Enthalpy of neutralisation and calorimetry explained: HSC Chemistry Module 6

A 5 mark answer needs the heat calculation, the moles of water formed, the molar enthalpy with sign, the accepted comparison, and a stated source of error.

Step 1: Heat absorbed by the solution. Assume the combined solution has the density and specific heat of water ($\rho = 1.00$ g/mL, $c = 4.18$ J/g/K). Total mass $m = 100.0$ g. Temperature change $\Delta T = 28.7 - 22.0 = 6.7$ K.

Step 2: Moles of water formed. $n(HCl) = n(NaOH) = 0.0500 \times 1.00 = 0.0500$ mol. Limiting reagent: stoichiometric, so $n(H_2O) = 0.0500$ mol.

Step 3: Molar enthalpy.

The sign is negative because the reaction is exothermic (the solution gained heat from the reaction).

Step 4: Comparison. The measured value (-56.0 kJ/mol) is slightly less exothermic than the accepted -57.6 kJ/mol. The discrepancy (about 3 percent) is due to heat loss from the polystyrene cup to the surroundings and to the cup itself, which a perfect calorimeter would prevent.

Markers reward (1) correct $q = mc\Delta T$ with stated assumptions, (2) correct moles and ratio, (3) the negative sign on $\Delta H$, (4) numerical comparison, (5) a sensible source of error.

The strong acid plus strong base reaction has the simple net ionic equation:

All of the energy released goes into forming water from already-ionised reactants.

With ethanoic acid (weak), the actual reaction is:

Before the proton can combine with $OH^-$, the $CH_3COOH$ must first ionise (break the $O-H$ bond), which is endothermic. The net release is therefore the strong-strong value minus the ionisation enthalpy of the weak acid, giving a smaller magnitude.

Markers reward (1) the strong-strong net ionic equation, (2) recognising that weak acid ionisation costs energy, (3) the conclusion that the magnitude of $\Delta H_{neut}$ is smaller for weak combinations.

The accepted value is $\Delta H_{neut} \approx -57.6$ kJ/mol.

This value is the same for any strong-strong combination because the only chemistry occurring is the net ionic reaction $H^+_{(aq)} + OH^-_{(aq)} \rightarrow H_2O_{(l)}$; the spectator ions (e.g. $Na^+$, $Cl^-$) do not take part energetically, so the identity of the strong acid or base does not change the heat released per mole of water formed.

Marking criteria: 1 mark for the correct value with correct sign and units, 1 mark for a correct explanation referencing the net ionic equation or spectator ions.

Step 1: total mass of solution.

Step 2: temperature change.

Step 3: heat absorbed by the solution.

Marking criteria: 1 mark for correct total mass, 1 mark for correct $\Delta T$, 1 mark for the correct $q$ value with correct units and 3 significant figures.

Step 1: moles of water formed.

Stoichiometric 1:1 ratio (from $HNO_3 + KOH \rightarrow KNO_3 + H_2O$), so $n(H_2O) = 0.0400$ mol.

Step 2: molar enthalpy of neutralisation. Using the unrounded heat value ($q = 2.20704$ kJ) to avoid compounding rounding error:

Step 3: round to 3 significant figures.

Step 4: comparison. This is slightly less exothermic than the accepted $-57.6$ kJ/mol for a strong-strong combination. Since $HNO_3$ and $KOH$ are both strong, the true value should match $-57.6$ kJ/mol; the discrepancy (about 4 percent) is most likely due to heat loss from the polystyrene cup to the surroundings during mixing, rather than a chemical cause.

Marking criteria: 1 mark for correct moles of each reactant, 1 mark for correctly identifying the 1:1 mole ratio to water, 1 mark for the correct $\Delta H_{neut}$ calculation with sign, 1 mark for rounding to 3 significant figures with units, 1 mark for a valid experimental (not chemical) explanation of the discrepancy.

What the extrapolation corrects for. After mixing, some heat is lost continuously to the surroundings (through the cup wall and to the air) before the maximum temperature is actually reached. Because the temperature has already begun falling due to heat loss by the time the true peak would occur, the observed maximum reading is slightly lower than the true temperature rise caused by the reaction. Plotting temperature against time for both before and after mixing and extrapolating the cooling line back to the moment of mixing ($t = 0$) recovers the true (corrected) maximum temperature, as if no heat had been lost.

Effect of using the uncorrected peak. Using the lower, uncorrected observed peak gives a smaller $\Delta T$, which gives a smaller calculated $q = mc\Delta T$, which in turn gives a smaller magnitude for $\Delta H_{neut}$ than the true value. In other words, failing to correct for heat loss makes the experimental enthalpy of neutralisation appear less exothermic than it actually is.

Marking criteria: 1 mark for identifying continuous heat loss to the surroundings as the cause, 1 mark for correctly describing the extrapolation-back-to-mixing-time method, 1 mark for stating the uncorrected $\Delta T$ is too small, 1 mark for correctly linking this to a $\Delta H_{neut}$ magnitude that is too small (less exothermic than true).

For most weak acid and strong base combinations, ionising the weak acid is endothermic, which reduces the magnitude of $\Delta H_{neut}$ below $-57.6$ kJ/mol (as seen with ethanoic acid, about $-55.2$ kJ/mol).

HF is the exception. Although HF is a weak acid (only partially ionised in solution), the overall process of HF ionising and the resulting ions being hydrated is actually exothermic overall, not endothermic. This is because the $F^-$ ion is very small and highly charge-dense, so its hydration (forming strong ion-dipole bonds with surrounding water molecules) releases an unusually large amount of energy, more than enough to outweigh the energy cost of breaking the $H-F$ bond during ionisation.

Because the ionisation step itself is exothermic (rather than the usual endothermic case), it ADDS to the $-57.6$ kJ/mol from the $H^+ + OH^-$ step instead of subtracting from it, giving a more exothermic overall $\Delta H_{neut}$ of about $-68$ kJ/mol.

Marking criteria: 1 mark for stating the usual rule (weak acid ionisation is normally endothermic, reducing the magnitude), 1 mark for identifying that HF is an exception, 1 mark for explaining the strong hydration of the small, highly charge-dense $F^-$ ion as the cause, 1 mark for correctly reasoning that an exothermic ionisation step ADDS to (rather than subtracts from) the base $-57.6$ kJ/mol value, 1 mark for the correct overall conclusion (more exothermic than strong-strong).

This is a 7-mark OUTLINE/PREDICT/EVALUATE, requiring a valid method, a justified prediction, and a genuine evaluation with an improvement.

Band 6 PLAN.

• (a) Method: equal volumes and equal, known concentrations of each acid mixed with the same volume and concentration of NaOH; both acids and the NaOH pre-equilibrated to the same starting temperature; same calorimeter, same stirring technique, temperature recorded to a consistent precision (ideally with a data logger) for both trials; repeat each combination at least twice and average.
• Controlled variables: concentration of acid and base, volume of acid and base, initial temperature of all solutions, type and size of calorimeter, stirring method.
• (b) Prediction: HCl (strong acid) will produce the LARGER temperature rise, because its net reaction is simply $H^+ + OH^- \rightarrow H_2O$ releasing the full $-57.6$ kJ/mol, whereas ethanoic acid must first ionise (an endothermic step) before the same neutralisation step occurs, so less net heat, and hence a smaller $\Delta T$, is produced for the same moles of water formed.
• (c) Limitation and improvement: heat loss to the surroundings through the polystyrene cup and lid gap is the main limitation, causing both $\Delta T$ values to be systematically UNDERESTIMATED (though the comparison between the two acids may still broadly hold since the error affects both similarly). Improvement: use a better-insulated calorimeter (e.g. a double-walled vacuum flask/Dewar) with a tight-fitting lid, and apply a graphical cooling correction (extrapolating the temperature-time trace back to the mixing time) to recover the true $\Delta T$ for both trials.

Model paragraph (excerpt). A fair comparison requires that both acid-base pairs be tested under identical conditions except for the identity of the acid: equal volumes and concentrations of HCl or $CH_3COOH$ mixed with the same volume and concentration of NaOH, all reactants pre-equilibrated to the same starting temperature, in the same insulated cup, with temperature recorded at the same precision. HCl is predicted to produce the larger temperature rise because its neutralisation releases the full $-57.6$ kJ/mol associated with the direct $H^+ + OH^-$ reaction, while ethanoic acid's weak-acid ionisation step absorbs some of that energy before the same water-forming step can occur, lowering the net heat released and therefore the observed $\Delta T$ for equal moles reacted. The main limitation of a simple polystyrene cup is uncontrolled heat loss to the surroundings, which would be reduced by upgrading to a vacuum-insulated flask and applying a cooling-curve extrapolation to both trials.

Marker's note: full marks require (1) explicit control of concentration, volume AND starting temperature, not just "keep it fair", (2) a prediction justified by the ionisation-energy argument rather than just asserting which is bigger, and (3) an evaluation that names a SPECIFIC limitation (not just "errors could occur") paired with a workable, specific improvement.