2025 IB DP Chemistry Practice Paper with Answers

Phosphorus trifluoride (\(\text{PF}_3\)) has 4 electron domains around the central phosphorus atom (3 bonding pairs and 1 lone pair), giving it a tetrahedral electron domain geometry and a trigonal pyramidal molecular geometry. Due to repulsion from the lone pair, the F-P-F bond angle is compressed to less than the tetrahedral angle of \(109.5^\circ\). In contrast, \(\text{BF}_3\), \(\text{SO}_3\), and \(\text{CO}_3^{2-}\) all have 3 electron domains around their central atoms, resulting in trigonal planar geometries with bond angles of approximately \(120^\circ\).

Award 1 mark for the correct option C. Reject all other options.

A conjugate acid-base pair consists of two species that differ by exactly one proton (\(\text{H}^+\)). Here, \(\text{H}_2\text{PO}_4^-\text{ (acid)}\) and \(\text{HPO}_4^{2-}\text{ (conjugate base)}\) differ by one proton, forming a conjugate pair.

Award 1 mark for the correct option C. Reject all other options.

The atomic number of iron (\(\text{Fe}\)) is 26, so its ground-state configuration is \([\text{Ar}] 4\text{s}^2 3\text{d}^6\). When forming the \(\text{Fe}^{3+}\) ion, three electrons are removed. Electrons are lost first from the outer \(4\text{s}\) orbital and then from the \(3\text{d}\) orbital. Therefore, removing 2 electrons from \(4\text{s}\) and 1 electron from \(3\text{d}\) yields the configuration \([\text{Ar}] 3\text{d}^5\).

Award 1 mark for the correct option C. Reject all other options.

Assume a \(100\text{ g}\) sample of the oxide. The mass of manganese is \(100 - 36.8 = 63.2\text{ g}\) and the mass of oxygen is \(36.8\text{ g}\). Convert these masses to moles: \(n(\text{Mn}) = 63.2 / 54.94 = 1.15\text{ mol}\) and \(n(\text{O}) = 36.8 / 16.00 = 2.30\text{ mol}\). Find the simplest whole-number ratio by dividing both by the smallest mole value: \(\text{Mn} = 1.15 / 1.15 = 1\) and \(\text{O} = 2.30 / 1.15 = 2\). Thus, the empirical formula is \(\text{MnO}_2\).

Award 1 mark for the correct option B. Reject all other options.

A catalyst provides an alternative reaction pathway with a lower activation energy, which shifts the activation energy barrier to a lower energy value (to the left on the kinetic energy axis). The Maxwell-Boltzmann distribution curve itself depends only on the temperature of the system and thus remains unchanged.

Award 1 mark for the correct option C. Reject all other options.

The structural formula \(\text{CH}_3\text{CH(OH)COOCH}_2\text{CH}_3\) (ethyl lactate) contains a hydroxyl group bonded to a saturated carbon atom (\(\text{-OH}\)), which is an alcohol group, and an ester group (\(\text{-COO-}\)) linking the carbonyl carbon to the ethyl group. Therefore, the present functional groups are alcohol and ester.

Award 1 mark for the correct option A. Reject all other options.

First ionization energy increases across a period (with minor exceptions) and decreases down a group. For option B, down group 1, first ionization energy decreases: \(\text{K} < \text{Na} < \text{Li}\). Comparing lithium to beryllium (next to it in Period 2), beryllium has a higher first ionization energy because of its greater nuclear charge. Thus, \(\text{K} < \text{Na} < \text{Li} < \text{Be}\) is in correct increasing order. Note: Option A is incorrect because aluminium has a lower first ionization energy than magnesium due to its outer electron residing in a higher-energy \(3\text{p}\) orbital.

Award 1 mark for the correct option B. Reject all other options.

A reducing agent is the species that is oxidized (loses electrons) in a redox reaction. In \(\text{SO}_2\), sulfur has an oxidation state of \(+4\). In \(\text{SO}_4^{2-}\), sulfur has an oxidation state of \(+6\). Since the oxidation state of sulfur increases, \(\text{SO}_2\) is oxidized and therefore acts as the reducing agent, with the oxidation state of S changing from \(+4\) to \(+6\).

Award 1 mark for the correct option B. Reject all other options.

To determine the polar character and hybridization: 1. \(\text{BF}_3\) has 3 single bonds and 0 lone pairs on the boron atom (3 electron domains). It is \(sp^2\) hybridized, but because of its symmetric trigonal planar geometry, it is non-polar. 2. \(\text{SO}_2\) has 2 double bonds (or 1 double and 1 coordinate bond) and 1 lone pair on the sulfur atom (3 electron domains). It is \(sp^2\) hybridized. Due to the lone pair, it has a bent geometry and is polar. 3. \(\text{CO}_2\) has 2 double bonds and 0 lone pairs on the carbon atom (2 electron domains). It is \(sp\) hybridized and non-polar. 4. \(\text{CH}_4\) has 4 single bonds and 0 lone pairs on the carbon atom (4 electron domains). It is \(sp^3\) hybridized and non-polar. Therefore, \(\text{SO}_2\) is the only polar molecule with an \(sp^2\) hybridized central atom.

Award 1 mark for the correct answer (B). No partial marks.

According to Le Chatelier's principle, if the temperature of an equilibrium mixture is increased, the system will shift in the direction that absorbs heat (the endothermic direction). Since the forward reaction is exothermic (\(\Delta H < 0\)), the reverse reaction is endothermic. Therefore, the equilibrium shifts to the left, towards the reactants. This produces more brown \(\text{NO}_2\), making the mixture a darker brown. Because the equilibrium shifts to the left, the concentration of the product decreases relative to the reactants, resulting in a decrease in the value of the equilibrium constant, \(K_c\).

Award 1 mark for the correct answer (B). No partial marks.

A buffer solution consists of a weak acid and its conjugate base, or a weak base and its conjugate acid. In Option A, \(5.0 \text{ mmol}\) of weak base \(\text{NH}_3\) is mixed with \(2.5 \text{ mmol}\) of strong acid \(\text{HCl}\). The \(\text{HCl}\) reacts completely to convert \(2.5 \text{ mmol}\) of \(\text{NH}_3\) to its conjugate acid, \(\text{NH}_4^+\), leaving \(2.5 \text{ mmol}\) of unreacted \(\text{NH}_3\). This forms a basic buffer solution containing \(\text{NH}_3\) and \(\text{NH}_4^+\), which will have a pH greater than 7 at \(298\text{ K}\). Option B forms an acidic buffer with pH less than 7. Option C has an excess of strong acid. Option D represents complete neutralization to form a salt.

Award 1 mark for the correct answer (A). No partial marks.

First, calculate the moles of \(\text{Fe}^{2+}\): \(n(\text{Fe}^{2+}) = 0.0600 \text{ dm}^3 \times 0.500 \text{ mol dm}^{-3} = 0.0300 \text{ mol}\). Next, determine the required moles of \(\text{Cr}_2\text{O}_7^{2-}\) using the mole ratio from the balanced equation (1:6): \(n(\text{Cr}_2\text{O}_7^{2-}) = 0.0300 \text{ mol} / 6 = 0.00500 \text{ mol}\). Finally, calculate the volume of the \(\text{K}_2\text{Cr}_2\text{O}_7\) solution: \(V = n / C = 0.00500 \text{ mol} / 0.200 \text{ mol dm}^{-3} = 0.0250 \text{ dm}^3 = 25.0 \text{ cm}^3\).

Award 1 mark for the correct answer (B). No partial marks.

First, calculate the total mass of the mixture: \(m = 50.0 \text{ g} + 50.0 \text{ g} = 100.0 \text{ g}\). Next, calculate the temperature change: \(\Delta T = 26.8^\circ\text{C} - 20.0^\circ\text{C} = 6.8\text{ K}\). Calculate the heat released: \(q = m c \Delta T = 100.0 \text{ g} \times 4.18 \text{ J g}^{-1}\text{K}^{-1} \times 6.8 \text{ K} = 2842.4 \text{ J} = 2.8424 \text{ kJ}\). Determine the moles of water formed: \(n(\text{H}_2\text{O}) = 0.0500 \text{ dm}^3 \times 1.00 \text{ mol dm}^{-3} = 0.0500 \text{ mol}\). Finally, calculate the enthalpy of neutralization: \(\Delta H = -q / n = -2.8424 \text{ kJ} / 0.0500 \text{ mol} = -56.8 \text{ kJ mol}^{-1}\).

Award 1 mark for the correct answer (C). No partial marks.

For a spontaneous reaction, \(E^\theta_{\text{cell}}\) must be positive: \(E^\theta_{\text{cell}} = E^\theta_{\text{reduction}} - E^\theta_{\text{oxidation}}\). The species with the more positive reduction potential (\(\text{Fe}^{3+}/\text{Fe}^{2+}\) at \(+0.77 \text{ V}\)) undergoes reduction at the cathode. The species with the more negative standard reduction potential (\(\text{Zn}^{2+}/\text{Zn}\) at \(-0.76 \text{ V}\)) undergoes oxidation at the anode. Thus, \(E^\theta_{\text{cell}} = +0.77 \text{ V} - (-0.76 \text{ V}) = +1.53 \text{ V}\). By IUPAC convention, the anode (oxidation) is written on the left and the cathode (reduction) on the right. An inert platinum electrode is used for the iron half-cell because both Fe species are in aqueous solution. Therefore, the cell diagram is \(\text{Zn}(\text{s}) \mid \text{Zn}^{2+}(\text{aq}) \parallel \text{Fe}^{3+}(\text{aq}), \text{Fe}^{2+}(\text{aq}) \mid \text{Pt}(\text{s})\).

Award 1 mark for the correct answer (A). No partial marks.

To find the order with respect to \(\text{A}\), compare Exp 1 and Exp 2: while \([\text{B}]\) is constant, doubling \([\text{A}]\) increases the rate by a factor of 4. This indicates a second-order dependency: \(\text{Rate} \propto [\text{A}]^2\). To find the order with respect to \(\text{B}\), compare Exp 1 and Exp 3: while \([\text{A}]\) is constant, doubling \([\text{B}]\) does not change the rate, indicating a zero-order dependency. Thus, the rate expression is \(\text{Rate} = k[\text{A}]^2\). Substituting values from Exp 1: \(2.0 \times 10^{-4} = k(0.10)^2 \implies k = 2.0 \times 10^{-2} \text{ dm}^3\text{mol}^{-1}\text{s}^{-1}\).

Award 1 mark for the correct answer (B). No partial marks.

A reaction is spontaneous when the change in Gibbs free energy is negative (\(\Delta G^\theta < 0\)). The transition occurs when \(\Delta G^\theta = 0\). Using the relation \(\Delta G^\theta = \Delta H^\theta - T\Delta S^\theta\) and setting \(\Delta G^\theta = 0\), we get \(T = \Delta H^\theta / \Delta S^\theta\). Converting \(\Delta H^\theta\) to Joules gives \(-125000 \text{ J mol}^{-1}\). Thus, \(T = -125000 / -400 = 312.5 \text{ K}\), which is approximately \(313 \text{ K}\).

Award 1 mark for the correct answer (C). No partial marks.

\(\text{SF}_6\) has an octahedral molecular geometry. Because of its highly symmetrical octahedral shape, the individual dipoles of the six polar \(\text{S-F}\) bonds cancel each other out completely, resulting in a net dipole moment of zero (non-polar molecule). In contrast, \(\text{SF}_4\) (see-saw), \(\text{ICl}_3\) (T-shaped), and \(\text{PF}_3\) (trigonal pyramidal) are asymmetrical geometries, meaning their bond dipoles do not cancel, making them polar.

[1 mark] for selecting B. Award 0 marks for any other option.

A conjugate base is defined as the species formed after an acid donates a proton (\(\text{H}^+\)). When \(\text{H}_2\text{PO}_4^-\text{(aq)}\) acts as an acid, it donates a proton: \(\text{H}_2\text{PO}_4^-\text{(aq)} \rightarrow \text{HPO}_4^{2-}\text{(aq)} + \text{H}^+\text{(aq)}\). Thus, \(\text{HPO}_4^{2-}\) is its conjugate base.

[1 mark] for selecting B. Award 0 marks for any other option.

The cathode undergoes reduction, which corresponds to the half-cell with the more positive reduction potential (\(\text{Ce}^{4+}(\text{aq}) + e^- \rightarrow \text{Ce}^{3+}(\text{aq})\), \(E^\ominus = +1.61\text{ V}\)). The anode undergoes oxidation, corresponding to the other half-cell reversed (\(\text{Fe}^{2+}(\text{aq}) \rightarrow \text{Fe}^{3+}(\text{aq}) + e^-\), \(E^\ominus = +0.77\text{ V}\)). Therefore, oxidation of \(\text{Fe}^{2+}\) occurs at the anode.

The cell potential is calculated as:
\(E^\ominus_{\text{cell}} = E^\ominus_{\text{cathode}} - E^\ominus_{\text{anode}} = +1.61\text{ V} - (+0.77\text{ V}) = +0.84\text{ V}\).

[1 mark] for selecting A. Award 0 marks for any other option.

The rate equation is \(\text{Rate} = k[\text{A}][\text{B}]\), which is second-order overall (first order with respect to \(\text{A}\) and first order with respect to \(\text{B}\)).

To find the units of \(k\):
\(k = \frac{\text{Rate}}{[\text{A}][\text{B}]} = \frac{\text{mol dm}^{-3}\text{ s}^{-1}}{(\text{mol dm}^{-3})(\text{mol dm}^{-3})} = \text{dm}^3\text{ mol}^{-1}\text{ s}^{-1}\).

- A is incorrect because the overall order is \(1 + 1 = 2\).
- B is incorrect because \(\text{B}\) is first-order; doubling \([\text{B}]\) will double the rate.
- C is incorrect because if the reaction occurred in a single elementary step, the rate expression would match the stoichiometry: \(\text{Rate} = k[\text{A}][\text{B}]^2\).

[1 mark] for selecting D. Award 0 marks for any other option.

Analyzing the formula of ethyl lactate, \(\text{CH}_3\text{CH}(\text{OH})\text{COOCH}_2\text{CH}_3\):
- The \(-\text{OH}\) group is a hydroxyl group (characteristic of alcohols).
- The \(-\text{COO-}\) group is an ester linkage.
Therefore, the functional groups present are hydroxyl and ester.

[1 mark] for selecting B. Award 0 marks for any other option.

Spontaneity is determined by the Gibbs free energy change, given by \(\Delta G^\ominus = \Delta H^\ominus - T\Delta S^\ominus\).

- Here, \(\Delta H^\ominus < 0\) (exothermic, thermodynamically favorable).
- \(\Delta S^\ominus < 0\) (entropy decreases, thermodynamically unfavorable).

Substituting these signs into the equation gives:
\(\Delta G^\ominus = (\text{negative}) - T(\text{negative}) = (\text{negative}) + T(\text{positive})\).

At low temperatures, the absolute value of the positive \(-T\Delta S^\ominus\) term is small, making \(\Delta G^\ominus < 0\) (spontaneous).
At high temperatures, the positive \(-T\Delta S^\ominus\) term becomes larger than the negative \(\Delta H^\ominus\) term, making \(\Delta G^\ominus > 0\) (non-spontaneous).
Therefore, the reaction is spontaneous only at low temperatures.

[1 mark] for selecting D. Award 0 marks for any other option.

Generally, first ionization energy increases across Period 3 from left to right due to increasing nuclear charge and decreasing atomic radius. However, there is a well-known exception between \(\text{Mg}\) (Group 2) and \(\text{Al}\) (Group 13).
- \(\text{Mg}\) has the electron configuration \([\text{Ne}]3s^2\).
- \(\text{Al}\) has the electron configuration \([\text{Ne}]3s^2 3p^1\).

The single \(3p\) electron of \(\text{Al}\) is in a higher-energy subshell, further from the nucleus, and is shielded by the inner \(3s^2\) electrons. Consequently, less energy is required to remove this electron compared to a \(3s\) electron from \(\text{Mg}\). Thus, the first ionization energy of \(\text{Al}\) is lower than that of \(\text{Mg}\).

Therefore, the correct increasing order is: \(\text{Na} < \text{Al} < \text{Mg} < \text{Si}\).

[1 mark] for selecting B. Award 0 marks for any other option.

The balanced chemical equation for the reaction is:
\(\text{CaCO}_3(\text{s}) + 2\text{HCl}(\text{aq}) \rightarrow \text{CaCl}_2(\text{aq}) + \text{H}_2\text{O}(\text{l}) + \text{CO}_2(\text{g})\)

1. Calculate the amount of \(\text{CaCO}_3\) in moles:
\(n(\text{CaCO}_3) = \frac{\text{mass}}{M_{\text{r}}} = \frac{10.0\text{ g}}{100.09\text{ g mol}^{-1}} \approx 0.100\text{ mol}\)

2. Determine the moles of \(\text{CO}_2\) produced:
Since the mole ratio of \(\text{CaCO}_3\) to \(\text{CO}_2\) is 1:1, \(n(\text{CO}_2) = 0.100\text{ mol}\).

3. Calculate the volume of \(\text{CO}_2\) at STP:
\(V(\text{CO}_2) = n \times V_{\text{m}} = 0.100\text{ mol} \times 22.7\text{ dm}^3\text{ mol}^{-1} = 2.27\text{ dm}^3\).

[1 mark] for selecting A. Award 0 marks for any other option.

To find the number of \(\sigma\) and \(\pi\) bonds in acrylonitrile (\(\text{CH}_2=\text{CH}-\text{C}\equiv\text{N}\)), we count the bonds individually:
- Three \(\text{C}-\text{H}\) single bonds: 3 \(\sigma\) bonds
- One \(\text{C}=\text{C}\) double bond: 1 \(\sigma\) and 1 \(\pi\) bond
- One \(\text{C}-\text{C}\) single bond: 1 \(\sigma\) bond
- One \(\text{C}\equiv\text{N}\) triple bond: 1 \(\sigma\) and 2 \(\pi\) bonds

Summing these up:
- Total \(\sigma\) bonds = \(3 + 1 + 1 + 1 = 6\)
- Total \(\pi\) bonds = \(1 + 2 = 3\)

Therefore, there are 6 \(\sigma\) and 3 \(\pi\) bonds.

[1 mark] for selecting option A.
Award 0 marks for any other option.

Set up an ICE (Initial, Change, Equilibrium) table to find equilibrium concentrations:
- Initial concentration of A: \([\text{A}] = 1.0\text{ mol dm}^{-3}\)
- Initial concentrations of B and C: \([\text{B}] = 0\), \([\text{C}] = 0\)
- Change: \(-2x\) for A, \(+x\) for B, \(+x\) for C

Given \([\text{A}]_{\text{eq}} = 1.0 - 2x = 0.4\text{ mol dm}^{-3}\):
\(2x = 0.6 \implies x = 0.3\text{ mol dm}^{-3}\)

At equilibrium:
- \([\text{A}] = 0.4\text{ mol dm}^{-3}\)
- \([\text{B}] = 0.3\text{ mol dm}^{-3}\)
- \([\text{C}] = 0.3\text{ mol dm}^{-3}\)

The equilibrium constant expression is:
\(K_c = \frac{[\text{B}][\text{C}]}{[\text{A}]^2} = \frac{(0.3)(0.3)}{(0.4)^2} = \frac{0.09}{0.16} = 0.5625 \approx 0.56\)

[1 mark] for correct selection of Option A.
Award 0 marks for incorrect responses.

First, determine the hydrogen ion concentration from the pH:
\([\text{H}^+] = 10^{-\text{pH}} = 10^{-3.0} = 1.0 \times 10^{-3}\text{ mol dm}^{-3}\)

For a weak acid \(\text{HA} \rightleftharpoons \text{H}^+ + \text{A}^-\), the dissociation expression is:
\(K_a = \frac{[\text{H}^+][\text{A}^-]}{[\text{HA}]}\)

Assuming \([\text{H}^+] \approx [\text{A}^-]\) and the change in concentration of \(\text{HA}\) is negligible (\([\text{HA}] \approx 0.10\text{ mol dm}^{-3}\)):
\(K_a \approx \frac{(1.0 \times 10^{-3})^2}{0.10} = \frac{1.0 \times 10^{-6}}{0.10} = 1.0 \times 10^{-5}\)

[1 mark] for correct selection of Option B.

Assume a \(100\text{ g}\) sample of the compound:
- Mass of \(\text{N} = 30.4\text{ g}\)
- Mass of \(\text{O} = 100 - 30.4 = 69.6\text{ g}\)

Convert the masses to moles:
- Moles of \(\text{N} = \frac{30.4}{14.01} \approx 2.17\text{ mol}\)
- Moles of \(\text{O} = \frac{69.6}{16.00} \approx 4.35\text{ mol}\)

Divide each value by the smaller number of moles to find the simplest whole-number ratio:
- Ratio of \(\text{N} = \frac{2.17}{2.17} = 1\)
- Ratio of \(\text{O} = \frac{4.35}{2.17} \approx 2\)

Therefore, the empirical formula is \(\text{NO}_2\).

[1 mark] for selecting option B.

The equation for the formation of methane is:
\(\text{C(s)} + 2\text{H}_2\text{(g)} \rightarrow \text{CH}_4\text{(g)}\n\nAccording to Hess's Law, the enthalpy change of a reaction can be calculated from the enthalpies of combustion of the reactants and products:\n\)\Delta H_f^\ominus = \sum \Delta H_c^\ominus(\text{reactants}) - \sum \Delta H_c^\ominus(\text{products})\n\n\(\Delta H_f^\ominus = [\Delta H_c^\ominus(\text{C(s)}) + 2 \times \Delta H_c^\ominus(\text{H}_2\text{(g)})] - [\Delta H_c^\ominus(\text{CH}_4\text{(g)})]\n\n\)\Delta H_f^\ominus = [-394 + 2(-286)] - [-891]\n\n\(\Delta H_f^\ominus = [-394 - 572] + 891 = -966 + 891 = -75\text{ kJ mol}^{-1}\)

[1 mark] for selecting option A.

Balance using the half-reaction method:

1. Reduction half-reaction (manganese):
\(\text{MnO}_4^- + 8\text{H}^+ + 5e^- \rightarrow \text{Mn}^{2+} + 4\text{H}_2\text{O}\)

2. Oxidation half-reaction (carbon):
\(\text{H}_2\text{C}_2\text{O}_4 \rightarrow 2\text{CO}_2 + 2\text{H}^+ + 2e^-\)

3. Equalize the electrons transferred by multiplying the reduction equation by 2 and the oxidation equation by 5:
- Reduction: \(2\text{MnO}_4^- + 16\text{H}^+ + 10e^- \rightarrow 2\text{Mn}^{2+} + 8\text{H}_2\text{O}\)
- Oxidation: \(5\text{H}_2\text{C}_2\text{O}_4 \rightarrow 10\text{CO}_2 + 10\text{H}^+ + 10e^-\)

4. Add the two half-reactions together and simplify by canceling out \(\text{H}^+\) ions on both sides:
\(2\text{MnO}_4^- + 5\text{H}_2\text{C}_2\text{O}_4 + 6\text{H}^+ \rightarrow 2\text{Mn}^{2+} + 10\text{CO}_2 + 8\text{H}_2\text{O}\)

Thus, the balanced coefficient of \(\text{CO}_2\text{(g)}\) is 10.

[1 mark] for selecting option D.

Let the initial rate be \(\text{rate}_1 = k[\text{NO}]^2[\text{O}_2]\).

When the concentration of \(\text{NO}\) is doubled to \(2[\text{NO}]\) and \(\text{O}_2\) is halved to \(0.5[\text{O}_2]\), the new rate is:
\(\text{rate}_2 = k(2[\text{NO}])^2(0.5[\text{O}_2])\)
\(\text{rate}_2 = k \times 4[\text{NO}]^2 \times 0.5[\text{O}_2]\)
\(\text{rate}_2 = 2 \times k[\text{NO}]^2[\text{O}_2] = 2 \times \text{rate}_1\)

Thus, the rate increases by a factor of 2.

[1 mark] for selecting option B.

For an endothermic reaction, the change in enthalpy is positive (\(\Delta H^\ominus > 0\)). An increase in entropy means the change in entropy is positive (\(\Delta S^\ominus > 0\)).

The Gibbs free energy change equation is:
\(\Delta G^\ominus = \Delta H^\ominus - T\Delta S^\ominus\)

For the reaction to be spontaneous, \(\Delta G^\ominus\) must be negative (\(\Delta G^\ominus < 0\)).
- Since \(\Delta H^\ominus > 0\) (unfavorable) and \(\Delta S^\ominus > 0\) (favorable), the term \(-T\Delta S^\ominus\) must be larger in magnitude than \(\Delta H^\ominus\).
- This occurs only when the temperature, \(T\), is high enough so that \(T\Delta S^\ominus > \Delta H^\ominus\).

Therefore, the reaction is spontaneous only at high temperatures.

[1 mark] for selecting option D.

First, determine the number of moles of copper: \(n = \frac{3.01 \times 10^{22}}{6.02 \times 10^{23}\text{ mol}^{-1}} = 0.0500\text{ mol}\). Then, calculate the mass: \(m = n \times M = 0.0500\text{ mol} \times 63.55\text{ g mol}^{-1} = 3.18\text{ g}\).

Award 1 mark for correct calculation and selection of option A. Deduct 0 marks for incorrect options.

The central oxygen atom in ozone (\(\text{O}_3\)) has three electron domains (one lone pair, one single bond, and one double bond in its resonance forms). The electron domain geometry is trigonal planar, giving an ideal bond angle of 120°. The lone pair slightly compresses this angle, but it remains approximately 120° (about 117°). \(\text{H}_3\text{O}^+\) is trigonal pyramidal (\(\approx 107^\circ\)), \(\text{NH}_4^+\) is tetrahedral (\(109.5^\circ\)), and \(\text{OF}_2\) is bent (\(\approx 103^\circ\)).

Award 1 mark for identifying ozone (A) as the species with a bond angle closest to 120°.

For a first-order reaction, the half-life (\(t_{1/2}\)) is constant. The concentration decreases as follows: \(0.80\text{ mol dm}^{-3} \rightarrow 0.40\text{ mol dm}^{-3}\) (1st half-life) \(\rightarrow 0.20\text{ mol dm}^{-3}\) (2nd half-life). Two half-lives have elapsed in 20 minutes, so \(2 \times t_{1/2} = 20\text{ minutes}\), meaning \(t_{1/2} = 10\text{ minutes}\).

Award 1 mark for the correct calculation of half-life (B).

A conjugate acid-base pair consists of two species that differ by a single proton (\(\text{H}^+\)). Here, \(\text{H}_2\text{PO}_4^-\) acts as the acid, donating a proton to form its conjugate base, \(\text{HPO}_4^{2-}\). These two species differ by exactly one proton, so they constitute a conjugate pair.

Award 1 mark for selecting option A.

Write and balance the two half-reactions. Oxidation half-reaction: \(\text{Fe}^{2+} \rightarrow \text{Fe}^{3+} + \text{e}^-\). Reduction half-reaction: \(\text{MnO}_4^- + 8\text{H}^+ + 5\text{e}^- \rightarrow \text{Mn}^{2+} + 4\text{H}_2\text{O}\). To balance the electrons transferred, multiply the oxidation half-reaction by 5. This gives a combined reaction of: \(5\text{Fe}^{2+} + \text{MnO}_4^- + 8\text{H}^+ \rightarrow 5\text{Fe}^{3+} + \text{Mn}^{2+} + 4\text{H}_2\text{O}\). Thus, the coefficient of \(\text{Fe}^{2+}\) is 5.

Award 1 mark for the correct balanced coefficient (C).

Spontaneity is determined by the equation \(\Delta G = \Delta H - T\Delta S\). A reaction transitions from spontaneous to non-spontaneous (or vice-versa) when \(\Delta G = 0\). Setting \(\Delta H - T\Delta S = 0\) gives \(T = \frac{\Delta H}{\Delta S}\). Convert \(\Delta H\) to \(\text{J mol}^{-1}\): \(-92\text{ kJ mol}^{-1} = -92000\text{ J mol}^{-1}\). Therefore, \(T = \frac{-92000\text{ J mol}^{-1}}{-198\text{ J K}^{-1}\text{mol}^{-1}} \approx 464.6\text{ K}\), which rounds to 465 K.

Award 1 mark for the correct temperature calculation (A).

In a tertiary alcohol, the carbon atom bonded to the hydroxyl (\(-\text{OH}\)) group is directly bonded to three other carbon atoms. In 2-methylpropan-2-ol, the central carbon is bonded to three methyl groups and the hydroxyl group, making it a tertiary alcohol. 2-methylpropan-1-ol is a primary alcohol, whereas butan-2-ol and 3-methylbutan-2-ol are secondary alcohols.

Award 1 mark for identifying 2-methylpropan-2-ol as the tertiary alcohol (A).

As you descend Group 17, the number of electron shells increases, resulting in an increased atomic radius. The outer electrons are further from the nucleus and more shielded, leading to a weaker electrostatic attraction between the nucleus and bonding electrons (decreased electronegativity) as well as easier removal of the outermost electron (decreased first ionization energy).

Award 1 mark for identifying that both electronegativity and first ionization energy decrease (B).

**(a)** Zinc powder is used to increase the surface area of the solid reactant, which increases the rate of the reaction and ensures the maximum temperature is reached quickly before excessive heat is lost to the surroundings.

**(b)**
- Moles of \(\text{Zn} = \frac{1.25\text{ g}}{65.38\text{ g mol}^{-1}} = 0.0191\text{ mol}\)
- Moles of \(\text{Cu}^{2+} = 0.0500\text{ dm}^3 \times 0.200\text{ mol dm}^{-3} = 0.0100\text{ mol}\)
- Since the reaction ratio is \(1:1\), \(\text{Cu}^{2+}\) is the limiting reactant as it is present in a smaller molar amount.

**(c)**
- Mass of solution, \(m = 50.0\text{ g}\) (since \(50.0\text{ cm}^3 \times 1.00\text{ g cm}^{-3} = 50.0\text{ g}\))
- \(q = m \cdot c \cdot \Delta T = 50.0\text{ g} \times 4.18\text{ J g}^{-1}\text{ K}^{-1} \times 10.5\text{ K} = 2194.5\text{ J} = 2.19\text{ kJ}\)

**(d)**
- \(\Delta H = -\frac{q}{n_{\text{limiting}}} = -\frac{2.1945\text{ kJ}}{0.0100\text{ mol}} = -219\text{ kJ mol}^{-1}\) (the negative sign indicates an exothermic reaction).

**(e)**
- Since a temperature change requires two readings (initial and final), the absolute uncertainty is \(2 \times (\pm 0.1\,^{\circ}\text{C}) = \pm 0.2\,^{\circ}\text{C}\) (or if only considering one reading's uncertainty: \(\pm 0.1\,^{\circ}\text{C}\)).
- Percentage uncertainty \(= \frac{0.2}{10.5} \times 100\% = 1.9\%\) (or \(\frac{0.1}{10.5} \times 100\% = 0.95\%\)).

**(f)**
- **Systematic error:** Heat loss to the surroundings / polystyrene cup / thermometer / air.
- **Explanation:** Heat escaping to the surroundings results in a lower measured maximum temperature change (\(\Delta T\)), which leads to a lower calculated heat release (\(q\)) and thus a less negative (less exothermic) value for \(\Delta H\).
- **Minimization:** Use a lid on the polystyrene cup, wrap the cup in cotton wool/insulating material, or use a vacuum flask.

**(a)** [1 mark]
- State that powder increases surface area and increases rate / ensures rapid reaction / minimizes heat loss. [1]

**(b)** [2 marks]
- Correct calculation of moles of Zn (\(0.0191\text{ mol}\)) AND Cu2+ (\(0.0100\text{ mol}\)) [1]
- Correct identification of Cu2+ as limiting reactant with justification [1]

**(c)** [2 marks]
- Correct equation/values substituted: \(q = 50.0 \times 4.18 \times 10.5\) [1]
- Final value \(2.19\text{ kJ}\) (accept \(2.2\text{ kJ}\) or \(2190\text{ J}\)) [1]

**(d)** [2 marks]
- Calculation using limiting moles: \(\frac{q}{0.0100}\) [1]
- Final value with negative sign: \(-219\text{ kJ mol}^{-1}\) (accept range \(-219\) to \(-220\)) [1]

**(e)** [1 mark]
- Correct percentage uncertainty: \(1.9\%\) (accept \(0.95\%\)) [1]

**(f)** [3.6 marks]
- Identifies heat loss to surroundings as systematic error [1]
- Explains that heat loss decreases \(\Delta T\), making calculated \(\Delta H\) less negative / less exothermic [1.6]
- Suggests appropriate insulation method (e.g. use of lid / double cup / vacuum flask) [1]

**(a)** Carbon dioxide (\(\text{CO}_2\)) is a gaseous product. As it is formed, it escapes from the unstoppered flask into the laboratory, causing the mass of the flask and its contents to decrease.

**(b)**
- Average rate \(= \frac{\Delta \text{mass}}{\Delta t} = \frac{0.30\text{ g} - 0.00\text{ g}}{60\text{ s} - 0\text{ s}} = 0.0050\text{ g s}^{-1}\) (or \(5.0 \times 10^{-3}\text{ g s}^{-1}\)).

**(c)**
- As the reaction proceeds, reactants (hydrochloric acid/\(\text{H}^+\) ions) are consumed, meaning their concentration decreases over time.
- According to collision theory, a lower concentration of reactant particles leads to a lower frequency of collisions between reactant particles per unit time, resulting in a lower frequency of successful collisions and therefore a decreasing rate of reaction.

**(d)**
- **Independent variable:** Time (the variable manipulated or controlled by the experimenter).
- **Dependent variable:** Mass loss (the variable being measured as it responds to time).

**(e)**
- Maximum mass loss represents the mass of \(\text{CO}_2\) produced \(= 0.50\text{ g}\).
- Moles of \(\text{CO}_2 = \frac{0.50\text{ g}}{44.01\text{ g mol}^{-1}} = 0.01136\text{ mol}\).
- Volume of \(\text{CO}_2 = 0.01136\text{ mol} \times 24.0\text{ dm}^3\text{ mol}^{-1} = 0.2726\text{ dm}^3\).
- Converting to \(\text{cm}^3\): \(0.2726\text{ dm}^3 \times 1000\text{ cm}^3\text{ dm}^{-3} = 273\text{ cm}^3\) (or \(270\text{ cm}^3\) using 2 significant figures).

**(f)**
- **Difference in curve:** The curve would have a gentler/shallower initial slope (indicating a slower initial rate) but would eventually reach the same maximum mass loss plateau of \(0.50\text{ g}\).
- **Reasoning:** A single large marble chip has a much smaller surface area than the same mass of powder. A smaller surface area reduces the frequency of collisions between the acid particles and the solid reactant. However, since the same mass of reactants is used, the total quantity of gas produced remains the same.

**(a)** [1 mark]
- State that carbon dioxide is a gas and escapes from the flask [1]

**(b)** [1 mark]
- \(0.0050\text{ g s}^{-1}\) (or \(5.0 \times 10^{-3}\text{ g s}^{-1}\)) [1]

**(c)** [2 marks]
- State that reactant concentration (or amount of reactants) decreases [1]
- Explain that this reduces collision frequency / successful collision frequency [1]

**(d)** [2 marks]
- Identify Independent variable = Time [1]
- Identify Dependent variable = Mass loss (or mass of flask/contents) [1]

**(e)** [3 marks]
- Moles of \(\text{CO}_2 = \frac{0.50}{44.01} = 0.0114\text{ mol}\) [1]
- Multiply moles of \(\text{CO}_2\) by \(24.0\text{ dm}^3\text{ mol}^{-1}\) to get volume in \(\text{dm}^3\) (\(0.273\text{ dm}^3\)) [1]
- Final volume in \(\text{cm}^3 = 273\text{ cm}^3\) (or \(270\text{ cm}^3\) to 2 sig figs) [1]

**(f)** [2.6 marks]
- Curve has a gentler/shallower slope initially but levels off at the same mass loss (\(0.50\text{ g}\)) [1]
- Explanation: Smaller surface area of marble chip leads to lower collision frequency [1.6]

**(a)** Since the volume delivered is the difference between two burette readings (final reading \(-\) initial reading), the absolute uncertainty is the sum of the individual uncertainties:
$$\text{Uncertainty} = \pm (0.05 + 0.05)\text{ cm}^3 = \pm 0.10\text{ cm}^3$$

**(b)**
- Moles of KHP \(= \frac{\text{mass}}{M_r} = \frac{0.510\text{ g}}{204.22\text{ g mol}^{-1}} = 0.002497\text{ mol} \approx 2.50 \times 10^{-3}\text{ mol}\) (or \(0.00250\text{ mol}\)).

**(c)**
- Volume of \(\text{NaOH}\) used \(= 24.65\text{ cm}^3 - 0.15\text{ cm}^3 = 24.50\text{ cm}^3 = 0.02450\text{ dm}^3\).
- Since KHP is monoprotic, the reaction stoichiometry with \(\text{NaOH}\) is \(1:1\).
- Moles of \(\text{NaOH} = \text{moles of KHP} = 0.002497\text{ mol}\).
- Concentration of \(\text{NaOH} = \frac{0.002497\text{ mol}}{0.02450\text{ dm}^3} = 0.1019\text{ mol dm}^{-3} \approx 0.102\text{ mol dm}^{-3}\).

**(d)**
- **Indicator:** Phenolphthalein.
- **Color change at end-point:** Colorless to pale pink (accept colorless to permanent pink; do not accept dark pink or red).

**(e)**
- **Effect:** The calculated concentration of \(\text{NaOH}\) would be lower than the true concentration.
- **Explanation:** Water remaining in the burette dilutes the sodium hydroxide solution. A larger volume of this diluted solution is required to neutralize the KHP. Because concentration is calculated using the formula \(C = \frac{n}{V}\), a larger measured volume (\(V\)) in the denominator results in a lower calculated concentration.

**(f)**
- Moles of \(\text{NaOH}\) used \(= 0.01850\text{ dm}^3 \times 0.102\text{ mol dm}^{-3} = 0.001887\text{ mol}\) (or \(0.001850\text{ mol}\) if using exactly \(0.100\text{ mol dm}^{-3}\)).
- Reaction is \(1:1\), so moles of \(\text{CH}_3\text{COOH}\) in \(25.00\text{ cm}^3 = 0.001887\text{ mol}\).
- Moles of \(\text{CH}_3\text{COOH}\) in \(1.00\text{ dm}^3 = 0.001887\text{ mol} \times \frac{1000\text{ cm}^3}{25.00\text{ cm}^3} = 0.07548\text{ mol}\).
- Mass of \(\text{CH}_3\text{COOH}\) in \(1.00\text{ dm}^3 = 0.07548\text{ mol} \times 60.06\text{ g mol}^{-1} = 4.53\text{ g}\) (or \(4.54\text{ g}\)).
- *Note:* If using the unrounded concentration \(0.1019\text{ mol dm}^{-3}\), mass \(= 4.53\text{ g}\). If using \(0.100\text{ mol dm}^{-3}\), mass \(= 4.44\text{ g}\). All these pathways are accepted.

**(a)** [1 mark]
- \(\pm 0.10\text{ cm}^3\) (units must be included) [1]

**(b)** [2 marks]
- Show division by molar mass: \frac{0.510}{204.22} [1]
- Correct amount: \(0.00250\text{ mol}\) or \(2.50 \times 10^{-3}\text{ mol}\) (accept \(0.002497\text{ mol}\)) [1]

**(c)** [2 marks]
- Correct volume of \(\text{NaOH}\) calculated (\(24.50\text{ cm}^3\) or \(0.02450\text{ dm}^3\)) [1]
- Correct concentration of \(\text{NaOH}\): \(0.102\text{ mol dm}^{-3}\) (accept range \(0.1019\) to \(0.102\)) [1]

**(d)** [2 marks]
- Phenolphthalein [1]
- Colorless to pale/permanent pink [1]

**(e)** [2 marks]
- States that calculated concentration is lower than actual [1]
- Explains that dilution increases the volume of titrant needed, reducing the calculated concentration [1]

**(f)** [2.6 marks]
- Calculate moles of \(\text{NaOH}\)/acid used in titration: \(0.01850 \times 0.102 = 1.89 \times 10^{-3}\text{ mol}\) (or equivalent using own value/0.100) [1]
- Calculate moles in \(1.00\text{ dm}^3\): \(\text{moles} \times 40 = 0.0755\text{ mol}\) [0.6]
- Calculate mass of ethanoic acid: \(0.0755 \times 60.06 = 4.53\text{ g}\) (accept range \(4.44\text{ g}\) to \(4.54\text{ g}\)) [1]

(a)(i) Mass of water lost = \(2.379\text{ g} - 1.299\text{ g} = 1.080\text{ g}\). (ii) Moles of anhydrous \(\text{CoCl}_2 = \frac{1.299\text{ g}}{129.83\text{ g mol}^{-1}} = 0.0100\text{ mol}\). Moles of water lost = \frac{1.080\text{ g}}{18.02\text{ g mol}^{-1}} = 0.0600\text{ mol}\). (iii) Ratio of \(\text{H}_2\text{O}\) to \(\text{CoCl}_2 = \frac{0.0600}{0.0100} = 6\), so \(x = 6\). (b)(i) \(\text{Ag}^+\text{(aq)} + \text{Cl}^-\text{(aq)} \rightarrow \text{AgCl(s)}\). (ii) Concentration of anhydrous \(\text{CoCl}_2 = \frac{0.0100\text{ mol}}{0.1000\text{ dm}^3} = 0.100\text{ mol dm}^{-3}\). Since each formula unit of \(\text{CoCl}_2\) produces 2 chloride ions, \([\text{Cl}^-] = 2 \times 0.100 = 0.200\text{ mol dm}^{-3}\). (iii) In a \(10.0\text{ cm}^3\) aliquot: \(n(\text{Cl}^-) = 0.200\text{ mol dm}^{-3} \times 0.0100\text{ dm}^3 = 0.00200\text{ mol}\). Since \(\text{Ag}^+\) and \(\text{Cl}^-\) react in a 1:1 ratio, \(n(\text{AgNO}_3) = 0.00200\text{ mol}\). Volume of \(\text{AgNO}_3 = \frac{0.00200\text{ mol}}{0.100\text{ mol dm}^{-3}} = 0.0200\text{ dm}^3 = 20.0\text{ cm}^3\).

(a)(i) \(1.080\text{ g}\) [1] (ii) Moles of \(\text{CoCl}_2 = 0.0100\text{ mol}\) [1], Moles of \(\text{H}_2\text{O} = 0.0600\text{ mol}\) [1] (iii) \(x = 6\) [1] (b)(i) \(\text{Ag}^+\text{(aq)} + \text{Cl}^-\text{(aq)} \rightarrow \text{AgCl(s)}\) with correct state symbols [1] (ii) \([\text{CoCl}_2] = 0.100\text{ mol dm}^{-3}\) [1], \([\text{Cl}^-] = 0.200\text{ mol dm}^{-3}\) [1] (iii) Moles of \(\text{Cl}^- = 0.00200\text{ mol}\) [1], Volume = \(20.0\text{ cm}^3\) [1]

(a)(i) \(\text{HCOOH(aq)} + \text{H}_2\text{O(l)} \rightleftharpoons \text{HCOO}^-\text{(aq)} + \text{H}_3\text{O}^+\text{(aq)}\) (or with \(\text{H}^+\text{(aq)}\)). (ii) The conjugate base is \(\text{HCOO}^-\). (b) \(K_a = \frac{[\text{H}^+][\text{HCOO}^-]}{[\text{HCOOH}]} \approx \frac{[\text{H}^+]^2}{c_0}\). \([\text{H}^+] = \sqrt{1.77 \times 10^{-4} \times 0.150} = 5.15 \times 10^{-3}\text{ mol dm}^{-3}\). \(\text{pH} = -\log_{10}(5.15 \times 10^{-3}) = 2.29\). Assumptions: The dissociation of HCOOH is negligible compared to its initial concentration (or \([\text{HCOOH}]_{\text{eq}} \approx [\text{HCOOH}]_{\text{initial}}\)), and standard water autoionization is neglected. (c)(i) Volume of mixture = \(100.0\text{ cm}^3\). \([\text{HCOOH}] = \frac{0.150 \times 50.0}{100.0} = 0.0750\text{ mol dm}^{-3}\). \([\text{HCOO}^-] = \frac{0.100 \times 50.0}{100.0} = 0.0500\text{ mol dm}^{-3}\). (ii) \(\text{p}K_a = -\log_{10}(1.77 \times 10^{-4}) = 3.75\). \(\text{pH} = \text{p}K_a + \log_{10}\left(\frac{[\text{HCOO}^-]}{[\text{HCOOH}]}\right) = 3.75 + \log_{10}\left(\frac{0.0500}{0.0750}\right) = 3.57\).

(a)(i) Correct chemical equation with equilibrium arrow [1] (ii) \(\text{HCOO}^-\) [1] (b) \([\text{H}^+] = 5.15 \times 10^{-3}\text{ mol dm}^{-3}\) [1], \(\text{pH} = 2.29\) [1], State assumption (negligible dissociation of acid) [1] (c)(i) \([\text{HCOOH}] = 0.0750\text{ mol dm}^{-3}\) [1], \([\text{HCOO}^-] = 0.0500\text{ mol dm}^{-3}\) [1] (ii) \(\text{p}K_a = 3.75\) [1], \(\text{pH} = 3.57\) [1]

(a)(i) \(\text{SF}_4\): S is central, bonded to 4 F atoms with single bonds. F atoms have 3 lone pairs each. S has 1 lone pair (total of 34 valence electrons). \(\text{XeF}_4\): Xe is central, bonded to 4 F atoms with single bonds. F atoms have 3 lone pairs each. Xe has 2 lone pairs (total of 36 valence electrons). (ii) \(\text{SF}_4\) has 5 electron domains around S. Electron domain geometry: trigonal bipyramidal. Molecular geometry: see-saw. (iii) \(\text{XeF}_4\) has 6 electron domains around Xe. Electron domain geometry: octahedral. Molecular geometry: square planar. (b) Hybridization: S in \(\text{SF}_4\) is \(\text{sp}^3\text{d}\) and Xe in \(\text{XeF}_4\) is \(\text{sp}^3\text{d}^2\). (c) \(\text{SF}_4\) is polar because its asymmetrical see-saw geometry prevents the individual S-F dipoles from cancelling out. \(\text{XeF}_4\) is non-polar because its highly symmetrical square planar shape allows the four polar Xe-F bonds and the two lone pairs to oppose and cancel each other out completely, giving a net dipole of zero.

(a)(i) Correct Lewis structure for \(\text{SF}_4\) showing lone pair on sulfur [1], Correct Lewis structure for \(\text{XeF}_4\) showing 2 lone pairs on xenon [1] (ii) Trigonal bipyramidal [1], See-saw [1] (iii) Octahedral [1], Square planar [1] (b) \(\text{sp}^3\text{d}\) and \(\text{sp}^3\text{d}^2\) (both must be correct) [1] (c) \(\text{SF}_4\) polar with explanation of asymmetry/non-cancelling dipoles [1], \(\text{XeF}_4\) non-polar with explanation of symmetry/cancelling dipoles [1]

(a)(i) Comparing Experiment 1 and 2: \([\text{H}_2]\) is kept constant. \([\text{NO}]\) doubles, and the rate increases by a factor of 4 (\(\frac{5.00 \times 10^{-5}}{1.25 \times 10^{-5}} = 4\)). Since \(2^2 = 4\), the reaction is second order with respect to \(\text{NO}\). (ii) Comparing Experiment 2 and 3: \([\text{NO}]\) is kept constant. \([\text{H}_2]\) doubles, and the rate doubles (\(\frac{1.00 \times 10^{-4}}{5.00 \times 10^{-5}} = 2\)). Since \(2^1 = 2\), the reaction is first order with respect to \(\text{H}_2\). (b)(i) \(\text{Rate} = k[\text{NO}]^2[\text{H}_2]\). (ii) Using data from Experiment 1: \(1.25 \times 10^{-5} = k (5.00 \times 10^{-3})^2 (2.00 \times 10^{-3})\). \(1.25 \times 10^{-5} = k (5.00 \times 10^{-8})\) so \(k = 250\). The units of \(k\) for a third-order reaction are \(\text{dm}^6\text{ mol}^{-2}\text{ s}^{-1}\). (c) Proposed mechanism: Step 1: \(2\text{NO} + \text{H}_2 \rightarrow \text{N}_2\text{O} + \text{H}_2\text{O}\) (slow step) and Step 2: \(\text{N}_2\text{O} + \text{H}_2 \rightarrow \text{N}_2 + \text{H}_2\text{O}\) (fast step). The molecularity of the slow step matches the rate expression.

(a)(i) Identifies comparison between Exp 1 & 2 showing doubling of [NO] [1], concludes second order with respect to [NO] [1] (ii) Identifies comparison between Exp 2 & 3 showing doubling of [H2] [1], concludes first order with respect to [H2] [1] (b)(i) \(\text{Rate} = k[\text{NO}]^2[\text{H}_2]\) [1] (ii) \(k = 250\) [1], units \(\text{dm}^6\text{ mol}^{-2}\text{ s}^{-1}\) [1] (c) Correct slow step (must contain 2 NO and 1 H2 as reactants) [1], correct subsequent fast step leading to the overall reaction stoichiometry [1]

(a)(i) \(\text{Zn(s)} \mid \text{Zn}^{2+}\text{(aq)} \parallel \text{Ag}^+\text{(aq)} \mid \text{Ag(s)}\). (ii) The anode is the zinc electrode (where oxidation occurs). The half-equation is: \(\text{Zn(s)} \rightarrow \text{Zn}^{2+}\text{(aq)} + 2\text{e}^-\). (iii) Electrons flow from the anode (Zn) to the cathode (Ag) through the external circuit. (b)(i) \(\text{Zn(s)} + 2\text{Ag}^+\text{(aq)} \rightarrow \text{Zn}^{2+}\text{(aq)} + 2\text{Ag(s)}\). (ii) \(E^\theta_{\text{cell}} = E^\theta_{\text{cathode}} - E^\theta_{\text{anode}} = +0.80\text{ V} - (-0.76\text{ V}) = +1.56\text{ V}\). (c) \(\Delta G^\theta = -nFE^\theta_{\text{cell}}\). Here, \(n = 2\) moles of electrons are transferred. \(\Delta G^\theta = -2 \times 96485\text{ C mol}^{-1} \times 1.56\text{ V} = -301033.2\text{ J mol}^{-1} = -301\text{ kJ mol}^{-1}\).

(a)(i) \(\text{Zn(s)} \mid \text{Zn}^{2+}\text{(aq)} \parallel \text{Ag}^+\text{(aq)} \mid \text{Ag(s)}\) [1] (ii) Zinc (Zn) [1], \(\text{Zn(s)} \rightarrow \text{Zn}^{2+}\text{(aq)} + 2\text{e}^-\). [1] (iii) From Zn to Ag [1] (b)(i) \(\text{Zn(s)} + 2\text{Ag}^+\text{(aq)} \rightarrow \text{Zn}^{2+}\text{(aq)} + 2\text{Ag(s)}\) [1] (ii) \(+1.56\text{ V}\) [1] (c) Identification of \(n = 2\) [1], Substitution into \(\Delta G^\theta = -nFE^\theta\) [1], Correct calculation of \(-301\text{ kJ mol}^{-1}\) with negative sign [1]

(a)(i) Endothermic, as the temperature of the water decreased. (ii) \(\Delta T = 21.5 - 15.6 = 5.9^\circ\text{C} = 5.9\text{ K}\). Using \(m = 100.0\text{ g}\): \(q = m \cdot c \cdot \Delta T = 100.0\text{ g} \times 4.18\text{ J g}^{-1}\text{ K}^{-1} \times 5.9\text{ K} = 2466.2\text{ J}\). (If using total solution mass of \(108.0\text{ g}\), \(q = 108.0 \times 4.18 \times 5.9 = 2663.5\text{ J}\)). (b)(i) \(n(\text{NH}_4\text{NO}_3) = \frac{8.00\text{ g}}{80.06\text{ g mol}^{-1}} = 0.0999\text{ mol}\). (ii) \(\Delta H_{\text{sol}}^\theta = +\frac{2.466\text{ kJ}}{0.0999\text{ mol}} = +24.7\text{ kJ mol}^{-1}\) (or \(+26.7\text{ kJ mol}^{-1}\) if mass of 108.0 g is used). (c) Sources of systematic error: 1. Heat loss to surroundings: can be minimized by using a lid or vacuum flask. 2. The heat capacity of the calorimeter cup itself is ignored: can be minimized by calibrating the calorimeter to find its heat capacity.

(a)(i) Endothermic because temperature decreased [1] (ii) \(\Delta T = 5.9\text{ K}\) [1], substitution in \(q = mc\Delta T\) [1], \(q = 2466\text{ J}\) (or \(2664\text{ J}\)) [1] (b)(i) \(0.0999\text{ mol}\) [1] (ii) Divides \(q\) in kJ by moles [1], positive sign and correct value \(+24.7\text{ kJ mol}^{-1}\) (or \(+26.7\text{ kJ mol}^{-1}\)) [1] (c) Identifies heat loss/minimization (lid, jacket) [1], identifies heat capacity of cup/minimization (calibration) [1]

(a)(i) \(K_c = \frac{[\text{CH}_3\text{OH}]}{[\text{CO}][\text{H}_2]^2}\). (ii) The forward reaction is exothermic. An increase in temperature will shift the equilibrium position to the left (favoring the endothermic reverse reaction) to absorb heat. This increases the concentration of reactants and decreases the concentration of products, causing the value of \(K_c\) to decrease. (b)(i) Initial concentrations: \([\text{CO}]_0 = \frac{0.400}{2.00} = 0.200\text{ mol dm}^{-3}\); \([\text{H}_2]_0 = \frac{0.800}{2.00} = 0.400\text{ mol dm}^{-3}\). Equilibrium concentration of \(\text{CH}_3\text{OH}\) = \(\frac{0.150}{2.00} = 0.0750\text{ mol dm}^{-3}\). By stoichiometry: \([\text{CO}]_{\text{eq}} = 0.200 - 0.0750 = 0.125\text{ mol dm}^{-3}\); \([\text{H}_2]_{\text{eq}} = 0.400 - 2(0.0750) = 0.250\text{ mol dm}^{-3}\). (ii) \(K_c = \frac{0.0750}{0.125 \times (0.250)^2} = \frac{0.0750}{0.0078125} = 9.60\). Units: \(\text{dm}^6\text{ mol}^{-2}\).

(a)(i) \(K_c = \frac{[\text{CH}_3\text{OH}]}{[\text{CO}][\text{H}_2]^2}\) [1] (ii) State that \(K_c\) decreases [1], explain that the forward reaction is exothermic, hence raising temperature favors reverse reaction [1] (b)(i) \([\text{CH}_3\text{OH}]_{\text{eq}} = 0.0750\text{ mol dm}^{-3}\) [1], \([\text{CO}]_{\text{eq}} = 0.125\text{ mol dm}^{-3}\) [1], \([\text{H}_2]_{\text{eq}} = 0.250\text{ mol dm}^{-3}\) [1] (ii) Correct calculation of \(9.60\) [2], correct units \(\text{dm}^6\text{ mol}^{-2}\) [1]

(a)(i) Isomer 1: \(\text{CH}_3\text{CH}_2\text{CH}_2\text{CH}_2\text{Br}\) (1-bromobutane); Isomer 2: \(\text{CH}_3\text{CH(Br)CH}_2\text{CH}_3\) (2-bromobutane); Isomer 3: \((\text{CH}_3)_2\text{CHCH}_2\text{Br}\) (1-bromo-2-methylpropane); Isomer 4: \((\text{CH}_3)_3\text{CBr}\) (2-bromo-2-methylpropane). (ii) 1-bromobutane is primary; 2-bromobutane is secondary; 1-bromo-2-methylpropane is primary; 2-bromo-2-methylpropane is tertiary. (b)(i) Tertiary halogenoalkanes form highly stable tertiary carbocations due to the positive inductive effect of three alkyl groups, and they are sterically hindered, which prevents direct backside nucleophilic attack (favoring \(\text{S}_\text{N}1\)). Primary halogenoalkanes form highly unstable carbocations, but have little steric hindrance, allowing direct nucleophilic attack to occur easily (favoring \(\text{S}_\text{N}2\)). (ii) Mechanism: Step 1: Slow ionization. A curly arrow starts from the C-Br bond and points to the Br atom. This yields the carbocation intermediate \((\text{CH}_3)_3\text{C}^+\) and \(\text{Br}^-\). Step 2: Nucleophilic attack. A curly arrow starts from a lone pair on the oxygen of \(\text{OH}^-\)\ and points to the carbocation carbon atom, yielding \((\text{CH}_3)_3\text{COH}\).

(a)(i) Award [1] for each correct structure paired with its correct IUPAC name, up to [4] (ii) Correct classification of all four isomers [1] (b)(i) Tertiary: stable carbocation (inductive effect) / steric hindrance prevents SN2 [1], Primary: unstable carbocation / no steric hindrance allows direct backside attack [1] (ii) Correct curly arrow for C-Br fission [1], correct curly arrow for OH- nucleophile attack on the tertiary carbocation intermediate [1]

(a) (i) Propanoic acid is a weak acid and dissociates partially in water:
\(\text{CH}_3\text{CH}_2\text{COOH(aq)} + \text{H}_2\text{O(l)} \rightleftharpoons \text{CH}_3\text{CH}_2\text{COO}^-\text{(aq)} + \text{H}_3\text{O}^+\text{(aq)}\)
or:
\(\text{CH}_3\text{CH}_2\text{COOH(aq)} \rightleftharpoons \text{CH}_3\text{CH}_2\text{COO}^-\text{(aq)} + \text{H}^+\text{(aq)}\)

(ii) The expression for the acid dissociation constant is:
\(K_a = \frac{[\text{H}^+][\text{CH}_3\text{CH}_2\text{COO}^-]}{[\text{CH}_3\text{CH}_2\text{COOH}]}\)

Using the standard assumptions that \([\text{H}^+] \approx [\text{CH}_3\text{CH}_2\text{COO}^-]\) and that the amount of acid dissociated is negligible such that \([\text{CH}_3\text{CH}_2\text{COOH}]_{\text{equilibrium}} \approx 0.100\text{ mol dm}^{-3}\):
\(1.35 \times 10^{-5} = \frac{[\text{H}^+]^2}{0.100}\)
\([\text{H}^+]^2 = 1.35 \times 10^{-6}\)
\([\text{H}^+] = 1.16 \times 10^{-3}\text{ mol dm}^{-3}\)

\(\text{pH} = -\log_{10}(1.16 \times 10^{-3}) = 2.94\)

(b) (i) The reaction is a 1:1 neutralization:
\(\text{CH}_3\text{CH}_2\text{COOH(aq)} + \text{NaOH(aq)} \rightarrow \text{CH}_3\text{CH}_2\text{COONa(aq)} + \text{H}_2\text{O(l)}\)

Since concentration of acid = concentration of base = \(0.100\text{ mol dm}^{-3}\), the volume of \(\text{NaOH}\) required to reach the equivalence point is equal to the volume of acid:
\(V = 25.0\text{ cm}^3\)

At the equivalence point, the pH is greater than 7 (alkaline) because the conjugate base \(\text{CH}_3\text{CH}_2\text{COO}^-\text{(aq)}\) undergoes hydrolysis in water to form hydroxide ions:
\(\text{CH}_3\text{CH}_2\text{COO}^-\text{(aq)} + \text{H}_2\text{O(l)} \rightleftharpoons \text{CH}_3\text{CH}_2\text{COOH(aq)} + \text{OH}^-\text{(aq)}\)

(ii) At the half-equivalence point, a buffer solution containing roughly equal amounts of weak acid (\(\text{CH}_3\text{CH}_2\text{COOH}\)) and its conjugate base (\(\text{CH}_3\text{CH}_2\text{COO}^-\)) is formed.
When a small amount of acid (\(\text{H}^+\)) is added, it reacts with the conjugate base present in the buffer:
\(\text{CH}_3\text{CH}_2\text{COO}^-\text{(aq)} + \text{H}^+\text{(aq)} \rightarrow \text{CH}_3\text{CH}_2\text{COOH(aq)}\)
This removes the added \(\text{H}^+\) ions, thereby resisting a decrease in pH.

(a) (i) [1 mark]
- Reversible dissociation equation with correct reactants, products, and equilibrium arrow (\(\rightleftharpoons\)). State symbols are not required.

(ii) [3 marks]
- Correct formulation and calculation of \([\text{H}^+] = 1.16 \times 10^{-3}\text{ mol dm}^{-3}\) [1]
- Correct calculation of \(\text{pH} = 2.94\) (must be to 2 decimal places) [1]
- Any one of the following assumptions stated: [1]
* The dissociation of the acid is negligible compared to the initial concentration (i.e., \(0.100 - x \approx 0.100\)).
* The dissociation of water itself is negligible (does not contribute significantly to \([\text{H}^+]\)).
* Temperature is maintained constant at \(298\text{ K}\).

(b) (i) [2 marks]
- Volume of \(\text{NaOH(aq)} = 25.0\text{ cm}^3\) [1]
- Predicition of \(\text{pH} > 7\) AND justification that \(\text{CH}_3\text{CH}_2\text{COO}^-\text{(aq)}\) undergoes hydrolysis to produce \(\text{OH}^-\text{(aq)}\) ions [1]

(ii) [3 marks]
- Identification that the solution contains a high concentration of the conjugate base \(\text{CH}_3\text{CH}_2\text{COO}^-\text{(aq)}\) [1]
- Equation showing response to addition of acid: \(\text{CH}_3\text{CH}_2\text{COO}^-\text{(aq)} + \text{H}^+\text{(aq)} \rightarrow \text{CH}_3\text{CH}_2\text{COOH(aq)}\) [1]
- Explanation that the added \(\text{H}^+\) ions are consumed/neutralized, keeping the pH relatively constant [1]

(a) (i)
- Comparing Run 1 and Run 2: \([\text{S}_2\text{O}_8^{2-}]\) is kept constant, \([\text{I}^-]\) is doubled (from \(0.040\) to \(0.080\text{ mol dm}^{-3}\)), and the initial rate doubles (from \(1.25 \times 10^{-5}\) to \(2.50 \times 10^{-5}\text{ mol dm}^{-3}\text{ s}^{-1}\)). Therefore, the reaction is first-order with respect to \(\text{I}^-\).
- Comparing Run 1 and Run 3: \([\text{I}^-]\) is kept constant, \([\text{S}_2\text{O}_8^{2-}]\) is doubled (from \(0.080\) to \(0.160\text{ mol dm}^{-3}\)), and the initial rate doubles (from \(1.25 \times 10^{-5}\) to \(2.50 \times 10^{-5}\text{ mol dm}^{-3}\text{ s}^{-1}\)). Therefore, the reaction is first-order with respect to \(\text{S}_2\text{O}_8^{2-}\).

(ii) The rate expression is:
\(\text{Rate} = k[\text{S}_2\text{O}_8^{2-}][\text{I}^-]\)

To find \(k\), substitute values from Run 1:
\(1.25 \times 10^{-5}\text{ mol dm}^{-3}\text{ s}^{-1} = k (0.080\text{ mol dm}^{-3})(0.040\text{ mol dm}^{-3})\)
\(k = \frac{1.25 \times 10^{-5}}{3.2 \times 10^{-3}} = 3.91 \times 10^{-3}\text{ dm}^3\text{ mol}^{-1}\text{ s}^{-1}\)

Units calculation: \(\frac{\text{mol dm}^{-3}\text{ s}^{-1}}{(\text{mol dm}^{-3})^2} = \text{dm}^3\text{ mol}^{-1}\text{ s}^{-1}\).

(b) (i) A catalyst provides an alternative reaction pathway with a lower activation energy (\(E_a\)). On a Maxwell–Boltzmann distribution curve, lowering the activation energy shift barrier to the left, which means a larger area under the curve (a greater fraction of reactant molecules) possesses kinetic energy equal to or exceeding this new lower activation energy threshold. Consequently, there is a higher frequency of successful collisions.

(ii) Using the Arrhenius equation:
\(k = A e^{-E_a / RT}\)
Rearranging to solve for \(A\):
\(A = k e^{E_a / RT}\)

Converting activation energy to J/mol:
\(E_a = 52.0\text{ kJ mol}^{-1} = 52000\text{ J mol}^{-1}\)
Using \(R = 8.31\text{ J K}^{-1}\text{ mol}^{-1}\) and \(T = 298\text{ K}\):
\(\frac{E_a}{RT} = \frac{52000}{8.31 \times 298} = 21.00\)
\(e^{E_a / RT} = e^{21.00} = 1.32 \times 10^9\)

Using calculated \(k = 3.91 \times 10^{-3}\text{ dm}^3\text{ mol}^{-1}\text{ s}^{-1}\):
\(A = (3.91 \times 10^{-3}) \times (1.32 \times 10^9) = 5.15 \times 10^6\text{ dm}^3\text{ mol}^{-1}\text{ s}^{-1}\)

(If using the alternative value \(k = 4.00 \times 10^{-3}\text{ dm}^3\text{ mol}^{-1}\text{ s}^{-1}\)):
\(A = (4.00 \times 10^{-3}) \times (1.32 \times 10^9) = 5.26 \times 10^6\text{ dm}^3\text{ mol}^{-1}\text{ s}^{-1}\)
*(Note: If \(R = 8.314\text{ J K}^{-1}\text{ mol}^{-1}\) is used, \(A = 5.10 \times 10^6\text{ dm}^3\text{ mol}^{-1}\text{ s}^{-1}\) for the first case, and \(5.21 \times 10^6\text{ dm}^3\text{ mol}^{-1}\text{ s}^{-1}\) for the alternative case)*.

(a) (i) [3 marks]
- Order with respect to \(\text{I}^-\): First order AND reasoning referencing Run 1 and Run 2 (concentration of iodide doubles, rate doubles while peroxydisulfate concentration is constant) [1]
- Order with respect to \(\text{S}_2\text{O}_8^{2-}\): First order AND reasoning referencing Run 1 and Run 3 (concentration of peroxydisulfate doubles, rate doubles while iodide concentration is constant) [1]
- Correct final statement of both orders [1]

(ii) [2 marks]
- Correct rate expression: \(\text{Rate} = k[\text{S}_2\text{O}_8^{2-}][\text{I}^-]\) [0.5]
- Calculated value: \(k = 3.91 \times 10^{-3}\) (accept \(3.9 \times 10^{-3}\)) [1]
- Correct units: \(\text{dm}^3\text{ mol}^{-1}\text{ s}^{-1}\) (or \(\text{L mol}^{-1}\text{ s}^{-1}\) or \(\text{M}^{-1}\text{ s}^{-1}\)) [0.5]

(b) (i) [2 marks]
- States that a catalyst provides an alternative pathway with a lower activation energy [1]
- Explains that a larger fraction/proportion of molecules now have kinetic energy greater than or equal to this lower activation energy (leading to a higher frequency of successful collisions) [1]

(ii) [2 marks]
- Correct substitution of values into Arrhenius equation, with \(E_a\) correctly converted to \(\text{J mol}^{-1}\) (e.g., \(52000\)) [1]
- Correct calculation of \(A\) including matching units (accept values in range \(5.10 \times 10^6\) to \(5.30 \times 10^6\) depending on \(R\) used and whether the alternative \(k\) was used) [1]