Enthalpy of vaporisation of napthalene is 48.5 kJ/mol and enthalpy of fusion of napthalene is 22.5 kJ/mol, then enthalpy of subtlimation of naphthalene will be

Options

A. `-48.5" KJ mole"^(-1)`
B. `+22.5" KJ mole"^(-1)`
C. `26.0" KJ mole"^(-1)`
D. `71.0" KJ mole"^(-1)`

Answer

D

Explanation

Subliation of naphthalene is
`"Naphthalene"_((s)) rarr "Nephthalene"_((g))`
`"Naphthalene"_((s)) overset("fusion")underset(Delta H_(f))rarr"Naphthalene"_((l))overset("vapourisation")underset(Delta H_(v))rarr"Naphthalene"_((g))`
`"Nephthalene"_((s))overset("Sublimation")underset(Delta H_(s))rarr"Nephthalene"_((g))`
`Delta H_("sublimation") = Delta H_("fusion")+Delta H_("vapourisation")`
`=22.5 + 48.5`
`Delta H_("sublimation")=71.0 kJ mol^(-1)`