Q. Oxidising power of chlorine in aqueous solution can be determined by the parameters indicated below $\frac{1}{2}{Cl}_{2}(g)\overset{\frac{1}{2}\mathrm{\Delta}_{diss^{H{^\circ}}}^{\ }}{\rightarrow}Cl(g)\overset{\mathrm{\Delta}_{EA\ ^{H{^\circ}}}^{\ }}{\rightarrow}{Cl}^{-}(g)\overset{\mathrm{\Delta}_{hyd\ ^{H{^\circ}}}^{\ }}{\rightarrow}{Cl}^{-}(aq)$ The energy involved in the conversion of $\frac{1}{2}Cl_{2}(g)$ to ${Cl}^{-}(aq)$(Using the data) $\mathrm{\Delta}_{diss^{H{^\circ}}Cl_{2}}^{\ } = 240\ kJ\ mol^{- 1}$ $\mathrm{\Delta}_{{EA}^{H{^\circ}}Cl_{\ }}^{\ } = - 349\ kJ\ mol^{- 1}$ $\mathrm{\Delta}_{hyd\ ^{H{^\circ}}}^{\ }Cl = - 381\ kJ\ mol^{- 1}$ will be
A) $+ 152\ kJ\ mol^{- 1}$
B) $- 610\ kJ\ mol^{- 1}$
C) $- 850\ kJ\ mol^{- 1}$
D) $+ 120\ kJ\ mol^{- 1}$
Explanation
(b)
$\frac{1}{2}Cl_{2}(g) \longrightarrow Cl^{-}(aq)$
$\mathrm{\Delta}H = \frac{1}{2}\mathrm{\Delta}H_{diss}\left( Cl_{2} \right) + \mathrm{\Delta}H_{EA}\ Cl + \mathrm{\Delta}H_{hyd}(Cl^{-})$
$= \frac{240}{2} - 349 - 381$
$= - 610\ kJ\ mol^{- 1}$
Reviewed by Dr. Abubacker
Physics Educator & NEET/JEE Specialist
Ph.D. in Physics with 19+ years of teaching experience.