1.

The combustion of benzene (1) gives CO_(2)(g) and H_(2)O(1). 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=8.314 JK^(-1) mol^(-1))

Answer»

`- 452.46`
3260
`- 3267.6`
4152.6

Solution :`C_(6)H_(6("liq"))+7.5O_(2(g))rarr6CO_(2(g))+3H_(2)O_("liq")`
`Deltan_((g))=6-7.5=-15`
`DeltaH=DeltaE+Deltan_((g))RT`
`DeltaH=-3263.9kJ-(1.5xx8.314xx298)/(1000)KJ=-3267.6 kJ`


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