1. Enthalpy of Solution

2. HIGHER GRADE CHEMISTRY CALCULATIONS Enthalpy of Solution. The enthalpy of solution of a substance is the energy change when one mole of a substance dissolves in water. Worked example 1. 5g of ammonium chloride, NH 4 Cl, is completely dissolved in 100cm 3 of water. The water temperature falls from 21 o C to 17.7 o C. Use  H = -cm  T  H = -4.18 x 0.1 x -3.3  H = 1.38 kJ ( c is specific heat capacity of water, 4.18 kJ kg -1 o C -1 ) m is mass of water in kg, 0.1 kg  T is change in temperature in o C, -3.3 o C) Use proportion to find the enthalpy change for 1 mole of ammonium chloride, 53.5g, dissolving. 5g  1.38 kJ So 53.5 g  53.5 / 5 x 1.38 = 17.77 kJ mol -1. Note:- As the temperature falls the reaction is endothermic (takes in heat) – this is shown by the positive value for the enthalpy change.

3. Higher Grade Chemistry Calculations for you to try. 1.8g of ammonium nitrate, NH 4 NO 3, is dissolved in 200cm 3 of water. The temperature of the water falls from 20 o C to 17.1 o C. Use  H = -cm  T  H = -4.18 x 0.2 x -2.9  H = 2.42 kJ Use proportion to find the enthalpy change for 1 mole, 80g, of ammonium nitrate dissolving. 8g  2.42 kJ So 80g  80 / 8 x 2.42 = 24.2 kJ mol -1. 2. When 0.1 mol of a compound dissolves in 100cm 3 of water the temperature of the water rises from 19 o C to 22.4 o C. Calculate the enthalpy of solution of the compound. Use  H = -cm  T  H = -4.18 x 0.1 x 3.4  H = -1.42 kJ Use proportion to find the enthalpy change for 1 mole of the compound. 0.1 mol  - 1.42 kJ So 1 mol  1 / 0.1 x -1.42 = -14.2 kJ mol -1.

4. Higher Grade Chemistry Calculations for you to try. 3.The enthalpy of solution of potassium chloride, KCl, is + 16.75kJ mol -1. What will be the temperature change when 14.9g of potassium chloride is dissolved in 150cm 3 of water? Use proportion to find the enthalpy change for 14.9g of potassium chloride dissolving. 74.5g (1 mol)  16.75 kJ So 14.9g  14.9 / 74.5 x 16.75 = 3.35 kJ Rearranging  H = -cm  T Gives  T =  H -cm  T = = -5.34 o C 3.35 -4.18 x 0.15