1.

The combustion of benzene (l) gives CO_(2)(g) and H_(2)O(l). Given that heat of combustion of benzene at constant volume is -3263.9 kJ*mol^(-1) at 25^(@)C, heat of combustion (in kJ*mol^(-1)) of benzene at constant pressure will be (R=8.314J*K^(-1)*mol^(-1))-

Answer»

4152.6
`-452.46`
`3260`
`-3267.6`

Solution :`C_(6)H_(6)(l)+(15)/(2)O_(2)(g)to6CO_(2)(g)+3H_(2)O(l)`
`Deltan=6-(15)?(2)=-(3)/(2)`
`DeltaH=DeltaU+DeltanRT`
`DeltaH=[-3263.9-(3)/(2)xx8.314xx10^(-3)xx298]KJ*MOL^(-1)`
`=-3267.6kJ*mol^(-1)`.


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