1 g of graphite is burnt in a bomb calorimeter in excess of `O_(2)` at 298 K and 1 atm. Pressure according to the equations.
`C_("graphite")+O_(2(g)) to CO_(2(g))`
During the reaction the temperature rises from 298 K to 200K. Heat capacity of the bomb calorimeter is `20.7KJK^(-1)`. What is the enthalpy change for the above reaction at 298 K 1 atm?

Explanation

Given equation `C_("graphite")+O_(2(g)) to CO_(2(g))`
`q=C_(v) xx triangleT=-20.7 xx 1=-20.7kJ`
Given 1 gm of graphite `triangleH=triangleU (triangleh=0)`
`triangleU=-20.7KJ//K`
For 1 mole
`triangleU=12 xx -20.7="-248 KJ/mole"`