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ivann1987 [24]
3 years ago
11

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 tempera

ture 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.
Chemistry
1 answer:
vredina [299]3 years ago
5 0

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

Explanation :

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

q=[q_1+q_2]

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

where,

q = heat released by the reaction

q_1 = heat absorbed by the calorimeter

q_2 = heat absorbed by the water

c_1 = specific heat of calorimeter = 1.55J/^oC

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

m_2 = mass of water = 107.70 g

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

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

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

q=1112.3J=1.1123kJ

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

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

where,

\Delta H = enthalpy change = ?

q = heat released = 1.1123 kJ

m = mass of KClO_4 = 2.84 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{2.84g}{138.55g/mole}=0.0205mole

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

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

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