Question

When 0.455 g of anthracene, C14H10, is combusted in a bomb calorimeter that has a water jacket containing 500.0 g of water, the temperature of the water increases by 8.63 degrees C. Assuming that the specific heat of water is 4.18 J/(g degrees C) and that the heat absorption by the calorimeter is negligible, estimate the enthalpy of combustion per mole of anthracene.

Answer 1

Expression for heat in terms of specific heat can be written as follows:

Q = c m ΔT

Where, Q = heat added or removed

c = specific heat = 4.18 J.g^-1.^oC^-1

m = mass = 500.0 g

ΔT = 8.63 ^oC

Thus, heat absorbed by water can be calculated as follows:

Q = (4.18 J.g^-1.^oC^-1) (500.0 g) (8.63 ^oC)

Number of moles of anthracene in 0.455g can be calculated as follows:

No. of moles = Wt. of anthracene / molar mass of anthracene

No. of moles = 0.455 g / (178.23 g/mol)

Thus, heat absorbed per mole = heat absorbed /no. of moles.

Heat absorbed per mole = [(4.18 J.g^-1.^oC^-1) (500.0 g) (8.63 ^oC)] / [0.455 g / (178.23 g/mol)]

Heat absorbed per mole = 7065233 J = 7065kJ

Since, heat absorbed by water from combustion of per mole of anthracene is 7065kJ

Thus, enthalpy of combustion per mole of anthracene is -7065 kJ