Question

A student determines the heat of dissolution of solid potassium perchlorate using a coffee-cup calorimeter of negligible heat capacity. When 1.69 g of KClO4(s) is dissolved in 102.00 g of water, the temperature of the solution drops from 25.00 to 23.48 °C. Based on the student's observation, calculate the enthalpy of dissolution of KClO4(s) in kJ/mol. Assume the specific heat of the solution is 4.184 J/g°C. ΔHdissolution = kJ/mol

Answer 1

Answer : The enthalpy change of dissolution of
`KClO_4` is -54.0 kJ/mole

Explanation :

`q=m* c* \Delta T`

where,

q = heat released by the solution

c = specific heat of water =
`4.184J/g^oC`

m = mass of solution = mass of water + mass of
`KClO_4` = 102.00 + 1.69 = 103.69 g

`\Delta T` = change in temperature =
`T_2-T_1=(25.00-23.48)=1.52^oC`

Now put all the given values in the above formula, we get:

`q=103.69g* 4.184J/g^oC* 1.52^oC`

`q=659.4J=0.659kJ`

Now we have to calculate the enthalpy change of dissolution of
`KClO_4`

`\Delta H=-(q)/(n)`

where,

`\Delta H` = enthalpy change of dissolution = ?

q = heat released = 0.659 kJ

m = mass of
`KClO_4` = 1.69 g

Molar mass of
`KClO_4` = 138.55 g/mol

`\text{Moles of }KClO_4=\frac{\text{Mass of }KClO_4}{\text{Molar mass of }KClO_4}=(1.69g)/(138.55g/mole)=0.0122mole`

`\Delta H=-(0.659kJ)/(0.0122mole)=-54.0kJ/mole`

Therefore, the enthalpy change of dissolution of
`KClO_4` is -54.0 kJ/mole