Question

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:

Answer 1

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.