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) + O2 (g) ? CO2 (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? ?

suppose q is the quantity of heat from the reaction mixture and Cv is the heat capacity of the calorimeter..then the quantity of heat absorbed by the calorimeter q=Cv*deltaTquantity of heat from the reaction will have the same magnitude but opposite sign because the heat lost by the system that is reaction mixture is equal to the heat gained by the caloriemeterq=-Cv*deltaT =-20.7kJ/K*(299-298)K = -20.7kJ/KFor combustion of one mole of Graphite= (12.0 g/mol)*(-20.7kJ)divided by 1g=-2.48*10"2(ten square) deltaH=enthalpy change=-2.48*10"2kJ/mole