JEE MAIN - Chemistry (2008 - No. 10)
Oxidising power of chlorine in aqueous solution can be determined by the parameters indicated
below:
$${1 \over 2}C{l_2}(g)$$ $$\buildrel {{1 \over 2}{\Delta _{diss}}{H^\Theta }} \over \longrightarrow $$ $$Cl(g)$$ $$\buildrel {{\Delta _{eg}}{H^\Theta }} \over \longrightarrow $$ $$C{l^ - }(g)$$ $$\buildrel {{\Delta _{Hyd}}{H^\Theta }} \over \longrightarrow $$ $$C{l^ - }(aq)$$
(Using the data, $${\Delta _{diss}}H_{C{l_2}}^\Theta $$ = 240 kJ/mol, $${\Delta _{eg}}H_{Cl}^\Theta $$ = -349 kJ/mol, $${\Delta _{hyd}}H_{C{l^ - }}^\Theta $$ = - 381 kJ/mol) will be :
$${1 \over 2}C{l_2}(g)$$ $$\buildrel {{1 \over 2}{\Delta _{diss}}{H^\Theta }} \over \longrightarrow $$ $$Cl(g)$$ $$\buildrel {{\Delta _{eg}}{H^\Theta }} \over \longrightarrow $$ $$C{l^ - }(g)$$ $$\buildrel {{\Delta _{Hyd}}{H^\Theta }} \over \longrightarrow $$ $$C{l^ - }(aq)$$
(Using the data, $${\Delta _{diss}}H_{C{l_2}}^\Theta $$ = 240 kJ/mol, $${\Delta _{eg}}H_{Cl}^\Theta $$ = -349 kJ/mol, $${\Delta _{hyd}}H_{C{l^ - }}^\Theta $$ = - 381 kJ/mol) will be :
+152 kJ mol−1
−610 kJ mol−1
−850 kJ mol−1
+120 kJ mol−1
Explanation
The energy involved in the conversion of
$${1 \over 2}C{{\rm}_2}\left( g \right)\,\,$$ to $${\mkern 1mu} {\mkern 1mu} {\mkern 1mu} C{l^{ - 1}}\left( {aq} \right)$$ is given by
$${\mkern 1mu} \Delta H = {1 \over 2}{\Delta _{diss}}H_{C{l_2}}^{\left( - \right)} + {\Delta _{eg}}H_{Cl}^{\left( - \right)} + {\Delta _{hy{\rm{I}}}}H_{Cl}^{\left( - \right)}$$
Substituting various values from given data, we get
$$\Delta H = \left( {{1 \over 2} \times 240} \right) + \left( { - 349} \right) + \left( { - 381} \right)kJ\,mo{l^{ - 1}}$$
$$ = \left( {120 - 349 - 381} \right)kJmo{l^{ - 1}}$$
$$ = - 610\,kJ\,mo{l^{ - 1}}$$
i.e., the correct answer is $$(b)$$
$${1 \over 2}C{{\rm}_2}\left( g \right)\,\,$$ to $${\mkern 1mu} {\mkern 1mu} {\mkern 1mu} C{l^{ - 1}}\left( {aq} \right)$$ is given by
$${\mkern 1mu} \Delta H = {1 \over 2}{\Delta _{diss}}H_{C{l_2}}^{\left( - \right)} + {\Delta _{eg}}H_{Cl}^{\left( - \right)} + {\Delta _{hy{\rm{I}}}}H_{Cl}^{\left( - \right)}$$
Substituting various values from given data, we get
$$\Delta H = \left( {{1 \over 2} \times 240} \right) + \left( { - 349} \right) + \left( { - 381} \right)kJ\,mo{l^{ - 1}}$$
$$ = \left( {120 - 349 - 381} \right)kJmo{l^{ - 1}}$$
$$ = - 610\,kJ\,mo{l^{ - 1}}$$
i.e., the correct answer is $$(b)$$
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