2025 IB DP Chemistry 模擬試題連答案詳解

Ozone (\(\text{O}_3\)) has a central oxygen atom bonded to two other oxygen atoms, with one lone pair. This gives rise to a trigonal planar electron domain geometry (three electron domains). Due to lone pair-bonding pair repulsion, the bond angle is slightly less than the ideal \(120^\circ\), measuring approximately \(117^\circ\). \(\text{NF}_3\), \(\text{H}_2\text{O}\), and \(\text{OF}_2\) all have four electron domains (tetrahedral electron domain geometry), resulting in bond angles significantly closer to \(109.5^\circ\) or less (e.g., \(\text{H}_2\text{O}\) is \(104.5^\circ\)).

[1 mark] for selecting B (\(\text{O}_3\)).

The equilibrium constant, \(K_c\), is only affected by changes in temperature. For an endothermic reaction (\(\Delta H > 0\)), increasing the temperature shifts the equilibrium position to the right (products side) to absorb the added heat energy. Since more products are formed, the value of \(K_c\) increases. Changes in volume, pressure, or adding a catalyst alter the position of equilibrium or the rate of reaction, but do not affect the value of \(K_c\).

[1 mark] for selecting B.

The pH of a salt solution depends on the hydrolysis of its ions:
- \(\text{NH}_4\text{Cl}\) is the salt of a weak base (\(\text{NH}_3\)) and a strong acid (\(\text{HCl}\)). The \(\text{NH}_4^+\) ion undergoes hydrolysis to produce \(\text{H}_3\text{O}^+\) ions, resulting in an acidic solution (\(\text{pH} < 7\)).
- \(\text{CH}_3\text{COOH}\) is a weak acid, resulting in an acidic solution (\(\text{pH} < 7\)).
- \(\text{Na}_2\text{CO}_3\) is the salt of a strong base (\(\text{NaOH}\)) and a weak acid (\(\text{H}_2\text{CO}_3\)). The \(\text{CO}_3^{2-}\) ion undergoes hydrolysis to produce \(\text{OH}^-\), resulting in a basic solution (\(\text{pH} > 7\)).
- \(\text{NaCl}\) is the salt of a strong base and strong acid, and does not undergo hydrolysis (\(\text{pH} \approx 7\)).
Therefore, \(\text{Na}_2\text{CO}_3\text{(aq)}\) has the highest pH.

[1 mark] for selecting C.

The half-cell with the more positive standard reduction potential is the cathode where reduction takes place. Thus, the cathode reaction is:
\(\text{Fe}^{3+}\text{(aq)} + \text{e}^- \rightarrow \text{Fe}^{2+}\text{(aq)}\), and \(\text{Fe}^{3+}\text{(aq)}\) is reduced.

The anode reaction is where oxidation takes place:
\(\text{Cr}\text{(s)} \rightarrow \text{Cr}^{3+}\text{(aq)} + 3\text{e}^-\).

The standard cell potential is:
\(E^\theta_{\text{cell}} = E^\theta_{\text{cathode}} - E^\theta_{\text{anode}} = +0.77\text{ V} - (-0.74\text{ V}) = +1.51\text{ V}\).

[1 mark] for selecting A.

The rate-determining step is the slow step (Step 2). From Step 2, the rate expression is:
\(\text{Rate} = k_2[\text{N}_2\text{O}_2][\text{O}_2]\)

Since \(\text{N}_2\text{O}_2\) is an intermediate, we express its concentration using the fast pre-equilibrium of Step 1:
\(K_{\text{eq}} = \frac{[\text{N}_2\text{O}_2]}{[\text{NO}]^2} \Rightarrow [\text{N}_2\text{O}_2] = K_{\text{eq}}[\text{NO}]^2\)

Substituting this back into the rate expression of the rate-determining step:
\(\text{Rate} = k_2 K_{\text{eq}}[\text{NO}]^2[\text{O}_2] = k[\text{NO}]^2[\text{O}_2]\).

[1 mark] for selecting C.

The compound has two functional groups: an aldehyde (\(-\text{CHO}\)) and a hydroxyl group (\(-\text{OH}\)).
According to IUPAC naming priorities, the aldehyde group has a higher priority than the hydroxyl group and is designated as the principal functional group (suffix '-al').
The carbon chain is numbered starting from the aldehyde carbon (C1):
- C1: \(-\text{CHO}\)
- C2: \(-\text{CH}_2-\)
- C3: \(-\text{CH(OH)}-\)
- C4: \(-\text{CH}_3\)

This gives a 4-carbon parent chain (butanal) with a hydroxyl substituent on carbon 3 (referred to as a 'hydroxy' prefix). Therefore, the correct name is 3-hydroxybutanal.

[1 mark] for selecting A.

First, calculate the empirical formula:
- Amount of C: \(40.0\text{ g} / 12.01\text{ g mol}^{-1} \approx 3.33\text{ mol}\)
- Amount of H: \(6.7\text{ g} / 1.01\text{ g mol}^{-1} \approx 6.63\text{ mol}\)
- Amount of O: \(53.3\text{ g} / 16.00\text{ g mol}^{-1} \approx 3.33\text{ mol}\)

Divide each mole value by the smallest value (3.33):
- C: \(3.33 / 3.33 = 1\)
- H: \(6.63 / 3.33 \approx 2\)
- O: \(3.33 / 3.33 = 1\)

The empirical formula is \(\text{CH}_2\text{O}\).

The formula mass of the empirical unit is:
\(12.01 + 2(1.01) + 16.00 = 30.03\text{ g mol}^{-1}\)

Find the ratio of the molar mass of the compound to the empirical formula mass:
\(\text{Ratio} = 180 / 30.03 \approx 6\)

Therefore, the molecular formula is \(\text{C}_6\text{H}_{12}\text{O}_6\), which has 6 carbon atoms.

[1 mark] for selecting C.

The neutral iron atom (Fe) has an atomic number of 26. Its electron configuration is:
\([\text{Ar}] 4\text{s}^2 3\text{d}^6\)

When transition metals form cations, they lose electrons from the outer \(s\) subshell (the \(4\text{s}\) orbital) first, before losing any from the \(d\) subshell (the \(3\text{d}\) orbital).

To form the \(\text{Fe}^{2+}\) ion, two electrons are removed from the neutral atom. Removing the two \(4\text{s}\) electrons results in the configuration:
\([\text{Ar}] 3\text{d}^6\).

[1 mark] for selecting B.

Step 1: Determine the mass of water lost. Mass of water = 5.00 g - 2.44 g = 2.56 g. Step 2: Calculate the amount in moles of anhydrous magnesium sulfate. n(MgSO4) = 2.44 g / 120.37 g/mol = 0.0203 mol. Step 3: Calculate the amount in moles of water lost. n(H2O) = 2.56 g / 18.02 g/mol = 0.142 mol. Step 4: Determine the ratio of water to anhydrous salt. x = n(H2O) / n(MgSO4) = 0.142 / 0.0203 = 7.00. Therefore, x = 7.

Award 1 mark for the correct answer C. Method: Calculate moles of water and anhydrous salt, find the simplest ratio.

The terminal carbon atoms (C1 and C3) form two single bonds to hydrogen and one double bond to the central carbon (C2). This represents three electron domains, corresponding to sp2 hybridization. The central carbon atom (C2) forms two double bonds (one to C1 and one to C3). This represents two electron domains, corresponding to sp hybridization. Therefore, the hybridizations are sp2, sp, and sp2 respectively.

Award 1 mark for the correct answer B. Method: Count electron domains around each carbon to deduce hybridization (3 domains = sp2, 2 domains = sp).

The overall order is the sum of the powers of the concentration terms in the rate expression: 1 + 2 = 3. To find the units of the rate constant: k = Rate / ([A][B]^2). Substituting units: [mol dm^-3 s^-1] / ([mol dm^-3] * [mol dm^-3]^2) = [mol dm^-3 s^-1] / [mol^3 dm^-9] = mol^-2 dm^6 s^-1, which is written as dm^6 mol^-2 s^-1.

Award 1 mark for the correct answer B. Method: Sum exponents for overall order, rearrange rate law to solve for the units of k.

When equal volumes are mixed, the sodium hydroxide reacts with half of the ethanoic acid to produce sodium ethanoate: CH3COOH + NaOH -> CH3COONa + H2O. Let the volume of each solution be V. Initial moles of CH3COOH = 0.20*V, initial moles of NaOH = 0.10*V. After reaction, moles of CH3COOH remaining = 0.20*V - 0.10*V = 0.10*V. Moles of CH3COO- formed = 0.10*V. Because [CH3COOH] = [CH3COO-], the pH of the buffer equals the pKa of the weak acid. pH = pKa = 4.76.

Award 1 mark for the correct answer C. Method: Determine the neutralization stoichiometry, identify that the buffer is at the half-equivalence point where pH = pKa.

The half-cell with the more positive reduction potential (Fe3+/Fe2+, E = +0.77 V) will undergo reduction. The other half-cell (Cr3+/Cr, E = -0.74 V) will undergo oxidation (Cr -> Cr3+ + 3e-). E_cell = E_reduction - E_oxidation = +0.77 V - (-0.74 V) = +1.51 V. Combining the half-equations to balance electrons: 3(Fe3+ + e- -> Fe2+) and Cr -> Cr3+ + 3e- yields 3Fe3+(aq) + Cr(s) -> 3Fe2+(aq) + Cr3+(aq).

Award 1 mark for the correct answer B. Method: Determine which species is oxidized/reduced, compute cell potential as E(cathode) - E(anode), and balance the electrons for the overall equation.

The formula contains a -OH group attached to a carbon chain, which is an alcohol (hydroxyl group). It also contains a -COOCH3 group, which is an ester group. Thus, the correct functional groups are alcohol and ester.

Award 1 mark for the correct answer B. Method: Identify -OH as alcohol and -COO- as ester from the condensed molecular formula.

Since the forward reaction is endothermic, increasing the temperature will shift the equilibrium to the right, which increases the value of Kc (Kc depends only on temperature). Halving the volume of the vessel increases the pressure, which shifts the equilibrium towards the side with fewer gas moles (to the left, towards N2O4). Thus, Kc increases and the equilibrium shifts to the left due to the volume change.

Award 1 mark for the correct answer A. Method: Analyze temperature effect on Kc for endothermic reaction, and pressure/volume effect on equilibrium position according to Le Chatelier's principle.

First, calculate the heat absorbed by the water: q = m * c * dT = 150.0 g * 4.18 J/(g K) * 25.0 K = 15675 J = 15.675 kJ. Next, calculate the moles of methanol burned: n = mass / molar mass = 0.80 g / 32.0 g/mol = 0.025 mol. Finally, calculate the enthalpy of combustion: dH = -q / n = -15.675 kJ / 0.025 mol = -627 kJ/mol. Since the reaction is exothermic, the sign must be negative.

Award 1 mark for the correct answer A. Method: Use q = mc dT to find heat change, divide by moles of fuel burned, and assign a negative sign for exothermic combustion.

The ground-state electron configuration of a neutral iron (Fe) atom is \([\text{Ar}] 4\text{s}^2 3\text{d}^6\). When transition metals form ions, they first lose electrons from the outermost s-subshell (4s) before losing electrons from the d-subshell (3d). Removing three electrons to form \(\text{Fe}^{3+}\) results in the loss of both 4s electrons and one 3d electron, giving \([\text{Ar}] 3\text{d}^5\).

Award [1] for the correct ground-state electron configuration option (B).

To find the empirical formula, divide the percentage of each element by its relative atomic mass: \(n(\text{C}) = \frac{85.7}{12.01} = 7.14\text{ mol}\); \(n(\text{H}) = \frac{14.3}{1.01} = 14.16\text{ mol}\). Find the simplest whole-number ratio by dividing by the smallest value (7.14): \(\text{C}: \frac{7.14}{7.14} = 1\); \(\text{H}: \frac{14.16}{7.14} \approx 2\). Therefore, the empirical formula is \(\text{CH}_2\).

Award [1] for the correct empirical formula option (B).

Chlorine has 7 valence electrons. It forms 3 covalent bonds with fluorine atoms, leaving 4 non-bonding valence electrons (2 lone pairs). The total number of electron domains is 5 (3 bonding domains and 2 lone pairs), which gives a trigonal bipyramidal electron domain geometry. To minimize repulsions, the two lone pairs occupy the equatorial positions, resulting in a T-shaped molecular geometry. The repulsions from the lone pairs compress the axial-equatorial bond angle to slightly less than \(90^\circ\).

Award [1] for the correct molecular geometry and bond angle option (C).

A Brønsted-Lowry base is a proton (\(\text{H}^+\)) acceptor. In the forward reaction, \(\text{HCO}_3^-\text{ accepts a proton from } \text{H}_2\text{PO}_4^- \text{ to form } \text{H}_2\text{CO}_3\), making \(\text{HCO}_3^-\text{ a base}. In the reverse reaction, \)\text{HPO}_4^{2-}\text{ accepts a proton from } \text{H}_2\text{CO}_3\text{ to form } \text{H}_2\text{PO}_4^-\), making \(\text{HPO}_4^{2-}\text{ a base}.

Award [1] for the correct identification of bases (B).

Comparing Run 1 and Run 2: \([\text{NO}]\) is kept constant, \([\text{Cl}_2]\) doubles, and the rate doubles (\(1.2 \times 10^{-3}\text{ to } 2.4 \times 10^{-3}\)). Therefore, the reaction is first-order with respect to \(\text{Cl}_2\). Comparing Run 1 and Run 3: \([\text{Cl}_2]\) is kept constant, \([\text{NO}]\) doubles, and the rate quadruples (\(1.2 \times 10^{-3}\text{ to } 4.8 \times 10^{-3}\)). Therefore, the reaction is second-order with respect to \(\text{NO}\). The overall order of reaction is the sum of the individual orders: \(1 + 2 = 3\).

Award [1] for the correct overall order of reaction (C).

The equilibrium constant, \(K_c\), is only affected by changes in temperature. Since the forward reaction is exothermic (\(\Delta H < 0\)), lowering the temperature shifts the position of the equilibrium to the right to produce more heat, resulting in a higher concentration of products relative to reactants at the new equilibrium state, which increases the value of \(K_c\).

Award [1] for the correct temperature change (B).

Determine the oxidation states of the transition metals: In \(\text{K}_2\text{Cr}_2\text{O}_7\): \(2(+1) + 2(x) + 7(-2) = 0 \implies x = +6\) (for Cr). In \(\text{KMnO}_4\): \((+1) + y + 4(-2) = 0 \implies y = +7\) (for Mn). In \(\text{K}_2\text{MnO}_4\): \(2(+1) + z + 4(-2) = 0 \implies z = +6\) (for Mn). In \(\text{CrO}_2\text{Cl}_2\): \(w + 2(-2) + 2(-1) = 0 \implies w = +6\) (for Cr). Thus, manganese in \(\text{KMnO}_4\) has the highest oxidation state of +7.

Award [1] for the correct option containing the highest oxidation state (B).

Methyl 2-hydroxybenzoate contains a carboxyl group that is esterified to a methyl group (\(-\text{COOCH}_3\)), representing an ester functional group. It also contains a hydroxyl group (\(-\text{OH}\)) directly attached to a benzene ring, which represents a phenol functional group.

Award [1] for the correct functional groups (B).

First, calculate the amount (in moles) of carbon and hydrogen from the products:
- \(n(\text{C}) = n(\text{CO}_2) = \frac{2.20\text{ g}}{44.01\text{ g mol}^{-1}} \approx 0.050\text{ mol}\)
- Mass of \(\text{C} = 0.050\text{ mol} \times 12.01\text{ g mol}^{-1} = 0.60\text{ g}\)
- \(n(\text{H}) = 2 \times n(\text{H}_2\text{O}) = 2 \times \frac{0.90\text{ g}}{18.02\text{ g mol}^{-1}} \approx 0.10\text{ mol}\)
- Mass of \(\text{H} = 0.10\text{ mol} \times 1.01\text{ g mol}^{-1} = 0.10\text{ g}\)

Next, determine the mass and moles of oxygen by subtraction:
- Mass of \(\text{O} = 1.50\text{ g} - (0.60\text{ g} + 0.10\text{ g}) = 0.80\text{ g}\)
- \(n(\text{O}) = \frac{0.80\text{ g}}{16.00\text{ g mol}^{-1}} = 0.050\text{ mol}\)

Finally, find the simplest whole number ratio of the elements:
- \(\text{C} : \text{H} : \text{O} = 0.050 : 0.10 : 0.050 = 1 : 2 : 1\)
- Therefore, the empirical formula is \(\text{CH}_2\text{O}\).

[1 mark] for identifying the correct empirical formula (A).
- [0 marks] for any incorrect response.

- \(\text{CF}_4\): Tetrahedral molecular geometry. The individual polar C-F bonds cancel out symmetrically, resulting in a non-polar molecule.
- \(\text{SF}_4\): Seesaw molecular geometry (5 electron domains, 4 bonding pairs, 1 lone pair on the sulfur atom). Due to the lone pair, the molecule is asymmetrical and the polar bond dipoles do not cancel, giving a net dipole moment (polar).
- \(\text{XeF}_4\): Square planar molecular geometry (6 electron domains, 4 bonding pairs, 2 lone pairs). The dipoles cancel out symmetrically, resulting in a non-polar molecule.
- \(\text{BF}_3\): Trigonal planar molecular geometry. Symmetrical distribution of polar bonds cancels out the dipoles, resulting in a non-polar molecule.

[1 mark] for choosing B.
- [0 marks] for any other option.

- A is correct because both hydrochloric acid (strong monoprotic) and ethanoic acid (weak monoprotic) have the same concentration and volume, meaning they contain the same total amount (moles) of neutralizable hydrogen ions. Thus, they require the exact same volume of \(\text{NaOH(aq)}\) for complete neutralization.
- B is incorrect because \(\text{HCl}\) is a strong acid and dissociates fully, releasing a higher concentration of \(\text{H}^+\) ions and therefore having a lower pH than the weakly dissociated \(\text{CH}_3\text{COOH}\).
- C is incorrect because fully dissociated \(\text{HCl(aq)}\) has a higher concentration of mobile ions and thus a higher electrical conductivity.
- D is incorrect because the lower concentration of hydrogen ions in \(\text{CH}_3\text{COOH(aq)}\) leads to a slower initial rate of reaction with magnesium compared to \(\text{HCl(aq)}\).

[1 mark] for selecting statement A.
- [0 marks] for any incorrect statement.

For mechanism A:
- Step 1 is a fast equilibrium, so we can write the equilibrium constant: \(K_c = \frac{[\text{NOCl}_2]}{[\text{NO}][\text{Cl}_2]}\), which gives \([\text{NOCl}_2] = K_c [\text{NO}][\text{Cl}_2]\).
- Step 2 is the rate-determining step (slow), so the rate is determined by: \(\text{Rate} = k_2 [\text{NOCl}_2][\text{NO}]\).
- Substituting the expression for \([\text{NOCl}_2]\) into the rate equation gives: \(\text{Rate} = k_2 K_c [\text{NO}][\text{Cl}_2][\text{NO}] = k [\text{NO}]^2[\text{Cl}_2]\).
- This matches the experimental rate expression.

For mechanism B: \(\text{Rate} = k[\text{NO}]^2\).
For mechanism C: \(\text{Rate} = k[\text{Cl}_2]\).
For mechanism D: \(\text{Rate} = k[\text{NO}][\text{Cl}_2]\).

[1 mark] for selecting mechanism A.
- [0 marks] for any other mechanism.

A spontaneous reaction must have a positive overall standard cell potential (\(E^\ominus_{\text{cell}} > 0\)).

- For reaction B:
- Reduction half-reaction: \(\text{Sn}^{4+}\text{(aq)} + 2\text{e}^- \rightarrow \text{Sn}^{2+}\text{(aq)}\quad E^\ominus_{\text{red}} = +0.15\text{ V}\)
- Oxidation half-reaction: \(\text{Cr(s)} \rightarrow \text{Cr}^{3+}\text{(aq)} + 3\text{e}^-\quad E^\ominus_{\text{ox}} = -(-0.74\text{ V}) = +0.74\text{ V}\)
- \(E^\ominus_{\text{cell}} = E^\ominus_{\text{red}} + E^\ominus_{\text{ox}} = +0.15\text{ V} + 0.74\text{ V} = +0.89\text{ V}\).
- Since \(E^\ominus_{\text{cell}} > 0\), the reaction is spontaneous.

- For other reactions, \(E^\ominus_{\text{cell}}\) values are negative, indicating non-spontaneous reactions under standard conditions.

[1 mark] for identifying the correct spontaneous reaction (B).
- [0 marks] for any incorrect response.

- The atomic number of cobalt (\(\text{Co}\)) is 27.
- The ground-state electron configuration of a neutral \(\text{Co}\) atom is \([\text{Ar}] 4\text{s}^2 3\text{d}^7\).
- When transition metals form ions, electrons are lost first from the highest energy occupied main shell, which is the outer \(4\text{s}\) subshell before the \(3\text{d}\) subshell.
- Therefore, to form \(\text{Co}^{2+}\), the two \(4\text{s}\) electrons are removed, resulting in the configuration \([\text{Ar}] 3\text{d}^7\).

[1 mark] for selecting electron configuration A.
- [0 marks] for any other response.

(a) Extrapolation corrects for heat loss to the surroundings that occurs during the mixing process and before the maximum temperature is reached. It estimates the temperature change that would have occurred if the reaction had been instantaneous with no heat loss.

(b) Total volume of solution = \(50.0\text{ cm}^3 + 50.0\text{ cm}^3 = 100.0\text{ cm}^3\)
Mass of solution, \(m = 100.0\text{ g}\) (using density \(1.00\text{ g cm}^{-3}\))
Temperature change, \(\Delta T = 26.80 - 20.20 = 6.60^\circ\text{C}\) (or \(6.60\text{ K}\))
\(q = mc\Delta T = 100.0\text{ g} \times 4.18\text{ J g}^{-1}\text{ K}^{-1} \times 6.60\text{ K} = 2758.8\text{ J}\) (or \(2.76\text{ kJ}\))

(c) Number of moles of \(\text{HCl}\) used = \(1.00\text{ mol dm}^{-3} \times 0.0500\text{ dm}^3 = 0.0500\text{ mol}\)
Number of moles of \(\text{NaOH}\) used = \(1.00\text{ mol dm}^{-3} \times 0.0500\text{ dm}^3 = 0.0500\text{ mol}\)
Number of moles of water formed = \(0.0500\text{ mol}\)
\(\Delta H_{\text{neu}} = -\frac{q}{\text{moles of water}} = -\frac{2.7588\text{ kJ}}{0.0500\text{ mol}} = -55.2\text{ kJ mol}^{-1}\)
(Note: the sign must be negative as the reaction is exothermic.)

(d) Percentage uncertainty = \(\frac{0.06}{50.0} \times 100\% = 0.12\%\)

Award 8.33 marks as follows:

(a) [2 marks]
- Correctly states that heat is lost to the surroundings during the reaction [1]
- Explains that extrapolation estimates the temperature rise if the reaction were instantaneous / corrects for this heat loss [1]

(b) [2 marks]
- Calculates \(\Delta T = 6.60\text{ K}\) and \(m = 100.0\text{ g}\) [1]
- Calculates \(q = 2758.8\text{ J}\) or \(2.76\text{ kJ}\) [1]

(c) [2.33 marks]
- Calculates moles of \(\text{H}_2\text{O}\) formed = \(0.0500\text{ mol}\) [1]
- Calculates \(\Delta H_{\text{neu}} = -55.2\text{ kJ mol}^{-1}\) [1.33] (deduct 0.5 marks if negative sign is missing; accept answers in range -55.1 to -55.2 depending on rounding of \(q\))

(d) [2 marks]
- Shows calculation: \(\frac{0.06}{50.0} \times 100\) [1]
- Correct percentage uncertainty: \(0.12\%\) [1]

(a) A suitable diagram must show:
- A reaction vessel (conical flask or boiling tube) containing the reactants (\(\text{CaCO}_3\) and \(\text{HCl}\)), sealed with a stopper/bung.
- A delivery tube connecting the reaction vessel to a gas collection apparatus.
- A gas syringe OR an inverted measuring cylinder filled with water in a trough of water.
- All key components correctly labeled.

(b) Draw a tangent to the curve at time \(t = 0\text{ s}\) (the origin) and determine the gradient of this tangent (\(\text{gradient} = \frac{\Delta y}{\Delta x}\)), which represents the initial rate of reaction.

(c) The rate of reaction decreases. A single marble chip has a smaller total surface area than the same mass of powder. This results in fewer exposed particles, leading to a lower frequency of successful collisions between reactant particles.

(d) The measured volume of gas will be lower than the actual volume produced because some \(\text{CO}_2\) dissolves in the water. This systematic error can be minimized by using a gas syringe (which does not collect gas over water) OR by saturating the water with carbon dioxide (or using warm water/acidified water) before starting the collection.

Award 8.33 marks as follows:

(a) [3 marks]
- Diagram shows a closed system with a reaction flask and delivery tube [1]
- Diagram shows a gas syringe OR an inverted measuring cylinder in a water bath [1]
- All major apparatus components clearly labeled (reactants, delivery tube, syringe/measuring cylinder) [1]

(b) [2 marks]
- Mentions drawing a tangent at the origin / at \(t = 0\) [1]
- States that the gradient of this tangent is calculated to find the initial rate [1]

(c) [2 marks]
- States that the rate decreases [1]
- Explains that a single chip has a smaller surface area, resulting in a lower collision frequency / fewer successful collisions per unit time [1]

(d) [1.33 marks]
- Identifies that the measured volume is lower/underestimated [0.33]
- Suggests using a gas syringe OR pre-saturating the water with \(\text{CO}_2\) [1.00]

(a) Heating to constant mass ensures that all the water of crystallization has been completely driven off and the decomposition/dehydration reaction is complete.

(b)
(i) Mass of anhydrous \(\text{MgSO}_4\) residue = \(25.55\text{ g} - 24.35\text{ g} = 1.20\text{ g}\)
(ii) Mass of water lost = \(26.81\text{ g} - 25.55\text{ g} = 1.26\text{ g}\)

(c)
- Moles of \(\text{MgSO}_4 = \frac{1.20\text{ g}}{120.38\text{ g mol}^{-1}} = 0.00997\text{ mol}\)
- Moles of \(\text{H}_2\text{O} = \frac{1.26\text{ g}}{18.02\text{ g mol}^{-1}} = 0.0699\text{ mol}\)
- Ratio \(x = \frac{\text{moles of H}_2\text{O}}{\text{moles of MgSO}_4} = \frac{0.0699}{0.00997} \approx 7.01\)
Therefore, the value of \(x = 7\).

(d) If the crucible is cooled without a lid, the anhydrous \(\text{MgSO}_4\) (which is highly hygroscopic) can reabsorb moisture from the atmosphere. This increases the measured mass of the anhydrous residue, which decreases the calculated mass of water lost, resulting in a lower calculated value of \(x\).

Award 8.33 marks as follows:

(a) [1.33 marks]
- Explains that it ensures all water of crystallization has been completely removed / reaction is complete [1.33]

(b) [2 marks]
- Calculates mass of anhydrous residue = \(1.20\text{ g}\) [1]
- Calculates mass of water lost = \(1.26\text{ g}\) [1]

(c) [3 marks]
- Calculates moles of \(\text{MgSO}_4 = 0.010\text{ mol}\) (or \(0.00997\text{ mol}\)) [1]
- Calculates moles of \(\text{H}_2\text{O} = 0.070\text{ mol}\) (or \(0.0699\text{ mol}\)) [1]
- Finds ratio \(x = 7.0\) / shows division of moles [1]

(d) [2 marks]
- Identifies that the anhydrous salt reabsorbs water/moisture from the air [1]
- Explains that this makes the mass of anhydrous residue appear larger (or mass of water lost appear smaller), reducing the calculated \(x\) value [1]

(a) Calculate masses and moles of C, H, and O in the 1.500 g sample:
- Mass of C = \( 2.998 \text{ g CO}_2 \times \frac{12.01 \text{ g mol}^{-1}}{44.01 \text{ g mol}^{-1}} = 0.8182 \text{ g C} \)
- Moles of C = \( \frac{0.8182 \text{ g}}{12.01 \text{ g mol}^{-1}} = 0.06813 \text{ mol C} \)
- Mass of H = \( 1.227 \text{ g H}_2\text{O} \times \frac{2.016 \text{ g mol}^{-1}}{18.02 \text{ g mol}^{-1}} = 0.1373 \text{ g H} \)
- Moles of H = \( \frac{0.1373 \text{ g}}{1.008 \text{ g mol}^{-1}} = 0.1362 \text{ mol H} \)
- Mass of O = \( 1.500 \text{ g} - (0.8182 \text{ g} + 0.1373 \text{ g}) = 0.5445 \text{ g O} \)
- Moles of O = \( \frac{0.5445 \text{ g}}{16.00 \text{ g mol}^{-1}} = 0.03403 \text{ mol O} \)
Divide by the smallest value (0.03403):
- C: \( \frac{0.06813}{0.03403} \approx 2 \)
- H: \( \frac{0.1362}{0.03403} \approx 4 \)
- O: \( \frac{0.03403}{0.03403} = 1 \)
Thus, the empirical formula is \( \text{C}_2\text{H}_4\text{O} \).

(b) Empirical formula mass \( (\text{C}_2\text{H}_4\text{O}) = 2(12.01) + 4(1.01) + 16.00 = 44.06 \text{ g mol}^{-1} \).
Scaling factor = \( \frac{88.1 \text{ g mol}^{-1}}{44.06 \text{ g mol}^{-1}} \approx 2 \).
Therefore, the molecular formula is \( \text{C}_4\text{H}_8\text{O}_2 \).

(c) The carboxylic acid with formula \( \text{C}_4\text{H}_8\text{O}_2 \) is butanoic acid, \( \text{C}_3\text{H}_7\text{COOH} \).
Equation: \( \text{C}_3\text{H}_7\text{COOH(aq)} + \text{NaOH(aq)} \rightarrow \text{C}_3\text{H}_7\text{COONa(aq)} + \text{H}_2\text{O(l)} \)
(Accept: \( \text{C}_4\text{H}_8\text{O}_2\text{(aq)} + \text{NaOH(aq)} \rightarrow \text{C}_4\text{H}_7\text{O}_2\text{Na(aq)} + \text{H}_2\text{O(l)} \))

(d) Moles of acid dissolved: \( n = \frac{2.20 \text{ g}}{88.12 \text{ g mol}^{-1}} = 0.0250 \text{ mol} \).
Volume of solution = \( 250.0 \text{ cm}^3 = 0.2500 \text{ dm}^3 \).
Concentration = \( \frac{0.0250 \text{ mol}}{0.2500 \text{ dm}^3} = 0.100 \text{ mol dm}^{-3} \).

(e) Moles of acid in titration sample: \( n_{\text{acid}} = 0.100 \text{ mol dm}^{-3} \times 0.02500 \text{ dm}^3 = 2.50 \times 10^{-3} \text{ mol} \).
Using 1:1 stoichiometry, moles of NaOH required = \( 2.50 \times 10^{-3} \text{ mol} \).
Volume of NaOH solution = \( \frac{2.50 \times 10^{-3} \text{ mol}}{0.100 \text{ mol dm}^{-3}} = 0.0250 \text{ dm}^3 = 25.0 \text{ cm}^3 \).

(a) [5 marks]
- 1 mark for correct mass or moles of carbon.
- 1 mark for correct mass or moles of hydrogen.
- 1 mark for calculating oxygen mass by difference.
- 1 mark for correct moles of oxygen.
- 1 mark for correct simplest ratio leading to empirical formula \( \text{C}_2\text{H}_4\text{O} \).

(b) [1 mark]
- 1 mark for showing comparison of masses and deducing molecular formula \( \text{C}_4\text{H}_8\text{O}_2 \).

(c) [1.5 marks]
- 1 mark for correct chemical species in the equation.
- 0.5 marks for correct state symbols.

(d) [2 marks]
- 1 mark for calculating moles of acid.
- 1 mark for correct final concentration of \( 0.100 \text{ mol dm}^{-3} \) (accept range 0.099–0.100).

(e) [3 marks]
- 1 mark for calculating moles of acid used in the titration.
- 1 mark for identifying 1:1 molar ratio.
- 1 mark for correct final volume of \( 25.0 \text{ cm}^3 \) (or \( 0.0250 \text{ dm}^3 \)).

(a) The rate-determining step is the slowest elementary step in a reaction mechanism, which controls/determines the overall rate of the chemical reaction.

(b) Comparing Run 1 and Run 2: \( [\text{H}_2] \) is held constant, while \( [\text{NO}] \) doubles. The initial rate increases by a factor of \( \frac{4.8 \times 10^{-3}}{1.2 \times 10^{-3}} = 4 \). Since \( 2^2 = 4 \), the reaction is second order with respect to \( \text{NO} \).
Comparing Run 1 and Run 3: \( [\text{NO}] \) is held constant, while \( [\text{H}_2] \) doubles. The rate increases by a factor of \( \frac{2.4 \times 10^{-3}}{1.2 \times 10^{-3}} = 2 \). Since \( 2^1 = 2 \), the reaction is first order with respect to \( \text{H}_2 \).
Rate expression: \( \text{Rate} = k[\text{NO}]^2[\text{H}_2] \).

(c) Using data from Run 1:
\( 1.2 \times 10^{-3} \text{ mol dm}^{-3} \text{ s}^{-1} = k (0.100 \text{ mol dm}^{-3})^2 (0.100 \text{ mol dm}^{-3}) \)
\( k = \frac{1.2 \times 10^{-3}}{0.00100} = 1.2 \)
Units calculation: \( \frac{\text{mol dm}^{-3} \text{ s}^{-1}}{(\text{mol dm}^{-3})^3} = \text{dm}^6 \text{ mol}^{-2} \text{ s}^{-1} \).
Value: \( 1.2 \text{ dm}^6 \text{ mol}^{-2} \text{ s}^{-1} \).

(d) Step 2 is the slow (rate-determining) step, so the rate law based on this step is:
\( \text{Rate} = k_2[\text{N}_2\text{O}_2][\text{H}_2] \)
Since \( \text{N}_2\text{O}_2 \) is an unstable intermediate, we express its concentration using Step 1 (fast equilibrium):
\( K_c = \frac{[\text{N}_2\text{O}_2]}{[\text{NO}]^2} \Rightarrow [\text{N}_2\text{O}_2] = K_c[\text{NO}]^2 \)
Substituting this into the rate law gives:
\( \text{Rate} = k_2 K_c [\text{NO}]^2 [\text{H}_2] = k'[\text{NO}]^2[\text{H}_2] \)
This matches the experimentally derived rate law, so the mechanism is consistent.

(e) Sketch requirements:
- Y-axis labeled "Number of molecules / fraction of particles with kinetic energy E"; X-axis labeled "Kinetic energy".
- Curve \( T_1 \) starts at origin, rises to a peak, and asymptotes to the x-axis.
- Curve \( T_2 \) (higher temperature) has a lower peak height, peak shifted to the right, and is flatter/broader than \( T_1 \), crossing \( T_1 \) only once.
- Vertical line labeled "Activation Energy \( E_a \)" on x-axis.
- Explanation: At \( T_2 \), a significantly larger fraction of molecules possess kinetic energy greater than or equal to \( E_a \), leading to more frequent successful collisions per unit time.

(a) [1 mark]
- 1 mark for defining rate-determining step as the slowest step in the mechanism.

(b) [3 marks]
- 1 mark for correct working and deduction that order with respect to NO is 2.
- 1 mark for correct working and deduction that order with respect to H2 is 1.
- 1 mark for correct rate expression: \( \text{Rate} = k[\text{NO}]^2[\text{H}_2] \).

(c) [2.5 marks]
- 1.5 marks for correct numerical calculation of \( k = 1.2 \).
- 1 mark for correct units of \( \text{dm}^6 \text{ mol}^{-2} \text{ s}^{-1} \).

(d) [3 marks]
- 1 mark for identifying that rate-determining step expression depends on \( [\text{N}_2\text{O}_2] \) and \( [\text{H}_2] \).
- 1 mark for expressing \( [\text{N}_2\text{O}_2] \) in terms of reactants from Step 1 equilibrium.
- 1 mark for substituting and matching the overall experimental rate expression.

(e) [3 marks]
- 1 mark for correct axis labels and general shape of the Maxwell-Boltzmann curves at \( T_1 \) and \( T_2 \).
- 1 mark for indicating \( E_a \) and highlighting the larger fraction of molecules with \( E \ge E_a \) at \( T_2 \).
- 1 mark for explaining that higher temperature increases the rate primarily because a larger fraction of collisions are successful (have energy \( \ge E_a \)) (accept higher frequency of overall collisions as secondary factor).

(a) Lewis structures:
- For \( \text{PO}_4^{3-} \): Central P atom bonded to 4 oxygen atoms. The structure showing 1 double bond (\( \text{P}=\text{O} \)) and 3 single bonds (\( \text{P}-\text{O}^- \)) is preferred to minimize formal charge, with P having an expanded octet (10 valence electrons). Single-bonded O atoms have 3 lone pairs each; double-bonded O has 2 lone pairs. Alternatively, drawing 4 single bonds to O with P satisfying octet is accepted if overall charge of \( 3- \) is shown.
- For \( \text{PO}_3^{3-} \): Central P atom with 1 lone pair, single-bonded to 3 oxygen atoms (each O with 3 lone pairs, giving P an octet and minimizing formal charge), overall charge \( 3- \) shown.

(b) Geometries:
- \( \text{PO}_4^{3-} \): 4 electron domains around P. Electron domain geometry: tetrahedral. Molecular geometry: tetrahedral.
- \( \text{PO}_3^{3-} \): 4 electron domains (3 bonding domains, 1 lone pair) around P. Electron domain geometry: tetrahedral. Molecular geometry: trigonal pyramidal.

(c) Compare and justify bond angles:
- In \( \text{PO}_4^{3-} \), the bond angle is \( 109.5^\circ \) (or \( 109^\circ \)). There are no lone pairs on the P atom, so repulsion between all bonding domains is equal.
- In \( \text{PO}_3^{3-} \), the bond angle is less than \( 109.5^\circ \) (approx. \( 107^\circ \)).
- Justification: The presence of the non-bonding lone pair on phosphorus in \( \text{PO}_3^{3-} \) results in stronger lone pair-bonding pair repulsion compared to bonding pair-bonding pair repulsion. This compresses the \( \text{O}-\text{P}-\text{O} \) bond angles.

(d) (i) Delocalization and bond lengths:
- In \( \text{CO}_3^{2-} \), the pi electrons are delocalized across all three carbon-oxygen bonds, resulting in three equivalent resonance structures. The bond order of each carbon-oxygen bond is equal to \( \frac{4 \text{ bonding pairs}}{3 \text{ positions}} = 1.33 \).
- In \( \text{CO}_2 \), the molecule contains two localized double bonds, meaning each carbon-oxygen bond has a bond order of 2.0.
- Since higher bond order means a stronger and shorter bond, the bonds in \( \text{CO}_3^{2-} \) (bond order 1.33) are weaker and therefore longer (129 pm) than the localized double bonds in \( \text{CO}_2 \) (116 pm).

(d) (ii) Hybridization:
- Carbon in \( \text{CO}_2 \): \( \text{sp} \) hybridized.
- Carbon in \( \text{CO}_3^{2-} \): \( \text{sp}^2 \) hybridized.

(a) [3 marks]
- 1.5 marks for correct Lewis structure of \( \text{PO}_4^{3-} \) showing all valence electrons, correct bonding, and overall charge.
- 1.5 marks for correct Lewis structure of \( \text{PO}_3^{3-} \) showing a lone pair on the phosphorus, correct bonding, and overall charge.

(b) [2.5 marks]
- 1 mark for correct geometries of \( \text{PO}_4^{3-} \) (tetrahedral / tetrahedral).
- 1.5 marks for correct geometries of \( \text{PO}_3^{3-} \) (tetrahedral / trigonal pyramidal).

(c) [3 marks]
- 1 mark for stating both angles correctly (\( 109.5^\circ \) and \( < 109.5^\circ \)).
- 1 mark for identifying that \( \text{PO}_3^{3-} \) contains a lone pair on phosphorus.
- 1 mark for applying VSEPR repulsion rules (lone pair-bonding pair repulsion > bonding pair-bonding pair repulsion).

(d) [4 marks]
- (i) [3 marks]: 1 mark for noting resonance/delocalization in carbonate; 1 mark for determining bond orders (1.33 for carbonate vs 2.0 for carbon dioxide); 1 mark for relating higher bond order to shorter/stronger bonds.
- (ii) [1 mark]: 0.5 marks for sp in carbon dioxide and 0.5 marks for \( \text{sp}^2 \) in carbonate.

(a) (i)
- Anode (oxidation occurs here): \( \text{Zn(s)} \rightarrow \text{Zn}^{2+}\text{(aq)} + 2\text{e}^- \)
- Cathode (reduction occurs here): \( \text{Cu}^{2+}\text{(aq)} + 2\text{e}^- \rightarrow \text{Cu(s)} \)

(a) (ii)
- Overall equation: \( \text{Zn(s)} + \text{Cu}^{2+}\text{(aq)} \rightarrow \text{Zn}^{2+}\text{(aq)} + \text{Cu(s)} \)
- Standard cell potential: \( E^\theta_{\text{cell}} = E^\theta_{\text{cathode}} - E^\theta_{\text{anode}} = (+0.34 \text{ V}) - (-0.76 \text{ V}) = +1.10 \text{ V} \).

(a) (iii)
- The salt bridge completes the electrical circuit and maintains charge neutrality in the half-cells.
- During discharge, cations (e.g., \( \text{K}^+ \)) migrate from the salt bridge into the copper half-cell (cathode compartment) to balance the loss of positive charge as \( \text{Cu}^{2+} \) is reduced to Cu.
- Anions (e.g., \( \text{NO}_3^- \)) migrate into the zinc half-cell (anode compartment) to balance the excess positive charge of newly formed \( \text{Zn}^{2+} \) ions.

(b) (i) Molten NaCl electrolysis:
- Cathode (reduction): \( \text{Na}^+\text{(l)} + \text{e}^- \rightarrow \text{Na(l)} \) (Product: liquid sodium metal)
- Anode (oxidation): \( 2\text{Cl}^-\text{(l)} \rightarrow \text{Cl}_2\text{(g)} + 2\text{e}^- \) (Product: chlorine gas)

(b) (ii) Aqueous NaCl electrolysis:
- Product at cathode: Hydrogen gas, \( \text{H}_2\text{(g)} \) (along with hydroxide ions, \( \text{OH}^-\text{(aq)} \)).
- Explanation: In aqueous solution, water molecules (or \( \text{H}^+ \) ions) compete with sodium ions (\( \text{Na}^+ \)) for reduction at the cathode. Since water is more easily reduced (has a more positive/less negative reduction potential) than sodium ions (\( E^\theta(\text{Na}^+/\text{Na}) = -2.71 \text{ V} \)), hydrogen gas is produced preferentially over sodium metal.

(a) [8 marks]
- (i) [3 marks]: 1.5 marks for correct anode equation and identification as oxidation; 1.5 marks for correct cathode equation and identification as reduction.
- (ii) [2 marks]: 1 mark for correct calculation showing working (\( +1.10 \text{ V} \)); 1 mark for overall reaction equation.
- (iii) [3 marks]: 1 mark for functional description of salt bridge (maintaining neutrality/completing circuit); 1 mark for correct direction of cation movement; 1 mark for correct direction of anion movement.

(b) [4.5 marks]
- (i) [2.5 marks]: 1 mark for correct cathode reaction and product (Na); 1 mark for correct anode reaction and product (\( \text{Cl}_2 \)); 0.5 marks for correct state symbols (l and g).
- (ii) [2 marks]: 1 mark for identifying the product at the cathode as hydrogen gas (\( \text{H}_2 \)); 1 mark for explaining that water is more easily reduced / has a higher (more positive) reduction potential than \( \text{Na}^+ \).