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From the balanced chemical equation, 4 moles of nitroglycerin decompose to produce a total of 29 moles of gas (\(12\text{ CO}_2 + 10\text{ H}_2\text{O} + 6\text{ N}_2 + 1\text{ O}_2 = 29\text{ moles of gas}\)). Therefore, 0.10 mol of nitroglycerin produces \(0.10 \times \frac{29}{4} = 0.725\text{ mol}\) of gas. The volume at STP is \(0.725\text{ mol} \times 22.7\text{ dm}^3\text{ mol}^{-1} = 16.5\text{ dm}^3\).

[1 mark] for correct choice B.

The hydronium ion, \(H_3O^+\), has 3 bonding pairs and 1 lone pair of electrons on the oxygen atom, giving it a trigonal pyramidal geometry. The repulsion of the lone pair reduces the ideal tetrahedral bond angle from \(109.5^\circ\) to approximately \(107^\circ\). \(BF_3\) and \(CO_3^{2-}\) are trigonal planar (\(120^\circ\)), and \(NH_4^+\) is tetrahedral (\(109.5^\circ\)).

[1 mark] for correct choice C.

A conjugate acid has one more proton (\(H^+\)) than its conjugate base. \(H_3O^+\) is the conjugate acid of \(H_2O\) because adding a proton to \(H_2O\) yields \(H_3O^+\). In the other options, the first species is either the conjugate base of the second (B and D) or has fewer protons (C).

[1 mark] for correct choice A.

Letting the initial rate be \(\text{rate}_1 = k[A][B]^2\). Under the new conditions, \([A]\) becomes \(2[A]\) and \([B]\) becomes \(0.5[B]\). The new rate, \(\text{rate}_2\), is given by: \(\text{rate}_2 = k(2[A])(0.5[B])^2 = k \cdot 2[A] \cdot 0.25[B]^2 = 0.5 k[A][B]^2\). Therefore, the rate decreases by a factor of 2.

[1 mark] for correct choice C.

The molecule contains a carbonyl group bonded to two carbons (\(C-CO-C\)), which is a ketone, and a hydroxyl group (\(-OH\)) bonded to a saturated carbon, which is an alcohol.

[1 mark] for correct choice B.

Calculate the heat absorbed by the water: \(q = m c \Delta T = 50.0\text{ g} \times 4.18\text{ J g}^{-1}\ ^\circ\text{C}^{-1} \times (44.0 - 20.0)\ ^\circ\text{C} = 50.0 \times 4.18 \times 24.0 = 5016\text{ J} = 5.016\text{ kJ}\). Calculate moles of ethanol burned: \(n = \frac{0.46\text{ g}}{46.0\text{ g mol}^{-1}} = 0.010\text{ mol}\). The enthalpy of combustion is: \(\Delta H_c = -\frac{q}{n} = -\frac{5.016\text{ kJ}}{0.010\text{ mol}} = -501.6\text{ kJ mol}^{-1} \approx -502\text{ kJ mol}^{-1}\).

[1 mark] for correct choice A.

For \(S_2O_3^{2-}\): let the oxidation state of sulfur be \(x\). \(2(x) + 3(-2) = -2 \Rightarrow 2x - 6 = -2 \Rightarrow 2x = +4 \Rightarrow x = +2\). For \(S_4O_6^{2-}\): let the oxidation state of sulfur be \(y\). \(4(y) + 6(-2) = -2 \Rightarrow 4y - 12 = -2 \Rightarrow 4y = +10 \Rightarrow y = +2.5\).

[1 mark] for correct choice A.

The equilibrium constant, \(K_c\), is only affected by changes in temperature. Because the reaction is exothermic (\(\Delta H < 0\)), lowering the temperature shifts the equilibrium position to the right to produce more products, which increases the value of \(K_c\).

[1 mark] for correct choice B.

First, calculate the moles of \(CO_2\) produced: \(n(CO_2) = 13.2\text{ g} / 44.01\text{ g mol}^{-1} = 0.300\text{ mol}\). Each mole of \(CO_2\) contains 1 mole of carbon atoms, so there are 0.300 mol of C in the sample. Next, calculate the moles of \(H_2O\) produced: \(n(H_2O) = 7.20\text{ g} / 18.02\text{ g mol}^{-1} = 0.400\text{ mol}\). Each mole of \(H_2O\) contains 2 moles of hydrogen atoms, so there are \(0.400 \times 2 = 0.800\text{ mol}\) of H in the sample. To find the molecular formula, determine the number of moles of each element per mole of hydrocarbon: Moles of C = \(0.300\text{ mol} / 0.100\text{ mol} = 3\). Moles of H = \(0.800\text{ mol} / 0.100\text{ mol} = 8\). The molecular formula is \(C_3H_8\).

Award [1] for the correct option (B). [0] for all other options.

The hydronium ion, \(H_3O^+\), has 3 bonding pairs and 1 lone pair on the oxygen atom. According to VSEPR theory, this results in a trigonal pyramidal molecular geometry with a bond angle of approximately \(107^\circ\). Ammonia, \(NH_3\), also has 3 bonding pairs and 1 lone pair on the central nitrogen atom, giving it the same trigonal pyramidal shape and a bond angle of \(107^\circ\). \(BF_3\), \(CH_3^+\), and \(SO_3\) all have trigonal planar geometries with bond angles of \(120^\circ\).

Award [1] for the correct option (C). [0] for all other options.

First, calculate the heat absorbed by the water: \(q = m c \Delta T = 100.0\text{ g} \times 4.18\text{ J g}^{-1}\text{ K}^{-1} \times 25.0\text{ K} = 10450\text{ J} = 10.45\text{ kJ}\). Next, calculate the moles of fuel burned: \(n = 0.500\text{ g} / 58.1\text{ g mol}^{-1} = 0.008606\text{ mol}\). Finally, calculate the enthalpy of combustion: \(\Delta H_c = -q / n = -10.45\text{ kJ} / 0.008606\text{ mol} \approx -1214\text{ kJ mol}^{-1}\), which rounds to -1210 to 3 significant figures.

Award [1] for the correct option (B). [0] for all other options.

A Brønsted-Lowry base is defined as a proton (\(H^+\)) acceptor. In reaction I, \(HPO_4^{2-}\) donates a proton to \(OH^-\), acting as an acid. In reaction II, \(HPO_4^{2-}\) accepts a proton from \(HF\) to form \(H_2PO_4^-\), acting as a base. In reaction III, \(HPO_4^{2-}\) accepts a proton from \(H_3O^+\) to form \(H_2PO_4^-\), acting as a base. Therefore, it acts as a base in reactions II and III only.

Award [1] for the correct option (D). [0] for all other options.

The compound contains two functional groups: a hydroxyl group (alcohol) and a carbonyl group (ketone). Since ketones have a higher priority in IUPAC nomenclature than alcohols, the main chain suffix is '-one' and the hydroxyl group is designated with the prefix 'hydroxy-'. The carbon chain must be numbered from the right end to give the carbonyl group the lowest possible number (C2 instead of C4). This puts the hydroxyl group at C4. The correct IUPAC name is therefore 4-hydroxypentan-2-one.

Award [1] for the correct option (B). [0] for all other options.

The rate law is represented by: \(\text{Rate} = k[A]^2[B]^0 = k[A]^2\). Since the reaction is second-order with respect to \(A\), doubling the concentration of \(A\) (multiplying it by 2) will increase the rate by a factor of \(2^2 = 4\). Because the reaction is zero-order with respect to \(B\), changing its concentration has no effect on the initial rate.

Award [1] for the correct option (B). [0] for all other options.

First, the forward reaction is exothermic (\(\Delta H < 0\)). According to Le Chatelier's principle, decreasing the temperature will shift the equilibrium to favor the exothermic direction (the right side). Second, there are 3 moles of gaseous reactants and 2 moles of gaseous products. Increasing the pressure will shift the equilibrium towards the side with fewer moles of gas to relieve the pressure (the right side). Therefore, decreasing temperature and increasing pressure will shift the equilibrium to the right.

Award [1] for the correct option (C). [0] for all other options.

Copper has a more positive standard reduction potential (\(+0.34\text{ V}\)) than zinc (\(-0.76\text{ V}\)), which means that \(Cu^{2+}\) is more readily reduced than \(Zn^{2+}\). Therefore, reduction occurs at the copper electrode (cathode): \(Cu^{2+}(aq) + 2e^- \rightarrow Cu(s)\). Electrons flow from the zinc electrode (anode) to the copper electrode (cathode) in the external circuit. Zinc is oxidized, causing the mass of the zinc electrode to decrease. Anions in the salt bridge migrate towards the anode (zinc half-cell) to balance the build-up of positive charge.

Award [1] for the correct option (C). [0] for all other options.

Assume \(100.0\text{ g}\) of the metal oxide is present.

1. Calculate the mass of each element:
- Mass of \(M = 70.0\text{ g}\)
- Mass of \(O = 100.0 - 70.0 = 30.0\text{ g}\)

2. Calculate the moles of oxygen:
- \(n(O) = \frac{30.0\text{ g}}{16.00\text{ g mol}^{-1}} = 1.875\text{ mol}\)

3. Determine the moles of metal \(M\) using the empirical formula ratio (\(M : O = 2 : 3\)):
- \(n(M) = 1.875\text{ mol} \times \frac{2}{3} = 1.25\text{ mol}\)

4. Calculate the molar mass of metal \(M\):
- \(M_r(M) = \frac{70.0\text{ g}}{1.25\text{ mol}} = 56.0\text{ g mol}^{-1}\)

[1 mark] awarded for selecting option A.
[0 marks] for incorrect answers.

Let's analyze the electron geometry and molecular geometry of each species:

- \(XeF_4\): Xenon has 8 valence electrons + 4 electrons from the fluorine atoms = 12 electrons (6 electron domains). This gives an octahedral electron domain geometry. With 4 bonding pairs and 2 lone pairs, the molecular geometry is square planar.
- \(SF_4\): Sulfur has 6 valence electrons + 4 from fluorine = 10 electrons (5 electron domains). This gives a trigonal bipyramidal electron domain geometry. With 4 bonding pairs and 1 lone pair, the molecular geometry is see-saw.
- \(CF_4\): Carbon has 4 valence electrons + 4 from fluorine = 8 electrons (4 electron domains). This is tetrahedral.
- \([BF_4]^-\): Boron has 3 valence electrons + 1 (from the charge) + 4 from fluorine = 8 electrons (4 electron domains). This is tetrahedral.

[1 mark] awarded for selecting option A.
[0 marks] for incorrect answers.

1. Calculate the total mass of the mixture:
- \(m = 50.0\text{ cm}^3 + 50.0\text{ cm}^3 = 100.0\text{ cm}^3 \Rightarrow 100.0\text{ g}\)

2. Calculate the temperature change:
- \(\Delta T = 26.8^\circ\text{C} - 20.0^\circ\text{C} = 6.8\text{ K}\)

3. Calculate the heat released (\(q\)):
- \(q = m c \Delta T = 100.0\text{ g} \times 4.18\text{ J g}^{-1}\text{ K}^{-1} \times 6.8\text{ K} = 2842.4\text{ J} = 2.84\text{ kJ}\)

4. Calculate the moles of limiting reactant (\(H^+\) or \(OH^-\)):
- \(n = 0.0500\text{ dm}^3 \times 1.00\text{ mol dm}^{-3} = 0.0500\text{ mol}\)

5. Calculate the enthalpy change of neutralization (\(\Delta H\)):
- \(\Delta H = -\frac{q}{n} = -\frac{2.8424\text{ kJ}}{0.0500\text{ mol}} = -56.8\text{ kJ mol}^{-1}\)

[1 mark] awarded for selecting option A.
[0 marks] for incorrect answers.

A conjugate acid-base pair consists of two species that differ by exactly one hydrogen ion (\(\text{H}^+\)).
In the given reaction:
- \(\text{H}_2\text{PO}_4^-\) is the acid, and \(\text{HPO}_4^{2-}\) is its conjugate base (differing by one proton).
- \(\text{HCO}_3^-\) is the base, and \(\text{H}_2\text{CO}_3\) is its conjugate acid (differing by one proton).

[1 mark] awarded for selecting option A.
[0 marks] for incorrect answers.

Let the initial rate be \(\text{Rate}_1 = k[A][B]^2\).

If \([A]\) is doubled, the new concentration is \(2[A]\).
If \([B]\) is halved, the new concentration is \(0.5[B]\).

Substitute these values into the rate equation to find the new rate (\(\text{Rate}_2\)):
\(\text{Rate}_2 = k(2[A])(0.5[B])^2 = k(2[A])(0.25[B]^2) = 0.5 \times k[A][B]^2 = 0.5 \times \text{Rate}_1\).

Therefore, the rate of reaction is halved.

[1 mark] awarded for selecting option C.
[0 marks] for incorrect answers.

Let's analyze the formula \(\text{CH}_3\text{CH(OH)COOCH}_2\text{CH}_3\):

- The group \(\text{-OH}\) attached to a carbon atom represents an alcohol (hydroxyl group).
- The group \(\text{-COO-}\) linked between two alkyl groups represents an ester group.
- Therefore, the functional groups present are ester and alcohol.

[1 mark] awarded for selecting option B.
[0 marks] for incorrect answers.

Let's calculate the oxidation state of sulfur (\(x\)) in each option:

- \(\text{H}_2\text{SO}_3\): \(2(+1) + x + 3(-2) = 0 \Rightarrow 2 + x - 6 = 0 \Rightarrow x = +4\).
- \(\text{H}_2\text{SO}_4\): \(2(+1) + x + 4(-2) = 0 \Rightarrow 2 + x - 8 = 0 \Rightarrow x = +6\).
- \(\text{S}_2\text{O}_3^{2-}\): \(2x + 3(-2) = -2 \Rightarrow 2x - 6 = -2 \Rightarrow 2x = +4 \Rightarrow x = +2\).
- \(\text{SF}_6\): \(x + 6(-1) = 0 \Rightarrow x = +6\).

[1 mark] awarded for selecting option A.
[0 marks] for incorrect answers.

Construct an ICE table:
- Initial: \([\text{H}_2] = 1.0\text{ M}\), \([\text{CO}_2] = 1.0\text{ M}\), \([\text{H}_2\text{O}] = 0\), \([\text{CO}] = 0\)
- Change: \(-x\), \(-x\), \(+x\), \(+x\)
- Equilibrium: \(1.0-x\), \(1.0-x\), \(x\), \(x\)

Set up the equilibrium expression:
\(K_c = \frac{[\text{H}_2\text{O}][\text{CO}]}{[\text{H}_2][\text{CO}_2]} = \frac{x^2}{(1.0 - x)^2} = 4.0\)

Take the square root of both sides:
\(\frac{x}{1.0 - x} = 2.0\)

Solve for \(x\):
\(x = 2.0 - 2x \Rightarrow 3x = 2.0 \Rightarrow x = 0.67\text{ mol dm}^{-3}\)

Therefore, the equilibrium concentration of \(\text{CO}\) is \(0.67\text{ mol dm}^{-3}\).

[1 mark] awarded for selecting option B.
[0 marks] for incorrect answers.

Step 1: Calculate the theoretical moles of reactant: \(n(\text{CaCO}_3) = \frac{10.0\text{ g}}{100.1\text{ g mol}^{-1}} \approx 0.100\text{ mol}\). Step 2: Since the molar ratio of \(\text{CaCO}_3\) to \(\text{CO}_2\) is 1:1, the theoretical yield of \(\text{CO}_2\) is 0.100 mol. Step 3: Calculate the theoretical volume of \(\text{CO}_2\): \(0.100\text{ mol} \times 24.0\text{ dm}^3\text{ mol}^{-1} = 2.40\text{ dm}^3\). Step 4: Calculate the percentage yield: \(\text{Percentage Yield} = \frac{1.80\text{ dm}^3}{2.40\text{ dm}^3} \times 100\% = 75.0\%\).

Award 1 mark for the correct option C.

Iron (Fe) has an atomic number of 26, giving it a ground-state electron configuration of \([\text{Ar}] 4\text{s}^2 3\text{d}^6\). When transition metals form cations, electrons are first removed from the outer \(4s\) subshell, followed by the \(3d\) subshell. Removing 3 electrons to form \(\text{Fe}^{3+}\) results in the loss of both \(4s\) electrons and one \(3d\) electron, leaving the electron configuration as \([\text{Ar}] 3\text{d}^5\).

Award 1 mark for the correct option C.

\(\text{BF}_3\) has 3 bonding electron domains and no lone pairs on the boron atom, resulting in a trigonal planar geometry with a bond angle of \(120^\circ\). \(\text{NH}_3\) has 3 bonding domains and 1 lone pair on the nitrogen atom, resulting in a trigonal pyramidal shape with a bond angle of approximately \(107^\circ\). \(\text{H}_2\text{O}\) has 2 bonding domains and 2 lone pairs on the oxygen atom, resulting in a bent shape with a bond angle of approximately \(104.5^\circ\). Therefore, the correct order of decreasing bond angles is \(\text{BF}_3 > \text{NH}_3 > \text{H}_2\text{O}\).

Award 1 mark for the correct option A.

To obtain the target equation using Hess's Law: Reverse equation (1): \(2\text{CO(g)} \rightarrow 2\text{C(s)} + \text{O}_2\text{(g)} \quad \Delta H^\theta = +221\text{ kJ mol}^{-1}\). Multiply equation (2) by 2: \(2\text{C(s)} + 2\text{O}_2\text{(g)} \rightarrow 2\text{CO}_2\text{(g)} \quad \Delta H^\theta = 2 \times (-394\text{ kJ mol}^{-1}) = -788\text{ kJ mol}^{-1}\). Summing these two equations gives the target reaction: \(2\text{CO(g)} + \text{O}_2\text{(g)} \rightarrow 2\text{CO}_2\text{(g)}\). \Delta H^\theta = +221 + (-788) = -567\text{ kJ mol}^{-1}\).

Award 1 mark for the correct option B.

A Brønsted-Lowry base is defined as a proton (\(\text{H}^+\)) acceptor. In the forward reaction, \(\text{HCO}_3^-\text{(aq)}\) accepts a proton to become \(\text{H}_2\text{CO}_3\text{(aq)}\), acting as the base. In the reverse reaction, \(\text{HPO}_4^{2-}\text{(aq)}\) accepts a proton to become \(\text{H}_2\text{PO}_4^-\text{(aq)}\), acting as the base. Thus, the forward base is \(\text{HCO}_3^-\) and the reverse base is \(\text{HPO}_4^{2-}\).

Award 1 mark for the correct option B.

By identifying the functional groups in the given structure: 1) The \(\text{CH}_2=\text{CH}-\) group represents an alkene functional group. 2) The \(-\text{COO}-\) group linking the alkyl sections represents an ester functional group. 3) The terminal \(-\text{OH}\) group represents an alcohol functional group. Therefore, the compound contains alkene, ester, and alcohol groups.

Award 1 mark for the correct option B.

\( \text{(a) }\Delta T = 27.8 - 21.3 = 6.5\text{ K (or }^\circ\text{C)} \)
\( m = 50.0 + 50.0 = 100.0\text{ g} \)
\( q = m \cdot c \cdot \Delta T = 100.0\text{ g} \times 4.18\text{ J g}^{-1}\text{ K}^{-1} \times 6.5\text{ K} = 2717\text{ J} = 2.717\text{ kJ} \)
\( n(\text{HCl}) = 1.00\text{ mol dm}^{-3} \times 0.0500\text{ dm}^3 = 0.0500\text{ mol} \)
\( \Delta H_{\text{n}} = -\frac{q}{n} = -\frac{2.717\text{ kJ}}{0.0500\text{ mol}} = -54.3\text{ kJ mol}^{-1} \text{ (accept -54 kJ mol}^{-1}\text{ due to 2 sig figs in temperature change)} \)

\( \text{(b) Absolute uncertainty in }\Delta T = 0.2 + 0.2 = 0.4\text{ K} \)
\( \text{Percentage uncertainty in }\Delta T = \frac{0.4}{6.5} \times 100\% = 6.15\% \)
\( \text{Absolute uncertainty in mass } m = 0.5 + 0.5 = 1.0\text{ g} \)
\( \text{Percentage uncertainty in mass } m = \frac{1.0}{100.0} \times 100\% = 1.00\% \)

\( \text{(c) Moles of HCl is calculated using } n = C \times V \text{, where uncertainty in } V_\text{HCl} \text{ is } 1.00\% \text{ (from } \frac{0.5}{50.0} \times 100\% \text{).} \)
\( \text{Total percentage uncertainty in }\Delta H_{\text{n}} = \%\text{ uncertainty in } m + \%\text{ uncertainty in }\Delta T + \%\text{ uncertainty in } V_{\text{HCl}} = 1.00\% + 6.15\% + 1.00\% = 8.15\% \text{ (accept 7.15\% if the uncertainty in volume/moles was not added).} \)

\( \text{(d) Glass is a better thermal conductor than polystyrene, meaning more heat is lost to the surroundings. This results in a smaller measured temperature change and a less exothermic (less negative/smaller) calculated }\Delta H_{\text{n}}\text{.} \)

\( \text{Part (a): [3 marks]} \)
- Award [1] for calculating \(\Delta T = 6.5\text{ K}\) and \(m = 100.0\text{ g}\).
- Award [1] for calculating \(q = 2.717\text{ kJ}\) or \(2.72\text{ kJ}\).
- Award [1] for \(\Delta H_{\text{n}} = -54.3\text{ kJ mol}^{-1}\) (or \(-54\text{ kJ mol}^{-1}\)). Accept negative sign (essential for enthalpy of neutralization).

\( \text{Part (b): [2 marks]} \)
- Award [1] for percentage uncertainty in \(\Delta T = 6.15\%\) (accept \(6.2\%\) or \(6\%\)).
- Award [1] for percentage uncertainty in mass = \(1.00\%\).

\( \text{Part (c): [2 marks]} \)
- Award [1] for finding percentage uncertainty in the volume of \(\text{HCl}\) used to determine moles is \(1.00\%\).
- Award [1] for correct sum: \(1.00\% + 6.15\% + 1.00\% = 8.15\%\) (accept \(8.2\%\) or \(8\%\); accept \(7.15\%\) if volume uncertainty for mole calculation was omitted but mass and temperature uncertainties were summed correctly).

\( \text{Part (d): [1 mark]} \)
- Award [1] for stating that temperature change would be smaller AND the calculated \(\Delta H_{\text{n}}\) would be less exothermic/less negative because glass allows more heat loss to the surroundings.

\( \text{(a) Comparing Experiments 1 and 2: } [\text{I}^-] \text{ is constant. } [\text{S}_2\text{O}_8^{2-}] \text{ doubles (from 0.0100 to 0.0200) and the rate doubles (from } 1.10 \times 10^{-5} \text{ to } 2.20 \times 10^{-5}\text{). Therefore, the order with respect to } \text{S}_2\text{O}_8^{2-} \text{ is 1 (first-order).}\)

\( \text{Comparing Experiments 2 and 3: } [\text{S}_2\text{O}_8^{2-}] \text{ doubles (from 0.0200 to 0.0400) and } [\text{I}^-] \text{ is halved (from 0.100 to 0.0500), but the rate remains constant (} 2.20 \times 10^{-5}\text{).}\)
\( \text{Since doubling } [\text{S}_2\text{O}_8^{2-}] \text{ alone would double the rate, the halving of } [\text{I}^-] \text{ must have halved the rate to keep it constant. Therefore, the order with respect to } \text{I}^- \text{ is 1 (first-order).}\)

\( \text{(b) Rate expression: Rate} = k[\text{S}_2\text{O}_8^{2-}][\text{I}^-] \)
\( \text{Using Experiment 1 data to calculate } k: \)
\( 1.10 \times 10^{-5}\text{ mol dm}^{-3}\text{ s}^{-1} = k \times (0.0100\text{ mol dm}^{-3}) \times (0.100\text{ mol dm}^{-3}) \implies k = 1.10 \times 10^{-2}\text{ (or } 0.0110\text{) dm}^3\text{ mol}^{-1}\text{ s}^{-1} \)

\( \text{(c) Since the reaction is first-order with respect to both } \text{S}_2\text{O}_8^{2-} \text{ and } \text{I}^- \text{, the rate-determining step must involve one } \text{S}_2\text{O}_8^{2-} \text{ ion and one } \text{I}^- \text{ ion. This matches the stoichiometry of the proposed first step, so the first step must be the rate-determining step.}\)

\( \text{Part (a): [3 marks]} \)
- Award [1] for deducing first-order for \(\text{S}_2\text{O}_8^{2-}\) with valid reasoning comparing Exps 1 & 2.
- Award [1] for deducing first-order for \(\text{I}^-\) with valid reasoning comparing Exps 2 & 3 (or calculating using rate equation ratios).
- Award [1] for clear presentation of comparison calculations or logical statements.

\( \text{Part (b): [3 marks]} \)
- Award [1] for correct rate expression: \(\text{Rate} = k[\text{S}_2\text{O}_8^{2-}][\text{I}^-]\) (accept alternative notation, but reactant formulas must be correct).
- Award [1] for correct numerical value of \(k = 1.10 \times 10^{-2}\) (or \(0.0110\)).
- Award [1] for correct units: \(\text{dm}^3\text{ mol}^{-1}\text{ s}^{-1}\) (accept \(\text{dm}^3\text{ mol}^{-1}\text{ s}^{-1}\) or similar standard unit representations).

\( \text{Part (c): [2 marks]} \)
- Award [1] for identifying that the first step is the rate-determining step.
- Award [1] for explaining that the molecularity/reactants of the first step (one of each reactant) matches the species in the experimental rate expression (first-order in both reactants).

\( \text{(a) } P(\text{H}_2) = P_{\text{total}} - P_{\text{water}} = 101.3\text{ kPa} - 2.6\text{ kPa} = 98.7\text{ kPa} \)

\( \text{(b) } n(\text{Mg}) = \frac{0.0389\text{ g}}{24.31\text{ g mol}^{-1}} = 1.60 \times 10^{-3}\text{ mol} \)
\( \text{Since } \text{Mg} : \text{H}_2 = 1:1\text{, } n(\text{H}_2) = 1.60 \times 10^{-3}\text{ mol} \)

\( \text{(c) } P = 98.7\text{ kPa} = 98.7 \times 10^3\text{ Pa} \)
\( V = 39.5\text{ cm}^3 = 39.5 \times 10^{-6}\text{ m}^3 \)
\( n = 1.60 \times 10^{-3}\text{ mol} \)
\( T = 295\text{ K} \)
\( R = \frac{PV}{nT} = \frac{(98.7 \times 10^3\text{ Pa}) \times (39.5 \times 10^{-6}\text{ m}^3)}{(1.60 \times 10^{-3}\text{ mol}) \times 295\text{ K}} = \frac{3.89865}{0.472} = 8.26\text{ J K}^{-1}\text{ mol}^{-1} \)

\( \text{(d) If water vapor pressure is not subtracted, a higher pressure value (101.3 kPa instead of 98.7 kPa) would be substituted into the ideal gas equation. Since } R = \frac{PV}{nT}\text{, a higher value of } P \text{ would lead to a higher calculated value of } R\text{.} \)

\( \text{Part (a): [1 mark]} \)
- Award [1] for \(98.7\text{ kPa}\).

\( \text{Part (b): [2 marks]} \)
- Award [1] for calculating \(n(\text{Mg}) = 1.60 \times 10^{-3}\text{ mol}\).
- Award [1] for relating moles of \(\text{Mg}\) to moles of \(\text{H}_2\) in a \(1:1\) ratio, yielding \(1.60 \times 10^{-3}\text{ mol}\).

\( \text{Part (c): [3 marks]} \)
- Award [1] for converting pressure to \(\text{Pa}\) (or keeping in \(\text{kPa}\) but converting volume to \(\text{dm}^3\) to match unit dimensions).
- Award [1] for correct substitution of values into \(R = \frac{PV}{nT}\).
- Award [1] for final answer of \(8.26\text{ J K}^{-1}\text{ mol}^{-1}\) (accept \(8.3\text{ J K}^{-1}\text{ mol}^{-1}\) if rounded to two significant figures, but the question specifies three significant figures).

\( \text{Part (d): [2 marks]} \)
- Award [1] for stating the calculated value of \(R\) would be higher.
- Award [1] for explaining that a higher pressure value of \(101.3\text{ kPa}\) would be used, and since \(R\) is directly proportional to \(P\) in the rearranged equation \(R = \frac{PV}{nT}\), the calculated \(R\) would increase.

(i)
\(\Delta T = 45.6\text{ }^\circ\text{C} - 20.2\text{ }^\circ\text{C} = 25.4\text{ }^\circ\text{C}\) (or \(25.4\text{ K}\))
\(\Delta m = 120.45\text{ g} - 119.85\text{ g} = 0.60\text{ g}\)

(ii)
\(q = m \cdot c \cdot \Delta T = 150.0\text{ g} \times 4.18\text{ J g}^{-1}\text{ K}^{-1} \times 25.4\text{ K} = 15925.8\text{ J} = 15.9\text{ kJ}\)

(iii)
\(n(\text{propan-1-ol}) = \frac{0.60\text{ g}}{60.11\text{ g mol}^{-1}} = 0.00998\text{ mol}\)
\(\Delta H_c = -\frac{q}{n} = -\frac{15.9258\text{ kJ}}{0.00998\text{ mol}} = -1596\text{ kJ mol}^{-1}\)
Since \(\Delta m\) has only 2 significant figures, the final value should be rounded to 2 significant figures: \(-1600\text{ kJ mol}^{-1}\) (or \(-1.6 \times 10^3\text{ kJ mol}^{-1}\)).

(iv)
- Error 1: Heat lost to the surroundings / draft.
Improvement: Use draft shields around the spirit burner and calorimeter, or use a bomb calorimeter.
- Error 2: Incomplete combustion of the alcohol (evidenced by soot/black carbon depositing on the beaker base).
Improvement: Provide a pure oxygen stream / feed to ensure complete combustion.

(v)
The experimental standard enthalpy of combustion is determined under standard conditions where water is in the liquid state, whereas bond enthalpies or gaseous products assume water vapor. Alternatively, standard conditions (298 K, 100 kPa) are not strictly maintained, or incomplete combustion leads to a less exothermic output.

(vi)
Balanced equation: \(\text{CH}_3\text{CH}_2\text{CH}_2\text{OH}(g) + 4.5\text{O}_2(g) \rightarrow 3\text{CO}_2(g) + 4\text{H}_2\text{O}(g)\)

Bonds broken (Reactants):
- 2 \(\text{C}-\text{C}\): \(2 \times 346 = 692\text{ kJ}\)
- 7 \(\text{C}-\text{H}\): \(7 \times 414 = 2898\text{ kJ}\)
- 1 \(\text{C}-\text{O}\): \(1 \times 358 = 358\text{ kJ}\)
- 1 \(\text{O}-\text{H}\): \(1 \times 463 = 463\text{ kJ}\)
- 4.5 \(\text{O}=\text{O}\): \(4.5 \times 498 = 2241\text{ kJ}\)
Total bonds broken = \(692 + 2898 + 358 + 463 + 2241 = 6652\text{ kJ}\)

Bonds formed (Products):
- 6 \(\text{C}=\text{O}\): \(6 \times 804 = 4824\text{ kJ}\)
- 8 \(\text{O}-\text{H}\): \(8 \times 463 = 3704\text{ kJ}\)
Total bonds formed = \(4824 + 3704 = 8528\text{ kJ}\)

\(\Delta H_c = \Sigma\text{Bonds Broken} - \Sigma\text{Bonds Formed} = 6652 - 8528 = -1876\text{ kJ mol}^{-1}\)

(i) [2 marks]
- 1 mark for temperature change \(25.4\text{ }^\circ\text{C}\) (or \(\text{K}\))
- 1 mark for mass change \(0.60\text{ g}\)

(ii) [2 marks]
- 1 mark for substituting values correctly into \(q = mc\Delta T\)
- 1 mark for correct calculation of heat \(15.9\text{ kJ}\) (accept \(15.93\text{ kJ}\))

(iii) [3 marks]
- 1 mark for calculation of moles of propan-1-ol (\(0.00998\text{ mol}\))
- 1 mark for correct division of \(q\) by moles with a negative sign
- 1 mark for correct significant figures (2 SF: \(-1600\text{ kJ mol}^{-1}\) or \(-1.6 \times 10^3\text{ kJ mol}^{-1}\))

(iv) [4 marks]
- 1 mark for each valid error identified (max 2 marks)
- 1 mark for each corresponding valid improvement (max 2 marks)

(v) [2 marks]
- 1 mark for stating that experimental combustion yields liquid water while standard bond enthalpies assume gas phase products
- 1 mark for linking this state difference to the heat required to vaporize water (less exothermic)

(vi) [3.67 marks]
- 1 mark for correctly balancing the combustion equation or listing correct number of bonds
- 1 mark for calculating correct total bond energy broken (\(6652\text{ kJ}\))
- 1 mark for calculating correct total bond energy formed (\(8528\text{ kJ}\))
- 0.67 marks for subtracting formed from broken to get \(-1876\text{ kJ mol}^{-1}\) (must include correct negative sign)

(i)
- A Brønsted–Lowry acid is a proton/\(\text{H}^+\) donor.
- Equation: \(\text{HCOOH}(aq) + \text{H}_2\text{O}(l) \rightleftharpoons \text{HCOO}^-(aq) + \text{H}_3\text{O}^+(aq)\)
- Conjugate pairs: \(\text{HCOOH}\) (acid) and \(\text{HCOO}^-\rightleftharpoons\) (conjugate base); \(\text{H}_2\text{O}\) (base) and \(\text{H}_3\text{O}^+\) (conjugate acid).

(ii)
\(K_a = \frac{[\text{H}^+][\text{HCOO}^-]}{[\text{HCOOH}]}\)
Since \([\text{H}^+] \approx [\text{HCOO}^-]\) and assuming \([\text{HCOOH}]_{\text{eq}} \approx [\text{HCOOH}]_{\text{initial}} = 0.150\text{ mol dm}^{-3}\):
\(1.77 \times 10^{-4} = \frac{[\text{H}^+]^2}{0.150}\)
\([\text{H}^+]^2 = 2.655 \times 10^{-5} \Rightarrow [\text{H}^+] = 5.15 \times 10^{-3}\text{ mol dm}^{-3}\)
\(\text{pH} = -\log_{10}(5.15 \times 10^{-3}) = 2.29\)
Assumption: The dissociation of methanoic acid is negligible (so equilibrium concentration is approximately equal to initial concentration) or autoionization of water is negligible.

(iii)
\(n(\text{HCOOH})_{\text{initial}} = 0.0500\text{ dm}^3 \times 0.150\text{ mol dm}^{-3} = 7.50 \times 10^{-3}\text{ mol}\)
\(n(\text{NaOH})_{\text{added}} = 0.0250\text{ dm}^3 \times 0.150\text{ mol dm}^{-3} = 3.75 \times 10^{-3}\text{ mol}\)
Reaction: \(\text{HCOOH} + \text{OH}^- \rightarrow \text{HCOO}^- + \text{H}_2\text{O}\)
Since \(\text{NaOH}\) is the limiting reactant, it reacts completely:
\(n(\text{HCOO}^-)_{\text{formed}} = 3.75 \times 10^{-3}\text{ mol}\)
\(n(\text{HCOOH})_{\text{remaining}} = 7.50 \times 10^{-3} - 3.75 \times 10^{-3} = 3.75 \times 10^{-3}\text{ mol}\)
Since the moles of weak acid and conjugate base are equal, this is the half-equivalence point:
\([\text{HCOOH}] = [\text{HCOO}^-]\)
Using the Henderson-Hasselbalch equation:
\(\text{pH} = \text{p}K_a + \log_{10}\left(\frac{[\text{HCOO}^-]}{[\text{HCOOH}]}\right) = \text{p}K_a = -\log_{10}(1.77 \times 10^{-4}) = 3.75\)

(iv)
When \(\text{H}^+\) (from \(\text{HCl}\)) is added, it reacts with the conjugate base (\(\text{HCOO}^-\)) present in the buffer:
\(\text{HCOO}^-(aq) + \text{H}^+(aq) \rightarrow \text{HCOOH}(aq)\)
This removes the added hydronium ions, keeping the free \([\text{H}^+]\) and thus the \(\text{pH}\) relatively constant.

(v)
Phenolphthalein. The titration involves a weak acid and a strong base, so the salt formed hydrolyzes to produce a basic solution at the equivalence point (\(\text{pH} > 7\)). The transition range of phenolphthalein (8.3 – 10.0) coincides with the steep vertical region of the titration curve.

(i) [3 marks]
- 1 mark for defining Brønsted–Lowry acid as a proton donor.
- 1 mark for correct balanced equilibrium equation.
- 1 mark for correctly matching both conjugate pairs.

(ii) [4 marks]
- 1 mark for setting up the approximation \(K_a = \frac{[\text{H}^+]^2}{[\text{HA}]}\).
- 1 mark for calculating \([\text{H}^+] = 5.15 \times 10^{-3}\text{ mol dm}^{-3}\).
- 1 mark for pH calculation of \(2.29\).
- 1 mark for stating a valid assumption (e.g., \([\text{HCOOH}]\) at equilibrium is equal to initial concentration).

(iii) [4.67 marks]
- 1 mark for finding initial moles of \(\text{HCOOH}\) and \(\text{NaOH}\).
- 1 mark for determining remaining moles of acid and formed moles of salt.
- 1 mark for identifying that it is a half-equivalence point where \(\text{pH} = \text{p}K_a\).
- 1.67 marks for calculating the final \(\text{pH}\) as \(3.75\).

(iv) [3 marks]
- 1 mark for writing the correct ionic equation: \(\text{HCOO}^- + \text{H}^+ \rightarrow \text{HCOOH}\).
- 1 mark for explaining that the conjugate base neutralizes the added \(\text{H}^+\).
- 1 mark for stating that the concentration of free hydrogen ions remains relatively unchanged.

(v) [2 marks]
- 1 mark for selecting phenolphthalein.
- 1 mark for explaining that the equivalence point is basic (pH > 7) due to a weak acid-strong base titration.

(i)
- **Order with respect to NO**: Compare Experiments 1 and 2, where \([\text{Cl}_2]\) is kept constant at \(0.10\text{ mol dm}^{-3}\).
\([\text{NO}]\) doubles from \(0.10\) to \(0.20\text{ mol dm}^{-3}\).
The rate quadruples: \(\frac{1.00 \times 10^{-4}}{2.50 \times 10^{-5}} = 4\).
Since \(2^x = 4\), the order with respect to \(\text{NO}\) is **2**.

- **Order with respect to \(\text{Cl}_2\)**: Compare Experiments 2 and 3, where \([\text{NO}]\) is kept constant at \(0.20\text{ mol dm}^{-3}\).
\([\text{Cl}_2]\) doubles from \(0.10\) to \(0.20\text{ mol dm}^{-3}\).
The rate doubles: \(\frac{2.00 \times 10^{-4}}{1.00 \times 10^{-4}} = 2\).
Since \(2^y = 2\), the order with respect to \(\text{Cl}_2\) is **1**.

(ii)
Rate expression: \(\text{Rate} = k[\text{NO}]^2[\text{Cl}_2]\)
To find \(k\), substitute values from Experiment 1:
\(2.50 \times 10^{-5}\text{ mol dm}^{-3}\text{ s}^{-1} = k (0.10\text{ mol dm}^{-3})^2 (0.10\text{ mol dm}^{-3})\)
\(2.50 \times 10^{-5} = k (0.0010)\)
\(k = 2.50 \times 10^{-2}\)
Units: \(\frac{\text{mol dm}^{-3}\text{ s}^{-1}}{(\text{mol dm}^{-3})^3} = \text{dm}^6\text{ mol}^{-2}\text{ s}^{-1}\)

(iii)
A plausible mechanism must sum to the overall stoichiometry: \(2\text{NO} + \text{Cl}_2 \rightarrow 2\text{NOCl}\)
Step 1 (fast equilibrium):
\(\text{NO}(g) + \text{Cl}_2(g) \rightleftharpoons \text{NOCl}_2(g)\)
Step 2 (slow, rate-determining step):
\(\text{NOCl}_2(g) + \text{NO}(g) \rightarrow 2\text{NOCl}(g)\)
Overall: \(2\text{NO} + \text{Cl}_2 \rightarrow 2\text{NOCl}\)

(iv)
The energy profile should show:
- Y-axis: Potential Energy, X-axis: Reaction Coordinate / Progress of Reaction.
- Two humps (transition states) since it is a two-step mechanism, with an intermediate valley (for \(\text{NOCl}_2\)).
- The second step is the rate-determining step, so the second hump peak must be higher than the first peak relative to their starting states, representing a larger activation energy for Step 2.
- Products are at a lower potential energy level than the reactants (exothermic).
- Labels for \(\Delta H\) (negative distance between reactants and products) and \(E_a\) for the RDS.

(v)
- An increase in temperature increases the average kinetic energy of the reacting species.
- In terms of the Maxwell-Boltzmann distribution, the curve shifts to the right and flattens, meaning a significantly larger fraction of reactant particles have kinetic energy greater than or equal to the activation energy (\(E \ge E_a\)).
- This leads to a higher frequency of successful, productive collisions, thereby increasing the rate constant \(k\) and the overall reaction rate.

(i) [4 marks]
- 1 mark for identifying Experiments 1 and 2 to find NO order.
- 1 mark for showing mathematical step or deduction that order wrt NO is 2.
- 1 mark for identifying Experiments 2 and 3 to find \(\text{Cl}_2\) order.
- 1 mark for showing mathematical step or deduction that order wrt \(\text{Cl}_2\) is 1.

(ii) [3.67 marks]
- 1 mark for correct rate expression: \(\text{Rate} = k[\text{NO}]^2[\text{Cl}_2]\).
- 1 mark for correct calculation of numerical value \(2.50 \times 10^{-2}\).
- 1.67 marks for correct units: \(\text{dm}^6\text{ mol}^{-2}\text{ s}^{-1}\).

(iii) [3 marks]
- 1 mark for proposing a valid Step 1 and Step 2 which add up to the overall reaction.
- 1 mark for identifying the correct slow step (Step 2) that yields the rate expression.
- 1 mark for identifying \(\text{NOCl}_2\) (or another species) as a transient intermediate.

(iv) [4 marks]
- 1 mark for correctly labeled axes (Potential Energy vs. reaction progress).
- 1 mark for showing two distinct peaks with the second transition state higher than the first.
- 1 mark for indicating products lower in energy than reactants (exothermic) and labeling \(\Delta H\).
- 1 mark for correctly labeling the overall activation energy \(E_a\) from the reactant energy level to the highest transition state peak.

(v) [2 marks]
- 1 mark for stating that a higher temperature increases the average kinetic energy of the molecules.
- 1 mark for stating that a larger fraction of molecules have \(E \ge E_a\), leading to more successful collisions.