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?

Options

A. `-2.48xx10^(2)kJ" "mol^(-1)`
B. `2.48xx10^(2)"kJ "mol^(-1)`
C. `-5.46xx10^(2)"kJ "mol^(-1)`
D. `5.46xx10^(2)"kJ "mol^(-1)`

Answer

A

Explanation

Given, in bom calorimeter, volume remains the constant, thus, the heat involved is internal energy i.e., `DeltaU`. We know that,
Since, heat is lost be the system.
`thereforeq_(V)=-C_(V)DeltaT=-20.7kJ//Kxx(299-298)K=-20.7kJ`
(Here, negative sign indicates the exothermic nature of the reaction.)
Thus, `DeltaU` for the combustion of the 1 g of graphite`=-20.7kJ" "K^(-1)`
For cumbusion of 1 mole (12.0 g) of graphite
`=(12.0" g "mol^(-1)xx(-20.7kJ))/(1g)=-2.48xx10^(2)" kJ "mol^(-1)`
Since, `Deltan_(g)=0,DeltaH=DeltaU=-2.48xx10^(2)"kJ "mol^(-1)`