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 ?

1 g of graphite is burnt in a bomb calorimeter in excess of oxygen at

1.	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 ?
Answer» Solution :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)`