At constant volume, the heat of combustion of a particular compound is -3952.0 kJ/mol. When 1.211 g of this compound (molar mass = 130.08 g/mol) was burned in a bomb calorimeter, the temperature of the calorimeter (including its contents) rose by 4.453 ◦C. What is the heat capacity (calorimeter constant) of the calorimeter?-qrxn = qcalorimeterWhat is the formula for heat (q) for the calorimeter?___________________Did the calorimeter absorb/release heat? Circle your answer.Calculate the energy absorbed by the calorimeter.Solve:
Molar heat of combustion of the compound = -3952.0 kJ/mol
Mass of the compound = 1.211 g
Molar mass of the compound = 130.08 g/mol.
Mol(s) of the compound corresponding to 1.211 g = (1.211 g)/(130.08 g/mol)
= 0.009310 mol.
Heat of reaction, qrxn = (mols of the compound)*(molar heat of combustion of the compound)
= (0.009310 mol)*(-3952.0 kJ/mol)
= -36.79312 kJ
= (-36.79312 kJ)*(1000 J)/(1 kJ)
= -36793.12 J
The heat of the calorimeter is given by
qcal = (heat constant of the calorimeter)*(change in temperature of the calorimeter)
= Scal*(4.453ºC)
As per the problem,
-qrxn = qcal
=======> -(-36793.12 J) = Scal*(4.453ºC)
=======> Scal = (36793.12 J)/(4.453ºC)
=======> Scal = 8262.546 J/ºC
=======> Scal = (8262.546 J/ºC)*(1 kJ)/(1000 J)
=======> Scal = 8.262546 kJ/ºC ≈ 8.26 kJ/ºC
The heat capacity of the bomb calorimeter is 8.26 kJ/ºC (ans).
Since the temperature of the bomb calorimeter increased, hence, the calorimeter absorbs heat.
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