Question

A 1.435g of Naphthalene (C10H8) was burned in Constant Volume Bomb Calorimeter. Consequently temperature of the water rose from 20.17°C to 25.84°C. If the mass of water surrounding the calorimeter was exactly 2.0 X 103 kg and the heat capacity of the bob calorimeter was 1.80 Kj/°C, Calculate the molar heat of reaction (combustion) in KJ of Naphthalene.

Answer 1

$$-q_{naphthalene}=Q_{water}+Q_{calorimeter},$$

$$-n\Delta H=cm\Delta t+C\Delta t,$$

$$\Delta H=-\frac{\Delta t(cm+C)}{n}=-\frac{\Delta t(cm+C)}{m/\mu}=\\ =-\frac{(25.84-20.17)(4182\cdot2000+1800)}{0.001435/[(12\cdot10+1\cdot8)\cdot10^{-3}]}=-4.231\cdot10^9\text{ J/mol}.$$