\[\frac{1}{2}C{{l}_{2}}(g)\xrightarrow{\frac{1}{2}{{\Delta }_{diss}}{{H}^{\Theta }}}Cl(g)\xrightarrow{{{\Delta }_{eg}}{{H}^{\Theta }}}C{{l}^{-}}\]\[(g)\xrightarrow{\frac{1}{2}{{\Delta }_{hyd}}{{H}^{\Theta }}}C{{l}^{-}}(aq)\] |
The energy involved in the conversion of |
\[\frac{1}{2}C{{l}_{2}}(g)\] to \[C{{l}^{-}}(aq)\] |
(using the data, \[{{\Delta }_{diss}}H_{C{{l}_{2}}}^{\Theta }=240\,kJ\,mo{{l}^{-1}},\,{{\Delta }_{eg}}H_{Cl}^{\Theta }=-349\]\[kJ\,mo{{l}^{-1}},{{\Delta }_{hyd}}{{H}_{C{{l}^{-}}}}=-381\,kJ\,mo{{l}^{-1}})\] will be |
A) - 850 kJ \[mo{{l}^{-1}}\]
B) +120 kJ \[mo{{l}^{-1}}\]
C) +152 kJ \[mo{{l}^{-1}}\]
D) - 610 kJ \[mo{{l}^{-1}}\]
Correct Answer: D
Solution :
\[\Delta H=\left( \frac{1}{2}\times 240 \right)+(-349)+(-381)\] \[=-610\,kJ\,mo{{l}^{-1}}\].You need to login to perform this action.
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