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

`q = C_v xx Delta`
Quantity of heat from the reaction will have the same magnitude but opposite sign because the heat lost by the system (reaction mixture) is equal to the heat gained by the calorimeter
`q= - C_v xx DeltaT = -20.7kJ//K xx (299-298) K = -20.7kJ `
For combustion of 1 mol of graphite
`DeltaE = -2.48 xx 10^2 kJ mol^(-1)`
`DeltaH = DeltaE = -2.48 xx 10^2 kJmol^(-1)`, Since `Deltan = 0`.