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What is `Deltan` for combustion of 1 mole of benzene, when both the reactants and the products are gas at 298 K

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What is `Deltan` for combustion of 1 mole of benzene, when both the reactants and the products are gas at 298 K
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`C_(6)H_(6(g))+(15)/(2)O_(2(g))rarr6CO_(2(g))+3H_(2)O_((g))`
`Deltan=6+3-1-(15)/(2)=+(1)/(2)`.
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