2023 Cambridge IAS-Level Chemistry (9701) Practice Paper with Answers

Zinc is in excess, so the total amount of hydrogen gas produced depends entirely on the limiting reactant, hydrochloric acid. Since the amount of hydrochloric acid is unchanged, the final volume of hydrogen gas remains constant. Powdering the zinc granules increases the surface area of the zinc, which increases the frequency of successful collisions per unit time, thereby increasing the initial rate of reaction without changing the amount of limiting reactant. Option B reduces the concentration of the acid, which decreases the initial rate. Option C doubles the concentration of the acid while keeping volume the same, which increases the initial rate but also doubles the final volume of hydrogen. Option D does not change the initial rate as concentration remains constant, and using a larger mass of the already excess zinc will not alter the rate or the final volume.

Award 1 mark for the correct option A. Method: Identify that changing the physical state of the excess reactant (powdering) increases the surface area and hence the initial rate, while keeping the limiting reactant (and therefore the total volume of gas) identical.

The equation for the standard enthalpy change of formation of propane is: 3\text{C(graphite)} + 4\text{H}_{2}\text{(g)} \rightarrow \text{C}_{3}\text{H}_{8}\text{(g)}. Using Hess's law and standard enthalpies of combustion: \Delta H_{f}^{\ominus} = 3 \times \Delta H_{c}^{\ominus}[\text{C(graphite)}] + 4 \times \Delta H_{c}^{\ominus}[\text{H}_{2}\text{(g)}] - \Delta H_{c}^{\ominus}[\text{C}_{3}\text{H}_{8}\text{(g)}]. Substituting the values: \Delta H_{f}^{\ominus} = 3(-393.5) + 4(-285.8) - (-2220.0) = -1180.5 - 1143.2 + 2220.0 = -103.7 \text{ kJ mol}^{-1}.

Award 1 mark for the correct option A. Method: Use Hess's cycle relating enthalpies of combustion of reactants to products: \Delta H_f = \Sigma \Delta H_c(\text{reactants}) - \Sigma \Delta H_c(\text{products}) with correct stoichiometry.

The reaction occurring in the catalytic converter is: 2\text{NO(g)} + 2\text{CO(g)} \rightarrow \text{N}_{2}\text{(g)} + 2\text{CO}_{2}\text{(g)}. In nitrogen monoxide, the oxidation state of nitrogen is +2. In nitrogen gas, the oxidation state of nitrogen is 0. Therefore, the oxidation state of nitrogen changes from +2 to 0, representing reduction. Carbon monoxide is oxidized to carbon dioxide, meaning it acts as a reducing agent. Platinum acts as a heterogeneous catalyst because it is in a different phase (solid) than the reactants (gases).

Award 1 mark for the correct option B. Method: Determine the oxidation state of nitrogen in \text{NO} and \text{N}_2 to identify the change from +2 to 0.

The chloride of magnesium (\text{MgCl}_{2}) dissolves in water and undergoes slight hydrolysis to form a weakly acidic solution with a pH of approximately 6.5. The chloride of silicon (\text{SiCl}_{4}) reacts violently with water in a complete hydrolysis reaction to form silicon dioxide (\text{SiO}_{2}) and hydrogen chloride (\text{HCl}) gas, which appears as white fumes and forms a strongly acidic solution with a pH of approximately 2.

Award 1 mark for the correct option B. Method: Recall the reactions and pH values of the Period 3 chlorides in water.

Monochlorination of butane can occur at the C1 position or the C2 position. Substitution at C1 gives 1-chlorobutane, \text{CH}_{2}\text{ClCH}_{2}\text{CH}_{2}\text{CH}_{3}, which does not have any chiral centers and therefore has no optical isomers. Substitution at C2 gives 2-chlorobutane, \text{CH}_{3}\text{C}^*\text{HClCH}_{2}\text{CH}_{3}. The C2 carbon is chiral, meaning 2-chlorobutane exists as a pair of optical enantiomers: (R)-2-chlorobutane and (S)-2-chlorobutane. This gives a total of 1 + 2 = 3 monochlorinated isomers.

Award 1 mark for the correct option B. Method: Identify the two structural isomers (1-chlorobutane and 2-chlorobutane) and recognize that 2-chlorobutane contains a chiral carbon atom, leading to two stereoisomers.

The reaction of Z with acidified potassium dichromate(VI) with a color change from orange to green indicates that Z is a primary or secondary alcohol (this rules out the tertiary alcohol 2-methylpropan-2-ol). Dehydration of Z with hot concentrated sulfuric acid forms alkenes. For butan-1-ol, dehydration produces only but-1-ene, which cannot exhibit cis-trans isomerism. For 2-methylpropan-1-ol, dehydration produces only 2-methylpropene, which cannot exhibit cis-trans isomerism. For butan-2-ol, dehydration produces a mixture of but-1-ene and but-2-ene. While but-1-ene does not show cis-trans isomerism, but-2-ene exhibits cis-trans isomerism. Thus, butan-2-ol is the correct compound.

Award 1 mark for the correct option B. Method: Correlate the functional group chemical test (oxidation of primary/secondary alcohols) and the dehydration outcomes (elimination products and stereoisomerism of alkenes).

An increase in temperature by 10 K causes a very small increase (about 1-2%) in the average kinetic energy of the particles, which only slightly increases collision frequency. However, from the Maxwell-Boltzmann distribution, even a small shift in temperature significantly increases the proportion of particles with kinetic energy equal to or greater than the activation energy (\text{E}_{a}). Consequently, the frequency of successful collisions increases exponentially, leading to a much higher reaction rate. The activation energy itself remains unchanged.

Award 1 mark for the correct option D. Method: Apply collision theory and the Maxwell-Boltzmann distribution to explain the exponential effect of temperature on rate.

Total volume of the mixture = 50.0 + 50.0 = 100.0 \text{ cm}^{3}. Mass of the mixture = 100.0 \text{ g}. Temperature rise (\Delta T) = 26.8 - 20.0 = 6.8 \text{ K}. Heat released (q) = m \times c \times \Delta T = 100.0 \text{ g} \times 4.18 \text{ J g}^{-1} \text{ K}^{-1} \times 6.8 \text{ K} = 2842.4 \text{ J} = 2.8424 \text{ kJ}. Number of moles of water formed = moles of \text{HCl} used = 0.0500 \text{ dm}^{3} \times 1.00 \text{ mol dm}^{-3} = 0.0500 \text{ mol}. Enthalpy of neutralisation (\Delta H_n) = -q / n = -2.8424 \text{ kJ} / 0.0500 \text{ mol} = -56.8 \text{ kJ mol}^{-1}. Since neutralisation is exothermic, the sign must be negative.

Award 1 mark for the correct option C. Method: Calculate heat released using q = mc\Delta T, find moles of water formed, divide heat by moles, and apply correct sign.

When temperature increases, the average kinetic energy of the molecules increases. This causes the Maxwell–Boltzmann distribution curve to change in two key ways: the peak of the curve shifts to the right (higher energy) and downwards (to keep the total area under the curve constant, since the total number of molecules is unchanged). The activation energy, \(E_a\), of the reaction is a fixed constant for a given reaction and does not change with temperature.

1 mark for selecting correct option B.
- Option A is incorrect because activation energy does not decrease, and the peak shifts lower, not higher.
- Option C is incorrect because the total area under the curve represents the total number of molecules, which remains constant.
- Option D is incorrect because the activation energy remains constant.

The equation for the formation of propanoic acid is:
\(3C(s) + 3H_2(g) + O_2(g) \rightarrow CH_3CH_2CO_2H(l)\)

Using Hess's law and enthalpy changes of combustion:
\(\Delta H_f^\theta = \sum \Delta H_c^\theta(\text{reactants}) - \sum \Delta H_c^\theta(\text{products})\)
\(\Delta H_f^\theta = [3 \times \Delta H_c^\theta(C) + 3 \times \Delta H_c^\theta(H_2)] - [\Delta H_c^\theta(CH_3CH_2CO_2H)]\)
\(\Delta H_f^\theta = [3(-394) + 3(-286)] - [-1527]\)
\(\Delta H_f^\theta = [-1182 - 858] + 1527 = -2040 + 1527 = -513\text{ kJ mol}^{-1}\)

1 mark for selecting correct option A.
- Option B is incorrect due to a sign error (+513 instead of -513).
- Option C is incorrect due to incorrect stoichiometry in the cycle.
- Option D is incorrect because it is only the sum of reactant combustion enthalpies.

The properties described belong to phosphorus:
1. Phosphorus burns in oxygen to form phosphorus(V) oxide, \(P_4O_{10}\).
2. \(P_4O_{10}\) dissolves in water to form phosphoric(V) acid, \(H_3PO_4\), which is an acidic solution.
3. Being an acidic oxide, it reacts with bases like sodium hydroxide but does not react with acids like hydrochloric acid.
Sodium forms \(Na_2O\), which is a basic oxide. Aluminium forms \(Al_2O_3\), which is amphoteric (reacts with both acids and bases) and insoluble in water. Silicon forms \(SiO_2\), which is an acidic oxide but is completely insoluble in water.

1 mark for selecting correct option D.
- Option A is incorrect as sodium oxide is basic.
- Option B is incorrect as aluminium oxide is amphoteric and insoluble in water.
- Option C is incorrect as silicon dioxide is insoluble in water.

Nitrogen dioxide, \(NO_2\), acts as a catalyst in the atmospheric oxidation of sulfur dioxide to sulfur trioxide. The process is a two-step cycle:
1. \(SO_2(g) + NO_2(g) \rightarrow SO_3(g) + NO(g)\)
2. \(NO(g) + \frac{1}{2}O_2(g) \rightarrow NO_2(g)\)

- Option A is incorrect because \(NO\) in internal combustion engines is formed by the direct reaction of atmospheric nitrogen and oxygen at high temperatures, not from fuel-bound nitrogen.
- Option B is incorrect because \(SO_2\) is oxidized to \(SO_3\), meaning it acts as a reducing agent, not an oxidizing agent.
- Option D is incorrect because direct oxidation of \(SO_2\) by atmospheric oxygen is extremely slow without catalysts.

1 mark for selecting correct option C.
- Option A is incorrect (origin of nitrogen oxides in engines is atmospheric nitrogen).
- Option B is incorrect (sulfur dioxide is oxidized, so it is a reducing agent).
- Option D is incorrect (the direct uncatalysed atmospheric reaction is too slow).

1. The molecular formula \(C_4H_{10}O\) corresponds to an alcohol.
2. Acidified potassium dichromate(VI) remaining orange indicates that the alcohol is resistant to oxidation. This identifies \(W\) as a tertiary alcohol. The only tertiary alcohol with four carbon atoms is 2-methylpropan-2-ol.
3. Dehydration of 2-methylpropan-2-ol with concentrated sulfuric acid yields 2-methylpropene. Since one of the carbon atoms in the double bond is bonded to two identical methyl groups, 2-methylpropene cannot exhibit stereoisomerism (cis-trans isomerism), confirming the identity.

1 mark for selecting correct option D.
- Option A is incorrect because butan-1-ol is primary and would be oxidized (color change to green).
- Option B is incorrect because butan-2-ol is secondary and would be oxidized, plus its dehydration products include stereoisomers (but-2-ene).
- Option C is incorrect because 2-methylpropan-1-ol is primary and would be oxidized.

During the propagation phase of free-radical substitution of propane:
1. A chlorine radical abstracts a hydrogen atom from propane to form a propyl radical and hydrogen chloride:
\(CH_3CH_2CH_3 + Cl\bullet \rightarrow CH_3CH_2\dot{C}H_2 + HCl\) (or the isopropyl radical \(CH_3\dot{C}HCH_3\)).
2. The propyl radical then reacts with a chlorine molecule to form the halogenoalkane product and regenerate the chlorine radical:
\(CH_3CH_2\dot{C}H_2 + Cl_2 \rightarrow CH_3CH_2CH_2Cl + Cl\bullet\).

Therefore, \(CH_3CH_2\dot{C}H_2\) is a propagation radical intermediate.
- \(Cl^-\) is a chloride ion (not a radical).
- \(H\bullet\) is not formed in this reaction because the C-H bond cleavage results in hydrogen combining with chlorine to form stable HCl.
- \(CH_3CH_2CH_2^+\) is a carbocation, which is not involved in free-radical mechanisms.

1 mark for selecting correct option B.
- Option A is incorrect as it is an ion, not a radical.
- Option C is incorrect as hydrogen atoms are not formed as free radicals in this mechanism.
- Option D is incorrect as it is a carbocation.

The mechanism of this homogeneous catalytic pathway involves two steps:
Step 1: The oxidation of \(Fe^{2+}\) to \(Fe^{3+}\) by peroxodisulfate ions:
\(S_2O_8^{2-} + 2Fe^{2+} \rightarrow 2SO_4^{2-} + 2Fe^{3+}\)

Step 2: The subsequent reduction of \(Fe^{3+}\) back to \(Fe^{2+}\) by iodide ions, which are oxidized to iodine:
\(2Fe^{3+} + 2I^- \rightarrow 2Fe^{2+} + I_2\)

This confirms that \(Fe^{3+}\) is formed as an intermediate and oxidizes iodide ions to iodine (Option C). Since all reactants and the catalyst are in the aqueous phase, this is homogeneous catalysis (making Option B incorrect). \(Fe^{2+}\) oxidizes, not reduces, the other species in the first step (making Option A incorrect). The catalyst lowers the activation energy by offering an alternative pathway, not by changing collision frequencies (making Option D incorrect).

1 mark for selecting correct option C.
- Option A is incorrect because \(Fe^{2+}\) reduces peroxodisulfate, not iodide.
- Option B is incorrect because this is a homogeneous catalysis reaction.
- Option D is incorrect because catalysts do not lower the activation energy by changing the physical collision frequency of the original uncatalysed pathway.

- \(BF_3\): Boron has 3 valence electrons and forms 3 single covalent bonds with fluorine with 0 lone pairs. This gives a trigonal planar shape with bond angles of exactly \(120^\circ\).
- \(ClF_3\): Chlorine has 7 valence electrons, forms 3 bonding pairs and has 2 lone pairs. This is a T-shaped molecule with bond angles of less than \(90^\circ\) (approx. \(87.5^\circ\)).
- \(NF_3\): Nitrogen has 5 valence electrons, forms 3 bonding pairs and has 1 lone pair. This is a trigonal pyramidal molecule with a bond angle of approximately \(102^\circ\).
- \(SO_2\): Sulfur has 6 valence electrons, forms 2 double bonds (or equivalent) and has 1 lone pair. This gives a bent shape with a bond angle of approximately \(119^\circ\) (less than \(120^\circ\) due to lone pair-bonding pair repulsion).

1 mark for selecting correct option B.
- Option A is incorrect as \(ClF_3\) is T-shaped.
- Option C is incorrect as \(NF_3\) is trigonal pyramidal.
- Option D is incorrect as \(SO_2\) has a lone pair which compresses the angle to slightly less than \(120^\circ\).

A catalyst provides an alternative reaction pathway with a lower activation energy. Therefore, the value of the activation energy, \(E_{\text{a}}\), shifts to the left on the energy axis. Since the temperature of the gas remains constant, the Maxwell-Boltzmann distribution curve representing the molecular energies of the gas particles remains entirely unchanged.

1 mark: Correctly identify that introducing a catalyst decreases the activation energy (shifts \(E_{\text{a}}\) to the left) without modifying the shape of the molecular energy distribution curve itself.

The skeletal structure of 2-methylpropane is \((CH_3)_3CH\). It contains nine primary hydrogen atoms (in the three methyl groups) and one tertiary hydrogen atom (attached to the central carbon). Replacing any of the nine primary hydrogen atoms yields 1-chloro-2-methylpropane, while replacing the single tertiary hydrogen atom yields 2-chloro-2-methylpropane. Under the assumption that all hydrogen atoms are equally reactive, the yield of each product is directly proportional to the number of hydrogen atoms available for substitution, giving an expected ratio of \(9:1\).

1 mark: Identify the number of primary (9) and tertiary (1) hydrogen atoms in 2-methylpropane and deduce the ratio of the corresponding monochloroalkane products as \(9:1\).

The equation for the formation of liquid ethanol is:
\(2C(\text{s}) + 3H_2(\text{g}) + \frac{1}{2}O_2(\text{g}) \rightarrow C_2H_5OH(\text{l})\)

According to Hess's Law, using the enthalpy changes of combustion:
\(\Delta H_{\text{f}}^{\ominus}[C_2H_5OH(\text{l})] = 2 \Delta H_{\text{c}}^{\ominus}[C(\text{s})] + 3 \Delta H_{\text{c}}^{\ominus}[H_2(\text{g})] - \Delta H_{\text{c}}^{\ominus}[C_2H_5OH(\text{l})]\)

Substitute the given values:
\(\Delta H_{\text{f}}^{\ominus} = 2(-394) + 3(-286) - (-1367)\)
\(\Delta H_{\text{f}}^{\ominus} = -788 - 858 + 1367 = -1646 + 1367 = -279\text{ kJ mol}^{-1}\)

1 mark: Correctly apply Hess's Law to relate enthalpy of formation to combustion data, and compute the value of \(-279\text{ kJ mol}^{-1}\).

Phosphorus(V) oxide and sulfur trioxide are acidic non-metal oxides. When reacted with excess water, they undergo vigorous reactions to form strongly acidic solutions: \(P_4O_{10}(\text{s}) + 6H_2O(\text{l}) \rightarrow 4H_3PO_4(\text{aq})\) (phosphoric(V) acid) and \(SO_3(\text{g}) + H_2O(\text{l}) \rightarrow H_2SO_4(\text{aq})\) (sulfuric(VI) acid). Both solutions have low pH values.

1 mark: Identify that both Period 3 oxides dissolve and react with water to form strongly acidic solutions of \(H_3PO_4\) and \(H_2SO_4\).

The atmospheric oxidation of sulfur dioxide is catalyzed by nitrogen oxides via a two-step mechanism:
1) Oxidation of nitrogen monoxide: \(2NO(\text{g}) + O_2(\text{g}) \rightarrow 2NO_2(\text{g})\)
2) Oxidation of sulfur dioxide by nitrogen dioxide: \(NO_2(\text{g}) + SO_2(\text{g}) \rightarrow NO(\text{g}) + SO_3(\text{g})\)
In this second step, sulfur dioxide is oxidized to sulfur trioxide while nitrogen dioxide is reduced back to nitrogen monoxide, which can repeat the cycle.

1 mark: Identify the correct equation showing the role of \(NO_2\) in oxidizing \(SO_2\) to \(SO_3\) with the regeneration of \(NO\).

Let's analyze the products of dehydration for each option:
- Pentan-1-ol yields only pent-1-ene (1 alkene).
- Pentan-2-ol, \(CH_3CH_2CH_2CH(OH)CH_3\), can undergo elimination of hydrogen from C1 or C3. Eliminating from C1 yields pent-1-ene. Eliminating from C3 yields pent-2-ene. Since pent-2-ene contains a double bond where each carbon has different groups attached, it exhibits cis-trans stereoisomerism, giving cis-pent-2-ene and trans-pent-2-ene. Therefore, pentan-2-ol produces exactly three isomeric alkenes in total.
- 2-Methylbutan-2-ol yields 2-methylbut-1-ene and 2-methylbut-2-ene (neither has stereoisomers, so only 2 alkenes).
- 3-Methylbutan-2-ol yields 3-methylbut-1-ene and 2-methylbut-2-ene (neither has stereoisomers, so only 2 alkenes).

1 mark: Identify that dehydration of pentan-2-ol yields pent-1-ene, cis-pent-2-ene, and trans-pent-2-ene, totaling three isomers.

Let's determine the electronic configuration of each ion:
- Cr atom has configuration \([\text{Ar}] 3\text{d}^5 4\text{s}^1\). \(\text{Cr}^{3+}\) is formed by removing three electrons (one 4s and two 3d), giving \([\text{Ar}] 3\text{d}^3\).
- Mn atom has configuration \([\text{Ar}] 3\text{d}^5 4\text{s}^2\). \(\text{Mn}^{2+}\) is formed by removing the two 4s valence electrons first, giving \([\text{Ar}] 3\text{d}^5\).
- Fe atom has configuration \([\text{Ar}] 3\text{d}^6 4\text{s}^2\). \(\text{Fe}^{2+}\) is formed by removing the two 4s valence electrons, giving \([\text{Ar}] 3\text{d}^6\).
- Co atom has configuration \([\text{Ar}] 3\text{d}^7 4\text{s}^2\). \(\text{Co}^{3+}\) is formed by removing two 4s and one 3d electrons, giving \([\text{Ar}] 3\text{d}^6\).

1 mark: Identify that \(\text{Mn}^{2+}\) has the correct electronic configuration of \([\text{Ar}] 3\text{d}^5\) by first removing 4s electrons.

First, find the order of reaction with respect to each reactant:
- Doubling \([P]\) quadruples the rate: \(2^x = 4 \Rightarrow x = 2\) (second order with respect to \(P\)).
- Doubling \([Q]\) doubles the rate: \(2^y = 2 \Rightarrow y = 1\) (first order with respect to \(Q\)).

The rate equation is: \(\text{rate} = k[P]^2[Q]\).

When both concentrations are tripled:
\(\text{new rate} = k(3[P])^2(3[Q]) = 9 \times 3 \times k[P]^2[Q] = 27 \times \text{original rate}\).
Thus, the rate increases by a factor of 27.

1 mark: Deduce the reaction orders (2 for P and 1 for Q) and calculate the combined effect of tripling both concentrations to find the factor of 27.

A catalyst provides an alternative reaction pathway with a lower activation energy, so the activation energy barrier decreases (shifts to the left). However, because the temperature is kept constant at \(300\text{ K}\), the average kinetic energy of the molecules does not change, meaning the shape and position of the peak of the Boltzmann distribution curve remain completely unchanged. Therefore, the distribution curve is identical, but the activation energy threshold is lower.

1 mark for the correct option (B).
- Reject A: The Boltzmann distribution peak only shifts to higher energy when temperature increases.
- Reject C: The activation energy must decrease in the presence of a catalyst.
- Reject D: The Boltzmann peak does not shift to a lower energy.

By Hess's Law, the standard enthalpy change of combustion can be calculated from standard enthalpies of formation:
\(\Delta H_c^\ominus = \sum \Delta H_f^\ominus(\text{products}) - \sum \Delta H_f^\ominus(\text{reactants})\)

For the reaction:
\(\text{N}_2\text{H}_4(\text{l}) + \text{O}_2(\text{g}) \rightarrow \text{N}_2(\text{g}) + 2\text{H}_2\text{O}(\text{l})\)

We have:
\(-622 = [\Delta H_f^\ominus(\text{N}_2(\text{g})) + 2 \times \Delta H_f^\ominus(\text{H}_2\text{O}(\text{l}))] - [\Delta H_f^\ominus(\text{N}_2\text{H}_4(\text{l})) + \Delta H_f^\ominus(\text{O}_2(\text{g}))]\)

Since \(\Delta H_f^\ominus\) for elements in their standard states is zero:
\(-622 = [0 + 2(-286)] - [\Delta H_f^\ominus(\text{N}_2\text{H}_4(\text{l})) + 0]\)
\(-622 = -572 - \Delta H_f^\ominus(\text{N}_2\text{H}_4(\text{l}))\)
\(\Delta H_f^\ominus(\text{N}_2\text{H}_4(\text{l})) = -572 - (-622) = +50\text{ kJ mol}^{-1}\)

1 mark for the correct option (B).
- Award 1 mark for correct Hess's Law application and calculation resulting in +50 kJ mol^-1.
- Reject A (-50 kJ mol^-1) due to sign error.
- Reject C and D (-908 and +908 kJ mol^-1) due to incorrect multipliers or cycle construction.

Magnesium (Mg) burns with a characteristic bright white flame to form magnesium oxide (MgO), which is a white solid. MgO is sparingly soluble in water and forms a basic suspension of magnesium hydroxide, \(\text{Mg(OH)}_2\), which has a pH of approximately 10. Sodium burns with a yellow flame and forms a highly soluble oxide that yields a pH of 14. Aluminium oxide is insoluble in water. Phosphorus burns to form an oxide that dissolves to form an acidic solution (pH 1-2).

1 mark for the correct option (B).
- Reject A: Sodium burns with a yellow flame, and its oxide forms a highly alkaline solution (pH 14).
- Reject C: Aluminium oxide is insoluble and does not change the pH of water.
- Reject D: Phosphorus oxide reacts with water to form an acidic solution (pH 1-2).

In the atmosphere, nitrogen dioxide (\(\text{NO}_2\)) oxidizes sulfur dioxide (\(\text{SO}_2\)) to sulfur trioxide (\(\text{SO}_3\)), being reduced to nitrogen monoxide (\(\text{NO}\)) in the process: \(\text{SO}_2 + \text{NO}_2 \rightarrow \text{SO}_3 + \text{NO}\). The \(\text{NO}\) is then reoxidized back to \(\text{NO}_2\) by atmospheric oxygen (\(2\text{NO} + \text{O}_2 \rightarrow 2\text{NO}_2\)), regenerating the catalyst.

1 mark for the correct option (B).
- Reject A: This is not a balanced or valid catalytic step.
- Reject C: This is the reverse of the actual oxidation step.
- Reject D: This represents thermal decomposition of nitrogen dioxide, not the pathway for catalytic oxidation of SO2.

1. Compound \(\text{W}\) is an alcohol with molecular formula \(\text{C}_4\text{H}_{10}\text{O}\).
2. Since it does not react with acidified potassium dichromate(VI) (the solution remains orange), \(\text{W}\) must be a tertiary alcohol.
3. The only tertiary alcohol with four carbon atoms is 2-methylpropan-2-ol, \((\text{CH}_3)_3\text{COH}\).
4. Dehydration of 2-methylpropan-2-ol with concentrated \(\text{H}_2\text{SO}_4\) yields 2-methylpropene, \((\text{CH}_3)_2\text{C}=\text{CH}_2\).
5. 2-Methylpropene does not exhibit stereoisomerism because one double-bonded carbon has two identical methyl groups, and the other has two identical hydrogen atoms. Therefore, only 1 alkene product is formed.

1 mark for the correct option (A).
- Correctly identifying W as 2-methylpropan-2-ol: 0.5 marks.
- Correctly determining that only 1 alkene (2-methylpropene) is formed with no cis-trans isomers: 0.5 marks.

To find the number of structurally isomeric dichloropropanes, we look at the possible positions for the two chlorine atoms on a propane skeleton (\(\text{C}-\text{C}-\text{C}\)):
1. Both chlorines on carbon-1: 1,1-dichloropropane (\(\text{CH}_3-\text{CH}_2-\text{CHCl}_2\))
2. One chlorine on carbon-1 and one on carbon-2: 1,2-dichloropropane (\(\text{CH}_3-\text{CHCl}-\text{CH}_2\text{Cl}\))
3. One chlorine on carbon-1 and one on carbon-3: 1,3-dichloropropane (\(\text{CH}_2\text{Cl}-\text{CH}_2-\text{CH}_2\text{Cl}\))
4. Both chlorines on carbon-2: 2,2-dichloropropane (\(\text{CH}_3-\text{CCl}_2-\text{CH}_3\))

This gives exactly 4 structural isomers.

1 mark for the correct option (C).
- Reject A, B, D: Incorrect counts of structural isomers. Stereoisomers are excluded by the term 'structurally isomeric'.

Group 2 nitrates and carbonates both undergo thermal decomposition when heated strongly.
1. Calcium nitrate decomposes to form calcium oxide (solid), nitrogen dioxide (gas), and oxygen (gas):
\[2\text{Ca(NO}_3)_2(\text{s}) \rightarrow 2\text{CaO}(\text{s}) + 4\text{NO}_2(\text{g}) + \text{O}_2(\text{g})\]
2. Calcium carbonate decomposes to form calcium oxide (solid) and carbon dioxide (gas):
\[\text{CaCO}_3(\text{s}) \rightarrow \text{CaO}(\text{s}) + \text{CO}_2(\text{g})\]

Therefore, the gases evolved are \(\text{CO}_2\), \(\text{NO}_2\), and \(\text{O}_2\).

1 mark for the correct option (D).
- Reject A, B, C: These options fail to include one or more of the three evolved gases.

Let's determine the shape and bond angles using VSEPR theory:
- \(\text{NH}_2^-\): Nitrogen has 2 bonding pairs and 2 lone pairs. This is a non-linear (bent) shape with a bond angle of approximately \(104.5^\circ\) due to strong lone pair-lone pair repulsion.
- \(\text{NH}_3\): Nitrogen has 3 bonding pairs and 1 lone pair. This is a trigonal pyramidal shape with a bond angle of approximately \(107^\circ\).
- \(\text{NH}_4^+\): Nitrogen has 4 bonding pairs and 0 lone pairs. This is a regular tetrahedral shape with a bond angle of \(109.5^\circ\).
- \(\text{BF}_3\): Boron has 3 bonding pairs and 0 lone pairs. This is a trigonal planar shape with a bond angle of \(120^\circ\).

Thus, \(\text{NH}_2^-\) has the smallest bond angle.

1 mark for the correct option (A).
- Reject B (107 degrees), C (109.5 degrees), and D (120 degrees) as they all have larger bond angles than NH2^- (104.5 degrees).

At a higher temperature, the activation energy, \(E_a\), of the reaction remains unchanged, so option A is incorrect. The collision frequency increases only slightly (by approximately \(1.6\%\) as it is proportional to \(\sqrt{T}\)), making option B incorrect. The average kinetic energy of the reactant molecules is directly proportional to the temperature in Kelvin, so it increases by only about \(3.3\%\) (\(310/300\)), making option C incorrect. The main reason for the exponential increase in reaction rate is that the Boltzmann distribution curve shifts, significantly increasing the fraction of molecules with kinetic energy greater than or equal to the activation energy (\(E \ge E_a\)), leading to a much higher frequency of successful collisions.

1 mark for selecting the correct explanation of temperature effect on rates (D).

Using Hess's Law, we can construct an enthalpy cycle where both the anhydrous and hydrated salts dissolve in excess water to form the same final aqueous solution:

\(\Delta H_{\text{hydration}} + \Delta H_{\text{solution}}(\text{CaCl}_2\cdot 6\text{H}_2\text{O}\text{(s)}) = \Delta H_{\text{solution}}(\text{CaCl}_2\text{(s)})\)

Substituting the values:

\(\Delta H_{\text{hydration}} + (+19.1) = -82.9\)

\(\Delta H_{\text{hydration}} = -82.9 - 19.1 = -102.0\text{ kJ mol}^{-1}\).

1 mark for setting up the Hess cycle correctly and obtaining the correct hydration enthalpy value (A).

In the free-radical substitution of ethane, ethyl radicals (\(\text{C}_2\text{H}_5^\bullet\)) are formed during the propagation steps. During the termination phase, two ethyl radicals can collide and combine to form butane: \(\text{C}_2\text{H}_5^\bullet + \text{C}_2\text{H}_5^\bullet \rightarrow \text{C}_4\text{H}_{10}\). Butane is a hydrocarbon with a higher molecular mass (\(M_r = 58\)) than the reactant ethane (\(M_r = 30\)). Option A is incorrect because hydrogen radicals do not feature in this mechanism. Option B produces ethane, which does not have a higher molecular mass. Option D produces chloroethane, which is a halogenoalkane rather than a hydrocarbon.

1 mark for identifying the correct termination step that produces a higher molecular mass hydrocarbon (C).

\(\text{NaCl}\) is an ionic chloride that simply dissolves in water to form hydrated ions, without undergoing hydrolysis; hence, the solution remains neutral (pH 7). \(\text{MgCl}_2\) contains the magnesium ion which has a relatively high charge density, leading to very slight hydrolysis and forming a weakly acidic solution with a pH of approximately 6.5. \(\text{SiCl}_4\) is a covalent chloride that reacts vigorously and completely with water to undergo complete hydrolysis, producing hydrochloric acid (\(\text{HCl}\)) and silicon dioxide (\(\text{SiO}_2\)); this forms a strongly acidic solution with a pH around 2.

1 mark for correctly identifying the pH of all three Period 3 chloride solutions (B).

Inside a catalytic converter, nitrogen oxides (such as \(\text{NO}\) or \(\text{NO}_2\)) are reduced to harmless \(\text{N}_2\) gas by reacting with reducing agents like carbon monoxide (CO) or unburnt hydrocarbons (such as \(\text{CH}_4\)). Therefore, reactions A, C, and D occur on the catalyst surface. Reaction B, however, is the oxidation of atmospheric sulfur dioxide by nitrogen dioxide, which acts as a catalyst in the troposphere for the formation of acid rain; it does not take place within the vehicle's catalytic converter.

1 mark for identifying the atmospheric reaction that does not occur in a catalytic converter (B).

Since compound X is oxidized by acidified dichromate(VI), it cannot be 2-methylpropan-2-ol, which is a tertiary alcohol (D). The oxidation product Y does not react with Tollens' reagent, which indicates that Y is a ketone, not an aldehyde. This rules out the primary alcohols butan-1-ol (A) and 2-methylpropan-1-ol (C), which oxidize to aldehydes. Therefore, X must be the secondary alcohol, butan-2-ol, which is oxidized to butanone (Y). Butanone contains a methyl ketone group (\(\text{CH}_3\text{CO}-\)) and thus reacts with alkaline aqueous iodine to form a yellow precipitate of tri-iodomethane.

1 mark for correctly determining the structure of X from the chemical properties of its oxidation product Y (B).

The catalysis of this reaction by \(\text{Fe}^{3+}\) is homogeneous. The mechanism consists of two redox steps. Because both steps involve reactions between oppositely charged ions, they have lower activation energies than the direct reaction:

Step 1: \(2\text{Fe}^{3+}\text{(aq)} + 2\text{I}^-\text{(aq)} \rightarrow 2\text{Fe}^{2+}\text{(aq)} + \text{I}_2\text{(aq)}\) (where \(\text{Fe}^{3+}\) oxidizes \(\text{I}^-\))

Step 2: \(2\text{Fe}^{2+}\text{(aq)} + \text{S}_2\text{O}_8^{2-}\text{(aq)} \rightarrow 2\text{Fe}^{3+}\text{(aq)} + 2\text{SO}_4^{2-}\text{(aq)}\) (where \(\text{Fe}^{2+}\) reduces \(\text{S}_2\text{O}_8^{2-}\) and is regenerated back to \(\text{Fe}^{3+}\)).

This perfectly matches option A.

1 mark for identifying the correct redox steps in the homogeneous catalysis mechanism (A).

During nucleophilic hydrolysis of halogenoalkanes, the carbon-halogen bond must be broken. The ease of breaking this bond is determined by the bond enthalpy (bond strength) rather than bond polarity. The average bond enthalpies decrease in the order \(\text{C}-\text{Cl} > \text{C}-\text{Br} > \text{C}-\text{I}\). Because the \(\text{C}-\text{I}\) bond is the weakest, it is broken most easily, making 1-iodobutane undergo hydrolysis at the fastest rate to form a yellow silver iodide precipitate. 1-chlorobutane has the strongest bond and reacts the slowest. Therefore, the rate increases in the order 1-chlorobutane < 1-bromobutane < 1-iodobutane due to the decreasing C-halogen bond strength.

1 mark for identifying both the correct order of hydrolysis and the bond-strength justification (A).

(a) Activation energy is the minimum energy colliding particles must possess [1] for a collision to result in a successful reaction [1]. (b) (i) Both \(\text{S}_2\text{O}_8^{2-}\) and \(\text{I}^-\right.\) are negatively charged anions [1]. They repel each other, which leads to a very high activation energy [1]. (ii) Step 1: \(\text{S}_2\text{O}_8^{2-} + 2\text{Fe}^{2+} \rightarrow 2\text{SO}_4^{2-} + 2\text{Fe}^{3+}\) [1]. Step 2: \(2\text{Fe}^{3+} + 2\text{I}^- \rightarrow 2\text{Fe}^{2+} + \text{I}_2\) [1]. (iii) Diagram requirements: y-axis labeled 'Energy' and x-axis labeled 'Progress of reaction' [1]. Uncatalyzed pathway shows a single high curve starting at reactants and ending at products [1]. Catalyzed pathway shows a two-step mechanism (two smaller peaks/humps) with lower activation energies, starting and ending at the same energy levels as the uncatalyzed pathway [1]. (c) (i) Rate = \(k[\text{S}_2\text{O}_8^{2-}][\text{I}^-]\) [1]. (ii) Rearranging rate equation: \(k = \text{Rate} / ([\text{S}_2\text{O}_8^{2-}][\text{I}^-])\) [1]. Substituting units: \(k = (\text{mol dm}^{-3} \text{s}^{-1}) / (\text{mol dm}^{-3} \times \text{mol dm}^{-3}) = \text{dm}^3 \text{mol}^{-1} \text{s}^{-1}\) [1].

(a) M1: Minimum energy colliding particles must possess. M2: In order to react / undergo a successful collision. (b) (i) M1: Both reactant species are negatively charged ions / anions. M2: Like charges repel each other, resulting in a high activation energy. (ii) M1: Balanced equation for reduction of peroxodisulfate by iron(II). M2: Balanced equation for oxidation of iodide by iron(III). (iii) M1: Correctly labeled axes (Energy on y-axis, Progress/Coordinate on x-axis) and reactants/products. M2: Uncatalyzed path represented by a single high barrier. M3: Catalyzed path represented by a lower energy pathway featuring two transition states/peaks. (c) (i) M1: Correct rate equation containing rate constant, peroxodisulfate concentration, and iodide concentration. (ii) M1: Shows substitution of units into the rearranged rate equation. M2: Correct units: dm3 mol-1 s-1 (accept in any order).

(a) The enthalpy change when 1 mole of a compound [1] is formed from its constituent elements [1] under standard conditions (298 K and 100 kPa / 1 bar) with all reactants and products in their standard states [1]. (b) \(3\text{C}(\text{s}) + 3\text{H}_2(\text{g}) + \frac{1}{2}\text{O}_2(\text{g}) \rightarrow \text{CH}_3\text{COCH}_3(\text{l})\). Correct formulas and balancing [1], correct state symbols [1]. (c) The Hess cycle should show: Top-left: \(3\text{C}(\text{s}) + 3\text{H}_2(\text{g}) + \frac{1}{2}\text{O}_2(\text{g})\) (or just elements); Top-right: \(\text{CH}_3\text{COCH}_3(\text{l})\); Bottom: \(3\text{CO}_2(\text{g}) + 3\text{H}_2\text{O}(\text{l})\). Arrow from Elements to Propanone labeled \(\Delta H_f^\ominus\) [1]. Arrow from Elements to Combustion Products labeled \(3\Delta H_c^\ominus[\text{C}] + 3\Delta H_c^\ominus[\text{H}_2]\) [1]. Arrow from Propanone to Combustion Products labeled \(\Delta H_c^\ominus[\text{propanone}]\) [1]. (d) From Hess's Law: \(\Delta H_f^\ominus + \Delta H_c^\ominus[\text{propanone}] = 3\Delta H_c^\ominus[\text{C}] + 3\Delta H_c^\ominus[\text{H}_2]\) [1] \(\Delta H_f^\ominus = 3(-393.5) + 3(-285.8) - (-1786.0)\) [1] \(\Delta H_f^\ominus = -1180.5 - 857.4 + 1786.0 = -2037.9 + 1786.0 = -251.9 \text{ kJ mol}^{-1}\) [1]. (e) Gaseous propanone has a less exothermic enthalpy of formation because energy is required (endothermic process) to convert liquid propanone to gaseous propanone / intermolecular forces must be overcome [1].

(a) M1: Enthalpy change when 1 mole of a substance is formed. M2: From its constituent elements. M3: Under standard conditions (298 K, 1 atm/100 kPa) with elements and products in standard states. (b) M1: 3C + 3H2 + 1/2O2 -> CH3COCH3. M2: All state symbols correct: C(s), H2(g), O2(g), CH3COCH3(l). (c) M1: Elements and propanone in top tier, and 3CO2 + 3H2O in bottom tier. M2: Arrow from elements to propanone labeled DHf and arrow from elements to combustion products labeled with correct multipliers. M3: Arrow from propanone to combustion products labeled DHc(propanone). (d) M1: Hess's Law equation: DHf = 3DHc(C) + 3DHc(H2) - DHc(propanone). M2: Correct substitution of values: 3(-393.5) + 3(-285.8) - (-1786.0). M3: Correct final answer: -251.9 kJ mol-1 (must include minus sign and correct units). (e) M1: Less exothermic because vaporisation/conversion from liquid to gas is an endothermic process.

(a) (i) Type of reaction: Substitution [1]. Mechanism: Free-radical [1]. (ii) Initiation: \(\text{Cl}_2 \rightarrow 2\text{Cl}^\bullet\) [1]. Propagation step 1: \(\text{CH}_3\text{CH}_2\text{CH}_2\text{CH}_3 + \text{Cl}^\bullet \rightarrow \text{CH}_3\text{CH}_2\text{CH}_2\text{CH}_2^\bullet + \text{HCl}\) [1]. Propagation step 2: \(\text{CH}_3\text{CH}_2\text{CH}_2\text{CH}_2^\bullet + \text{Cl}_2 \rightarrow \text{CH}_3\text{CH}_2\text{CH}_2\text{CH}_2\text{Cl} + \text{Cl}^\bullet\) [1]. (iii) Reason 1: Further substitution can occur, replacing more hydrogen atoms to form di-, tri-, or poly-chlorinated alkanes [1]. Reason 2: Chlorine radicals can attack different carbon atoms along the chain, forming positional isomers (e.g., 2-chlorobutane as well as 1-chlorobutane) [1]. (b) (i) \(\text{C}_4\text{H}_{10} + 6.5\text{O}_2 \rightarrow 4\text{CO}_2 + 5\text{H}_2\text{O}\) (or doubled: \(2\text{C}_4\text{H}_{10} + 13\text{O}_2 \rightarrow 8\text{CO}_2 + 10\text{H}_2\text{O}\)) [1]. (ii) Carbon monoxide binds irreversibly / very strongly to the haemoglobin in red blood cells [1], preventing it from binding and transporting oxygen around the body [1]. (c) The C-H and C-C bonds in alkanes are very strong / have high bond energies, meaning they require a lot of energy to break [1]. Furthermore, carbon and hydrogen have very similar electronegativities, making these bonds non-polar and thus less susceptible to attack by polar reagents [1].

(a) (i) M1: Substitution. M2: Free radical. (ii) M1: Correct initiation step showing homolytic fission of Cl2 with radical dot on Cl. M2: First propagation step showing formation of butyl radical and HCl. M3: Second propagation step showing reaction of butyl radical with Cl2 to form 1-chlorobutane and regenerate Cl radical. (iii) M1: Multiple substitution / further substitution occurs (forming dichlorobutanes, etc.). M2: Positional isomerism / substitution can occur on carbon-2 (forming 2-chlorobutane). (b) (i) M1: Correctly balanced equation for complete combustion of butane (fractional coefficients accepted). (ii) M1: CO binds irreversibly / coordinates strongly to haemoglobin. M2: Reduces the oxygen-carrying capacity of blood. (c) M1: C-C and C-H bonds are strong / have high bond enthalpies. M2: Electronegativities of C and H are very similar, so the bonds are non-polar.

(a) (i) Nitrogen molecules have a triple covalent bond between the nitrogen atoms (\(\text{N}\equiv\text{N}\)) [1]. This bond has an extremely high bond energy and requires a large amount of energy (high temperature) to break [1]. (ii) Any one of: causes acid rain / photochemical smog / respiratory problems [1]. (iii) \(2\text{CO} + 2\text{NO} \rightarrow 2\text{CO}_2 + \text{N}_2\) [1]. (b) (i) Step 1: \(\text{SO}_2 + \text{NO}_2 \rightarrow \text{SO}_3 + \text{NO}\) [1]. Step 2: \(2\text{NO} + \text{O}_2 \rightarrow 2\text{NO}_2\) (or \(\text{NO} + \frac{1}{2}\text{O}_2 \rightarrow \text{NO}_2\)) [1]. (ii) Homogeneous catalysis [1] because both the reactants (\(\text{SO}_2\) and \(\text{O}_2\)) and the catalyst (\(\text{NO}_2\)) are in the same physical state / are all gases [1]. (c) (i) \(\text{SO}_3 + \text{H}_2\text{O} \rightarrow \text{H}_2\text{SO}_4\) [1]. (ii) \(\text{CaCO}_3 + \text{H}_2\text{SO}_4 \rightarrow \text{CaSO}_4 + \text{CO}_2 + \text{H}_2\text{O}\) [1]. (iii) Any two of: Acidification of lakes/rivers, killing aquatic life [1]; Leaching of toxic metal ions (e.g., \(\text{Al}^{3+}\)) from soils into water bodies [1]; Damaging leaves/foliage on trees and crops [1].

(a) (i) M1: Nitrogen has a triple covalent bond. M2: Very high bond energy, requiring substantial energy input to break. (ii) M1: Acid rain / photochemical smog / respiratory irritation. (iii) M1: Correctly balanced equation: 2CO + 2NO -> 2CO2 + N2. (b) (i) M1: SO2 + NO2 -> SO3 + NO. M2: 2NO + O2 -> 2NO2 (or NO + 0.5O2 -> NO2). (ii) M1: Homogeneous (catalysis). M2: Catalyst and reactants are in the same physical state / same phase / all are gases. (c) (i) M1: SO3 + H2O -> H2SO4. (ii) M1: CaCO3 + H2SO4 -> CaSO4 + CO2 + H2O. (iii) M1: Acidifies lakes / harms aquatic life. M2: Damages trees/crops / leaches toxic ions from soil (accept any two distinct effects, 1 mark each).

(a) (i) Skeletal structure of butan-1-ol: a zig-zag chain of 4 carbon vertices with -OH attached to terminal carbon [1]. Skeletal structure of 2-methylpropan-2-ol: 3 methyl groups attached to a central carbon which is also bonded to an -OH group [1]. (ii) Isomer name: butan-2-ol [1]. 3D representation showing the central carbon bonded to -H, -OH, -CH3, and -CH2CH3, using wedge-and-dash representation [1]. Mirror image drawn correctly with corresponding groups aligned [1]. (b) (i) \(\text{CH}_3\text{CH}_2\text{CH}_2\text{COOH}\) (butanoic acid) [1]. (ii) \(\text{CH}_3\text{COCH}_2\text{CH}_3\) (butanone) [1]. (iii) Observation: The solution remains orange / no color change is seen [1]. Explanation: 2-Methylpropan-2-ol is a tertiary alcohol and cannot be oxidized because the carbon atom bonded to the -OH group is not bonded to any hydrogen atoms [1]. (c) (i) Reagent: Concentrated sulfuric acid (\(\text{H}_2\text{SO}_4\)) or concentrated phosphoric acid (\(\text{H}_3\text{PO}_4\)) [1]. Conditions: Heat (at approx. 170 °C) [1]. (Alternative: Reagent: Aluminium oxide (\(\text{Al}_2\text{O}_3\)) [1]. Conditions: Heat/high temperature [1]). (ii) \(\text{CH}_3\text{CH}_2\text{CH}=\text{CH}_2\) (but-1-ene) [1].

(a) (i) M1: Correct skeletal formula of butan-1-ol. M2: Correct skeletal formula of 2-methylpropan-2-ol. (ii) M1: Identifies the isomer as butan-2-ol. M2: Draws one 3D tetrahedral representation of the chiral center with 4 different groups (H, OH, CH3, C2H5) using wedges/dashes. M3: Draws the second 3D enantiomer as a clear mirror image of the first. (b) (i) M1: CH3CH2CH2COOH (or C3H7COOH). (ii) M1: CH3COCH2CH3 (or C2H5COCH3). (iii) M1: Remains orange / no reaction. M2: It is a tertiary alcohol, which does not have a hydrogen on the C-OH carbon. (c) (i) M1: Concentrated H2SO4 or concentrated H3PO4 (accept hot Al2O3). M2: Heat (or high temperature for Al2O3). (ii) M1: CH3CH2CH=CH2.

1. Average titre from concordant results is \(21.50\text{ cm}^3\). 2. Moles of \(\text{NaOH} = 0.120 \times (21.50 / 1000) = 2.58 \times 10^{-3}\text{ mol}\). 3. Since the reaction equation at the first equivalence point is \(\text{H}_3\text{PO}_4 + \text{NaOH} \rightarrow \text{NaH}_2\text{PO}_4 + \text{H}_2\text{O}\), the moles of \(\text{H}_3\text{PO}_4\) in \(25.0\text{ cm}^3\) of FA 1 is \(2.58 \times 10^{-3}\text{ mol}\). The concentration of \(\text{H}_3\text{PO}_4\) in FA 1 is \((2.58 \times 10^{-3}) / 0.0250 = 0.1032\text{ mol dm}^{-3}\). 4. The concentration of \(\text{H}_3\text{PO}_4\) in the original commercial rust remover is \(0.1032 \times 10 = 1.03\text{ mol dm}^{-3}\) (3 s.f.). 5. If phenolphthalein is used, it detects the second equivalence point corresponding to \(\text{H}_3\text{PO}_4 + 2\text{NaOH} \rightarrow \text{Na}_2\text{HPO}_4 + 2\text{H}_2\text{O}\). This requires a 1:2 molar ratio, so the titre volume would double to \(43.00\text{ cm}^3\).

Table of results: 4 marks (1 mark for table structure with initial and final burette readings and titres; 1 mark for correct units; 1 mark for all readings recorded to 0.05 cm3; 1 mark for obtaining two concordant titres within 0.10 cm3). Calculations: 6 marks (1 mark for correct average titre calculation; 1 mark for moles of NaOH; 2 marks for concentration of FA 1; 2 marks for concentration of original rust remover). Analysis: 3.3 marks (1.3 marks for stating the titre volume doubles to 43.00 cm3; 1 mark for stating the 1:2 reaction stoichiometry; 1 mark for stating that phenolphthalein changes colour in the pH range of the second equivalence point).

1. Plotting the temperature-time graph and extrapolating to the 4th minute gives a maximum temperature change of \(dT = 9.5\text{ }^{\circ}\text{C}\). 2. Heat released \(q = m \times c \times dT = 50.0\text{ g} \times 4.18\text{ J g}^{-1}\text{ K}^{-1} \times 9.5\text{ K} = 1985.5\text{ J} = 1.986\text{ kJ}\). 3. Moles of glycolic acid \(= 1.50 \times 0.0250 = 0.0375\text{ mol}\). Moles of \(\text{NaOH} = 2.00 \times 0.0250 = 0.0500\text{ mol}\). Since glycolic acid is the limiting reactant, the moles of water formed is \(0.0375\text{ mol}\). 4. Enthalpy change of neutralization \(\Delta H_{\text{neut}} = -q / \text{moles} = -1.9855\text{ kJ} / 0.0375\text{ mol} = -53.0\text{ kJ mol}^{-1}\). 5. Heat loss to the surroundings can be minimized by adding a lid to the polystyrene cup or lagging it.

Table of temperatures: 2 marks (1 mark for recording all temperatures to 1 decimal place; 1 mark for complete data). Graph and extrapolation: 3 marks (1 mark for axes and plotting; 2 marks for drawing two straight lines and extrapolating to t = 4 min to get dT). Heat calculation: 2 marks (1 mark for using m = 50.0 g; 1 mark for correct q value). Moles: 2 marks (1 mark for identifying glycolic acid as limiting; 1 mark for calculating 0.0375 mol). Enthalpy change: 2.3 marks (1.3 marks for value with negative sign and 3 significant figures; 1 mark for correct units). Improvement: 1 mark (suggesting a lid or extra lagging).

1. Keeping total volume constant ensures that the volume of FA 1 added is directly proportional to its initial concentration in the mixture. 2. Initial concentration of \(\text{S}_2\text{O}_8^{2-}\) in Experiment 1 is \(0.050 \times (20.0 / 55.0) = 0.0182\text{ mol dm}^{-3}\). 3. Typical times are t1 = 45 s, t2 = 90 s. Rate 1 = \(1/45 = 0.0222\text{ s}^{-1}\), Rate 2 = \(1/90 = 0.0111\text{ s}^{-1}\). The ratio of Rate 1 / Rate 2 = 2.0, which matches the ratio of concentrations (2.0). Therefore, the order of reaction with respect to peroxodisulfate is 1. 4. Thiosulfate ions instantly reduce \(\text{I}_2\) back to \(\text{I}^{-}\). Once all thiosulfate is consumed, free \(\text{I}_2\) accumulates and reacts with starch to turn blue-black.

Table of times: 2 marks (1 mark for recording times to the nearest second; 1 mark for clear headings and units). Total volume explanation: 1 mark (for stating that volume of FA 1 is directly proportional to concentration). Concentration calculation: 2 marks (1 mark for correct dilution formula; 1 mark for 0.0182 mol dm-3). Rate and Order deduction: 4 marks (2 marks for calculating 1/t for both experiments; 2 marks for showing that halving concentration halves the rate, ducing first-order kinetics). Thiosulfate mechanism: 2.3 marks (1.3 marks for explaining the reduction of iodine by thiosulfate; 1 mark for explaining starch color change after thiosulfate is completely consumed).