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 :

$q=m\times c\times \Delta T$

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:

$q=103.69g\times 4.184J/g^oC\times 1.52^oC$

$q=659.4J=0.659kJ$

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

$\Delta H=-\frac{q}{n}$

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

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

$\Delta H=-\frac{0.659kJ}{0.0122mole}=-54.0kJ/mole$

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