1 g of graphite is burnt in a bomb calorimeter in excess of `O_(2)` at 298 K and 1 atm. Pressure according to the equations.
`C_("graphite")+O_(2(g)) to CO_(2(g))`
During the reaction the temperature rises from 298 K to 200K. Heat capacity of the bomb calorimeter is `20.7KJK^(-1)`. What is the enthalpy change for the above reaction at 298 K 1 atm?

Explanation

Suppose q is the quantity of heat from the reaction mixture and `C_v` is the heat capacity of the calorimeter, then the quantity of heat absorbed by the calorimeter.
`q=C_v xx triangleT`
Quantity of heat from the reaction will have the same magnitude but opposite sign be cause the heat lost by the system (reaction mixture) is equal to the heat gained by the calorimeter.
`q=-C_(v) xx triangleT=-20.7"kJ/K" xx (299-298)K`
=-20.7kJ
(Here, negative sign indicates the exothermic nature of the reaction)
Thus, `triangleU` for the combustion of the 1g of graphite = -20.7 kJ
For combustion of 1 mol of graphite, 12.0g moll ( 20.7kJ).
`=(12.0"g mol"^(-1) (20.7 kJ))/(1g)`
`=-2.48 xx 10^2" kJ mol"^(-1)," Since "trianglen_(g)=0`
`triangleH=triangleU=-2.48 xx 10^(2)"kJ mol"^(-1)`