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)") + O_(2(g)) to CO_(2(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 ?

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 Delta T`
Quantity of heat from the reaction will have the same magnitude but opposite sign because the heat lost by the system is equal to the heat gained by the calorimeter.
`q=-C_(v) xx Delta T`
`=-20.7 "kJ/K" K xx (299- 298) K = -20.7 "kJ"`
Here, negative sign indicates the exothermic nature of the reaction.
Thus, `Delta U` for the combustion of the 1g of graphite `= -20.7 "kJ K"^(-1)`
For combustion of 1 mol of graphite,
`= (12.0 g "mol"^(-1) xx (-20.7 "kJ") )/(1g)`
`= -2.48 xx 10^(2) "kJ mol"^(-1)` since `Deltan_(g) =0`
`therefore Delta H = Delta U = - 2.48 xx 10^(2) "kJ mol"^(-1)`