1 g of graphite is burnt in a bomb calorimeter in excess of oxygen at 298 K and 1 atmospheric pressure according to the

equation  C (graphite) + O₂ (g) → CO₂ (g) 

During the reaction, temperature rises from 298 K to 299 K. If the heat capacity of the bomb calorimeter is 20.7 kJ/K, what is the enthalpy change for the above reaction at 298 K and 1 atm?

Rise in temperature of the calorimeter = 299 – 298 K = 1 K 

Heat capacity of the calorimeter = 20.7 kJ K^–1 

∴ Heat absorbed by the calorimeter = C_v × ∆T = (20.7kjK^–1) (1 K) = 20.7 kJ

This is the heat evolved in the combustion of 1 g of graphite. 

∴ Heat evolved in the combustion of 1 mole of graphite, i.e., 12 g f graphite = 20.7 × 12 kJ = 248.4 kJ 

As this is the heat evolved and the vessel is closed, therefore, enthalpy change of the reaction (∆U)

= – 248.4 kJ mol^–1