1. 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, NH4Cl, is completely dissolved in 100cm3 of water. The water temperature falls from 21oC to 17.7oC.
Use DH = -cmDT
DH = x x -3.3
DH = kJ
( c is specific heat capacity of water, kJ kg-1 oC-1)
m is mass of water in kg, 0.1 kg
DT is change in temperature in oC, -3.3oC)
Use proportion to find the enthalpy change for 1 mole of ammonium chloride, 53.5g, dissolving.
5g  1.38 kJ
So g  53.5/5 x = 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.

2. Calculations for you to try.
8g of ammonium nitrate, NH4NO3, is dissolved in 200cm3 of water.
The temperature of the water falls from 20oC to 17.1oC.
Use DH = -cmDT
DH = x x -2.9
DH = 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 = kJ mol-1.
When 0.1 mol of a compound dissolves in 100cm3 of water the
temperature of the water rises from 19oC to 22.4oC .
Calculate the enthalpy of solution of the compound.
Use DH = -cmDT
DH = x x 3.4
DH = kJ
Use proportion to find the enthalpy change for 1 mole of the compound.
0.1 mol  kJ
So 1 mol  1/ x = kJ mol-1.
Higher Grade Chemistry

3. Calculations for you to try.
The enthalpy of solution of potassium chloride, KCl, is kJ mol-1.
What will be the temperature change when 14.9g of potassium chloride
is dissolved in 150cm3 of water?
Use proportion to find the enthalpy change for 14.9g of potassium chloride dissolving.
74.5g (1 mol)  kJ
So g  /74.5 x = kJ
Rearranging DH = -cmDT
Gives DT = DH
-cm
DT = = oC
3.35
-4.18 x 0.15
Higher Grade Chemistry