1) At constant volume, the heat of combustion of a particular compound is –3717.
ID: 914847 • Letter: 1
Question
1) At constant volume, the heat of combustion of a particular compound is –3717.0 kJ/mol. When 1.339 g of this compound (molar mass = 186.00 g/mol) was burned in a bomb calorimeter, the temperature of the calorimeter (including its contents) rose by 3.323 °C. What is the heat capacity (calorimeter constant) of the calorimeter?
2) When a 3.00-g sample of KSCN is dissolved in water in a calorimeter that has a total heat capacity of 2.019 kJ·K–1, the temperature decreases by 0.370 K. Calculate the molar heat of solution of KSCN.
Explanation / Answer
1.No of moles of the compound = (1.339)/(186g/mol) = 0.0072 mole
Heat released to calorimeter = 0.0072 moles * 3717 kJ/mole = 26.76 kJ
Calorimeter constant, CV=26.76 kJ/3.323 oC = 8.05 kJ/oC
2. Heat released to the calorimeter, qcal = 2.019 kJ.K-1 * (-0.370 K) =-0.747 kJ
Heat of the reacton(dissolution) = -qcal = 0.747 kJ
moles of KSCN = 3/97.181 moles = 0.031 moles
dHdissolution = 0.747 kJ/0.031mole = 24.2 kJ/mol
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