`0.562g` of graphite kept in a bomb calorimeter in excess of oxygen at `298K` and 1 atmospheric pressure was burnt according to the equation,
`C_(Graph ite)+O_(2(g))rarr CO_(2(g))`
durgin the reaction, temperature rises from `298K` o `298.89K`. If the heat capacity of the calorimeter and its contents is `20.7 kJ//K`, what is the enthalpy change for the above reaction at `298K` and `1 atm`?

Explanation

Let `q` amount of heat is produced during the course of reaction, when `0.562g` of graphite in a bomb calorimeter is burnt in excess of `O_(2)`. This is condition of constant volume `(` volume of bomb calorimeter is constant `)` Thus heat produced `(q_(v))` is `:`
Thus `q_(v)=-C_(v)xxDeltaT`
Where `C_(v)` heat capacity of calorimeter and its content, `DeltaT` is change in temperature. The negative sign indicates for exothermic reaction.
`q_(v)=20.7xx10^(3)xx0.89`
`=-18.42xx10^(3)J per 0.562g ` carbon
`:. `Amount of heat liberated during burning of 1 mole carbon
`=(18.42xx10^(3)xx12)/(0.562)=-3.93xx10^(5)J`
Thus heat is used to change the internal energy `U i.e., DeltaU` as the conditions are of constant volume.
Now `DeltaH=Delatu+DeltanRT`
`[C_((s))+O_(2(g))=CO_(2(g)):. Deltan=0]`
`:. Delta n=1-1=0`
`:. DeltaH=DeltaU=-3.93xx10^(5)J`