Given standard enthalpy of formation of ...

Given standard enthalpy of formation of `CO(-110 "KJ mol"^(-1))` and `CO_(2)(-394 "KJ mol"^(-1))`. The heat of combustion when one mole of graphite burns is

`-504` kJ

`-394` kJ

`-284` kJ

`-110` kJ

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The correct Answer is:

Graphite burns to give `CO_2`. So , standard enthalpy of formation of `CO_2` and the heat of combustion when one mole of graphite burns are identical

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