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?

Answer:

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: \[\Delta {{n}_{g}}=0\] \[\therefore \]\[\Delta H=\Delta U=-2.48\times {{10}^{2}}kJ\,mo{{l}^{-1}}\]