \(\text{CuO}\) (1)
*Reject CuO2 / copper oxide (must be the formula)*

(a)(iii)
- Carbonate / \(\text{CO}_3^{2-}\) (1)

(b)(i)
- Add sodium hydroxide solution / aqueous ammonia (1)
- Blue precipitate (1)
*Reject other colours*

(b)(ii)
Any two from:
- Brown / pink / red-brown solid / deposit (1)
- Blue solution turns colourless / fades (1)
- Magnesium ribbon gets smaller / disappears / dissolves (1)
- Bubbles / effervescence / mixture gets warm (1)

(b)(iii)
- Magnesium is more reactive than copper / magnesium is higher in the reactivity series (1)

(a) (i) Heating a carbonate thermally decomposes it, producing carbon dioxide gas which turns limewater cloudy.

(ii) Copper(II) carbonate decomposes to form copper(II) oxide, which is a black solid. The chemical formula of copper(II) oxide is \(\text{CuO}\).

(iii) Since carbon dioxide gas was released, the anion present in the green solid (copper(II) carbonate) is the carbonate ion (\(\text{CO}_3^{2-}\)).

(b) (i) To test for transition metal cations such as copper(II), add sodium hydroxide solution (or ammonia solution) dropwise. A blue precipitate of copper(II) hydroxide, \(\text{Cu(OH)}_2\), will form.

(ii) When magnesium is added to the copper(II) nitrate solution:
- Magnesium is more reactive than copper and displaces it, forming a brown/pink copper solid deposit.
- As copper(II) ions are displaced and leave the solution, the blue colour fades and the solution eventually becomes colourless.
- The magnesium ribbon dissolves / gets smaller.
- Since the green solid was dissolved in dilute nitric acid, any excess acid reacts with the magnesium to produce bubbles of hydrogen gas.

(iii) Magnesium is higher in the reactivity series (more reactive) than copper, so it displaces copper from its salt in solution.

(a)(i)
- Carbon dioxide (1)
*Accept \(\text{CO}_2\)*

(a)(ii)
- \(\text{CuO}\) (1)
*Reject CuO2 / copper oxide (must be the formula)*

(a)(iii)
- Carbonate / \(\text{CO}_3^{2-}\) (1)

(b)(i)
- Add sodium hydroxide solution / aqueous ammonia (1)
- Blue precipitate (1)
*Reject other colours*

(b)(ii)
Any two from:
- Brown / pink / red-brown solid / deposit (1)
- Blue solution turns colourless / fades (1)
- Magnesium ribbon gets smaller / disappears / dissolves (1)
- Bubbles / effervescence / mixture gets warm (1)

(b)(iii)
- Magnesium is more reactive than copper / magnesium is higher in the reactivity series (1)

(a)(i) The two most common straight-chain alkene isomers of \(\text{C}_4\text{H}_8\) are but-1-ene and but-2-ene:
- But-1-ene: \(\text{CH}_2=\text{CH}-\text{CH}_2-\text{CH}_3\) (fully displayed with all single and double bonds shown).
- But-2-ene: \(\text{CH}_3-\text{CH}=\text{CH}-\text{CH}_3\) (fully displayed with all single and double bonds shown).
Alternatively, methylpropene: \(\text{CH}_2=\text{C}(\text{CH}_3)_2\) (fully displayed).

(a)(ii) IUPAC names corresponding to the drawn structures: 'but-1-ene' and 'but-2-ene' (or 'methylpropene').

(b) Add bromine water to both samples. Bromine water has an initial color of orange/yellow/brown. With butane (an alkane), there is no reaction, so the solution remains orange/yellow/brown. With but-1-ene (an alkene), addition reaction occurs and the solution becomes colorless.

(c)(i) The repeat unit of poly(but-1-ene) has a backbone of two single-bonded carbon atoms with extension bonds on either side. One carbon is bonded to two H atoms, and the other carbon is bonded to one H atom and an ethyl group (\(-\text{CH}_2\text{CH}_3\) or \(-\text{C}_2\text{H}_5\)):
\(-\text{[CH}_2-\text{CH}(\text{C}_2\text{H}_5)]_n-\)

(c)(ii) Addition polymers contain strong carbon-carbon (\(\text{C}-\text{C}\)) single bonds which make them highly stable, inert, and unreactive. Decomposers/microorganisms do not possess the necessary enzymes to break down these bonds. A key environmental issue is that they fill up landfill sites, taking up valuable space because they persist indefinitely.

(d) Calculate moles of products:
- \(\text{Moles of CO}_2 = \frac{17.6\text{ g}}{44\text{ g/mol}} = 0.40\text{ mol}\)
Each mole of \(\text{CO}_2\) contains \(1\text{ mol}\) of \(\text{C}\) atoms, so there are \(0.40\text{ mol}\) of \(\text{C}\) atoms in \(0.10\text{ mol}\) of \(Y\).

- \(\text{Moles of H}_2\text{O} = \frac{7.2\text{ g}}{18\text{ g/mol}} = 0.40\text{ mol}\)
Each mole of \(\text{H}_2\text{O}\) contains \(2\text{ mol}\) of \(\text{H}\) atoms, so there are \(2 \times 0.40 = 0.80\text{ mol}\) of \(\text{H}\) atoms in \(0.10\text{ mol}\) of \(Y\).

Now, scale up to find the moles of atoms in \(1.0\text{ mol}\) of \(Y\):
- \(\text{Moles of C} = \frac{0.40}{0.10} = 4\)
- \(\text{Moles of H} = \frac{0.80}{0.10} = 8\)
Therefore, the molecular formula of \(Y\) is \(\text{C}_4\text{H}_8\).

(a)(i)
- 1 mark for correct displayed structure of but-1-ene showing all atoms and all bonds (including C-H bonds).
- 1 mark for correct displayed structure of but-2-ene (or methylpropene) showing all atoms and all bonds.

(a)(ii)
- 1 mark for correct name matching the first drawn isomer (e.g., but-1-ene).
- 1 mark for correct name matching the second drawn isomer (e.g., but-2-ene).

(b)
- 1 mark for stating the test reagent: bromine water / aqueous bromine (Reject liquid bromine / bromine in UV light).
- 1 mark for stating the initial color: orange / yellow / brown.
- 1 mark for stating both correct observations: butane remains orange/yellow/brown (or no change) AND but-1-ene decolourises / becomes colorless (Reject 'clear').

(c)(i)
- 1 mark for drawing a two-carbon backbone with a single bond between them and open extension bonds on each side through brackets.
- 1 mark for correctly showing the four attached groups: three H atoms and one ethyl group (\(-\text{CH}_2\text{CH}_3\) or \(-\text{C}_2\text{H}_5\)).

(c)(ii)
- 1 mark for stating they are inert / unreactive because they have strong carbon-carbon (\(\text{C}-\text{C}\)) single bonds.
- 1 mark for stating they cannot be broken down by microorganisms / decomposers / bacteria (or they lack the enzymes to do so).
- 1 mark for stating they take up space in landfill sites / persist for hundreds of years / do not decay.

(d)
- 1 mark for calculating moles of carbon atoms (\(0.40\text{ mol}\)) and moles of hydrogen atoms (\(0.80\text{ mol}\)).
- 1 mark for dividing both by \(0.10\) (or scaling the ratio) to show that \(1\text{ mol}\) of \(Y\) contains \(4\text{ mol}\) of \(\text{C}\) and \(8\text{ mol}\) of \(\text{H}\), leading to \(\text{C}_4\text{H}_8\).

(a) (i) Any two from: the metal floats on the water surface / moves around on the surface; the metal melts into a spherical ball; there is effervescence / bubbles of gas / fizzing; the metal disappears / dissolves / gets smaller. (ii) An alkaline solution / potassium hydroxide solution is formed (which has a pH > 7) due to the presence of hydroxide ions / \(\text{OH}^-\). (b) (i) Clean a platinum or nichrome wire by dipping it in concentrated hydrochloric acid and placing it in a Bunsen burner flame until no color is observed. Dip the wire into the solid sample of the salt, then place the wire into the hot, non-luminous (blue) Bunsen burner flame and observe the color of the flame. (ii) Potassium / \(\text{K}\) (since potassium ions produce a lilac flame). (c) (i) A white precipitate is formed. (ii) The ionic equation is: \(\text{Ag}^+(\text{aq}) + \text{Cl}^-(\text{aq}) \rightarrow \text{AgCl}(\text{s})\).

(a) (i) 2 marks: Award 1 mark for each correct observation up to a maximum of 2. Acceptable observations: floats / moves on water; melts into a ball; effervescence / fizzing / bubbles; metal disappears / gets smaller. Reject: burns with a lilac flame (this only occurs if potassium ignites, which is not guaranteed for any generic group 1 metal without specifying). (ii) 2 marks: 1 mark for stating that an alkaline / basic solution is formed / pH increases; 1 mark for identifying hydroxide ions / \(\text{OH}^-\) as the cause. (b) (i) 3 marks: 1 mark for cleaning a platinum/nichrome wire using (concentrated) hydrochloric acid; 1 mark for dipping the wire in the solid sample; 1 mark for placing the wire in a non-luminous / blue Bunsen burner flame. (ii) 1 mark: Potassium / \(\text{K}\). (c) (i) 1 mark: White precipitate / white solid. Reject: cream/yellow precipitate. (ii) 2 marks: 1 mark for correct formulae of reactants and products: \(\text{Ag}^+ + \text{Cl}^- \rightarrow \text{AgCl}\); 1 mark for correct state symbols: \((\text{aq}) + (\text{aq}) \rightarrow (\text{s})\) (dependent on correct formulae or near-misses such as incorrect charges).

(a) The balanced equation is: \(\text{FeCl}_{3}(\text{aq}) + 3\text{KOH}(\text{aq}) \rightarrow \text{Fe(OH)}_{3}(\text{s}) + 3\text{KCl}(\text{aq})\)

(b)(i) The graph consists of two intersecting straight lines: a rising straight line starting from the origin, and a horizontal line. The exact volume of KOH needed is found at the point where these two lines intersect (which is at 3.0 cm³).

(b)(ii) Initially, the height of the precipitate increases because iron(III) chloride is in excess, and adding more KOH produces more iron(III) hydroxide precipitate. After 3.0 cm³ of KOH has been added, all the iron(III) chloride has completely reacted. Therefore, adding further KOH does not produce any more precipitate, so the height remains constant.

(c) Step 1: Calculate the amount, in moles, of \(\text{FeCl}_{3}\) used:
\(\text{Moles of FeCl}_{3} = \text{concentration} \times \text{volume in dm}^{3}\)
\(\text{Moles of FeCl}_{3} = 0.20 \text{ mol/dm}^{3} \times \frac{5.0}{1000} \text{ dm}^{3} = 0.0010 \text{ mol}\)

Step 2: Use the reaction stoichiometry to find the moles of \(\text{KOH}\) required:
From the equation, 1 mole of \(\text{FeCl}_{3}\) reacts with 3 moles of \(\text{KOH}\).
\(\text{Moles of KOH required} = 3 \times 0.0010 \text{ mol} = 0.0030 \text{ mol}\)

Step 3: Calculate the volume of 1.0 mol/dm³ \(\text{KOH}\) solution containing this number of moles:
\(\text{Volume of KOH in dm}^{3} = \frac{\text{moles}}{\text{concentration}} = \frac{0.0030 \text{ mol}}{1.0 \text{ mol/dm}^{3}} = 0.0030 \text{ dm}^{3}\)

Step 4: Convert this volume into \(\text{cm}^{3}\):
\(\text{Volume} = 0.0030 \text{ dm}^{3} \times 1000 = 3.0 \text{ cm}^{3}\).

(d) The precipitate may not have settled completely, there may be air gaps or water trapped in the solid, or the internal diameter of the test tubes may not be completely uniform.

Part (a): 2 marks
- 1 mark for correct chemical formulae of all reactants and products: \(\text{FeCl}_{3}\), \(\text{KOH}\), \(\text{Fe(OH)}_{3}\), \(\text{KCl}\).
- 1 mark for correct balancing (1, 3, 1, 3) AND correct state symbols: \((\text{aq})\), \((\text{aq})\), \((\text{s})\), \((\text{aq})\).

Part (b)(i): 2 marks
- 1 mark for describing two straight lines (one rising/increasing, one horizontal).
- 1 mark for stating that the volume is determined from the point where the two lines intersect/meet.

Part (b)(ii): 3 marks
- 1 mark for stating that initially \(\text{FeCl}_{3}\) is in excess / KOH is limiting, so adding more KOH produces more precipitate.
- 1 mark for stating that at the intersection point (3.0 cm³ of KOH), all \(\text{FeCl}_{3}\) has reacted / reactants are in exact stoichiometric ratio.
- 1 mark for stating that after this point, KOH is in excess / no more iron ions left to react, so no more precipitate forms.

Part (c): 4 marks
- 1 mark for calculating moles of \(\text{FeCl}_{3}\) as 0.0010 mol.
- 1 mark for identifying the 1:3 mole ratio and calculating the required moles of KOH as 0.0030 mol.
- 1 mark for using the concentration to calculate volume of KOH in \(\text{dm}^{3}\) (0.0030 \(\text{dm}^{3}\)).
- 1 mark for converting the volume to 3.0 \(\text{cm}^{3}\).

Part (d): 1 mark
- 1 mark for any acceptable practical error: precipitate not fully settled, solid not compacting uniformly, presence of trapped water/air pockets, variations in tube width/geometry. Reject: general 'human error' or 'incorrect measurement'.

(a) The balanced equation is: \(\text{FeCl}_{3}(\text{aq}) + 3\text{KOH}(\text{aq}) \rightarrow \text{Fe(OH)}_{3}(\text{s}) + 3\text{KCl}(\text{aq})\)

(b)(i) The graph consists of two intersecting straight lines: a rising straight line starting from the origin, and a horizontal line. The exact volume of KOH needed is found at the point where these two lines intersect (which is at 3.0 cm³).

(b)(ii) Initially, the height of the precipitate increases because iron(III) chloride is in excess, and adding more KOH produces more iron(III) hydroxide precipitate. After 3.0 cm³ of KOH has been added, all the iron(III) chloride has completely reacted. Therefore, adding further KOH does not produce any more precipitate, so the height remains constant.

(c) Step 1: Calculate the amount, in moles, of \(\text{FeCl}_{3}\) used:
\(\text{Moles of FeCl}_{3} = \text{concentration} \times \text{volume in dm}^{3}\)
\(\text{Moles of FeCl}_{3} = 0.20 \text{ mol/dm}^{3} \times \frac{5.0}{1000} \text{ dm}^{3} = 0.0010 \text{ mol}\)

Step 2: Use the reaction stoichiometry to find the moles of \(\text{KOH}\) required:
From the equation, 1 mole of \(\text{FeCl}_{3}\) reacts with 3 moles of \(\text{KOH}\).
\(\text{Moles of KOH required} = 3 \times 0.0010 \text{ mol} = 0.0030 \text{ mol}\)

Step 3: Calculate the volume of 1.0 mol/dm³ \(\text{KOH}\) solution containing this number of moles:
\(\text{Volume of KOH in dm}^{3} = \frac{\text{moles}}{\text{concentration}} = \frac{0.0030 \text{ mol}}{1.0 \text{ mol/dm}^{3}} = 0.0030 \text{ dm}^{3}\)

Step 4: Convert this volume into \(\text{cm}^{3}\):
\(\text{Volume} = 0.0030 \text{ dm}^{3} \times 1000 = 3.0 \text{ cm}^{3}\).

(d) The precipitate may not have settled completely, there may be air gaps or water trapped in the solid, or the internal diameter of the test tubes may not be completely uniform.

Part (a): 2 marks
- 1 mark for correct chemical formulae of all reactants and products: \(\text{FeCl}_{3}\), \(\text{KOH}\), \(\text{Fe(OH)}_{3}\), \(\text{KCl}\).
- 1 mark for correct balancing (1, 3, 1, 3) AND correct state symbols: \((\text{aq})\), \((\text{aq})\), \((\text{s})\), \((\text{aq})\).

Part (b)(i): 2 marks
- 1 mark for describing two straight lines (one rising/increasing, one horizontal).
- 1 mark for stating that the volume is determined from the point where the two lines intersect/meet.

Part (b)(ii): 3 marks
- 1 mark for stating that initially \(\text{FeCl}_{3}\) is in excess / KOH is limiting, so adding more KOH produces more precipitate.
- 1 mark for stating that at the intersection point (3.0 cm³ of KOH), all \(\text{FeCl}_{3}\) has reacted / reactants are in exact stoichiometric ratio.
- 1 mark for stating that after this point, KOH is in excess / no more iron ions left to react, so no more precipitate forms.

Part (c): 4 marks
- 1 mark for calculating moles of \(\text{FeCl}_{3}\) as 0.0010 mol.
- 1 mark for identifying the 1:3 mole ratio and calculating the required moles of KOH as 0.0030 mol.
- 1 mark for using the concentration to calculate volume of KOH in \(\text{dm}^{3}\) (0.0030 \(\text{dm}^{3}\)).
- 1 mark for converting the volume to 3.0 \(\text{cm}^{3}\).

Part (d): 1 mark
- 1 mark for any acceptable practical error: precipitate not fully settled, solid not compacting uniformly, presence of trapped water/air pockets, variations in tube width/geometry. Reject: general 'human error' or 'incorrect measurement'.

(a) Protons = 35, neutrons = 44, electrons = 35. (b) Let x be the percentage abundance of \(^{79}\text{Br}\). The equation is: \(\frac{79x + 81(100 - x)}{100} = 79.9\). Multiplying by 100 gives: \(79x + 8100 - 81x = 7990\). Simplifying gives: \(-2x = -110\), which results in \(x = 55\). Thus, the abundance of \(^{79}\text{Br}\) is 55%. (c)(i) The solution changes from colourless to orange. (ii) \(\text{Cl}_2 + 2\text{KBr} \rightarrow 2\text{KCl} + \text{Br}_2\) (iii) \(\text{Cl}_2(\text{g}) + 2\text{Br}^-(\text{aq}) \rightarrow 2\text{Cl}^-(\text{aq}) + \text{Br}_2(\text{aq})\) (iv) Chlorine has fewer electron shells and a smaller atomic radius than bromine, so its outer shell is closer to the nucleus. This leads to a stronger electrostatic attraction between the positive nucleus and incoming electrons, which means chlorine gains an electron more easily than bromine.

(a) 1 mark for 35 protons, 1 mark for 44 neutrons, 1 mark for 35 electrons. (b) 1 mark for setting up the equation correctly: \(\frac{79x + 81(100 - x)}{100} = 79.9\). 1 mark for correct algebraic simplification, e.g. \(2x = 110\). 1 mark for correct final value of 55%. (c)(i) 1 mark for starting color being colourless. 1 mark for final color being orange / yellow-brown / brown (reject yellow alone, red alone, or green). (c)(ii) 1 mark for correct formulae of reactants and products. 1 mark for correct balancing. (c)(iii) 1 mark for correct ionic equation: \(\text{Cl}_2 + 2\text{Br}^- \rightarrow 2\text{Cl}^- + \text{Br}_2\). 1 mark for correct state symbols: \(\text{Cl}_2(\text{g})\) [or \(\text{Cl}_2(\text{aq})\)], \(2\text{Br}^-(\text{aq})\), \(2\text{Cl}^-(\text{aq})\), \(\text{Br}_2(\text{aq})\). (c)(iv) 1 mark for stating that chlorine has fewer electron shells / outer shell closer to nucleus. 1 mark for stating that there is stronger attraction to incoming electrons / gains an electron more easily.

(a) Polystyrene is a good thermal insulator, which reduces heat loss to the surroundings during the reaction.

(b) Step 1: Calculate temperature change (\(\Delta T\))
\(\Delta T = 38.7 - 19.5 = 19.2\ ^\circ\text{C}\)

Step 2: Calculate mass of solution (\(m\))
\(m = 50.0\text{ cm}^3 \times 1.00\text{ g/cm}^3 = 50.0\text{ g}\)

Step 3: Calculate \(Q\)
\(Q = m \cdot c \cdot \Delta T = 50.0\text{ g} \times 4.18\text{ J/g/}^\circ\text{C} \times 19.2\ ^\circ\text{C} = 4012.8\text{ J}\) (or \(4010\text{ J}\) to 3 sig figs)

(c) \(\text{Moles of } \text{CuSO}_4 = \text{concentration} \times \text{volume in dm}^3\)
\(\text{Moles} = 0.500\text{ mol/dm}^3 \times \frac{50.0}{1000}\text{ dm}^3 = 0.0250\text{ mol}\)

(d) Step 1: Convert heat energy \(Q\) to kilojoules
\(Q = \frac{4012.8}{1000} = 4.0128\text{ kJ}\)

Step 2: Calculate molar enthalpy change (\(\Delta H\))
\(\Delta H = -\frac{Q}{\text{moles}} = -\frac{4.0128\text{ kJ}}{0.0250\text{ mol}} = -160.512\text{ kJ/mol}\)

Step 3: Round to 3 significant figures and include correct negative sign
\(\Delta H = -161\text{ kJ/mol}\)

(e) Any two valid reasons:
1. Heat is lost to the surroundings (air, cup, thermometer).
2. The reaction may be incomplete.
3. The specific heat capacity of the mixture is assumed to be that of water, and the mass of zinc is ignored.

(f) Using double the volume of the same concentration of copper(II) sulfate means there are double the moles of copper(II) ions reacting, which releases double the amount of heat energy (\(2 \times Q\)). However, there is also double the mass of solution to be heated (\(2 \times m\)). Since \(\Delta T = \frac{Q}{m \cdot c}\), the temperature rise remains constant because both the heat energy and mass are scaled by the same factor.

Part (a):
- 1 mark: Good insulator / reduces heat loss to surroundings (DO NOT accept 'prevents heat loss').

Part (b):
- 1 mark: Calculating \(\Delta T = 19.2\ ^\circ\text{C}\)
- 1 mark: Correct substitution into equation: \(50.0 \times 4.18 \times 19.2\)
- 1 mark: Correct value of \(4012.8\text{ J}\) (accept \(4010\text{ J}\) or \(4013\text{ J}\))

Part (c):
- 1 mark: Correct conversion of volume to dm\(^3\) (e.g., \(0.050\text{ dm}^3\))
- 1 mark: Correct evaluation of moles = \(0.0250\text{ mol}\)

Part (d):
- 1 mark: Division of \(Q\) by moles (ECF applies from (b) and (c))
- 1 mark: Value of \(161\) (or correct evaluation of ECF to 3 sig figs)
- 1 mark: Negative sign included with units (\(\text{kJ/mol}\))

Part (e):
- 1 mark each for any two valid reasons (max 2 marks):
- Heat loss (to surroundings/cup/thermometer)
- Incomplete reaction
- Specific heat capacity of solution is not exactly 4.18 J/g/\(^\circ\)C (or density is not 1.00 g/cm\(^3\))

Part (f):
- 1 mark: Double the moles react, releasing double the heat energy.
- 1 mark: There is double the mass/volume of solution to be heated (so they cancel out, keeping \(\Delta T\) the same).

Part (a):
1. Weigh the empty crucible and its lid.
2. Add the hydrated cobalt(II) chloride to the crucible and reweigh.
3. Heat the crucible and its contents strongly using a Bunsen burner.
4. Allow the crucible to cool down and then weigh it again.
5. Repeat the process of heating, cooling, and weighing until the mass remains constant. This ensures all the water has been completely removed.

Part (b):
(i) The lid prevents any solid from spitting out and being lost, while placing it loosely allows the water vapour (steam) to escape.
(ii) The color changes from pink to blue.

Part (c):
- Mass of anhydrous \(\text{CoCl}_2 = 21.10 - 18.50 = 2.60\text{ g}\)
- Mass of water lost = \(23.26 - 21.10 = 2.16\text{ g}\)
- Moles of anhydrous \(\text{CoCl}_2 = \frac{2.60}{130} = 0.02\text{ mol}\)
- Moles of \(\text{H}_2\text{O} = \frac{2.16}{18} = 0.12\text{ mol}\)
- Mole ratio of \(\text{H}_2\text{O} : \text{CoCl}_2 = \frac{0.12}{0.02} = 6\)
- Therefore, \(x = 6\).

Part (a) [5 marks total]:
- M1: Weigh the empty crucible and lid (1)
- M2: Weigh the crucible, lid, and hydrated salt (1)
- M3: Heat the crucible strongly (with a Bunsen burner) (1)
- M4: Cool and reweigh (1)
- M5: Repeat heating, cooling, and weighing until a constant mass is reached (1)

Part (b) [2 marks total]:
- M1 (i): To prevent loss of solid/spitting AND allow steam/water vapour to escape (1)
- M2 (ii): Pink to blue (1)

Part (c) [6 marks total]:
- M1: Find mass of anhydrous salt = \(2.60\text{ g}\) (1)
- M2: Find mass of water lost = \(2.16\text{ g}\) (1)
- M3: Moles of \(\text{CoCl}_2 = \frac{2.60}{130} = 0.02\text{ mol}\) (1)
- M4: Moles of \(\text{H}_2\text{O} = \frac{2.16}{18} = 0.12\text{ mol}\) (1)
- M5: Find the ratio of moles \(\frac{0.12}{0.02} = 6\) (1)
- M6: Deduce \(x = 6\) (1)
*(Note: Award full marks for correct final answer with working. Allow error carried forward (ECF) from calculation errors in M1 and M2)*

(a) When temperature increases:
1. The reactant particles gain kinetic energy and move faster.
2. This results in more frequent collisions (more collisions per unit time).
3. A greater proportion of particles have energy greater than or equal to the activation energy.
4. Therefore, there are more successful (or effective) collisions per unit time.

(b)(i):
1. The rate of reaction increases (the reaction is faster).
2. Small marble chips have a larger total surface area (for the same mass).
3. This leads to more frequent collisions (more collisions per unit time) between the acid particles and the surface of the marble chips.

(b)(ii):
1. Both curves start at the origin (0,0).
2. Curve S (small chips) has a steeper initial gradient (slope) than Curve L (large chips) because the initial rate of reaction is faster.
3. Both curves eventually level off (reach a horizontal plateau) at the exact same final volume of carbon dioxide gas, because the limiting reactant is used up and the same mass of reactants is used.

(c):
1. Dilute hydrochloric acid is the limiting reactant and is completely used up.
2. Therefore, there are no more acid particles left to collide with the remaining calcium carbonate.

Part (a) [4 marks]:
- MP1: Particles gain kinetic energy / move faster [1]
- MP2: More frequent collisions / more collisions per unit time [1] (reject 'more collisions' without time reference, unless MP4 is awarded)
- MP3: More particles have energy greater than or equal to the activation energy [1]
- MP4: More successful / effective collisions per unit time [1]

Part (b)(i) [3 marks]:
- MP1: Rate increases / reaction is faster [1]
- MP2: Small chips have a larger surface area [1]
- MP3: More frequent collisions / more collisions per unit time [1]

Part (b)(ii) [3 marks]:
- MP1: Both curves start at the origin [1]
- MP2: Curve S has a steeper initial gradient than Curve L [1]
- MP3: Both curves level off at the same final horizontal level / volume [1]

Part (c) [2 marks]:
- MP1: Hydrochloric acid is the limiting reactant / is completely used up [1]
- MP2: No more acid particles available to collide with the calcium carbonate [1]

**(a) Electronic configurations across Period 3:**
- The number of occupied electron shells remains the same (all have 3 shells).
- The number of outer-shell electrons increases by 1 for each successive element (from 1 in sodium to 8 in argon).

**(b)(i) Trend in boiling points down Group 7:**
- Boiling point increases down the group.
- This is because the molecules get larger / have more electrons / have higher relative molecular mass, which leads to stronger intermolecular forces that require more thermal energy to overcome.

**(ii) Trend in reactivity down Group 7:**
- As you go down the group, the atoms get larger / have more shells, meaning the outer shell is further from the positive nucleus.
- There is also more shielding by inner electron shells.
- Therefore, there is a weaker attraction between the nucleus and the incoming electron, making it harder for the atom to gain an electron.

**(a)**
- **M1**: Number of occupied electron shells remains the same / all have three shells (1)
- **M2**: Number of outer shell electrons increases / increases from 1 to 8 (1)

**(b)(i)**
- **M1**: Boiling points increase (1)
- **M2**: Because intermolecular forces become stronger / require more energy to overcome (due to larger molecules / more electrons) (1)
*Note: Reject any mention of breaking covalent bonds.*

**(b)(ii)**
- **M1**: Atomic radius increases / outer shell is further from the nucleus / there are more shells (1)
- **M2**: There is more shielding by inner shells (1)
- **M3**: There is a weaker attraction between the nucleus and the incoming electron / it is harder to gain an electron (1)

**(a) Electronic configurations across Period 3:**
- The number of occupied electron shells remains the same (all have 3 shells).
- The number of outer-shell electrons increases by 1 for each successive element (from 1 in sodium to 8 in argon).

**(b)(i) Trend in boiling points down Group 7:**
- Boiling point increases down the group.
- This is because the molecules get larger / have more electrons / have higher relative molecular mass, which leads to stronger intermolecular forces that require more thermal energy to overcome.

**(ii) Trend in reactivity down Group 7:**
- As you go down the group, the atoms get larger / have more shells, meaning the outer shell is further from the positive nucleus.
- There is also more shielding by inner electron shells.
- Therefore, there is a weaker attraction between the nucleus and the incoming electron, making it harder for the atom to gain an electron.

**(a)**
- **M1**: Number of occupied electron shells remains the same / all have three shells (1)
- **M2**: Number of outer shell electrons increases / increases from 1 to 8 (1)

**(b)(i)**
- **M1**: Boiling points increase (1)
- **M2**: Because intermolecular forces become stronger / require more energy to overcome (due to larger molecules / more electrons) (1)
*Note: Reject any mention of breaking covalent bonds.*

**(b)(ii)**
- **M1**: Atomic radius increases / outer shell is further from the nucleus / there are more shells (1)
- **M2**: There is more shielding by inner shells (1)
- **M3**: There is a weaker attraction between the nucleus and the incoming electron / it is harder to gain an electron (1)

a) Copper is a pink-brown metal that reacts with oxygen to form black copper(II) oxide. Therefore, the color change is from pink-brown to black.

b) The air is passed back and forth several times to ensure all the oxygen present in the sample has reacted completely with the excess copper. The apparatus must be allowed to cool because gases expand when heated. Reading the volume while hot would give an incorrectly high gas volume, leading to an underestimation of the volume of oxygen that reacted.

c) First, calculate the volume of oxygen that reacted:
Volume of oxygen = 80.0 cm³ - 63.6 cm³ = 16.4 cm³
Then, calculate the percentage of oxygen in the initial sample:
Percentage of oxygen = (16.4 cm³ / 80.0 cm³) * 100 = 20.5%

d) The main gas remaining is nitrogen. It is unreactive under these conditions because of the strong triple covalent bond between the nitrogen atoms, which requires a very high amount of energy to break.

a)
- M1: Pink / brown / pink-brown / red-brown (1)
- M2: to black (1)
[Reject: copper-colored to green/blue]

b)
- M1: (Passed back and forth) to ensure all the oxygen has reacted / complete reaction (1)
- M2: Gases expand when heated / contract when cooled (1)
- M3: (Cooling ensures) the volume is measured at the same temperature as the start / prevents an underestimation of the volume of oxygen reacted (1)

c)
- M1: Calculates volume of oxygen reacted: 80.0 - 63.6 = 16.4 (cm³) (1)
- M2: Shows division by initial volume and multiplication by 100: (16.4 / 80.0) x 100 (1)
- M3: Correct evaluation: 20.5 (%) (1)
[Note: Correct final answer with no working scores 3 marks. If student calculates percentage of gas remaining (79.5%), award max 1 mark for M1]

d)
- M1: Nitrogen (1)
- M2: It is unreactive / has a strong triple covalent bond (1)

Part (a)(i): Fermentation requires anaerobic conditions (absence of oxygen) and a temperature between \(30^\circ\text{C}\) and \(40^\circ\text{C}\) (or presence of yeast). Part (a)(ii): The equation is \(\text{C}_2\text{H}_4\text{(g)} + \text{H}_2\text{O(g)} \rightarrow \text{C}_2\text{H}_5\text{OH(g)}\). The catalyst used is phosphoric acid (\(\text{H}_3\text{PO}_4\)). Part (a)(iii): Hydration is a continuous process, which is faster and produces pure ethanol with no by-products. Part (b)(i): Molar mass of glucose (\(\text{C}_6\text{H}_{12}\text{O}_6\)) = \(6 \times 12 + 12 \times 1 + 6 \times 16 = 180\text{ g/mol}\). Moles of glucose = \(45.0 / 180 = 0.250\text{ mol}\). From the balanced equation, \(1\text{ mol}\) of glucose produces \(2\text{ mol}\) of ethanol. Moles of ethanol = \(0.250 \times 2 = 0.500\text{ mol}\). Molar mass of ethanol (\(\text{C}_2\text{H}_5\text{OH}\)) = \(2 \times 12 + 6 \times 1 + 16 = 46\text{ g/mol}\). Maximum theoretical mass of ethanol = \(0.500 \times 46 = 23.0\text{ g}\). Part (b)(ii): Percentage yield = \(\text{(Actual Yield / Theoretical Yield)} \times 100 = (14.4 / 23.0) \times 100 = 62.6\%\). Part (b)(iii): Moles of ethanol actually produced = \(14.4 / 46 = 0.31304\text{ mol}\). From the chemical equation, the mole ratio of \(\text{C}_2\text{H}_5\text{OH}\) to \(\text{CO}_2\) is \(2 : 2\) (or \(1 : 1\)). Thus, moles of \(\text{CO}_2\) produced = \(0.31304\text{ mol}\). Volume of \(\text{CO}_2\) gas at rtp = \(0.31304 \times 24 = 7.51\text{ dm}^3\).

Part (a)(i): M1: yeast (or zymase enzyme) AND anaerobic / absence of oxygen (1). M2: temperature range \(30 - 40^\circ\text{C}\) (1). Part (a)(ii): M1: correct reactants and products (\(\text{C}_2\text{H}_4 + \text{H}_2\text{O} \rightarrow \text{C}_2\text{H}_5\text{OH}\)) (1). M2: correct state symbols: \(\text{(g)}\) for reactants, and \(\text{(g)}\) or \(\text{(l)}\) for product (1). M3: phosphoric acid (\(\text{H}_3\text{PO}_4\)) catalyst (1). Part (a)(iii): M1: any one of: continuous process, faster rate of reaction, purer product, 100% atom economy (1). Part (b)(i): M1: calculation of glucose moles: \(45.0 / 180 = 0.25\text{ mol}\) (1). M2: determination of ethanol moles: \(0.25 \times 2 = 0.50\text{ mol}\) (1). M3: calculation of theoretical mass: \(0.50 \times 46 = 23.0\text{ g}\) (1). Part (b)(ii): M1: percentage yield expression: \((14.4 / 23.0) \times 100\) (1) [allow ECF from b(i)]. M2: evaluation to \(62.6\%\) (1) [must be 3 sig figs]. Part (b)(iii): M1: calculation of actual moles of ethanol: \(14.4 / 46 = 0.313\text{ mol}\) (1). M2: state that moles of \(\text{CO}_2\) equals moles of ethanol (1). M3: multiply moles of \(\text{CO}_2\) by 24 (1). M4: final answer of \(7.51\text{ dm}^3\) (1) [accept range \(7.50 - 7.52\); allow ECF from previous parts].

(a) At the start of the titration, the conical flask contains ethanoic acid (acidic), so phenolphthalein is colourless. At the end-point, as soon as the acid is neutralised by the added sodium hydroxide, the indicator turns pink.

(b) (i) Subtract the initial burette reading from the final burette reading for each column:
- Rough: \(21.20 - 0.00 = 21.20\text{ cm}^3\)
- Titration 1: \(21.40 - 1.00 = 20.40\text{ cm}^3\)
- Titration 2: \(41.90 - 21.20 = 20.70\text{ cm}^3\)
- Titration 3: \(20.90 - 0.50 = 20.40\text{ cm}^3\)

(ii) Concordant titres are within \(0.20\text{ cm}^3\) of each other. Titration 1 and Titration 3 are identical (\(20.40\text{ cm}^3\)), so these should be used.
\(\text{Mean titre} = \frac{20.40 + 20.40}{2} = 20.40\text{ cm}^3\)

(c) (i) \(\text{Moles of NaOH} = \text{concentration} \times \text{volume in dm}^3\)
\(\text{Moles} = 0.125\text{ mol/dm}^3 \times \frac{20.40}{1000}\text{ dm}^3 = 0.00255\text{ mol}\)

(ii) From the equation, the reacting ratio of \(\text{CH}_3\text{COOH}\) to \(\text{NaOH}\) is 1:1.
Therefore, \(\text{moles of CH}_3\text{COOH} = 0.00255\text{ mol}\).

(iii) \(\text{Concentration of diluted vinegar} = \frac{\text{moles}}{\text{volume in dm}^3}\)
\(\text{Concentration} = \frac{0.00255\text{ mol}}{0.0250\text{ dm}^3} = 0.102\text{ mol/dm}^3\)

(iv) The vinegar was diluted by taking 25.0 cm³ and making it up to 250 cm³, which is a 10-fold dilution:
\(\text{Dilution factor} = \frac{250}{25.0} = 10\)
\(\text{Original concentration} = 0.102\text{ mol/dm}^3 \times 10 = 1.02\text{ mol/dm}^3\)

(a)
- Colourless [1]
- To pink / pale pink [1] (REJECT red / purple)

(b) (i)
- All four titres correct [1]: Rough = 21.20, Titration 1 = 20.40, Titration 2 = 20.70, Titration 3 = 20.40

(b) (ii)
- Identifies Titrations 1 and 3 [1]
- Calculates mean of 20.40 (cm³) [1] (ALLOW ECF from incorrect titres in (i) for correct calculation of mean from chosen concordant values)

(c) (i)
- Calculation showing division by 1000: \(0.125 \times \frac{20.40}{1000}\) [1]
- \(0.00255\) / \(2.55 \times 10^{-3}\) (mol) [1] (ALLOW ECF from (b)(ii))

(c) (ii)
- \(0.00255\) (mol) [1] (ALLOW ECF from (c)(i))

(c) (iii)
- \(\frac{0.00255}{0.0250}\) or \(0.102\) (mol/dm³) [1] (ALLOW ECF from (c)(ii))

(c) (iv)
- Multiply diluted concentration by 10 [1]
- \(1.02\) (mol/dm³) [1] (ALLOW ECF from (c)(iii))

**(a)** Electrolysis is the chemical decomposition of an ionic compound (either molten or in aqueous solution) using electricity.

**(b) (i)** The product at the anode is **chlorine** gas.
- Test: Hold damp blue litmus paper in the gas.
- Result: The paper turns red then bleaches white.

**(b) (ii)** Chloride ions (\(\text{Cl}^-\)) migrate to the positive anode, where they lose electrons (oxidation) to form chlorine gas (\(\text{Cl}_2\)):
\[2\text{Cl}^-(\text{aq}) \rightarrow \text{Cl}_2(\text{g}) + 2\text{e}^-\]
*(Alternatively: \(2\text{Cl}^-(\text{aq}) - 2\text{e}^- \rightarrow \text{Cl}_2(\text{g})\))*

**(c) (i)** The gas produced at the negative electrode (cathode) is **hydrogen** (\(\text{H}_2\)).

**(c) (ii)** Both sodium ions (\(\text{Na}^+\)) and hydrogen ions (\(\text{H}^+\)) from the water are attracted to the cathode. Since sodium is more reactive than hydrogen on the reactivity series, sodium ions are more stable and remain in solution, while hydrogen ions are more easily reduced (gain electrons) to form hydrogen gas.

**(d)** Phenolphthalein turns **pink**. This is because hydrogen ions (\(\text{H}^+\)) are discharged at the cathode, leaving an excess of hydroxide ions (\(\text{OH}^-\)) in the solution, making it alkaline.

**Part (a):**
- **M1:** Decomposition / breakdown of an ionic compound / electrolyte (1)
- **M2:** Using electricity / electrical energy / direct current (1)

**Part (b)(i):**
- **M1:** Chlorine / \(\text{Cl}_2\) (1)
- **M2:** Damp blue litmus paper turns red then bleaches / turns white (1) *(Accept damp starch-iodide paper turns blue-black)*

**Part (b)(ii):**
- **M1:** Correct formulae of reactants and products: \(\text{Cl}^-\rightarrow \text{Cl}_2 + \text{e}^-\), or equivalent (1)
- **M2:** Correct balancing and correct state symbols: \(2\text{Cl}^-(\text{aq}) \rightarrow \text{Cl}_2(\text{g}) + 2\text{e}^-\) (1)

**Part (c)(i):**
- **M1:** Hydrogen / \(\text{H}_2\) (1)

**Part (c)(ii):**
- **M1:** Sodium is more reactive than hydrogen (1)
- **M2:** Hydrogen ions / \(\text{H}^+\) are preferentially discharged / gain electrons more easily than sodium ions / \(\text{Na}^+\) (1)

**Part (d):**
- **M1:** Pink AND because hydroxide ions (\(\text{OH}^-\)) are formed / left behind / solution becomes alkaline (1)

**(a)** Electrolysis is the chemical decomposition of an ionic compound (either molten or in aqueous solution) using electricity.

**(b) (i)** The product at the anode is **chlorine** gas.
- Test: Hold damp blue litmus paper in the gas.
- Result: The paper turns red then bleaches white.

**(b) (ii)** Chloride ions (\(\text{Cl}^-\)) migrate to the positive anode, where they lose electrons (oxidation) to form chlorine gas (\(\text{Cl}_2\)):
\[2\text{Cl}^-(\text{aq}) \rightarrow \text{Cl}_2(\text{g}) + 2\text{e}^-\]
*(Alternatively: \(2\text{Cl}^-(\text{aq}) - 2\text{e}^- \rightarrow \text{Cl}_2(\text{g})\))*

**(c) (i)** The gas produced at the negative electrode (cathode) is **hydrogen** (\(\text{H}_2\)).

**(c) (ii)** Both sodium ions (\(\text{Na}^+\)) and hydrogen ions (\(\text{H}^+\)) from the water are attracted to the cathode. Since sodium is more reactive than hydrogen on the reactivity series, sodium ions are more stable and remain in solution, while hydrogen ions are more easily reduced (gain electrons) to form hydrogen gas.

**(d)** Phenolphthalein turns **pink**. This is because hydrogen ions (\(\text{H}^+\)) are discharged at the cathode, leaving an excess of hydroxide ions (\(\text{OH}^-\)) in the solution, making it alkaline.

**Part (a):**
- **M1:** Decomposition / breakdown of an ionic compound / electrolyte (1)
- **M2:** Using electricity / electrical energy / direct current (1)

**Part (b)(i):**
- **M1:** Chlorine / \(\text{Cl}_2\) (1)
- **M2:** Damp blue litmus paper turns red then bleaches / turns white (1) *(Accept damp starch-iodide paper turns blue-black)*

**Part (b)(ii):**
- **M1:** Correct formulae of reactants and products: \(\text{Cl}^-\rightarrow \text{Cl}_2 + \text{e}^-\), or equivalent (1)
- **M2:** Correct balancing and correct state symbols: \(2\text{Cl}^-(\text{aq}) \rightarrow \text{Cl}_2(\text{g}) + 2\text{e}^-\) (1)

**Part (c)(i):**
- **M1:** Hydrogen / \(\text{H}_2\) (1)

**Part (c)(ii):**
- **M1:** Sodium is more reactive than hydrogen (1)
- **M2:** Hydrogen ions / \(\text{H}^+\) are preferentially discharged / gain electrons more easily than sodium ions / \(\text{Na}^+\) (1)

**Part (d):**
- **M1:** Pink AND because hydroxide ions (\(\text{OH}^-\)) are formed / left behind / solution becomes alkaline (1)

Part (a):
To draw the displayed formula of methyl methanoate (\(\text{HCOOCH}_3\)):
1. Draw the ester group: a carbon atom double-bonded to one oxygen atom (\(\text{C=O}\)) and single-bonded to a second oxygen atom (\(\text{C-O}\)).
2. The carbon atom from the ester group must be bonded to a single hydrogen atom (\(\text{H-C=O}\)).
3. The single-bonded oxygen atom must be bonded to a methyl carbon atom (\(\text{-CH}_3\)), which has three hydrogen atoms single-bonded to it.
4. Ensure every single bond and double bond is represented as a line (\(-\) or \(=\)).

Part (b):
1. Calculate the energy needed to break all bonds in the reactants:
- Bonds broken in 1 mole of \(\text{HCOOCH}_3\):
- \(4 \times \text{C-H}\) bonds = \(4 \times 413 = 1652\text{ kJ}\)
- \(1 \times \text{C=O}\) bond = \(1 \times 743 = 743\text{ kJ}\)
- \(2 \times \text{C-O}\) bonds = \(2 \times 358 = 716\text{ kJ}\)
- Total in ester = \(1652 + 743 + 716 = 3111\text{ kJ}\)
- Bonds broken in 2 moles of \(\text{O}_2\):
- \(2 \times \text{O=O}\) bonds = \(2 \times 495 = 990\text{ kJ}\)
- Total energy input (bonds broken) = \(3111 + 990 = 4101\text{ kJ}\)

2. Calculate the energy released when new bonds are formed in the products:
- Bonds formed in 2 moles of \(\text{CO}_2\):
- \(4 \times \text{C=O}\) bonds = \(4 \times 743 = 2972\text{ kJ}\)
- Bonds formed in 2 moles of \(\text{H}_2\text{O}\):
- \(4 \times \text{O-H}\) bonds = \(4 \times 463 = 1852\text{ kJ}\)
- Total energy output (bonds formed) = \(2972 + 1852 = 4824\text{ kJ}\)

3. Calculate the enthalpy change (\(\Delta H\)):
\(\Delta H = \text{Energy of bonds broken} - \text{Energy of bonds formed}\)
\(\Delta H = 4101 - 4824 = -723\text{ kJ/mol}\)

Part (a) [2 marks]:
- M1: Correctly drawn ester functional group showing \(\text{C=O}\) and \(\text{C-O}\) with all bonds shown (1 mark).
- M2: Rest of the molecule drawn correctly, including the \(\text{H}\) atom on the carbonyl carbon and the \(\text{-CH}_3\) group on the single-bonded oxygen (1 mark).
- Note: Deduct 1 mark if any bond is missing (e.g., writing \(\text{-CH}_3\) instead of showing all three \(\text{C-H}\) bonds separately).

Part (b) [4 marks]:
- M1: Calculation of energy required to break bonds in reactants: \(4101\text{ kJ}\) (1 mark).
- M2: Calculation of energy released making bonds in products: \(4824\text{ kJ}\) (1 mark).
- M3: Subtracting the energy released from the energy required (\(4101 - 4824\)) (1 mark).
- M4: Correct final value of \(-723\text{ kJ/mol}\) (or \(-723\text{ kJ}\)) with the negative sign (1 mark).
- Note: Award 4 marks for a correct final answer of \(-723\text{ kJ/mol}\). Award 3 marks for a final answer of \(+723\text{ kJ/mol}\) (omitting the negative sign).

**(a)**
Dynamic equilibrium means:
1. The rate of the forward reaction is equal to the rate of the reverse reaction.
2. The concentrations of reactants and products remain constant (in a closed system).

**(b)**
- Initial moles of \(\text{N}_2\text{O}_4 = 2.00\text{ mol}\).
- Moles of \(\text{N}_2\text{O}_4\) that dissociate: \(35.0\% \text{ of } 2.00 = 0.350 \times 2.00 = 0.70\text{ mol}\).
- Moles of \(\text{N}_2\text{O}_4\) remaining at equilibrium: \(2.00 - 0.70 = 1.30\text{ mol}\).
- Moles of \(\text{NO}_2\) formed at equilibrium: Since 1 mole of \(\text{N}_2\text{O}_4\) produces 2 moles of \(\text{NO}_2\), moles of \(\text{NO}_2\) formed = \(2 \times 0.70 = 1.40\text{ mol}\).
- Total moles at equilibrium: \(1.30 + 1.40 = 2.70\text{ mol}\).

**(c)**
- Increasing the temperature causes the position of equilibrium to shift to the right (in the forward direction).
- This is because the forward reaction is endothermic (absorbs heat), which acts to oppose the increase in temperature.
- Since more \(\text{NO}_2\) (brown gas) is formed, the colour of the gas mixture becomes darker brown.

**(d)**
- First, calculate the amount in moles of \(\text{N}_2\text{O}_4\) at RTP:
\(\text{moles} = \frac{\text{volume}}{\text{molar volume}} = \frac{0.54\text{ dm}^3}{24\text{ dm}^3\text{/mol}} = 0.0225\text{ mol}\)
- Next, calculate the mass of \(\text{N}_2\text{O}_4\):
\(\text{mass} = \text{moles} \times M_r = 0.0225\text{ mol} \times 92\text{ g/mol} = 2.07\text{ g}\)

**(a)**
- **M1**: Rate of forward reaction equals rate of reverse reaction (1)
- **M2**: Concentrations of reactants and products remain constant (1) *(Reject "concentrations are equal")*

**(b)**
- **M1**: Calculation of moles of \(\text{N}_2\text{O}_4\) reacted (\(0.70\text{ mol}\)) OR moles of \(\text{N}_2\text{O}_4\) remaining (\(1.30\text{ mol}\)) (1)
- **M2**: Calculation of moles of \(\text{NO}_2\) produced: \(2 \times 0.70 = 1.40\text{ mol}\) (1)
- **M3**: Calculation of total moles: \(1.30 + 1.40 = 2.70\text{ mol}\) (1)

**(c)**
- **M1**: Equilibrium shifts to the right / forward direction (1)
- **M2**: Because the forward reaction is endothermic / absorbs heat / to decrease the temperature (1)
- **M3**: The mixture becomes darker brown / more brown (1)

**(d)**
- **M1**: Relationship to find moles: \(\frac{0.54}{24}\) (1)
- **M2**: Moles = \(0.0225\text{ mol}\) (1)
- **M3**: Mass = \(2.07\text{ g}\) (1)
*(Allow full marks for correct final answer with no working shown. Allow ECF from incorrect M2 to M3)*