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In the laboratory a general chemistry student finds that when 2.84 g of KClO4(s) are dissolved in 107.70 g of water, the temperature of the solution drops from 22.82 to 20.34 °C. The heat capacity of the calorimeter (sometimes referred to as the calorimeter constant) was determined in a separate experiment to be 1.55 J/°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 equal to the specific heat of water.

Respuesta :

Answer : The enthalpy change of dissolution of [tex]KClO_4[/tex] is 54.3 kJ/mole

Explanation :

Heat released by the reaction = Heat absorbed by the calorimeter + Heat absorbed by the water

[tex]q=[q_1+q_2][/tex]

[tex]q=[c_1\times \Delta T+m_2\times c_2\times \Delta T][/tex]

where,

q = heat released by the reaction

[tex]q_1[/tex] = heat absorbed by the calorimeter

[tex]q_2[/tex] = heat absorbed by the water

[tex]c_1[/tex] = specific heat of calorimeter = [tex]1.55J/^oC[/tex]

[tex]c_2[/tex] = specific heat of water = [tex]4.184J/g^oC[/tex]

[tex]m_2[/tex] = mass of water = 107.70 g

[tex]\Delta T[/tex] = change in temperature = [tex]T_2-T_1=(22.80-20.34)=2.46^oC[/tex]

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

[tex]q=[(1.55J/^oC\times 2.46^oC)+(107.70g\times 4.184J/g^oC\times 2.46^oC)][/tex]

[tex]q=1112.3J=1.1123kJ[/tex]

Now we have to calculate the enthalpy change of dissolution of [tex]KClO_4[/tex]

[tex]\Delta H=\frac{q}{n}[/tex]

where,

[tex]\Delta H[/tex] = enthalpy change = ?

q = heat released = 1.1123 kJ

m = mass of [tex]KClO_4[/tex] = 2.84 g

Molar mass of [tex]KClO_4[/tex] = 138.55 g/mol

[tex]\text{Moles of }KClO_4=\frac{\text{Mass of }KClO_4}{\text{Molar mass of }KClO_4}=\frac{2.84g}{138.55g/mole}=0.0205mole[/tex]

[tex]\Delta H=\frac{1.1123kJ}{0.0205mole}=54.3kJ/mole[/tex]

Therefore, the enthalpy change of dissolution of [tex]KClO_4[/tex] is 54.3 kJ/mole

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