• # question_answer 6) 1 g of graphite is burnt in a bomb calorimeter in excess of oxygen at 298 K and 1 atm pressure according to the equation. ${{C}_{(graphite)}}+{{O}_{2}}(g)\to C{{O}_{2}}(g)$ During the reaction, temperature rises from 298 K to 299K. If the heat capacity of the bomb calorimeter is$20.7kJ\,{{K}^{-1}}$, what is the enthalpy change for the above reaction at 298 K and 1 atm?

Heat evolved in the reaction $=-{{C}_{v}}\,\Delta T$ $=-20.7\times 1=-20.7kJ$ Heat evolved in the combustion of 1 mole graphite, i.e.,12 g graphite will be: $\Delta U=-20.7\times 12kJ\,mo{{l}^{-1}}$ $=-2.48\times {{10}^{2}}kJ\,mo{{l}^{-1}}$ In the given reaction: $\Delta {{n}_{g}}=0$ $\therefore$$\Delta H=\Delta U=-2.48\times {{10}^{2}}kJ\,mo{{l}^{-1}}$