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)`+O_(2)(g)rarrCO_(2)(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?

Answer

`-2.48 xx 10^(2) kJ "mol"^(-1)`

Explanation

`q = C_(v) xx DeltaT`
` = -20.7 xx (299 - 298) = -20.7 kJ`
`therefore DeltaU = -20.7 kJ "mol"^(-1)`
Heat of combustion for 1 mol = `(-20.7 xx 12.0)/(1)`
` = -2.48 xx 10^(2)kJ "mol"^(-1)`.