Question
Oxidising power of chlorine in aqueous solution can be determined by the parameters indicated below:
\[\frac{1}{2}C{l_2}\left( g \right)\xrightarrow{{\frac{1}{2}{\Delta _{diss}}{H^\Theta }}}Cl\left( g \right)\xrightarrow{{{\Delta _{cg}}{H^\Theta }}}C{l^{^ - }}\left( g \right)\xrightarrow{{{\Delta _{Hyd}}{H^\Theta }}}C{l^ - }\left( {aq} \right)\]
$$\eqalign{
& \left( {{\text{using}}\,{\text{the}}\,{\text{data}}} \right.{\text{,}} \cr
& {\Delta _{diss}}H_{C{L_2}}^\Theta = 240\,kJ\,mo{l^{ - 1}},\,{\Delta _{eg}}H_{Cl}^\Theta = - 349\,kJ\,mo{l^{ - 1}}, \cr
& {\Delta _{hyd}}H_{C{l^ - }}^\Theta = - 381\,kJ\,\left. {mo{l^{ - 1}}} \right),\,{\text{will}}\,{\text{be}} \cr} $$
A.
$${\text{ + 152}}\,{\text{kJ}}\,{\text{mo}}{{\text{l}}^{ - 1}}$$
B.
$$ - 610\,kJ\,mo{l^{ - 1}}$$
C.
$$ - 850\,kJ\,mo{l^{ - 1}}$$
D.
$$ + 120\,kJ\,mo{l^{ - 1}}$$
Answer :
$$ - 610\,kJ\,mo{l^{ - 1}}$$
Solution :
The energy involved in the conversion of
$$\eqalign{
& \frac{1}{2}C{l_2}\left( g \right)\,\,{\text{to}}\,\,C{l^{ - 1}}\left( {aq} \right)\,{\text{is}}\,{\text{given}}\,{\text{by}} \cr
& \Delta H = \frac{1}{2}{\Delta _{diss}}H_{C{l_2}}^{\left( - \right)} + {\Delta _{eg}}H_{Cl}^{\left( - \right)} + {\Delta _{hyl}}H_{Cl}^{\left( - \right)} \cr} $$
Substituting various values from given data, we get
$$\eqalign{
& \Delta H = \left( {\frac{1}{2} \times 240} \right) + \left( { - 349} \right) + \left( { - 381} \right)kJ\,mo{l^{ - 1}} \cr
& = \left( {120 - 349 - 381} \right)\,kJ\,mo{l^{ - 1}} \cr
& = - 610\,kJ\,mo{l^{ - 1}} \cr} $$
i.e., the correct answer is (B)