Q. The sublimation energy of $I_{2}(s)$ is 57.3 kJ/mol and the enthalpy of fusion is 15.5 kJ/mol. The enthalpy of vaporisation of $I_{2}$ is
A) 41.8 kJ/mol
B) -41.8 kJ/mol
C) 72.8 kJ/mol
D) -72.8 kJ/mol
Explanation
(a)
According to Hess’s law total heat changes during a chemical reaction are independent of path of reaction.
Given, $I_{2}(s) \rightarrow I_{2}(g),\mathrm{\Delta}H_{1} = 57.3\ kJ/mol$ …(i)
$I_{2}(s) \rightarrow I_{2}(l),\mathrm{\Delta}H_{2} = + 15.5\ kJ/mol$ …(ii)
Required equation $I_{2}(l) \rightarrow I_{2}(g),\mathrm{\Delta}H_{1} = ?$ subtract
Eq. (ii) from Eq. (i)
$\therefore$ $I_{2}(l){\rightarrow I}_{2}(g),\ \ \ \ \ \ \ \ \mathrm{\Delta}H = 57.3 + ( - 15.5)$
$\ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ = + 41.8\ kJ/mol$
Reviewed by Dr. Abubacker
Physics Educator & NEET/JEE Specialist
Ph.D. in Physics with 19+ years of teaching experience.