ESAT Paper 2 Chemistry Worked Solutions

Correct Answer: C. $Cr_2O_7^{2-} + SO_3^{2-} \rightarrow Cr^{3+} + SO_4^{2-}$

Fastest Approach (🚀):
Track the oxidation state of the metal (Cr, Cu, Fe) through each equation first — if it is unchanged, the reaction is ligand substitution or acid-base, not redox, and can be eliminated immediately.

Step-by-Step Breakdown:

1. Identify Redox Reactions

A redox reaction involves a change in the oxidation states of atoms. Let's analyse the oxidation states in each option:

• A. $Cr_2O_7^{2-} + 2OH^- \rightarrow 2CrO_4^{2-} + H_2O$
• Cr in $Cr_2O_7^{2-}$: $2x + 7(-2) = -2 \implies 2x = +12 \implies x = +6$.
• Cr in $CrO_4^{2-}$: $x + 4(-2) = -2 \implies x = +6$.
• No change in oxidation state; this is the base-mediated dichromate–chromate equilibrium (an acid-base reaction).
• B. $[Cu(H_2O)_6]^{2+} + 4Cl^- \rightarrow CuCl_4^{2-} + 6H_2O$
• Cu is $+2$ on both sides. This is a ligand substitution reaction.
• C. $Cr_2O_7^{2-} + SO_3^{2-} \rightarrow Cr^{3+} + SO_4^{2-}$
• Cr goes from $+6$ (in dichromate) to $+3$. It is reduced.
• S in $SO_3^{2-}$ is $+4$. S in $SO_4^{2-}$ is $+6$. It is oxidised.
• Since both oxidation and reduction occur, this is a redox reaction. (This is shown as a skeletal ionic summary, as in the source paper; a fully balanced version in acid solution is $Cr_2O_7^{2-} + 3SO_3^{2-} + 8H^+ \rightarrow 2Cr^{3+} + 3SO_4^{2-} + 4H_2O$.)
• D. $[Fe(H_2O)_6]^{3+} + 3OH^- \rightarrow [Fe(H_2O)_3(OH)_3] + 3H_2O$
• Fe is $+3$ on both sides. This is an acid-base/ligand substitution (precipitation) reaction.

Therefore, the correct answer is C.

Why the Other Options Are Wrong (❌):

• A. $Cr_2O_7^{2-} + 2OH^- \rightarrow 2CrO_4^{2-} + H_2O$ — Conceptual Misunderstanding
  Cr₂O₇²⁻ + 2OH⁻ → 2CrO₄²⁻ + H₂O is the well-known dichromate-to-chromate interconversion. Assigning oxidation states (2x + 7(−2) = −2 and x + 4(−2) = −2) gives Cr = +6 on both sides, so this is an acid-base equilibrium with no electron transfer, not a redox reaction.
• B. $[Cu(H_2O)_6]^{2+} + 4Cl^- \rightarrow CuCl_4^{2-} + 6H_2O$ — Conceptual Misunderstanding
  [Cu(H₂O)₆]²⁺ + 4Cl⁻ → CuCl₄²⁻ + 6H₂O only exchanges the ligands bound to copper (water for chloride); copper remains +2 throughout, so this is a ligand-substitution (complex-ion) reaction, not a redox reaction.
• D. $[Fe(H_2O)_6]^{3+} + 3OH^- \rightarrow [Fe(H_2O)_3(OH)_3] + 3H_2O$ — Conceptual Misunderstanding
  [Fe(H₂O)₆]³⁺ + 3OH⁻ → [Fe(H₂O)₃(OH)₃] + 3H₂O replaces three aqua ligands with hydroxide ligands (an acid-base/precipitation step); iron stays at +3 on both sides, so no oxidation or reduction occurs.

Common Mistake (⚠️):
Assuming that any reaction involving a transition-metal complex must be redox simply because a metal ion is present. Ligand substitution (B, D) and proton-transfer/acid-base equilibria (A) can involve colour changes or precipitate formation without any electron transfer at all.

Takeaway (📌):
Confirm redox character by assigning oxidation numbers, not by looking for a metal, a colour change, or a precipitate — only a genuine change in oxidation state defines a redox reaction.

Correct Answer: E. Sodium hydrogen carbonate

Fastest Approach (🚀):
Recognise the classic 'baking soda' decomposition $2NaHCO_3 \xrightarrow{\Delta} Na_2CO_3 + CO_2 + H_2O$ — solid + gas + condensable vapour — and confirm that none of the other four sodium salts decompose this way on ordinary heating.

Step-by-Step Breakdown:

1. Analyse the Decomposition Products

Compound X decomposes on heating to yield:
A white solid
A colourless gas

• A colourless vapour that condenses to a liquid (this is typically water, $H_2O$).

2. Evaluate the Options

• A. Sodium carbonate ($Na_2CO_3$): Very stable to heat; does not decompose at typical heating temperatures. It is in fact the solid product of compound X's decomposition, not compound X itself.
• B. Sodium hydroxide ($NaOH$): Melts (m.p. ≈ 318 °C) but does not decompose into a gas and vapour on heating.
• C. Sodium nitrate ($NaNO_3$): Decomposes to sodium nitrite ($NaNO_2$) and oxygen gas ($O_2$), but produces no liquid/vapour ($H_2O$) and no separate solid+gas+vapour trio.
• D. Sodium sulphate ($Na_2SO_4$): Stable to heat; does not decompose.
• E. Sodium hydrogen carbonate ($NaHCO_3$): Also known as baking soda, it decomposes on heating according to:

$2NaHCO_3(s) \xrightarrow{\Delta} Na_2CO_3(s) + CO_2(g) + H_2O(g)$

• $Na_2CO_3(s)$ is a white solid.
• $CO_2(g)$ is a colourless gas.
• $H_2O(g)$ is a colourless vapour that condenses to a liquid.

This perfectly matches the description, so Compound X is sodium hydrogen carbonate (Option E).

Why the Other Options Are Wrong (❌):

• A. Sodium carbonate — Conceptual Misunderstanding
  Sodium carbonate (Na₂CO₃) is thermally very stable — it does not itself decompose into a solid, a gas, and a condensable vapour on ordinary heating; it is instead the solid RESIDUE produced when sodium hydrogen carbonate decomposes.
• B. Sodium hydroxide — Conceptual Misunderstanding
  Sodium hydroxide (NaOH) simply melts on heating; it does not decompose to release a gas and a separate condensable vapour, so it cannot be compound X.
• C. Sodium nitrate — Conceptual Misunderstanding
  Sodium nitrate (NaNO₃) decomposes on strong heating to sodium nitrite and oxygen gas (2NaNO₃ → 2NaNO₂ + O₂); this releases only one gas and no condensable liquid/vapour, so it does not match the description.
• D. Sodium sulphate — Conceptual Misunderstanding
  Sodium sulphate (Na₂SO₄) is thermally stable and does not decompose into a solid, gas, and vapour on heating, so it cannot be compound X.

Common Mistake (⚠️):
Choosing sodium carbonate (Option A) because 'carbonate' sounds like it should release CO₂ on heating — but Na₂CO₃ is thermally very stable and is in fact the solid product left behind once compound X (the hydrogencarbonate) has decomposed, not compound X itself.

Takeaway (📌):
The three-way signature — white solid residue, colourless gas, and a condensable colourless vapour — from heating an anhydrous white solid is diagnostic of a hydrogencarbonate decomposing to a carbonate, CO₂, and H₂O; carbonates, hydroxides, nitrates and sulfates of sodium each behave differently on heating.

Correct Answer: A. [CONTENT MISSING - source scan does not contain this question]

Step-by-Step Breakdown:
Question 3's body was not captured in any of the 27 source scan images provided for this paper. The scan sequence shows Question 2 complete (in esat-0002-chemistry-q03-full.png, which also carries the cut-off 'Question 3' header at its very bottom), then jumps directly to Question 4 complete (at the top of esat-0002-chemistry-q04-full.png) with no Question 3 content in between; esat-0002-chemistry-q05-full.png (the next file) is a byte-for-byte duplicate of q04's scan and likewise shows only Question 4 again, not Question 3. This content genuinely cannot be restored from the material supplied and is left flagged rather than fabricated.

Why the Other Options Are Wrong (❌):

• B. [CONTENT MISSING - source scan does not contain this question] — Content Unavailable
  Not available — Question 3's options are not present in any of the 27 source scans.
• C. [CONTENT MISSING - source scan does not contain this question] — Content Unavailable
  Not available — Question 3's options are not present in any of the 27 source scans.
• D. [CONTENT MISSING - source scan does not contain this question] — Content Unavailable
  Not available — Question 3's options are not present in any of the 27 source scans.
• E. [CONTENT MISSING - source scan does not contain this question] — Content Unavailable
  Not available — Question 3's options are not present in any of the 27 source scans.

Correct Answer: B. Row B: $x=3,\ y=8,\ z=3,\ b=2,\ a=4$

Fastest Approach (🚀):
Balance electrons between $Cu \rightarrow Cu^{2+} + 2e^-$ (2e⁻ lost) and $NO_3^- + 4H^+ + 3e^- \rightarrow NO + 2H_2O$ (3e⁻ gained) first — this fixes the ratio at 3 Cu : 2 NO, which (together with $x=z$, since every Cu ends up as one Cu(NO₃)₂) only row B satisfies.

Step-by-Step Breakdown:

1. Set up the half-equations for the redox process

The reaction is the oxidation of copper by nitric acid to form copper(II) nitrate, nitrogen monoxide, and water.

• Oxidation half-equation: Copper metal is oxidised to $Cu^{2+}$.

$Cu \rightarrow Cu^{2+} + 2e^-$

• Reduction half-equation: Nitrate ion ($NO_3^-$) in acidic solution is reduced to nitrogen monoxide ($NO$).

$NO_3^- + 4H^+ + 3e^- \rightarrow NO + 2H_2O$

2. Balance the electrons

To balance the electrons (2 lost, 3 gained), multiply the oxidation equation by 3 and the reduction equation by 2:

• $3Cu \rightarrow 3Cu^{2+} + 6e^-$
• $2NO_3^- + 8H^+ + 6e^- \rightarrow 2NO + 4H_2O$

3. Combine and balance the full equation

Combining the half-equations gives:
$3Cu + 2NO_3^- + 8H^+ \rightarrow 3Cu^{2+} + 2NO + 4H_2O$
Notice that we have $8H^+$ on the left, which must come from $8HNO_3$. This means we need 8 nitrate ions in total on the left side. Two of them are reduced to $NO$, and the remaining 6 act as spectator ions that pair with the $3Cu^{2+}$ to form $3Cu(NO_3)_2$.

The full balanced equation is:
$3Cu + 8HNO_3 \rightarrow 3Cu(NO_3)_2 + 2NO + 4H_2O$

4. Match the coefficients

Comparing this to $xCu + yHNO_3 \rightarrow zCu(NO_3)_2 + bNO + aH_2O$:

• $x = 3$
• $y = 8$
• $z = 3$
• $b = 2$
• $a = 4$

Looking at the table, row B correctly lists $x=3$, $y=8$, $z=3$, $b=2$, and $a=4$.

Why the Other Options Are Wrong (❌):

• A. Row A: $x=2,\ y=4,\ z=2,\ b=1,\ a=2$ — Miscounted Stoichiometric Coefficients
  Row A ($x=2,y=4,z=2,b=1,a=2$) fails electron balance: 2 Cu lose $2\times2=4$ electrons but only 1 NO is formed, gaining just 3 electrons — the electrons don't balance (4 ≠ 3), and there are too few HNO₃ (4) to supply both 1 spectator-nitrate pair per Cu(NO₃)₂ and the H⁺ needed for the reduction.
• C. Row C: $x=1,\ y=2,\ z=1,\ b=1,\ a=1$ — Miscounted Stoichiometric Coefficients
  Row C ($x=1,y=2,z=1,b=1,a=1$) only has enough HNO₃ (y=2) to supply the H⁺ for reducing 1 NO₃⁻ to NO, with no nitrate left over to form the $Cu(NO_3)_2$ spectator ions — the nitrogen and oxygen atom counts don't balance.
• D. Row D: $x=1,\ y=2,\ z=2,\ b=1,\ a=1$ — Miscounted Stoichiometric Coefficients
  Row D ($x=1,y=2,z=2,b=1,a=1$) has $x \neq z$ (1 Cu atom cannot become 2 formula units of Cu(NO₃)₂), so copper is not conserved — this row is impossible regardless of the redox balance.
• E. Row E: $x=4,\ y=8,\ z=8,\ b=2,\ a=1$ — Miscounted Stoichiometric Coefficients
  Row E ($x=4,y=8,z=8,b=2,a=1$) also has $x \neq z$ (4 Cu atoms cannot become 8 Cu(NO₃)₂ units), so copper is not conserved, and the water coefficient (a=1) is far too small to balance the 8 hydrogens supplied by 8 HNO₃.

Common Mistake (⚠️):
Forgetting that the HNO₃ coefficient (y) must account for BOTH the nitrate reduced to NO AND the nitrate surviving unchanged as spectator ions in Cu(NO₃)₂ — using only the reduced nitrate (giving y=2 instead of y=8) undercounts y, and also produces a row (like C or D) where x ≠ z, which is impossible since each Cu atom ends up in exactly one Cu(NO₃)₂ unit.

Takeaway (📌):
In metal + dilute HNO₃ reactions, split the acid's role into 'oxidising agent' (the fraction reduced to NO) and 'spectator/salt-forming' (the fraction that ends up as NO₃⁻ in the metal nitrate) before balancing — and always check that the metal coefficient equals the metal-nitrate coefficient as a quick consistency check.

Correct Answer: A. Arrow A: the full span from the reactants energy level up to the peak of the solid (uncatalysed) curve

Fastest Approach (🚀):
Identify which curve is solid (uncatalysed, the taller peak) versus dashed (catalysed, the lower peak), then pick the arrow that spans from the reactants baseline up to the solid curve's peak — that is arrow A, not the shorter dashed arrow E, which belongs to the catalysed pathway.

Step-by-Step Breakdown:

1. Understand Activation Energy

The activation energy ($E_a$) of a reaction pathway is the minimum energy required to initiate it, represented on an energy profile diagram by the vertical distance from the energy level of the reactants up to the highest point (transition state) of THAT pathway's curve.

2. Identify the Two Curves

• The solid curve is the higher-peaked pathway: this is the reaction in the absence of a catalyst (uncatalysed).
• The dashed curve has a lower peak because a catalyst provides an alternative route with lower activation energy: this is the catalysed pathway.

3. Match the Arrows to Each Curve

• Arrow A starts at the reactants energy level and extends vertically to the peak of the solid curve — this is the activation energy of the uncatalysed reaction.
• Arrow E (dashed) starts at the same reactants level but only extends to the lower peak of the dashed curve — this is the activation energy of the catalysed reaction, not the uncatalysed one.
• Arrows C and B run the opposite way, from each curve's peak down to the products level; these represent the activation energies of the reverse reactions (uncatalysed and catalysed respectively), not the forward activation energy asked for.
• Arrows D and F mark small residual gaps near the products plateau and are not activation energies at all.

4. Conclude

Since the question asks for the activation energy in the absence of a catalyst, this is the full vertical rise from the reactants level to the peak of the solid curve — arrow A.

The correct letter is A.

Why the Other Options Are Wrong (❌):

• B. Arrow B: the span from the peak of the dashed (catalysed) curve down to the products energy level — Misread Diagram
  Arrow B spans from the peak of the CATALYSED (dashed) curve down to the products level — this is the reverse activation energy of the catalysed pathway, not the forward activation energy of the uncatalysed reaction asked for.
• C. Arrow C: the span from the peak of the solid (uncatalysed) curve down to the products energy level — Misread Diagram
  Arrow C spans from the peak of the solid (uncatalysed) curve down to the products level — this is the activation energy of the REVERSE uncatalysed reaction, not the forward activation energy asked for.
• D. Arrow D: a short span just above the products energy level, below where the dashed curve levels off — Misread Diagram
  Arrow D marks a small energy gap near the products plateau, not a rise from the reactants level to a peak — it is not an activation energy at all.
• E. Arrow E: the span from the reactants energy level up to the peak of the dashed (catalysed) curve — Misread Diagram
  Arrow E is dashed and only reaches the lower, CATALYSED peak — it is the activation energy WITH a catalyst. The question specifically asks for the activation energy in the ABSENCE of a catalyst, which is the taller, solid-curve arrow A.
• F. Arrow F: the shortest marked span, just above the products energy level — Misread Diagram
  Arrow F marks the smallest energy gap near the products plateau, not a rise from the reactants level to a peak — it is not an activation energy at all.

Common Mistake (⚠️):
Selecting E instead of A: E is drawn as a dashed arrow reaching only the lower, catalysed peak, so it is the activation energy WITH a catalyst — the exact opposite of what the question asks for. The solid/dashed styling of each curve must be matched to 'uncatalysed'/'catalysed' before reading off any arrow.

Takeaway (📌):
Always check which curve (solid or dashed) a given arrow belongs to before reading an activation energy off an energy-profile diagram — height alone can mislead you into picking the wrong pathway's value.

Correct Answer: C. Row C: Potassium Permanganate → c; Sodium + water → a; Sodium carbonate + HCl → b

Fastest Approach (🚀):
Recall the three gases directly from the reagents: Na + water always gives H₂ (squeaky pop), a carbonate + acid always gives CO₂ (cloudy limewater), and KMnO₄ with concentrated HCl is a standard laboratory generator of Cl₂ (bleaches litmus) — no calculation needed, just recall of standard tests.

Step-by-Step Breakdown:

1. Analyse the Reactions

• Reaction a: Produces a gas that gives a 'squeaky pop' with a lit splint. This is the classic test for hydrogen gas ($H_2$). Sodium reacting with water produces hydrogen: $2Na + 2H_2O \rightarrow 2NaOH + H_2$.
• Reaction b: Produces a gas that turns limewater cloudy. This is the classic test for carbon dioxide ($CO_2$). Sodium carbonate reacting with hydrochloric acid produces carbon dioxide: $Na_2CO_3 + 2HCl \rightarrow 2NaCl + H_2O + CO_2$.
• Reaction c: Produces a gas that bleaches damp litmus paper. This is the classic test for chlorine gas ($Cl_2$). Potassium permanganate ($KMnO_4$) is a strong oxidising agent and, when reacted with concentrated hydrochloric acid, oxidises chloride ions to chlorine gas.

2. Match with the Table Columns

Potassium Permanganate (Column 1) corresponds to Reaction c.
Sodium reacting with water (Column 2) corresponds to Reaction a.

• Sodium carbonate reacting with hydrochloric acid (Column 3) corresponds to Reaction b.

Looking at the table, row C correctly aligns these: $KMnO_4 \to c$, $Na+H_2O \to a$, and $Na_2CO_3+HCl \to b$.

The correct answer is C.

Why the Other Options Are Wrong (❌):

• A. Row A: Potassium Permanganate → a; Sodium + water → b; Sodium carbonate + HCl → c — Misassigned Gas Test
  Row A assigns Potassium Permanganate → a (squeaky pop/H₂) and Sodium + water → b (cloudy limewater/CO₂), but KMnO₄ + HCl generates Cl₂ (the litmus-bleaching gas, reaction c), and Na + water generates H₂ (the squeaky-pop gas, reaction a) — both assignments in this row are wrong.
• B. Row B: Potassium Permanganate → b; Sodium + water → c; Sodium carbonate + HCl → a — Misassigned Gas Test
  Row B assigns Sodium carbonate + HCl → a (squeaky pop/H₂), but a carbonate reacting with acid produces CO₂ (which turns limewater cloudy, reaction b), not H₂ — carbonates never release H₂ with a dilute acid.
• D. Row D: Potassium Permanganate → c; Sodium + water → b; Sodium carbonate + HCl → a — Misassigned Gas Test
  Row D assigns Sodium + water → b (cloudy limewater/CO₂), but sodium and water release hydrogen gas (the squeaky-pop test, reaction a), not carbon dioxide — sodium contains no carbon to form CO₂.
• E. Row E: Potassium Permanganate → a; Sodium + water → c; Sodium carbonate + HCl → b — Misassigned Gas Test
  Row E assigns Potassium Permanganate → a (squeaky pop/H₂) and Sodium carbonate + HCl → b (cloudy limewater/CO₂ is actually correct for this pairing, but the KMnO₄ assignment is wrong) — KMnO₄ with concentrated HCl generates chlorine gas (reaction c, bleaches litmus), not hydrogen.

Common Mistake (⚠️):
Mixing up the CO₂ test (turns limewater cloudy) with the Cl₂ test (bleaches litmus) — both are 'a gas is passed into/onto something and something visually changes', so it is easy to swap reaction b and reaction c if the specific observation is not read carefully.

Takeaway (📌):
Learn the three core gas tests by their exact observation, not just 'a gas is produced': squeaky pop with a lit splint → H₂; limewater turns cloudy/milky → CO₂; damp litmus paper is bleached (decolourised) → Cl₂.

Correct Answer: D. Graph D: pH starts low and rises only slightly at first, then rises almost vertically over a short volume range around the equivalence point, before levelling off at a high pH

Fastest Approach (🚀):
Eliminate any straight-line graph immediately, since pH = $-\log_{10}[H^+]$ is never linear in volume added; then pick the curve with a low start, a steep near-vertical section, and a high plateau.

Step-by-Step Breakdown:

1. Understand Titration Curves

The question asks for the titration curve of a strong acid (in the flask) titrated with a strong alkali (base).

• A strong acid initially has a very low pH (around 1-2).

As strong base is added, the pH rises slowly at first.
Near the equivalence point, the pH shoots up almost vertically, crossing pH 7.

• After the equivalence point, the solution is dominated by excess strong base, so the pH levels off at a high value (around 13-14).

2. Evaluate the Graphs

• Graph A: Shows a smooth, ever-steepening rise with no sharp vertical equivalence point and no high-pH plateau. Since pH is logarithmic, this shape does not match a real titration curve. Incorrect.
• Graph B: Shows a linear increase in pH before it plateaus. pH cannot rise linearly with volume added, because $[H^+]$ (and hence pH) changes non-linearly as acid is neutralised. Incorrect.

Graph C: Shows a purely linear increase throughout, with no plateau and no steep jump. Incorrect for the same reason.
Graph D: Shows the classic sigmoidal (S-shaped) curve: it starts low and nearly flat, has a steep, near-vertical region around the equivalence point, and plateaus high afterwards. This is characteristic of a strong acid–strong base titration.

The correct graph is D.

Why the Other Options Are Wrong (❌):

• A. Graph A: pH stays low and flat initially, then rises smoothly and continuously with an ever-increasing gradient, and does not level off (no plateau) — Misread Diagram
  Graph A rises smoothly and continuously without a sharp vertical jump or a high-pH plateau — real strong acid/strong alkali titrations have a near-vertical equivalence-point jump followed by levelling off at high pH, which this exponential-looking shape lacks.
• B. Graph B: pH rises in a straight line from a low starting value, then levels off abruptly at a high, constant pH — Conceptual Misunderstanding
  Graph B rises as a straight line before plateauing — but pH = $-\log_{10}[H^+]$ is a logarithmic function of the acid/base amounts present, so it cannot increase linearly with volume added; the rise should be gradual, then near-vertical, not straight.
• C. Graph C: pH rises steadily in a single straight line from a low starting value to a higher value, with no flat regions — Conceptual Misunderstanding
  Graph C is a single straight line with no plateau at either end and no steep equivalence-point jump — this does not reflect the logarithmic pH scale or the buffering/neutralisation behaviour of a real titration.

Common Mistake (⚠️):
Picking a graph that has a plateau at the end (like B) without checking that the START of the curve is also flat and low — a genuine strong acid/strong alkali curve is flat-low, then steep, then flat-high; a curve that is merely a straight ramp into a plateau (B) is not sigmoidal and does not reflect the logarithmic nature of pH.

Takeaway (📌):
Strong acid–strong alkali titration curves are always S-shaped (sigmoidal): low flat start, steep near-vertical jump at the equivalence point, high flat plateau — never a straight line or a smooth exponential without a plateau.

Correct Answer: D. $-64.\ kJ\ mol^{-1}$ (i.e. approximately $-64.8\ kJ\ mol^{-1}$, the molar enthalpy change)

Fastest Approach (🚀):
Compute $q = mc\Delta T$ using the total mass of solution (both volumes combined, density 1.00 g/cm³), find moles of the limiting reagent (here, equal moles of acid and base react 1:1), then divide $-q$ by that mole count and convert J to kJ.

Step-by-Step Breakdown:

1. Calculate the Heat Transferred ($q$)

• Volume of acid = $25.0 \text{ cm}^3$
• Volume of base = $25.0 \text{ cm}^3$
• Total volume = $50.0 \text{ cm}^3$

Assuming the density is $1.00 \text{ g cm}^{-3}$, the mass of the solution ($m$) is $50.0 \text{ g}$.
The temperature change ($\Delta T$) = $37.5^{\circ}\text{C} - 22.0^{\circ}\text{C} = 15.5 \text{ K}$.
Using the specific heat capacity ($c = 4.18 \text{ J g}^{-1}\text{K}^{-1}$):
$q = mc\Delta T = 50.0 \times 4.18 \times 15.5 = 3239.5 \text{ J}$

2. Calculate the Moles of Water Formed ($n$)

The reaction is $HNO_3 + KOH \rightarrow KNO_3 + H_2O$. It's a 1:1 molar ratio.

• Moles of $HNO_3 = \text{concentration} \times \text{volume} = 2.00 \text{ mol dm}^{-3} \times \frac{25.0}{1000} \text{ dm}^3 = 0.0500 \text{ mol}$.
• Moles of $KOH = 2.00 \times \frac{25.0}{1000} = 0.0500 \text{ mol}$.

Since both are 0.0500 mol, they neutralise completely to form 0.0500 mol of $H_2O$.

3. Calculate the Enthalpy Change of Neutralisation ($\Delta H$)

The reaction is exothermic (temperature increased), so $\Delta H$ is negative.
$\Delta H = -\frac{q}{n} = -\frac{3239.5 \text{ J}}{0.0500 \text{ mol}} = -64790 \text{ J mol}^{-1}$
Converting to kJ mol$^{-1}$:
$\Delta H = -64.79 \text{ kJ mol}^{-1} \approx -64.8 \text{ kJ mol}^{-1}$

4. Match to the Options

Option A is a distractor: pH is not needed, since both reagent concentrations and volumes are already given, fixing the moles reacting exactly.
Option B ($-3239.5\text{ J}$) is just $-q$ itself — the raw heat released, not divided by moles, so it is NOT a molar enthalpy change and has the wrong units for $\Delta H_{neut}$.

• Option C ($+3457.2\text{ J}$) has both the wrong sign (the reaction is exothermic, so $\Delta H$ must be negative) and does not correspond to any correct intermediate value in this calculation.
• Option D ($\approx -64.8\ kJ\,mol^{-1}$) matches our calculated $-64.79\ kJ\,mol^{-1}$ (which rounds to $-65\ kJ\,mol^{-1}$ to 2 s.f.).

The correct answer is D.

Why the Other Options Are Wrong (❌):

• A. We need to know the pH of the nitric acid to calculate the enthalpy change of neutralisation — Conceptual Misunderstanding
  The pH of the nitric acid is not needed — both the concentration and volume of acid and base are already given, so the moles reacting (0.0500 mol each) are already fully determined without any pH measurement.
• B. $-3239.5\ J$ — Incomplete Calculation
  $-3239.5\ J$ is the raw heat released, $q=mc\Delta T = 50.0 \times 4.18 \times 15.5$, but with the sign not yet applied and — critically — not yet divided by the 0.0500 mol of water formed. This is an intermediate result, not the molar enthalpy change of neutralisation.
• C. $+3457.2\ J$ — Sign Error
  $+3457.2\ J$ has the wrong sign for an exothermic reaction (temperature rose, so $\Delta H$ must be negative) and does not match $q=mc\Delta T=3239.5\ J$ from the correct data, suggesting an arithmetic slip (e.g. using the wrong $\Delta T$ or mass).

Common Mistake (⚠️):
Stopping the calculation at $q = -3239.5\ J$ (option B) and treating that as the final answer, without dividing by the moles of water formed — enthalpy change of neutralisation must always be expressed per mole of water produced ($kJ\,mol^{-1}$), not as a raw heat energy in joules.

Takeaway (📌):
Enthalpy change of neutralisation is always $\Delta H = -q/n$ in $kJ\,mol^{-1}$: compute the heat released with $q=mc\Delta T$ using the TOTAL solution mass, then divide by the moles of the limiting reagent (here, equal moles of both, so either works), and give the answer a negative sign for this exothermic process.

Correct Answer: C. Row C: Reaction 1 — Iron oxidised, Iodide reduced; Reaction 2 — Hydrogen oxidised, Carbon reduced

Fastest Approach (🚀):
In Reaction 1, track iron (goes +2 → +3, so it is oxidised) and note that some of the iodine is reduced (from HIO₃'s +5 down to +1) while other iodine is oxidised (from FeI₂'s −1 up to +1) — since iron's change is unambiguous, use it to anchor the row. In Reaction 2, an alkene C=C carbon (oxidation state −1) becomes an alkane carbon (oxidation state −2, so carbon is reduced) as H₂ (0) becomes C–H hydrogen (+1, so hydrogen is oxidised) — hydrogenation always reduces the carbon skeleton and oxidises the added hydrogen.

Step-by-Step Breakdown:

1. Analyse Reaction 1

$5HIO_3 + 4FeI_2 + 25HCl \rightarrow 4FeCl_3 + 13ICl + 15H_2O$

• Iron (Fe): In $FeI_2$, Fe is +2. In the product $FeCl_3$, Fe is +3. Iron loses an electron, so it is oxidised.
• Iodine (I): In $HIO_3$, I is +5. In $FeI_2$, I is -1. In the product $ICl$, I is +1.
• The I from $HIO_3$ goes from +5 to +1, gaining electrons, so it is reduced.
• The I from $FeI_2$ goes from -1 to +1, losing electrons, so it is also oxidised.

The answer table simplifies this by using iron (unambiguously oxidised) and the reduced iodine (labelled 'iodide', the species that started at a negative/low oxidation state family and ends up reduced from the +5 iodate) as the two headline species: Iron is oxidised, Iodide/iodine (from $HIO_3$) is reduced.

2. Analyse Reaction 2

$CH_3CH=CHCH_3 + H_2 \rightarrow CH_3CH_2CH_2CH_3$

• Carbon (C): In the alkene (C=C), the alkene carbons have an oxidation state of -1. In the alkane product, they have an oxidation state of -2. The carbon atoms gain electrons (bonding to an extra H each), so they are reduced.
• Hydrogen (H): In $H_2$, H is 0. In the alkane, the newly-added H is +1. Hydrogen loses electrons, so it is oxidised.

So for Reaction 2: Species oxidised = Hydrogen, Species reduced = Carbon.

3. Match to the Table

Reaction 1: Species oxidised = Iron, Species reduced = Iodide. Reaction 2: Species oxidised = Hydrogen, Species reduced = Carbon. This matches row C exactly.

The correct answer is C.

Why the Other Options Are Wrong (❌):

• A. Row A: Reaction 1 — Iodide oxidised, Iron reduced; Reaction 2 — Carbon oxidised, Hydrogen reduced — Reversed Oxidation/Reduction
  Row A swaps Reaction 1's assignment (says Iodide oxidised/Iron reduced) — but iron goes from +2 in FeI₂ to +3 in FeCl₃, an INCREASE in oxidation state, meaning iron is oxidised, not reduced.
• B. Row B: Reaction 1 — Iron oxidised, Iodide reduced; Reaction 2 — Carbon oxidised, Hydrogen reduced — Reversed Oxidation/Reduction
  Row B gets Reaction 1 correct (Iron oxidised, Iodide reduced) but reverses Reaction 2: carbon goes from -1 (alkene) to -2 (alkane), a DECREASE in oxidation state (reduction), while hydrogen goes from 0 to +1 (oxidation) — this row has carbon and hydrogen's roles swapped.
• D. Row D: Reaction 1 — Iodide oxidised, Iron reduced; Reaction 2 — Hydrogen oxidised, Carbon reduced — Reversed Oxidation/Reduction
  Row D reverses BOTH reactions: in Reaction 1, iron is oxidised (+2→+3), not reduced; in Reaction 2, carbon is reduced (-1→-2) and hydrogen is oxidised (0→+1), the opposite of what this row states.

Common Mistake (⚠️):
Assuming that because iodine appears on 'both sides' of the oxidation-state ledger in Reaction 1, iodine alone must be both oxidised AND reduced (a disproportionation/comproportionation-style reading) without also checking iron — iron's oxidation state increase (+2→+3) makes iron the oxidised species in the table's simplified two-species framing, not iodine.

Takeaway (📌):
For hydrogenation of an alkene (Reaction 2 pattern), the carbon skeleton is always reduced (lower, more negative oxidation state) and the added hydrogen is always oxidised (0 → +1) — this is the general rule for every C=C + H₂ → C–C addition.

Correct Answer: D. Neither reaction

Fastest Approach (🚀):
Track iodine's oxidation states in Reaction 1: it starts at two DIFFERENT values (+5 in $HIO_3$ and -1 in $FeI_2$) and converges on ONE value (+1 in $ICl$) — that is the reverse pattern (comproportionation), not disproportionation. In Reaction 2, carbon and hydrogen are different elements, so no single element is both oxidised and reduced. Both fail the test, so the answer is 'neither'.

Step-by-Step Breakdown:

1. Define Disproportionation

A disproportionation reaction is a specific type of redox reaction in which a single element, starting in a single oxidation state, is simultaneously oxidised and reduced to form two different products with different oxidation states.

2. Check Reaction 1

$5HIO_3 + 4FeI_2 + 25HCl \rightarrow 4FeCl_3 + 13ICl + 15H_2O$

• Iodine starts in two different oxidation states: $+5$ (in $HIO_3$) and $-1$ (in $FeI_2$).
• It ends up in a single oxidation state: $+1$ (in $ICl$).

This is the reverse of disproportionation, known as comproportionation (two oxidation states of the same element converging to one). Disproportionation does not occur here.
Iron also changes, from $+2$ (in $FeI_2$) to $+3$ (in $FeCl_3$) — a single, ordinary oxidation, not a disproportionation.

3. Check Reaction 2

$CH_3CH=CHCH_3 + H_2 \rightarrow CH_3CH_2CH_2CH_3$

• Carbon is reduced (from $-1$ in the alkene to $-2$ in the alkane).
• Hydrogen is oxidised (from $0$ in $H_2$ to $+1$ in the new C–H bonds).

Two different* elements are oxidised and reduced — disproportionation requires the SAME element to do both, so this does not qualify either.

4. Conclude

Since disproportionation occurs in neither reaction, the correct option is D.

Why the Other Options Are Wrong (❌):

• A. Yes, both — Conceptual Misunderstanding
  'Yes, both' would require a single element in each reaction to split from one oxidation state into two different ones. Reaction 1 shows the opposite pattern (two iodine oxidation states converging to one — comproportionation), and Reaction 2 involves two different elements (C and H) changing, not one element splitting — so disproportionation occurs in neither, not both.
• B. Only reaction 1 — Reversed Concept (Comproportionation vs Disproportionation)
  Reaction 1's iodine chemistry ($+5$ and $-1$ both converging to $+1$) is comproportionation, not disproportionation — the oxidation states are merging, not a single starting state splitting apart, so Reaction 1 does not qualify.
• C. Only reaction 2 — Conceptual Misunderstanding
  Reaction 2 has carbon reduced and hydrogen oxidised — two different elements — whereas disproportionation requires the SAME element to be simultaneously oxidised and reduced. Reaction 2 is simply a redox reaction (hydrogenation), not a disproportionation.

Common Mistake (⚠️):
Seeing that iodine's oxidation state changes in Reaction 1 and immediately calling it disproportionation, without checking the DIRECTION of the change — two starting oxidation states merging into one (as happens here) is comproportionation, the mirror image of disproportionation, not the same thing.

Takeaway (📌):
Disproportionation always means ONE element, ONE starting oxidation state, splitting into two DIFFERENT product oxidation states — check both the direction (converging vs diverging) and that it is a single element before answering 'yes'.

Correct Answer: A. But-2-ene

Fastest Approach (🚀):
$CH_3CH=CHCH_3$ has 4 carbons ("but-") and one C=C double bond ("-ene") starting at carbon 2 from either end — so it must be but-2-ene, without needing to consider any other numbering.

Step-by-Step Breakdown:

1. Identify the Molecule

The organic reactant in Reaction 2 is $CH_3CH=CHCH_3$.

• It is a hydrocarbon with four carbon atoms in a single chain, so the stem is "but-".
• It contains one carbon-carbon double bond, so the suffix is "-ene" (not "-ane", which denotes a saturated alkane).
• Numbering from either end of the chain, the double bond starts at the second carbon: $C1H_3-C2H=C3H-C4H_3$. Numbering from the other end gives the identical locant (2), so "but-2-ene" is the lowest and only sensible locant.

2. Determine the IUPAC Name

Combining the stem, locant, and suffix gives the name but-2-ene.

3. Eliminate the Other Options

• B. But-3-ene: Numbering a 4-carbon chain can only place a double bond at position 1 or 2 (positions 3 and 4 are equivalent to 2 and 1 counted from the other end) — "but-3-ene" is not a valid lowest-locant name.
• C. Di-methene: Not a real IUPAC name; it does not follow the stem/suffix system at all.
• D. But-2-ane: The "-ane" suffix denotes a saturated alkane with no double bond, which contradicts the C=C shown in the structure.

The correct answer is A.

Why the Other Options Are Wrong (❌):

• B. But-3-ene — Incorrect Locant / Numbering Error
  'But-3-ene' numbers the double bond from the wrong end of the chain. IUPAC rules require the lowest possible locant, and for $CH_3CH=CHCH_3$ that is always 2 (numbering from either end gives the double bond position 2, never 3) — so 'but-3-ene' is not a valid lowest-locant name for this molecule.
• C. Di-methene — Invalid Nomenclature
  'Di-methene' is not a real IUPAC name — it does not follow the stem-locant-suffix naming system (which requires a chain-length stem like 'but-', a locant, and the '-ene' suffix), so it cannot describe any structure correctly.
• D. But-2-ane — Incorrect Suffix (Alkane vs Alkene)
  'But-2-ane' mismatches the suffix to the structure: '-ane' denotes a fully saturated alkane with no double bond, but $CH_3CH=CHCH_3$ clearly contains a C=C double bond, which requires the '-ene' suffix instead.

Common Mistake (⚠️):
Numbering the chain from the wrong end and writing 'but-3-ene' instead of applying the IUPAC rule that the double bond must get the LOWEST possible locant — for a 4-carbon chain, the lowest locant for this double bond position is always 2, not 3.

Takeaway (📌):
When naming an alkene, count the longest chain for the stem, use '-ene' for the C=C double bond, and always number from whichever end gives the double bond the lowest locant.

Correct Answer: C. 261 g

Fastest Approach (🚀):
$M_r(C_4H_8) = 56$ and $M_r(C_4H_{10}) = 58$; $252\,\text{g} \div 56 = 4.50\,\text{mol}$ of but-2-ene, which (1:1) gives 4.50 mol of butane, i.e. $4.50 \times 58 = 261\,\text{g}$.

Step-by-Step Breakdown:

1. Identify the Reactant and Product

• Reactant: but-2-ene, $C_4H_8$
• Product: butane, $C_4H_{10}$
• Reaction: $C_4H_8 + H_2 \rightarrow C_4H_{10}$ (1:1 molar ratio)

2. Calculate Molar Masses

• $M_r(C_4H_8) = (4 \times 12.0) + (8 \times 1.0) = 56.0\ \text{g mol}^{-1}$
• $M_r(C_4H_{10}) = (4 \times 12.0) + (10 \times 1.0) = 58.0\ \text{g mol}^{-1}$

3. Calculate Moles of Reactant

$n(C_4H_8) = \dfrac{252\ \text{g}}{56.0\ \text{g mol}^{-1}} = 4.50\ \text{mol}$

4. Calculate Mass of Product

The stoichiometry is 1:1, so 4.50 mol of but-2-ene produces 4.50 mol of butane.
$\text{Mass of butane} = 4.50\ \text{mol} \times 58.0\ \text{g mol}^{-1} = 261\ \text{g}$

The correct answer is C.

Why the Other Options Are Wrong (❌):

• A. 198 g — Arithmetic Error
  198 g does not correspond to the correct calculation with either molar mass ($252/56=4.50\ \text{mol}$ or $252/58=4.34\ \text{mol}$); it likely results from a compounded arithmetic slip rather than a single clean mole-ratio error.
• B. 142 g — Incomplete Calculation
  142 g is roughly what you would get by mistakenly treating 252 g as already being in moles-equivalent terms or misapplying the molar masses in a division/subtraction rather than the correct mass$\to$mole$\to$mass route; it does not follow from $n=4.50\ \text{mol}$ of product.
• D. 342 g — Molar Mass Mix-up
  342 g comes from using but-2-ene's own molar mass (56) for the PRODUCT step instead of butane's (58), or otherwise swapping which molar mass belongs to reactant vs product, inflating the final mass.

Common Mistake (⚠️):
Using the wrong molar mass for one of the two compounds (e.g. dividing 252 g by butane's $M_r=58$ instead of but-2-ene's $M_r=56$ to find moles), which silently carries an error through the rest of the calculation even though the 1:1 stoichiometry step is done correctly.

Takeaway (📌):
For a simple 1:1 addition reaction like alkene hydrogenation, moles of product equal moles of reactant — the only real work is converting mass to moles (and back) with the CORRECT molar mass for each side of the equation.

Correct Answer: A. The catalysed reaction will have a steeper gradient

Fastest Approach (🚀):
Recognise that a catalyst only changes the RATE (gradient), not the final YIELD — both curves must level off at the same final mass, ruling out any option claiming a different amount of product; the catalysed curve simply gets there faster (steeper).

Step-by-Step Breakdown:

1. Understand the Effect of a Catalyst

A catalyst provides an alternative reaction pathway with a lower activation energy, which increases the rate of the reaction.
On a graph of product mass vs. time, a faster reaction corresponds to a steeper initial gradient (slope).

• Reaction 2 ($CH_3CH=CHCH_3 + H_2 \rightarrow CH_3CH_2CH_2CH_3$) is a one-way hydrogenation, not a reversible equilibrium — it goes to completion. A catalyst speeds up how quickly it gets there but cannot change the final mass of product, since that is fixed by the stoichiometry and the amount of reactant available.

2. Evaluate the Options

A. The catalysed reaction will have a steeper gradient: Correct. This reflects the faster initial rate of reaction, while both curves still plateau at the same final mass.
B. Both graphs will be the same...: Incorrect. The rate will be faster with a catalyst, so the shape of the graph before it plateaus is different (steeper), even though the final plateau height is the same.

• C. The graph... will show more product being made: Incorrect. A catalyst doesn't change the total yield — Reaction 2 goes to completion regardless, so both curves reach the same final mass.
• D. Both will reach their end points at the same time...: Incorrect. It is the opposite: the catalysed reaction reaches its end point SOONER (not at the same time), while both reach the SAME final mass (not different amounts) — this option has both halves backwards.

3. Conclude

Only the gradient (rate) changes with a catalyst, not the amount of product formed, so the correct answer is A.

Why the Other Options Are Wrong (❌):

• B. Both graphs will be the same as the catalyst does not affect the position of equilibrium — Conceptual Misunderstanding
  The two graphs will NOT be the same shape — the catalysed reaction reaches completion faster (a steeper initial gradient), even though it is true that both level off at the same final mass. The premise given here ('catalyst does not affect equilibrium') is also misapplied: Reaction 2 goes to completion and is not a reversible equilibrium at all.
• C. The graph of the reaction with the catalyst will show more product being made — Conceptual Misunderstanding
  A catalyst cannot increase the total mass of product formed from a fixed amount of reactant — it only speeds up how quickly that (unchanged) final mass is reached. Reaction 2 goes to completion either way, so both curves plateau at the same height.
• D. Both will reach their end points at the same time, but the catalysed reaction will show more product at this point — Reversed Concept
  This has both parts backwards: the catalysed reaction reaches its end point SOONER, not at the same time as the uncatalysed one, while the amount of product at that end point is the SAME for both, not greater with the catalyst.

Common Mistake (⚠️):
Confusing 'more product formed' with 'product formed faster' — a catalyst only changes how quickly the reaction reaches its maximum mass of product (steeper gradient, reaching the plateau sooner), it never increases the total amount of product formed from a fixed quantity of reactant.

Takeaway (📌):
On a mass/time graph, a catalyst always makes the curve steeper at the start and reach its plateau sooner, but the FINAL height of the plateau (total product formed) is exactly the same with or without a catalyst.

Correct Answer: C. $2.34 \times 10^{-6}\ mol\ dm^{-3}$

Fastest Approach (🚀):
$[H^+] = 10^{-5.63} = 10^{-6} \times 10^{0.37} \approx 2.34 \times 10^{-6}\ \text{mol dm}^{-3}$ — since $HNO_3$ is a strong acid, this equals $[HNO_3]$ directly.

Step-by-Step Breakdown:

1. Understand the Relationship Between pH and Concentration

Nitric acid ($HNO_3$) is a strong acid, meaning it fully dissociates in water: $HNO_3 \rightarrow H^+ + NO_3^-$.
Therefore, the concentration of hydrogen ions $[H^+]$ is equal to the concentration of the acid $[HNO_3]$.

The formula for pH is:
$\text{pH} = -\log_{10}[H^+]$

2. Calculate $[H^+]$

Rearranging the formula to solve for $[H^+]$:
$[H^+] = 10^{-\text{pH}} = 10^{-5.63}$
$[H^+] \approx 2.34 \times 10^{-6}\ \text{mol dm}^{-3}$

3. Identify the Correct Option

Since $[HNO_3] = [H^+]$ for this strong, monoprotic acid, the concentration of $HNO_3(aq)$ is $2.34 \times 10^{-6}\ \text{mol dm}^{-3}$, matching option C.

The correct answer is C.

Why the Other Options Are Wrong (❌):

• A. $2.43 \times 10^{-4}\ mol\ dm^{-3}$ — Arithmetic Error
  $2.43 \times 10^{-4}\ \text{mol dm}^{-3}$ corresponds to $10^{-3.61}$, roughly 2 pH units away from 5.63 — this looks like a digit-transposition or off-by-two-in-the-exponent slip rather than the correct evaluation of $10^{-5.63}$.
• B. $3.24 \times 10^{-5}\ mol\ dm^{-3}$ — Arithmetic Error
  $3.24 \times 10^{-5}\ \text{mol dm}^{-3}$ corresponds to $10^{-4.49}$, about one pH unit away from 5.63 — likely from rounding or misplacing the decimal point when splitting 5.63 into its integer and fractional parts.
• D. $2.34 \times 10^{6}\ mol\ dm^{-3}$ — Sign Error
  $2.34 \times 10^{6}\ \text{mol dm}^{-3}$ has the correct digits but the wrong sign on the exponent — forgetting the minus sign in $[H^+]=10^{-\text{pH}}$ turns a very dilute acid concentration into an impossibly huge (and physically meaningless) one.

Common Mistake (⚠️):
Getting the magnitude right but the sign of the exponent wrong (writing $2.34 \times 10^{6}$ instead of $2.34 \times 10^{-6}$) — a solution with pH 5.63 is acidic and dilute, so $[H^+]$ must be a small fraction (a NEGATIVE power of 10), never a huge positive number.

Takeaway (📌):
For any strong monoprotic acid, $[\text{acid}] = [H^+] = 10^{-\text{pH}}$ directly — no equilibrium expression or dissociation constant is needed, only the definition of pH.

Correct Answer: C. $0.14\ J\ g^{-1}\ ^{\circ}C^{-1}$

Fastest Approach (🚀):
$\Delta T = 39-25 = 14^{\circ}\text{C}$, so $c = \dfrac{q}{m\Delta T} = \dfrac{19.6}{10 \times 14} = 0.14\ \text{J g}^{-1}\ ^{\circ}\text{C}^{-1}$.

Step-by-Step Breakdown:

1. Identify the Given Values

• Mass ($m$) = $10\ \text{g}$
• Energy ($q$) = $19.6\ \text{J}$
• Initial temperature ($T_1$) = $25^{\circ}\text{C}$
• Final temperature ($T_2$) = $39^{\circ}\text{C}$
• Change in temperature ($\Delta T$) = $39^{\circ}\text{C} - 25^{\circ}\text{C} = 14^{\circ}\text{C}$ (equivalently $14\ \text{K}$, since the size of a degree Celsius equals the size of a kelvin)

2. Use the Specific Heat Capacity Formula

The formula relating energy, mass, specific heat capacity ($c$), and temperature change is:
$q = mc\Delta T$

Rearranging to solve for $c$:
$c = \dfrac{q}{m \Delta T}$

3. Calculate the Value

$c = \dfrac{19.6\ \text{J}}{10\ \text{g} \times 14^{\circ}\text{C}} = \dfrac{19.6}{140}\ \text{J g}^{-1}\ ^{\circ}\text{C}^{-1} = 0.14\ \text{J g}^{-1}\ ^{\circ}\text{C}^{-1}$

This matches the real specific heat capacity of mercury (approximately $0.14\ \text{J g}^{-1}\ ^{\circ}\text{C}^{-1}$), confirming the calculation.

The correct answer is C.

Why the Other Options Are Wrong (❌):

• A. $0.08\ J\ g^{-1}\ ^{\circ}C^{-1}$ — Wrong Variable Used
  $0.08\ \text{J g}^{-1}\ ^{\circ}\text{C}^{-1}$ would result from using too large a value in the denominator (e.g. $19.6/(10\times 25)=0.078$, mistakenly using the initial temperature reading instead of $\Delta T=14$).
• B. $0.11\ J\ g^{-1}\ ^{\circ}C^{-1}$ — Arithmetic Error
  $0.11\ \text{J g}^{-1}\ ^{\circ}\text{C}^{-1}$ does not fall out of $19.6/(10\times14)$ with any of the given numbers substituted directly; it is consistent with a rounding or transcription slip partway through the division.
• D. $0.17\ J\ g^{-1}\ ^{\circ}C^{-1}$ — Arithmetic Error
  $0.17\ \text{J g}^{-1}\ ^{\circ}\text{C}^{-1}$ is close to $19.6/(10\times 11.5)$ or similar — consistent with using an incorrect $\Delta T$ (too small), which inflates $c$ above the correct 0.14.
• E. $0.2\ J\ g^{-1}\ ^{\circ}C^{-1}$ — Arithmetic Error
  $0.2\ \text{J g}^{-1}\ ^{\circ}\text{C}^{-1}$ would follow from $19.6/(10\times 9.8)$-style substitutions or from using only part of the mass/temperature data — it does not correspond to the correct $q=mc\Delta T$ rearrangement with the given values.

Common Mistake (⚠️):
Dividing by only one of the temperature readings (25 or 39) instead of the temperature CHANGE $\Delta T = 14$, or forgetting to multiply by the mass at all — either error scales the answer to a different (wrong) option.

Takeaway (📌):
$q=mc\Delta T$ always uses the CHANGE in temperature, never a single temperature reading; rearrange carefully to $c = q/(m\Delta T)$ and keep track of units throughout.

Correct Answer: B. Tollens' reagent: no silver produced; Fehling's solution: deep blue colour

Fastest Approach (🚀):
Whenever a reflux oxidation of a mixed primary/secondary polyol leaves no aldehyde behind (primary $\rightarrow$ carboxylic acid, secondary $\rightarrow$ ketone under full/reflux oxidation), both Tollens' and Fehling's must be negative — no silver mirror, and the Fehling's solution stays its original deep blue.

Step-by-Step Breakdown:

1. Analyse the First Reaction

• The starting material is 1,2,3,4-tetrabromobutane: $BrCH_2-CHBr-CHBr-CH_2Br$.
• Reacting it with warm aqueous sodium hydroxide ($NaOH$) causes nucleophilic substitution. Each bromine atom is replaced by a hydroxyl ($-OH$) group.
• The product is butane-1,2,3,4-tetraol: $HOCH_2-CH(OH)-CH(OH)-CH_2OH$. This molecule has two primary alcohol groups (at the ends, C1 and C4) and two secondary alcohol groups (in the middle, C2 and C3).

2. Analyse the Oxidation Reaction

• The tetraol is heated under reflux with acidified potassium dichromate ($K_2Cr_2O_7/H^+$), a strong oxidising agent.

Under reflux, primary alcohols are oxidised all the way to carboxylic acids (not stopped at the aldehyde stage, which requires gentler distillation conditions).
Secondary alcohols are oxidised to ketones.

• The product is 2,3-dioxobutanedioic acid: $HOOC-C(=O)-C(=O)-COOH$ — two terminal $-COOH$ groups and two central $C=O$ (ketone-like) groups.

3. Evaluate the Forensic Tests

• Tollens' reagent: gives a positive result (a silver mirror) only with aldehydes.
• Fehling's solution: gives a positive result (a brick-red precipitate of $Cu_2O$) only with aldehydes; it remains its original deep blue colour if no aldehyde is present.
• The final product contains only carboxylic acid and ketone groups — no aldehyde groups at all.
• Therefore, it reacts with neither reagent: Tollens' produces no silver, and Fehling's solution stays deep blue.

4. Conclude

Matching these results (no silver; deep blue) to the table, option B is correct.

Why the Other Options Are Wrong (❌):

• A. Tollens' reagent: silver mirror produced; Fehling's solution: deep blue colour — Conceptual Misunderstanding
  A silver mirror with Tollens' reagent requires an aldehyde group, but reflux with acidified dichromate oxidises the tetraol's primary alcohols all the way to carboxylic acids (and its secondary alcohols to ketones) — no aldehyde survives, so no silver mirror forms.
• C. Tollens' reagent: no silver produced; Fehling's solution: brick red colour — Conceptual Misunderstanding
  A brick-red precipitate with Fehling's solution requires an aldehyde group. The final product (2,3-dioxobutanedioic acid) has only carboxylic acid and ketone groups, no aldehyde, so Fehling's solution stays its original deep blue colour rather than turning brick red.
• D. Tollens' reagent: silver mirror produced; Fehling's solution: brick red colour — Conceptual Misunderstanding
  Both a silver mirror (Tollens') AND a brick-red precipitate (Fehling's) require an aldehyde group to be present. Since reflux oxidation converts every alcohol in this molecule to either a carboxylic acid or a ketone — never stopping at an aldehyde — neither positive result should be observed.

Common Mistake (⚠️):
Assuming that because the reflux step involves an oxidation, SOME oxidation-sensitive test must turn positive — but Tollens' and Fehling's are specific to the ALDEHYDE functional group, and reflux with excess acidified dichromate drives primary alcohols all the way past the aldehyde stage to carboxylic acids, so no aldehyde ever remains to give a positive result.

Takeaway (📌):
Reflux (not gentle distillation) with acidified $K_2Cr_2O_7$ takes primary alcohols all the way to carboxylic acids and secondary alcohols to ketones — since neither product is an aldehyde, both Tollens' reagent and Fehling's solution stay negative (no silver mirror; solution stays deep blue) regardless of how many alcohol groups reacted.

Correct Answer: A. $-202.6\ kJ/mol$

Fastest Approach (🚀):
Use equation 1 as-is, REVERSE equation 2 (to consume $H_2O_2$ and release $H_2$ back), and DOUBLE equations 3 and 4 (to produce $2H_2O_{(l)}$ via gaseous water); sum: $177.4 + 191.2 + 2(-241.8) + 2(-43.8) = -202.6\ \text{kJ mol}^{-1}$.

Step-by-Step Breakdown:

1. Set up the Hess's Law Cycle

We want to find $\Delta H$ for the target reaction:
$C_6H_4(OH)_2 + H_2O_2 \rightarrow C_6H_4O_2 + 2H_2O_{(l)}$

We are given the following equations:
(1) $C_6H_4(OH)_2 \rightarrow C_6H_4O_2 + H_2 \quad \Delta H = +177.4\ \text{kJ mol}^{-1}$
(2) $H_2 + O_2 \rightarrow H_2O_2 \quad \Delta H = -191.2\ \text{kJ mol}^{-1}$
(3) $H_2 + \tfrac{1}{2}O_2 \rightarrow H_2O_{(g)} \quad \Delta H = -241.8\ \text{kJ mol}^{-1}$
(4) $H_2O_{(g)} \rightarrow H_2O_{(l)} \quad \Delta H = -43.8\ \text{kJ mol}^{-1}$

2. Manipulate the Equations to Match the Target

• Keep Eq (1) as is, to get $C_6H_4(OH)_2$ reacting and $C_6H_4O_2$ forming: $\Delta H_1 = +177.4$
• REVERSE Eq (2), so that $H_2O_2$ is a reactant (consumed) rather than a product: $H_2O_2 \rightarrow H_2 + O_2 \quad \Delta H = +191.2$
• We need $2H_2O_{(l)}$ on the product side. First, DOUBLE Eq (3) to get $2H_2O_{(g)}$: $2H_2 + O_2 \rightarrow 2H_2O_{(g)} \quad \Delta H = 2 \times (-241.8) = -483.6$
• Then DOUBLE Eq (4) to condense the $2H_2O_{(g)}$ into $2H_2O_{(l)}$: $2H_2O_{(g)} \rightarrow 2H_2O_{(l)} \quad \Delta H = 2 \times (-43.8) = -87.6$

3. Check the Species Cancel

Adding the four modified equations: the $H_2$ produced by (1) is consumed by reversed-(2) and doubled-(3); the $O_2$ produced by reversed-(2) is consumed by doubled-(3); the $2H_2O_{(g)}$ produced by doubled-(3) is consumed by doubled-(4). What remains is exactly the target equation: $C_6H_4(OH)_2 + H_2O_2 \rightarrow C_6H_4O_2 + 2H_2O_{(l)}$.

4. Sum the Enthalpy Changes

$\Delta H_{\text{total}} = 177.4 + 191.2 + (-483.6) + (-87.6)$
$\Delta H_{\text{total}} = 368.6 - 571.2 = -202.6\ \text{kJ mol}^{-1}$

The correct answer is A.

Why the Other Options Are Wrong (❌):

• B. $-256.2\ kJ/mol$ — Sign / Equation-Combination Error
  $-256.2\ \text{kJ/mol}$ does not fall out of the correct combination (reverse eq. 2, double eqs. 3 and 4); it is consistent with mismanaging which equation gets reversed or applying a sign flip to the wrong term while still scaling eqs. 3 and 4 by 2, rather than a single clean transcription slip.
• C. $39.6\ kJ/mol$ — Missing Stoichiometric Scaling
  $39.6\ \text{kJ/mol}$ is close to what results from forgetting to DOUBLE equation (3) (using $-241.8$ once instead of $-483.6$) while still doubling equation (4): $177.4+191.2-241.8-87.6 \approx +39$ — a positive result because too little of the strongly exothermic vaporisation step was subtracted.
• D. $-161.8\ kJ/mol$ — Missing Stoichiometric Scaling
  $-161.8\ \text{kJ/mol}$ is close to what results from forgetting to DOUBLE equation (4) (using $-43.8$ once instead of $-87.6$) while still doubling equation (3): $177.4+191.2-483.6-43.8 \approx -159$ — under-subtracting the condensation step shifts the total away from the fully-scaled answer.

Common Mistake (⚠️):
Forgetting to DOUBLE both equation (3) and equation (4) before summing — since the target reaction produces $2H_2O_{(l)}$ (not $1H_2O_{(l)}$), both the vaporisation-equivalent and condensation steps must be scaled by 2, or the final total lands on the wrong option entirely.

Takeaway (📌):
In a Hess's law cycle, match each given equation's direction and scale (reverse it and/or multiply it by a coefficient) to what is needed in the target equation, and apply that SAME sign flip or multiplier to its $\Delta H$ — then confirm every non-target species cancels exactly before trusting the sum.

Correct Answer: B. $2.5^\circ$

Fastest Approach (🚀):
Recall the standard values: $H_2O \approx 104.5^\circ$ (2 lone pairs) and $H_3O^+ \approx 107^\circ$ (1 lone pair, like $NH_3$) — the difference is $107 - 104.5 = 2.5^\circ$.

Step-by-Step Breakdown:

1. Determine the Geometry of $H_2O$

Oxygen has 6 valence electrons. It forms 2 single bonds with hydrogen, using 2 electrons, leaving 4 electrons as 2 lone pairs.
Total electron domains = 4 (2 bonding pairs, 2 lone pairs). The electron geometry is tetrahedral.

• Because lone pairs repel more strongly than bonding pairs, the ideal $109.5^{\circ}$ tetrahedral angle is compressed by TWO lone pairs to approximately $104.5^{\circ}$.

2. Determine the Geometry of $H_3O^+$

• Oxygen has 6 valence electrons, minus 1 for the positive charge = 5. It forms 3 single bonds with hydrogen, using 3 electrons, leaving 2 electrons as 1 lone pair.
• Total electron domains = 4 (3 bonding pairs, 1 lone pair). The electron geometry is tetrahedral.
• With only ONE lone pair repelling the bonds (isoelectronic with $NH_3$'s bonding arrangement), the angle is compressed less than in water, resulting in an angle of approximately $107^{\circ}$.

3. Calculate the Difference

Difference $= 107^{\circ} - 104.5^{\circ} = 2.5^{\circ}$

The correct answer is B.

Why the Other Options Are Wrong (❌):

• A. $73^\circ$ — Conceptual Misunderstanding
  $73^\circ$ does not correspond to the difference between two tetrahedral-family bond angles that are both close to $109.5^\circ$ — both $H_2O$ ($104.5^\circ$) and $H_3O^+$ ($107^\circ$) are compressed only slightly from the ideal tetrahedral angle, so their difference must be small, not tens of degrees.
• C. $17^\circ$ — Arithmetic / Magnitude Error
  $17^\circ$ overstates how much a single extra lone pair compresses the bond angle; going from 1 lone pair ($H_3O^+$, $107^\circ$) to 2 lone pairs ($H_2O$, $104.5^\circ$) only costs about $2.5^\circ$ of compression, not $17^\circ$.
• D. Impossible to tell — Conceptual Misunderstanding
  The bond angles are not 'impossible to tell' — both species have well-defined, standard VSEPR geometries (tetrahedral electron-pair arrangement with 2 lone pairs for $H_2O$, 1 lone pair for $H_3O^+$), giving well-established angles of approximately $104.5^\circ$ and $107^\circ$ respectively.

Common Mistake (⚠️):
Assuming $H_2O$ and $H_3O^+$ must have identical bond angles because both have a 'tetrahedral' electron geometry around oxygen — the electron geometry is the same, but the number of LONE PAIRS differs (2 vs 1), which is exactly what shifts the observed bond angle between them.

Takeaway (📌):
More lone pairs on the central atom compress the bond angle further below the ideal tetrahedral $109.5^{\circ}$: $H_2O$ (2 lone pairs) $\approx 104.5^{\circ}$, while $H_3O^+$ (1 lone pair, like $NH_3$) $\approx 107^{\circ}$ — a small but real and calculable difference, not something 'impossible to tell'.

Correct Answer: A. $46^\circ$

Fastest Approach (🚀):
The axial-equatorial F-Cl-F angle in a T-shaped molecule is close to $90^\circ$ but compressed slightly by lone-pair repulsion to about $87.5^\circ$; treating the Cl-F(axial)-F(equatorial) triangle as isosceles, the two equal base angles are $(180^\circ-87.5^\circ)/2 = 46.25^\circ \approx 46^\circ$.

Step-by-Step Breakdown:

1. Determine the Shape of $ClF_3$

Chlorine has 7 valence electrons. It forms 3 single bonds with fluorine, using 3 electrons, leaving 4 electrons as 2 lone pairs.
Total electron domains = 5 (3 bonding pairs, 2 lone pairs). The electron geometry is trigonal bipyramidal.

• The lone pairs occupy the equatorial positions to minimise repulsion, resulting in a T-shaped molecular geometry: two axial fluorines and one equatorial fluorine.

2. Analyse the Angles

• In a T-shape, the angle between an axial and the equatorial bond ($F_{ax}-Cl-F_{eq}$) is theoretically $90^{\circ}$, but is slightly compressed to about $87.5^{\circ}$ by the two equatorial lone pairs.
• The question asks for the angle between a Cl-F bond and the shortest line joining two adjacent fluorine atoms — i.e. the angle inside the triangle formed by $Cl$, $F_{ax}$, and $F_{eq}$, at the F corners rather than at Cl.
• The angle at $Cl$ in this triangle is the $F_{ax}-Cl-F_{eq}$ angle, $\approx 87.5^{\circ}$.
• Treating the two Cl-F bonds as approximately equal in length, the triangle is roughly isosceles, so its other two angles (at each fluorine) are equal.

3. Calculate the Angle

Since a triangle's angles sum to $180^{\circ}$:
$\text{base angle} = \dfrac{180^{\circ} - 87.5^{\circ}}{2} = \dfrac{92.5^{\circ}}{2} = 46.25^{\circ}$

Looking at the options, $46^{\circ}$ is the closest value.

The correct answer is A.

Why the Other Options Are Wrong (❌):

• B. $88^\circ$ — Wrong Angle Identified
  $88^\circ$ is close to the F-Cl-F bond angle itself ($\approx 87.5^\circ$) — the angle AT chlorine, between the two Cl-F bonds — not the angle asked for, which is at a fluorine corner of the Cl-F-F triangle.
• C. $23^\circ$ — Arithmetic Error
  $23^\circ$ is roughly half of the correct $46^\circ$ — consistent with an extra, unnecessary halving (e.g. dividing $46.25^\circ$ by 2 again) rather than stopping once the isosceles triangle's base angle has been found.
• D. Impossible to tell — Conceptual Misunderstanding
  The angle is not 'impossible to tell' — $ClF_3$'s T-shaped VSEPR geometry gives a well-defined (if slightly compressed) $F_{ax}-Cl-F_{eq}$ angle of about $87.5^\circ$, from which the requested angle follows directly using triangle geometry.

Common Mistake (⚠️):
Confusing the F-Cl-F bond angle itself ($\approx 87.5^\circ$, the angle AT the central chlorine) with the angle asked for (the angle AT a fluorine atom, between a Cl-F bond and the F...F line) — these are two different angles in the same triangle, related by the fact that the triangle's angles must sum to $180^\circ$.

Takeaway (📌):
When a question asks for an angle 'between a bond and the line joining two other atoms', picture the triangle those three atoms form: use the known central bond angle to find the other angles via $180^\circ$ minus that angle, then halved if the triangle is isosceles.

Correct Answer: C. Reaction 3 requires a high temperature, high pressure but does not require a catalyst

Fastest Approach (🚀):
Recall that the industrial catalytic hydration of ethene (Reaction 3) is a textbook example that explicitly REQUIRES a solid phosphoric acid ($H_3PO_4$) catalyst alongside high temperature and pressure — any statement claiming it needs 'no catalyst' is automatically false, without needing to evaluate the other three statements in detail.

Step-by-Step Breakdown:

1. Analyse the Three Reactions

• Reaction 1 (Fermentation): $C_6H_{12}O_6 \rightarrow 2CH_3CH_2OH + 2CO_2$. Yields are low (the ethanol concentration is limited to around 15% before yeast enzymes are denatured/killed by the alcohol), it is a slow batch process, and it is often described as carbon-neutral (though this is debated once fertiliser use, farming, and transport emissions are counted).
• Reaction 2 (Nucleophilic substitution): $CH_3CH_2Br + NaOH \rightarrow CH_3CH_2OH + NaBr$. Rarely used industrially for bulk ethanol production because bromoethane itself is costly to manufacture and hazardous to handle.
• Reaction 3 (Catalytic hydration of ethene): $H_2C=CH_2 + H_2O \rightarrow CH_3CH_2OH$. Used industrially. Requires high temperature ($\approx 300^{\circ}\text{C}$), high pressure (60-70 atm), AND a solid phosphoric acid ($H_3PO_4$) catalyst — the catalyst is essential to achieving a useful reaction rate.

2. Evaluate Each Statement

• A: Reasonably true — Reaction 2 is not used industrially, largely because bromoethane is itself a hazardous, costly-to-produce reagent.
• B: True — fermentation is self-limiting to a dilute (~15%) ethanol solution, the lowest yield of the three routes, which can each in principle approach much higher conversions.
• C: False. This statement claims Reaction 3 does not require a catalyst, but the industrial hydration of ethene specifically and famously uses a phosphoric acid catalyst — without it, the reaction is far too slow to be practical even at high temperature and pressure.
• D: True — fermentation is often marketed as carbon-neutral (the $CO_2$ released was originally absorbed by the growing crop), but this claim is disputed once the fossil-fuel inputs to farming, transport, and processing are included.

3. Conclude

Since C is the only factually incorrect statement, it is the answer.

Why the Other Options Are Wrong (❌):

• A. Reaction 2 is not widely used in industry to produce ethanol because of a toxic byproduct produced — N/A - True Statement
  This statement is TRUE, not the answer: bromoethane (the reagent for Reaction 2) is itself hazardous and expensive to produce from crude oil derivatives, which is the real reason this route is not used for bulk industrial ethanol production.
• B. Reaction 1 produces the lowest yield of ethanol compared to the other reactions — N/A - True Statement
  This statement is TRUE, not the answer: fermentation (Reaction 1) self-limits to a dilute ethanol solution (yeast is poisoned by ethanol concentrations above roughly 15%), giving it the lowest yield of the three routes, each of which can otherwise approach much higher conversion.
• D. Reaction 1 is often regarded as carbon-neutral, but this is disputed — N/A - True Statement
  This statement is TRUE, not the answer: fermentation is often described as carbon-neutral because the $CO_2$ released was recently absorbed by the crop as it grew, but this claim is genuinely disputed once the fossil fuels used in farming, transport, and distillation are taken into account.

Common Mistake (⚠️):
Assuming that because Reaction 3 needs high temperature AND high pressure, a catalyst must be unnecessary ('the extreme conditions alone must be doing the work') — in reality, industrial ethene hydration uses high temperature, high pressure, AND a phosphoric acid catalyst together; all three conditions are needed simultaneously.

Takeaway (📌):
The industrial hydration of ethene to ethanol always requires all three of: high temperature ($\sim 300^{\circ}\text{C}$), high pressure ($60$-$70$ atm), AND a solid $H_3PO_4$ catalyst — a statement omitting the catalyst is a factual error, not a simplification.

Correct Answer: B. a: Less packed; b: More packed; c: Less packed

Fastest Approach (🚀):
Read off the sign of each $\Delta H$ directly: negative (exothermic) reactions release heat into the water (heats up $\rightarrow$ less packed); the positive (endothermic) reaction absorbs heat from the water (cools down $\rightarrow$ more packed). With a (-30) and c (-45) both exothermic and b (+45) endothermic, the pattern is Less/More/Less — row B.

Step-by-Step Breakdown:

1. Relate Enthalpy Change to Temperature Change

• Reaction a ($\Delta H = -30\ \text{kJ/mol}$): Exothermic. It releases heat into the surroundings (the water), so the water temperature increases.
• Reaction b ($\Delta H = +45\ \text{kJ/mol}$): Endothermic. It absorbs heat from the surroundings (the water), so the water temperature decreases.
• Reaction c ($\Delta H = -45\ \text{kJ/mol}$): Exothermic. It releases heat into the water, so the water temperature increases.

2. Relate Temperature Change to Density/Packing

• When water (above its temperature of maximum density, $4^{\circ}\text{C}$, so this applies at room temperature) is heated, it expands slightly. Its density decreases, meaning the molecules become less packed.
• When water is cooled (towards room temperature from above, or further towards $4^{\circ}\text{C}$), it contracts. Its density increases, meaning the molecules become more packed.

3. Determine the Outcome for Each Reaction

• Reaction a: Water heats up $\rightarrow$ Less packed
• Reaction b: Water cools down $\rightarrow$ More packed
• Reaction c: Water heats up $\rightarrow$ Less packed

4. Conclude

Matching Less / More / Less (for a / b / c respectively) with the table, option B is correct.

Why the Other Options Are Wrong (❌):

• A. a: Less packed; b: More packed; c: More packed — Sign Error
  Row A gets reaction c wrong: c has $\Delta H=-45\ \text{kJ/mol}$ (exothermic, same sign as a), so it should heat the water and make it LESS packed, like reaction a — not 'more packed' as this row states.
• C. a: No change; b: No change; c: No change — Conceptual Misunderstanding
  'No change' for all three ignores that each reaction has a non-zero $\Delta H$ and is releasing or absorbing real heat into/from the water on the tripod above it — with three different, non-zero enthalpy changes given, some change in packing must occur for at least two of the three reactions.
• D. a: More packed; b: Less packed; c: More packed — Reversed Concept (Exothermic vs Endothermic)
  Row D has every sign reversed: a ($\Delta H=-30$, exothermic) should make water LESS packed (not more), b ($\Delta H=+45$, endothermic) should make it MORE packed (not less), and c ($\Delta H=-45$, exothermic) should make it LESS packed (not more) — this row swaps exothermic and endothermic throughout.
• E. a: Less packed; b: Less packed; c: Less packed — Sign Error
  Row E has all three as 'less packed', but reaction b has $\Delta H=+45\ \text{kJ/mol}$ (endothermic) — it absorbs heat FROM the water rather than releasing it, cooling (not heating) the water, so b should be 'more packed', not 'less packed' like a and c.

Common Mistake (⚠️):
Assuming that ANY reaction taking place must change the water's packing (or that only the MAGNITUDE of $\Delta H$ matters), rather than checking the SIGN of each $\Delta H$ individually to determine whether that specific reaction heats or cools the water.

Takeaway (📌):
Exothermic reactions ($\Delta H$ negative) heat surrounding water, expanding it (less packed); endothermic reactions ($\Delta H$ positive) cool surrounding water, contracting it (more packed) — read the sign of $\Delta H$ for each reaction independently rather than assuming a single uniform effect.

Correct Answer: C. 100 times

Fastest Approach (🚀):
The pH difference is $4-2=2$ units, so the ratio of $[H^+]$ is $10^2 = 100$ times, directly — no need to compute either concentration explicitly.

Step-by-Step Breakdown:

1. Understand the pH Scale

The pH scale is logarithmic, meaning each whole number change in pH represents a 10-fold change in the concentration of hydrogen ions ($[H^+]$).
The formula is: $[H^+] = 10^{-\text{pH}}$

2. Calculate $[H^+]$ for Each Solution

• For pH 2: $[H^+] = 10^{-2} = 0.01\ \text{mol dm}^{-3}$
• For pH 4: $[H^+] = 10^{-4} = 0.0001\ \text{mol dm}^{-3}$

3. Compare the Concentrations

To find out how many times more concentrated the pH 2 solution is, divide its concentration by the pH 4 solution's concentration:
$\text{Ratio} = \dfrac{10^{-2}}{10^{-4}} = 10^{2} = 100$

There are 100 times more hydrogen ions in the pH 2 solution.

The correct answer is C.

Why the Other Options Are Wrong (❌):

• A. 2 times — Conceptual Misunderstanding (Linear vs Logarithmic Scale)
  '2 times' treats the pH difference (2 units) itself as the answer, forgetting that pH is a LOGARITHMIC scale — a difference of 2 pH units means a factor of $10^2=100$, not simply the number 2 itself.
• B. 10 times — Incomplete Calculation
  '10 times' is the factor for a difference of only ONE pH unit — since the difference here is 2 units (pH 4 to pH 2), the concentration ratio must be $10^2=100$, not $10^1=10$.
• D. 200 times — Misapplied Formula
  '200 times' looks like the pH difference (2) multiplied by 100, rather than used as the EXPONENT of 10 — the correct relationship is $10^{(\text{pH difference})}=10^2=100$, not $(\text{pH difference}) \times 100$.

Common Mistake (⚠️):
Treating the pH scale as if it were linear and simply multiplying the pH DIFFERENCE by some small factor (e.g. thinking a difference of 2 pH units means '2 times' or '20 times' more concentrated) — because pH is a LOGARITHMIC (base-10) scale, a difference of $n$ pH units always means a factor of $10^n$ in concentration.

Takeaway (📌):
Every single unit of pH difference corresponds to a factor of exactly 10 in $[H^+]$ — a 2-unit difference is $10^2=100$ times, a 3-unit difference is $10^3=1000$ times, and so on; never treat pH units as adding or multiplying linearly with concentration.

Correct Answer: B. It forms a polymer with the same repeating unit as Z-1,2-dichloroethene

Fastest Approach (🚀):
Sort each statement into 'depends on shape/polarity' (differs between E and Z: boiling point, fingerprint IR) versus 'depends only on formula' (same for E and Z: molecular ion peak, addition-polymer repeating unit) — only the polymer repeating-unit statement is both true AND about a formula-only property.

Step-by-Step Breakdown:

1. Analyse the Isomers

E-1,2-dichloroethene (trans) and Z-1,2-dichloroethene (cis) are geometric (E/Z) stereoisomers: same molecular formula ($C_2H_2Cl_2$) and connectivity, differing only in the spatial arrangement of the two Cl atoms across the C=C double bond.

2. Evaluate the Statements

• A. Boiling Point: Incorrect. The Z (cis) isomer has a net molecular dipole moment, because its two polar C-Cl bonds point in roughly the same overall direction; this gives it stronger dipole-dipole intermolecular forces and a higher boiling point. The E (trans) isomer's bond dipoles cancel by symmetry, making it far less polar overall and lowering its boiling point. The two isomers do NOT share the same boiling point.
• B. Polymer: Correct. When either isomer undergoes addition polymerisation, the C=C double bond breaks and each monomer forms single bonds to its neighbours. The stereochemistry along the resulting chain (tacticity) may differ, but the basic repeating unit — $-[CHCl-CHCl]-$ — is exactly the same for both, since it depends only on molecular formula/connectivity, not on E/Z geometry.
• C. IR Spectrum: Incorrect. The fingerprint region ($400$-$1500\ \text{cm}^{-1}$) is highly sensitive to the overall shape and symmetry of a molecule. Since E- and Z-1,2-dichloroethene have different molecular shapes, they give distinct (though perhaps partially overlapping) fingerprint region spectra.
• D. Molecular Ion Peak: Incorrect. The molecular ion peak ($M^+$) in a mass spectrum reflects the mass of the intact molecule. Since both isomers have the identical molecular formula ($C_2H_2Cl_2$) and therefore the identical molar mass, they give the SAME molecular ion peak — not a different one.

3. Conclude

Only option B correctly identifies a property (the addition-polymer repeating unit) that is genuinely the same between the two isomers, and correctly states that it IS the same. The correct answer is B.

Why the Other Options Are Wrong (❌):

• A. It has the same boiling point as Z-1,2-dichloroethene — Conceptual Misunderstanding
  E- and Z-1,2-dichloroethene do NOT share the same boiling point: the Z (cis) isomer has a net dipole moment (its C-Cl bond dipoles do not cancel), giving stronger intermolecular dipole-dipole forces and a higher boiling point than the E (trans) isomer, whose bond dipoles cancel by symmetry.
• C. It has the same IR spectrum as Z-1,2-dichloroethene in the range $400$-$1500\ cm^{-1}$ — Conceptual Misunderstanding
  The IR fingerprint region ($400$-$1500\ \text{cm}^{-1}$) is specifically sensitive to a molecule's overall shape and symmetry, so E- and Z-isomers — which have different shapes — give distinguishable fingerprint spectra, not identical ones.
• D. It has a molecular ion peak different from that of Z-1,2-dichloroethene in its mass spectrum — Conceptual Misunderstanding
  The molecular ion peak depends only on molar mass, and E- and Z-1,2-dichloroethene share the exact same molecular formula ($C_2H_2Cl_2$) and hence the same molar mass — so their molecular ion peaks are the SAME, not different as this option claims.

Common Mistake (⚠️):
Assuming that because E- and Z-1,2-dichloroethene are 'different compounds', EVERY physical or spectroscopic property must differ between them — properties that depend only on molecular formula (mass, and hence molecular ion peak and addition-polymer repeating unit) are identical, while properties that depend on 3D shape (boiling point, IR fingerprint region) differ.

Takeaway (📌):
For E/Z isomers, sort each property into 'formula-dependent' (same for both: molar mass, molecular ion peak, empirical/repeating unit) versus 'shape-or-polarity-dependent' (different for both: boiling point, melting point, dipole moment, IR fingerprint region) before deciding whether a statement about them is true.

Correct Answer: A. [CONTENT MISSING - source scan does not contain this question]

Step-by-Step Breakdown:
Question 24's body was not captured in any of the 27 source scan images provided for this paper. The scan sequence shows Question 23 complete (in esat-0002-chemistry-q23-full.png, which also carries the cut-off 'Question 24' header at its very bottom), then jumps directly to Question 25 complete (at the top of esat-0002-chemistry-q25-full.png) with no Question 24 content in between. The file esat-0002-chemistry-q24-full.png that does exist is a byte-for-byte duplicate of esat-0002-chemistry-q23-full.png (both show Question 22 and Question 23, not Question 24), which is consistent with a capture-sequence gap around this question rather than a rendering or legibility problem. This content genuinely cannot be restored from the material supplied and is left flagged rather than fabricated.

Why the Other Options Are Wrong (❌):

• B. [CONTENT MISSING - source scan does not contain this question] — Content Unavailable
  Not available — Question 24's options are not present in any of the 27 source scans.
• C. [CONTENT MISSING - source scan does not contain this question] — Content Unavailable
  Not available — Question 24's options are not present in any of the 27 source scans.
• D. [CONTENT MISSING - source scan does not contain this question] — Content Unavailable
  Not available — Question 24's options are not present in any of the 27 source scans.
• E. [CONTENT MISSING - source scan does not contain this question] — Content Unavailable
  Not available — Question 24's options are not present in any of the 27 source scans.

Correct Answer: D. $CH_3CH_2CH(OH_2^+)CH_3 \rightarrow CH_3CH=CHCH_3$ (acid-catalysed dehydration): one arrow runs from the C–O bond to the leaving $OH_2$ group, forming a secondary carbocation; a second arrow is drawn starting AT the hydrogen atom on the adjacent carbon (rather than at the C–H bond) and curving to the C–C bond, forming the C=C double bond of the alkene product.

Fastest Approach (🚀):
Scan each mechanism for any arrow tail that starts on an atom's SYMBOL rather than on a bond or an explicit lone pair; a proton ($H$) being removed has no electrons of its own to donate, so the arrow forming a new bond or double bond from an adjacent C–H must start on that C–H BOND, not on the H atom.

Step-by-Step Breakdown:

1. Understand the Rules of Curly Arrows

Curly arrows in reaction mechanisms must ALWAYS show the movement of an electron pair. Therefore, an arrow must ALWAYS start from a source of an electron pair, which is either a lone pair or a covalent bond — never from a bare atom that has no electrons of its own to give.

2. Analyse the Options

• A ($S_N2$ hydrolysis of 2-bromobutane): One arrow starts from the lone pair on $HO^-$ and points to the carbon bearing Br. A second arrow starts from the C–Br bond and points to Br, forming $Br^-$. Both arrows correctly originate from an electron source (a lone pair, then a bond). Correct.
• B (electrophilic addition of HBr to but-2-ene): One arrow starts from the C=C double bond and points to the H of H–Br. A second arrow starts from the H–Br bond and points to Br, forming the carbocation and $Br^-$. A third arrow starts from the lone pair on $Br^-$ and points to the carbocation carbon. All three arrows correctly originate from bonds or lone pairs. Correct.
• C (nucleophilic addition of NH$_3$ to butanone): One arrow starts from the lone pair on $NH_3$ and points to the carbonyl carbon. A second arrow starts from the C=O bond and points to O, forming the alkoxide. A third arrow starts from an N–H BOND (not the bare H) of the resulting ammonium group and points to the alkoxide oxygen, transferring a proton and giving a neutral –OH and –NH$_2$. Correct.
• D (E1 dehydration of butan-2-ol, second step): The first arrow (C–O bond to the leaving water molecule) is drawn correctly. But look closely at the SECOND arrow, the one forming the C=C double bond: it is drawn starting at the H atom itself, not at the C–H bond joining that hydrogen to the adjacent carbon. This is fundamentally incorrect — a hydrogen leaving as $H^+$ has no electron pair of its own to donate; the arrow must start on the C–H bond, showing those bonding electrons moving in to form the new C=C double bond (with $H^+$ released separately).

3. Conclude

Only option D has an arrow tail sitting on a bare atom instead of a bond, so the correct answer is D.

Why the Other Options Are Wrong (❌):

• A. $CH_3CH_2CHBrCH_3 \xrightarrow{HO^-} CH_3CH_2CH(OH)CH_3 + Br^-$ (nucleophilic substitution): one arrow runs from the lone pair on $HO^-$ to the carbon bearing Br; a second arrow runs from the C–Br bond to the Br atom, forming $Br^-$. — N/A - Correctly Drawn
  Both arrows in this $S_N2$ mechanism are drawn correctly: the nucleophile's lone pair (on $HO^-$) attacks the carbon, and the C–Br bond's electrons leave with the departing bromine to form $Br^-$ — each arrow starts on a genuine electron source (a lone pair, then a bond).
• B. $CH_3CH=CHCH_3 \xrightarrow{HBr} CH_3CH_2CHBrCH_3$ (electrophilic addition): one arrow runs from the C=C double bond to the H of H–Br; a second arrow runs from the H–Br bond to Br, forming a secondary carbocation and $Br^-$; a third arrow runs from the lone pair on $Br^-$ to the carbocation carbon. — N/A - Correctly Drawn
  All three arrows in this electrophilic addition are drawn correctly: the $\pi$ electrons of C=C attack the electrophilic H of H–Br, the H–Br bond's electrons leave onto Br, and the bromide lone pair then attacks the carbocation — every arrow starts on a bond or a lone pair.
• C. $CH_3CH_2COCH_3 \xrightarrow{NH_3} CH_3CH_2C(OH)(NH_2)CH_3$ (nucleophilic addition): one arrow runs from the lone pair on $NH_3$ to the carbonyl carbon; a second arrow runs from the C=O bond to the O, forming an alkoxide with an ammonium substituent; a third arrow runs from an N–H bond of that ammonium group to the alkoxide oxygen, giving a neutral –OH and –NH$_2$. — N/A - Correctly Drawn
  All three arrows in this nucleophilic addition (and subsequent proton transfer) are drawn correctly: the nitrogen lone pair attacks the carbonyl carbon, the C=O $\pi$ bond's electrons move onto oxygen, and the final proton transfer arrow starts on the N–H BOND (not the bare hydrogen) and moves to the alkoxide oxygen.

Common Mistake (⚠️):
Focusing only on WHERE an arrow points (checking that it ends up in a sensible place, like forming a new bond) while not checking WHERE the arrow starts — a curly arrow that points to the right place but originates on a bare atom (with no electron pair of its own) is still drawn incorrectly, because it fails to show a real pair of electrons moving.

Takeaway (📌):
Before trusting any curly arrow, check its TAIL, not just its head: it must sit on an explicit lone pair (often shown as two dots) or on a bond (a line between two atoms) — an arrow tail resting on a bare atomic symbol (like a lone $H$) is never valid, since that atom has no electrons of its own to donate.

Correct Answer: C. Graph C: yield starts high at low temperature and falls steadily as temperature increases, levelling off at a low yield at high temperature

Fastest Approach (🚀):
$\Delta H = 2(-290) - [(-286)+(-249)] = -580+535 = -45\ \text{kJ mol}^{-1}$ (exothermic forward reaction), so raising the temperature shifts equilibrium back towards $H_2O+D_2O$, meaning HDO yield DECREASES as temperature rises — a falling curve, graph C.

Step-by-Step Breakdown:

1. Calculate the Enthalpy Change ($\Delta H$)

The reaction is: $H_2O_{(l)} + D_2O_{(l)} \rightleftharpoons 2HDO_{(l)}$
Using standard enthalpies of formation:
$\Delta H_{\text{reaction}} = \Sigma \Delta H_f(\text{products}) - \Sigma \Delta H_f(\text{reactants})$
$\Delta H_{\text{reaction}} = [2 \times (-290)] - [(-286) + (-249)]$
$\Delta H_{\text{reaction}} = -580 - (-535)$
$\Delta H_{\text{reaction}} = -580 + 535 = -45\ \text{kJ mol}^{-1}$
Since $\Delta H$ is negative, the forward reaction (forming HDO) is exothermic.

2. Apply Le Chatelier's Principle

• For an exothermic forward reaction, heat can be thought of as an extra 'product'.
• Increasing the temperature adds heat to the system, so the equilibrium shifts to oppose this by favouring the ENDOTHERMIC direction — here, that is the reverse reaction (back towards $H_2O$ and $D_2O$).
• This shift towards the reactants means the yield of the product (HDO) at equilibrium decreases as temperature increases.

3. Select the Correct Graph

• Graph A shows yield increasing with temperature — this would describe an ENDOTHERMIC forward reaction, the opposite of what was calculated.
• Graph B shows yield remaining constant — this would only be true if $\Delta H = 0$, but it is $-45\ \text{kJ mol}^{-1}$, not zero.

Graph C shows yield decreasing steadily as temperature increases, matching our prediction for an exothermic forward reaction.
Graph D shows yield increasing then decreasing — there is no mechanism here (no change in dominant reaction pathway) that would produce a peak part-way through.

The correct answer is C.

Why the Other Options Are Wrong (❌):

• A. Graph A: yield starts low at low temperature and rises with an increasing gradient as temperature increases (an upward-curving increase, no plateau) — Sign Error
  Graph A (yield rising with temperature) would be correct only if the forward reaction were endothermic. The calculation gives $\Delta H=-45\ \text{kJ mol}^{-1}$ (exothermic), so raising temperature should DECREASE the HDO yield, not increase it.
• B. Graph B: yield stays constant (a flat horizontal line) at all temperatures — Conceptual Misunderstanding
  Graph B (constant yield) would only apply if $\Delta H=0$ for the reaction, meaning temperature has no effect on the equilibrium position. Here $\Delta H=-45\ \text{kJ mol}^{-1}$, a genuine non-zero exothermic value, so the yield must change with temperature.
• D. Graph D: yield starts low, rises to a peak at an intermediate temperature, then falls again as temperature increases further — Conceptual Misunderstanding
  Graph D (a peak in the middle) would require some additional mechanism causing the equilibrium to reverse its temperature-dependence partway through, which nothing in the given thermochemical data supports — a single reaction with a fixed sign of $\Delta H$ shifts consistently in one direction as temperature rises, giving a steadily rising or steadily falling curve, not a peak.

Common Mistake (⚠️):
Forgetting that increasing temperature favours the ENDOTHERMIC direction of a reaction, not automatically increasing the yield of whichever species is being asked about — since forming HDO here is exothermic, more heat pushes the equilibrium the OTHER way, lowering (not raising) the HDO yield.

Takeaway (📌):
Always calculate the sign of $\Delta H$ for the direction the question asks about (here, formation of HDO) using $\Delta H = \Sigma\Delta H_f(\text{products}) - \Sigma\Delta H_f(\text{reactants})$, then apply Le Chatelier's principle: exothermic forward reactions lose yield as temperature rises; endothermic forward reactions gain yield as temperature rises.

Correct Answer: A. When ozone molecules collide with nitrogen they may form nitrogen monoxide

Fastest Approach (🚀):
From $K_c$: $2NO_{(g)} + 2O_{2(g)} \rightleftharpoons N_{2(g)} + 2O_{3(g)}$. 'More ozone at higher temperature' means the forward reaction is endothermic (rules out B). Moles of gas: 4 (reactants) vs 3 (products), so higher pressure favours the products side (MORE ozone, ruling out C). $K_c$ is a true constant at fixed temperature regardless of concentration (ruling out D). That leaves A, which correctly describes the ever-present reverse reaction in a dynamic equilibrium.

Step-by-Step Breakdown:

1. Deduce the Reaction Equation

The equilibrium constant expression is $K_c = \dfrac{[N_2][O_3]^2}{[NO]^2[O_2]^2}$.
From this, we can write the balanced chemical equation: the terms in the numerator are products, and those in the denominator are reactants, with the exponents becoming stoichiometric coefficients:
$2NO_{(g)} + 2O_{2(g)} \rightleftharpoons N_{2(g)} + 2O_{3(g)}$

2. Analyse the Temperature Effect

We are told: "More ozone is formed as the temperature rises."
By Le Chatelier's principle, increasing the temperature favours the ENDOTHERMIC direction of a reversible reaction. Since raising the temperature shifts the equilibrium to the right (forming more $O_3$), the FORWARD reaction must be endothermic ($\Delta H$ is positive, not negative).

3. Evaluate the Statements

• A: "When ozone molecules collide with nitrogen they may form nitrogen monoxide." This describes the reverse reaction ($N_2 + 2O_3 \rightarrow 2NO + 2O_2$). In any dynamic equilibrium, both the forward and reverse reactions are continuously occurring at the molecular level, even while the NET (macroscopic) position sits at equilibrium — so product molecules genuinely are constantly colliding and reacting to reform reactants. This statement is true.
• B: "The enthalpy change for the reaction has a negative sign." False — we deduced the forward reaction is endothermic, so $\Delta H$ is positive, not negative.
• C: "Less ozone is formed at high pressure." False — there are 4 moles of gas on the left ($2NO+2O_2$) and only 3 moles on the right ($N_2+2O_3$). Increasing pressure shifts the equilibrium towards the side with FEWER moles of gas (the right), so MORE, not less, ozone would form.
• D: "At a fixed temperature, the magnitude of $K_c$ increases as the concentration of NO decreases." False — $K_c$ is a true constant at a given fixed temperature; it does not change as individual concentrations change (changing $[NO]$ simply shifts the position of equilibrium so that the SAME $K_c$ is recovered).

4. Conclude

Only statement A is true, so the correct answer is A.

Why the Other Options Are Wrong (❌):

• B. The enthalpy change for the reaction has a negative sign — Sign Error
  Since more ozone forms as temperature rises, the forward reaction is favoured by heating, which by Le Chatelier's principle means it is ENDOTHERMIC — so $\Delta H$ has a POSITIVE sign, not a negative one as this statement claims.
• C. Less ozone is formed at high pressure — Reversed Concept (Le Chatelier / Pressure)
  The reactant side ($2NO+2O_2$) has 4 moles of gas, while the product side ($N_2+2O_3$) has only 3 moles. Increasing pressure shifts equilibrium towards the side with FEWER gas moles — the product side — so high pressure produces MORE ozone, not less.
• D. At a fixed temperature, the magnitude of $K_c$ increases as the concentration of NO decreases — Conceptual Misunderstanding
  $K_c$ is a constant at a given fixed temperature and does not change when individual concentrations (like $[NO]$) change — changing $[NO]$ shifts the position of the equilibrium (the actual concentrations present), but the RATIO defining $K_c$ always settles back to the same value at that temperature.

Common Mistake (⚠️):
Assuming that because the NET reaction favours ozone formation as temperature rises, the REVERSE reaction (ozone + nitrogen forming NO) cannot be happening at all — in a true dynamic equilibrium, both directions occur continuously and simultaneously; it is only the NET, observable change that shifts one way or the other.

Takeaway (📌):
Read the $K_c$ expression directly to reconstruct the equation, use the given temperature trend (via Le Chatelier) to fix the sign of $\Delta H$, use mole counts on each side to predict the pressure effect, and remember that $K_c$ itself never changes with concentration at fixed temperature — and that 'equilibrium' never means the reverse reaction has stopped happening.