# Enthalpy of Neutralisation

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Enthalpy of Neutralisation

The enthalpy of neutralisation is the heat energy released when 1 mole of water is formed from neutralisation of an acid with a base. Neutralisation is an exothermic reaction. The reaction between NaOH with HCl can be written as: NaOH + HCl  NaCl + H2O By omitting the spectator ions the reaction becomes: H+(aq) + OH-(aq)  H2O(l)

To work out the enthalpy of neutralisation you need to record the following data:
The average starting temperature of acid + alkali The temperature of the final salt solution

Experiment 25 cm3 of HCl 1 mole l-1 25 cm3 of NaOH Na+OH- H+Cl-

 Measurements Taken Temperature acid-19oC Temperature alkali- 19oC
Final temperature- 25oC Calculate the enthalpy of neutralisation when 50cm3 of 1 mol l-1 NaOH is added to 50cm3 of 1 mol l-1 HCl. Specimen results

Solution 1. Calculate average starting temperature (19+19)/2= 19oC 2. Calculate average change in temperature 25-19= 6oC 3. Calculate the heat energy released EH = c x m x ΔT = 4.18x0.1x6 = 2.508kJ

4. Mole calculation NaOH + HCl  NaCl + H2O 1 mole 1 mole 1 mole 1 mole 0.05 moles 0.05 moles For both acid + alkali: n=c x v n=1 x 0.05= 0.05 moles So, 0.05 moles: 2.508kJ 1 mole : (2.508/0.05)= 50.16kJ mol-1 Enthalpy of neutralisation is kJ mol-1