Question

When a 0.245-g sample of benzoic acid is combusted in a bomb calorimeter, the temperature rises 1.644 ∘C . When a 0.275-g sample of caffeine, C8H10O2N4, is burned, the temperature rises 1.520 ∘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

The complete step wise calculation of the heat of combustion per mole of caffeine at constant volume is given as below:-

It is given that, mass of benzoic acid = 0.245 g = me Temperature rises = 1.644 oC Heat af Combustion of benzoic acid = 26.38 kJ/g. So, amo unt af heat released is given by Q = mB x 26.38 kJ/g = (0.2459)x(26.38 kJ19) on Qi = 6.4631 kJ - 0

Heat capacity (C) = a. -= 6.4631 Temperature rises 1.644 an ce - 6463) Ko/ •C = 3.9313 kJ/"c on C = 3.9313 kJ/•c) - © molecular weight of caffeine = 194.2 g/mol. mass af caffeine = 0,275 g

Number of males af Caffeine = 0.2759 194.2g/mol or n = 1.42 x 10-3 moles. - Heat rises = 1.520 °C Then amount of heat released from equation , Qą = (1.520 °C) x (3.9313 kJ 1°c) = 5.976kJ on as= 5.976 kJ 1 –

af caffeine Then the heat af Combustion per moles is given by 5.976 kJ Heat (a) = (ar) = 1.42 x103 moles (from equations, 29 and ③ we have) Heat (6) = 4.2885 x103 kJ/mole on Heat (61) = 4208.5 kJ/mole So, the heat of Combustion af Caffeine per mole is given by: - Ta = 4208. 5 kJ/mole