Coffee-Cup Calorimetry: Defending a Low Value for ΔH_neutralization
A student investigates the enthalpy of neutralization using a coffee-cup calorimeter. They combine 50.0 mL of 1.00 mol/L HCl(aq) with 50.0 mL of 1.00 mol/L NaOH(aq), both initially at 22.3 °C. The highest temperature reached after mixing is 28.4 °C. The student assumes the combined solution has a density of 1.00 g/mL and a specific heat capacity of 4.18 J/(g·°C), and that the calorimeter itself absorbs negligible heat. A second trial using 100.0 mL of each reactant at the same concentrations gives the same temperature change of 6.1 °C, which the student finds surprising. The accepted molar enthalpy of neutralization for a strong acid and strong base is −55.8 kJ/mol. The student reports −51.0 kJ/mol and must defend the result.
1. Calculate the molar enthalpy of neutralization from the student's data, including the sign, and determine the percent error relative to the accepted value. State whether the reported value is consistent with the raw data.
Read the worked answer
Heat absorbed by the solution
Total volume = 100.0 mL → mass = 100.0 g (density 1.00 g/mL)
q = mcΔT = (100.0 g)(4.18 J g⁻¹ °C⁻¹)(6.1 °C) = 2.55 × 10³ J = 2.55 kJ
Moles reacting
n(HCl) = (0.0500 L)(1.00 mol/L) = 0.0500 mol; n(NaOH) = 0.0500 mol. Neither is in excess, so 0.0500 mol of H₂O forms.
Molar enthalpy
ΔH = −q/n = −(2.55 kJ)/(0.0500 mol) = −51 kJ/mol
The negative sign is required: the mixture warms, so energy leaves the reacting system.
Percent error
|51.0 − 55.8| / 55.8 × 100% = 8.6%
The reported value follows directly from the data, so the student is not fabricating numbers - the question is why the measurement falls short of −55.8 kJ/mol.
2. Explain why doubling both reactant volumes leaves ΔT unchanged, and identify two specific systematic errors that would make the measured value less exothermic in magnitude than −55.8 kJ/mol.
Read the worked answer
Why ΔT does not double
Doubling both volumes doubles the heat released (2.55 kJ → 5.10 kJ), but it also doubles the mass being warmed (100 g → 200 g) and doubles the moles reacting (0.0500 → 0.100 mol).
ΔT = q/(mc) = (5.10 × 10³ J)/((200.0 g)(4.18 J g⁻¹ °C⁻¹)) = 6.1 °C
ΔH is an intensive ratio (energy per mole), so scaling the whole system cannot change it. The student is confusing total heat with temperature change.
Systematic errors that shrink |ΔH|
- Heat loss to the cup and surroundings - the recorded maximum temperature is lower than the true maximum, so q is underestimated. This is the dominant error in an uninsulated cup.
- Neglecting the calorimeter's own heat capacity - energy used to warm the cup and thermometer is not counted in mcΔT.
Other defensible answers: thermometer lag so the true peak is missed; using c = 4.18 J/(g·°C) for a solution containing 0.5 mol/L ions rather than for pure water; solutions not actually at the stated 1.00 mol/L.
3. Predict whether replacing the NaOH with 50.0 mL of 1.00 mol/L NH₃(aq) would give a larger or smaller temperature change. Justify with particle-level reasoning and estimate the new ΔT.
Read the worked answer
Prediction: smaller ΔT (ΔH_neutralization for NH₃ is about −51 kJ/mol, versus −55.8 kJ/mol for a strong base).
Reasoning
NH₃ is a weak base: in solution only a small fraction exists as OH⁻, with most present as un-ionized NH₃. Added H⁺ consumes the free OH⁻, and the NH₃ + H₂O ⇌ NH₄⁺ + OH⁻ equilibrium shifts right to replace it. That further ionization costs energy (breaking the N–H···O hydrogen-bonding arrangement with water), which partially offsets the energy released when H⁺ and OH⁻ combine. The net enthalpy released per mole is therefore smaller.
Estimate
q = (51 kJ/mol)(0.0500 mol) = 2.55 kJ → ΔT = 2550 J / ((100.0 g)(4.18 J g⁻¹ °C⁻¹)) ≈ 6.1 °C for the strong base; using −51.5 kJ/mol gives ΔT ≈ 5.6 °C.
So ΔT drops by roughly 0.5 °C. Note that the number of moles neutralized is still 0.0500 mol - every NH₃ molecule is eventually protonated - so the difference comes from the enthalpy per mole, not from a change in amount.