The heat of combustion of napthalene `{C_(10)H_(8)(s)}` at constant volume was measured to be `-5133 kJ mol^(-1)` at 298 K. Calculate the value of enthalpy change (Given `R=8.314 JK^(-1)mol^(-1)`)

Explanation

The combustion reaction of napthalene.
`C_(10)H_(8)(s)+12O_(2)(g)rarr 10CO_(2)(g)+4H_(2)O(l), " " Delta E=-5133 KJ`
`Delta n_(g)=10-12=-2` mol.
Now applying the relation. `Delta H = Delta H+Delta n_(g)RT=-5133xx10^(3)+(-2)(8.314)(298)`
`=-513000J-4955.14 J=-5137955.14` joule.