1g 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)`+O_(2)(g)rarrCO_(2)(g)`
During the reaction, temperature rises from 298 K to 299 K. If the heat capacity of the bomb calorimeter is 20.7kJ/K, what is the enthalpy change for the above reaction at 298 K and 1 atm?

Explanation

Rise in temperature of the calorimeter `= 299- 298K = 1K`
Heat capacity of the calorimeter `= 20.7 k J K^(-1)`
`:. `Heat absorbed by the calorimeter `= C_(v) xxDelta T = (20.7 k J K^(-1)) (1K) = 20.7 kJ`
This s the heat evolved in the combustionof1 g of graphite.
`:. ` Heat evolved inthe combustionof 1 mole of graphite , `i.e., 12g` of graphite `= 20.7 xx 12 kJ = 248.4 kJ`
As this is the heat evolved and the vessel is closed, therefore, enthalpy change of the reaction `( Delta U )`
`= - 248.4 kJ mol^(-1)`