Question

The sublimation energy of $I_2\left(s\right)$ is $57.3kJ/mol$ and the enthalpy of fusion is $15.5kJ/mol$. The enthalpy of vaporisation of $I_2$ is:

• A. $41.8kJ/mol$
• B. $-41.8kJ/mol$
• C. $72.8kJ/mol$
• D. $-72.8kJ/mol$

Answer 1

The correct option is A $41.8kJ/mol$

Enthalpy of Fusion :- It is the enthalpy change accompanied by melting of one mole of solid substance.

Enthalpy of Vapourisation :- It is the enthalpy change accompanied by conversion of one mole of liquid substance completely into vapours.

Enthalpy of Sublimation :- It is the enthalpy charge when one mole of a solid substance sublimes.

Enthalpy of sublimation of $I_2\left(s\right)$ $=$ Enthalpy of fusion of $I_2$ $+$ Enthalpy of vaparization of $I_2$

$\therefore 57.3=15.5+$ Enthalpy of vaparization of $I_2$

$\therefore$ Enthalpy of vapourization of $I_2$ $=57.3-15.5=41.8KJ/mol$