1. Tardigrade
  2. Question
  3. Chemistry
  4. Oxidising power of chlorine in aqueous solution can be determined by the parameters indicated below (1/2)(C l)2(g) overset(1/2) Δ d i s sH ° arrow Cl(g) oversetΔ E A ( )H ° arrow (C l)-(g) oversetΔ h y d ( )H ° arrow (C l)-(.aq.) The energy involved in the conversion of (1/2)Cl2(.g.) to (C l)-(.aq.) (Using the data) Δ d i s sH ° C l2 =240 kJ mol- 1 Δ E AH ° C l =-349 kJ mol- 1 Δ h y d H ° Cl=-381 kJ mol- 1 will be

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Q. Oxidising power of chlorine in aqueous solution can be determined by the parameters indicated below
$\frac{1}{2}\left(C l\right)_{2}\left(g\right)\overset{\frac{1}{2} \Delta _{d i s s^{H ^\circ }}^{ \, }}{ \rightarrow }Cl\left(g\right)\overset{\Delta _{E A \left( \, \right)^{H ^\circ }}^{ \, }}{ \rightarrow }\left(C l\right)^{-}\left(g\right)\overset{\Delta _{h y d \left( \, \right)^{H ^\circ }}^{ \, }}{ \rightarrow }\left(C l\right)^{-}\left(\right.aq\left.\right)$
The energy involved in the conversion of $\frac{1}{2}Cl_{2}\left(\right.g\left.\right)$ to $\left(C l\right)^{-}\left(\right.aq\left.\right)$ (Using the data)
$\Delta _{d i s s^{H ^\circ } C l_{2}}^{ \, }=240 \, kJ \, mol^{- 1}$
$\Delta _{E A^{H ^\circ } C l_{ \, }}^{ \, }=-349 \, kJ \, mol^{- 1}$
$\Delta _{h y d \, ^{H ^\circ }}^{ \, }Cl=-381 \, kJ \, mol^{- 1}$ will be

NTA AbhyasNTA Abhyas 2020

Solution:

$\frac{1}{2}Cl_{2}\left(\right.g\left.\right) \rightarrow Cl^{-}\left(\right.aq\left.\right)$
$\Delta H=\frac{1}{2}\Delta H_{d i s s}\left(C l_{2}\right)+\Delta H_{E A} \, Cl+\Delta H_{h y d}\left(\right.Cl^{-}\left.\right)$
$=\frac{240}{2}-349-381$
$=-610 \, kJ \, mol^{- 1}$