Calorimetry – Enthalpy of Neutralization of HCl + NaOH
ChemistryEnergetics & ThermodynamicsAges 15–16Virtual lab
Use it with your class: students open it from a link or a QR code, with no account needed.
Next in Energetics & Thermodynamics
A virtual experiment with a polystyrene-cup calorimeter: pour V₁ mL of C₁ M HCl and V₂ mL of C₂ M NaOH (both at 25 °C), stir, and follow the thermometer and the T(t) graph. From the maximum temperature, students calculate Q = m·c·ΔT, the moles of water formed (from the limiting reactant) and ΔrH ≈ −57.3 kJ/mol. Each run is recorded in a results table, and a heat-loss setting shows why measured values are usually smaller than the theoretical one.
Lesson: Enthalpy changes – measuring the enthalpy of neutralization
What it shows
Neutralization of a strong acid by a strong base is exothermic: H⁺(aq) + OH⁻(aq) → H₂O(l), with an enthalpy change of about −57.3 kJ per mole of water formed. In a polystyrene-cup calorimeter the heat released warms the mixed solution, so Q = m·c·ΔT, treating the solution as water (1 g/mL, 4.18 J/g·K). Dividing Q by the moles of water, which are set by the limiting reactant, gives ΔrH. Heat lost to the surroundings makes the measured value smaller than the theoretical one.
How to use
Set the volumes and concentrations of HCl and NaOH, choose a value for Heat loss from the cup, then press Pour and Stir. Watch the thermometer and the temperature–time graph; once the maximum has passed, the run is added to the table. Press New Run, change one variable at a time and compare ΔT and ΔrH across runs, for example by doubling both volumes or putting one reactant in excess.
Parameters you can change
- Volume of HCl solution 10–100 mL
- HCl concentration 0.1–2 M
- Volume of NaOH solution 10–100 mL
- NaOH concentration 0.1–2 M
- Heat loss from the cup 0–30 %
Questions to explore
- If both volumes are doubled at the same concentrations, how does ΔT change, and why?
- When NaOH is in excess, which reactant do you use to calculate the moles of water formed?
- Why is the measured value of |ΔrH| usually smaller than 57.3 kJ/mol?
Common questions
How do you calculate the enthalpy of neutralization from a temperature rise?
Calculate the heat released with Q = m·c·ΔT, then divide by the moles of water formed and add a minus sign: ΔrH = −Q/n(H₂O). In this formula m is the mass of the whole mixed solution and c is about 4.18 J/g·K, the value for water. Convert Q to kilojoules so the answer is in kJ/mol.
Why is the enthalpy of neutralization about the same for all strong acids and strong bases?
It is about the same because strong acids and strong bases are fully ionized, so the reaction is always H⁺(aq) + OH⁻(aq) → H₂O(l), about −57.3 kJ per mole of water. The other ions are spectators and do not react. Weak acids and bases usually give a less negative value, because some energy goes into ionizing them.
Why is a polystyrene cup used as a calorimeter?
A polystyrene cup is used because it is a good thermal insulator and takes up very little heat itself. Nearly all the heat from the reaction therefore warms the solution, and the heat taken up by the cup can be ignored in the calculation. A lid reduces further heat loss to the air.
Teaching ideas
All guides- Calorimetry Virtual Lab: Enthalpy of NeutralizationA calorimetry virtual lab for the temperature-change practical: mix HCl and NaOH in a polystyrene cup, find the peak temperature and calculate ΔH.
- Hess's Law Virtual Lab: Enthalpy of Hydration of CuSO₄A Hess's law virtual lab: dissolve anhydrous and hydrated copper(II) sulfate in a cup calorimeter, find both ΔH values and combine them in a cycle.
- Enthalpy of Combustion Virtual Lab: Alcohols and Heat LossAn enthalpy of combustion virtual lab: burn four alcohols under a copper can, calculate ΔcH from q = mcΔT, find the change per CH₂ and evaluate heat loss.
