Question

If the sublimation energy and enthalpy of fusion of $I _{2}$ are $57.3\, kJ\, mol ^{-1}$ and $15.5\, kJ\, mol ^{-1}$, respectively then, the enthalpy of vaporisation of $I _{2}$ will be

• A. $-72.8\, kJ\, mol ^{-1}$
• B. $72.8\, kJ\, mol ^{-1}$
• C. $-41.8\, kJ\, mol ^{-1}$
• D. $+41.8\, kJ\, mol ^{-1}$

Answer 1

Answer: (D)

(i) $I _{2}(s) \longrightarrow I _{2}(g) ; \Delta H_{1}=+57.3\, kJ\, mol ^{-1}$
(ii) $I _{2}(s) \longrightarrow I _{2}(l) ; \Delta H_{2}=+15.5\, kJ\, mol ^{-1}$
(iii) $I _{2}(l) \longrightarrow I _{2}(g) ; \Delta H=$ ?
$Eq$ (iii) will be obtained, if we substract Eq (ii) from (i), i.e (iii) = (i) - (ii)
Now, $\Delta H=57.3-15.5=+41.8\, kJ\, mol ^{-1}$