The molar enthalpy of neutralisation, why strong acid–strong base values are all about −57 kJ/mol and weak ones are smaller, and the practical investigation to measure it with a polystyrene-cup calorimeter, including temperature–time extrapolation, calculations and sources of error.
Mix 50 mL of 1 mol/L hydrochloric acid with 50 mL of 1 mol/L sodium hydroxide and the temperature jumps by almost 7 °C. Swap the hydrochloric acid for nitric acid, or the sodium hydroxide for potassium hydroxide, and you get the same jump to within the error of a school thermometer. That isn't a coincidence. It tells you that all four reactions are, at the particle level, the same reaction.
What the enthalpy of neutralisation measures
The molar enthalpy of neutralisation, Δ$_{\text{neut}}H$, is the enthalpy change when an acid and a base react to form one mole of water. Neutralisation is exothermic, so Δ$H$ is negative.
For any strong acid with any strong base, the net ionic equation is the same, so the enthalpy change is the same:
Typical data-book values. Strong–strong pairs cluster at about −57 kJ/mol; weak acids and bases release less.
Weak acids and weak bases give less negative values. A weak acid is mostly un-ionised molecules, so as the base removes H⁺, more of the acid has to ionise to replace it, and breaking the H–A bond absorbs energy. The difference is small for ethanoic acid (about −56 kJ/mol), larger for ammonia (about −52 kJ/mol), and huge for the very weak acid HCN (about −12 kJ/mol).
Notice the values are per mole of water, not per mole of acid. Sulfuric acid is diprotic, so one mole of H₂SO₄ reacting fully with NaOH forms two moles of water and releases about 2 × 57 = 114 kJ.
Calculating Δ$H$ from experimental data
The method is the same as for enthalpy of combustion. The heat released by the reaction is absorbed by the solution, so:
$$q = mc\Delta T \qquad \Delta H = -\frac{q}{n}$$
Make these standard assumptions (and state them in an answer): the solution has the density of water (1.00 g/mL) and the specific heat capacity of water (4.18 J g⁻¹ K⁻¹); the mass is the total volume of both solutions; no heat is lost to the cup or the surroundings; and $n$ is the moles of water formed, set by the limiting reagent.
Worked example: a strong acid and a strong base
50.0 mL of 1.00 mol/L HNO₃ is mixed with 50.0 mL of 1.00 mol/L KOH in a polystyrene cup. The temperature rises from 21.3 °C to 28.0 °C. Calculate the molar enthalpy of neutralisation.
0.0300 mol of water forms, so $q$ = 0.0300 × 57.1 = 1.713 kJ = 1713 J.
heat
$\Delta T = \dfrac{q}{mc} = \dfrac{1713}{100.0 \times 4.18} = 4.10$ °C.
ΔT
InteractiveCoffee-cup calorimeter
Try this
Run HCl + NaOH at 1.0 mol/L in polystyrene cups. What highest reading, extrapolated ΔT and Δ$H$ does the sim give? Compare the Δ$H$ with −57.1 kJ/mol.
Read the note at the end. What Δ$H$ would you get from the highest reading instead of the extrapolated ΔT?
Change to 2.0 mol/L and run it again. What happens to ΔT and to Δ$H$? Explain why one changes and the other doesn't.
Switch to the open glass beaker at 1.0 mol/L. How far is Δ$H$ from the accepted value now?
Run CH₃COOH + NaOH and HCl + NH₃ in polystyrene at 1.0 mol/L. Rank the three reactions by Δ$H$ and explain the order.
Test yourself
100.0 mL of 0.50 mol/L HCl is mixed with 100.0 mL of 0.50 mol/L NaOH. The temperature rises by 3.3 °C. The molar enthalpy of neutralisation is closest to
Why is the enthalpy of neutralisation of HCN with NaOH much smaller in magnitude than that of HCl with NaOH?
Both form one mole of water per mole of acid, but ionising HCN absorbs most of the energy released by H⁺ + OH⁻.
Which change would most improve the accuracy of a coffee-cup measurement?
Both reduce or correct for heat loss, the main systematic error.
Doubling the concentration of both solutions (same volumes) would
Twice as much water forms in the same mass of solution, so q and ΔT double, but q ÷ n stays the same.
Label the calorimeter.
Put the method in order.
The molar enthalpy of neutralisation is the enthalpy change per mole of formed. Neutralisation is , so ΔH is negative. For strong acids and strong bases the net ionic equation is H⁺ + OH⁻ → H₂O, so ΔH is always about kJ/mol. Heat to the surroundings makes experimental values less negative. Polystyrene is used because it is a poor of heat.
6 marks
A student mixed 25.0 mL of 2.00 mol/L NaOH with 25.0 mL of 2.00 mol/L HCl in a glass beaker. Both started at 20.5 °C and the highest temperature reached was 32.0 °C. (a) Calculate the molar enthalpy of neutralisation. (b) Compare your answer with the accepted value of −57.1 kJ/mol, and evaluate the student's method, suggesting two improvements.
q = 50.0 × 4.18 × 11.5 = 2404 J; n(H₂O) = 0.0500 mol; ΔH = −48.1 kJ/mol with sign and units; states the result is about 16% less negative than accepted; identifies heat loss through the conducting glass and open top and to the beaker as a systematic error; notes using the highest reading underestimates ΔT; two specific improvements (polystyrene cups with lid; record temperature over time and extrapolate; digital probe; repeat trials); judgement on validity.
Correct calculation with a sound evaluation missing one element, or a minor arithmetic slip with a full evaluation.
Correct q or n with a partial calculation, and one relevant source of error or improvement.
One correct step or relevant idea.
Sample answer
(a) Total mass = 50.0 g. q = mcΔT = 50.0 × 4.18 × (32.0 − 20.5) = 50.0 × 4.18 × 11.5 = 2.40 × 10³ J. n(HCl) = n(NaOH) = 2.00 × 0.0250 = 0.0500 mol, so 0.0500 mol of water forms. ΔH = −2.40 kJ ÷ 0.0500 mol = −48.1 kJ/mol.
(b) This is about 16% less negative than −57.1 kJ/mol, so the method has a large systematic error. Glass is a much better conductor of heat than polystyrene and the beaker had no lid, so heat escaped to the surroundings and warmed the glass itself. Using only the highest reading also underestimates ΔT, because the mixture was already losing heat before it reached its maximum. The method is therefore not valid for an accurate value, although the result is consistent with neutralisation being exothermic. Improvements: (1) use two nested polystyrene cups with a lid; (2) record the temperature every 30 seconds before and after mixing and extrapolate the cooling line back to the mixing time. Repeating the experiment and averaging would improve reliability.
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