Question

When a 0.225-g sample of benzoic acid is combusted in a bomb calorimeter, the temperature rises 1.643 ∘C . When a 0.265-g sample of caffeine, C8H10O2N4, is burned, the temperature rises 1.594 ∘C . Using the value 26.38 kJ/g for the heat of combustion of benzoic acid, calculate the heat of combustion per mole of caffeine at constant volume.

Answer 1

For benzoic acid:

Q=CdT

Q=26380*0.225=5935.5J

5935.5J=C*1.643

C=3612.6

For caffeine:

Q=3612.6*1.564=5650.1

moles=0.260/194=0.0013

Q(KJ/g)=21.73

Q(KJ/mol)=4346.2

Answer 2

Heat released by benzoic acid:
0.225 g benzoic acid * 26.38 kJ/g = 5.9355 kJ

Heat capacity of the calorimeter:
5.9355 kJ / 1.643 °C = 3.61 kJ/°C

Moles of a 0.265 g sample of caffeine, using molar mass:
0.265 g caffeine / 194.19 g/mol = 0.0013646 moles

Heat released by caffeine
the temperature rises 1.594 °C * 3.61 kJ/ °C = 5.75 kJ

Heat of combustion per mole of caffeine:
5.75 kJ / 0.0013646 moles = 4216.9 kJ/mol