1. Experiment Find the Enthalpy change of Dissolution

2. Starter What is the enthalpy change in Jmol -1 when 7g of sodium hydroxide is added to 150g of water. Initial temp of water 20 o C Final temp of water 35 o C Specific heat capacity of water = 4.18 kJ kg -1 K - 1

3. Calculation Energy change of water = m.c.∆T Conservation of energy says that thermal energy lost = thermal energy gained. Therefore ∆H = m.c.∆T Determine the number of moles of sodium hydroxide ∆H mol -1 = ∆H / moles

4. Plenary What results did you get?????? Lets look at some of your results.

5. hydrochloric acid-74.84 hydrochloric acid ammonium nitrate+25.69 ammonium nitrate sodium hydroxide -44.51 sodium hydroxide Change in enthalpy ΔH o in kJ/mol in water at 25°C [1]enthalpykJmolwater [1]

6. Enthalpy change of solution for some selected compounds hydrochloric acid-74.84 ammonium nitrate+25.69 ammonia-30.50 potassium hydroxide-57.61 caesium hydroxide-71.55 sodium chloride+3.88 potassium chlorate+41.38 acetic acid-1.51 sodium hydroxide-44.51 Change in enthalpy ΔH o in kJ/mol in water at 25°C [1]enthalpykJmolwater [1]