1.

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

Answer»

0
`3//2`
`-3//2`
`1//2`

Solution :`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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