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NSSCAS Chemistry · 8224/3 · Calorimetry
Enthalpy of neutralisation
Equal volumes of 1.0 M HCl and 1.0 M NaOH are mixed in a polystyrene cup. The temperature shoots up; the heat released per mole of water formed gives the standard enthalpy of neutralisation.
T = 22.00 °C
m (solution)
50 g
c
4.18 J/g/K
ΔT
0.00 K
n (water)
0.025 mol
Calculation
q = m c ΔT
ΔH = − q ÷ n
Net ionic equation: H⁺(aq) + OH⁻(aq) → H₂O(l)
Exam-style questions
1. Why is a styrofoam (polystyrene) cup used?
2. Why is ΔH of neutralisation roughly CONSTANT for strong acid + strong base?
3. Why is the experimental value usually a LITTLE less negative than −57 kJ/mol?
4. 25 cm³ 1.0 M HCl + 25 cm³ 1.0 M NaOH, ΔT = 6.6 °C. What is ΔH? (m=50 g, c=4.18, n=0.025)