Question

When a solid dissolves in water, heat may be evolved or absorbed. The heat of dissolution(dissolving) can be determined using a coffee cup calorimeter.

In the laboratory a general chemistry student finds that when 5.22 g of CsClO₄(s) are dissolved in 100.70 g of water, the temperature of the solution drops from 23.33 to 20.10 °C.

The heat capacity of the calorimeter (sometimes referred to as the calorimeter constant) was determined in a separate experiment to be 1.64 J/°C.

Based on the student's observation, calculate the enthalpy of dissolution of CsClO₄(s) in kJ/mol.

Assume the specific heat of the solution is equal to the specific heat of water.

ΔH_dissolution = _____kJ/mol

Answer 1

Heat lost by water Q = mcdT   c = heat capacity of water

                                 = (100.7 g) x (4.18 J/g·°C) x (23.33 - 20.10)°C

                                = 1359.6 J

Supposing the calorimeter to be at the same temperatures as the water:

Heat lost by calorimeter = heat capacity of calorimeter x temp difference
                                        = ( 1.64 J/°C) x ((23.33 - 20.10)°C

                                       = 5.3 J

Hence,

Heat gained by solute = total heat lost by combined apparatus

                                   = 1359.6 J + 5.3 J

                                  = 1364.9 J

Then,

enthalpy of dissolution of CsClO4(s) = 1364.9 J / moles of CsClO4(s)

= 1364.9 J/ 5.22 g / 232.36 g/mol [ moles = mass/ molar mass ]

= 60756 J

= 60.756 kJ/mol

Therefore,

ΔHdissolution = 60.756 kJ/mol