91Ó°ÊÓ

To solve the following problem, we have to use Hess’s law to calculate the change in enthalpy during the reaction.

\(\begin{array}{*{20}{l}}{{\rm{Hg}}\left( l \right){\rm{ + C}}{{\rm{l}}_{\rm{2}}}\left( {\rm{g}} \right) \to {\rm{HgC}}{{\rm{l}}_{\rm{2}}}\left( {\rm{s}} \right)..........{\rm{(1) \Delta H = }} - {\rm{224 kJ}}}\\{{\rm{Hg}}\left( l \right){\rm{ + HgC}}{{\rm{l}}_{\rm{2}}}\left( {\rm{s}} \right) \to {\rm{H}}{{\rm{g}}_{\rm{2}}}{\rm{C}}{{\rm{l}}_{\rm{2}}}\left( {\rm{s}} \right).....{\rm{(2) \Delta H = }} - {\rm{41}}{\rm{.2 kJ }}}\end{array}\)

Further, equations (1) and (2) are multiplied by 1 and then added up.

If we observe the given equation and equation(3), both are reverses of one another. So, we will follow the given steps to find the change in enthalpy of the correct equation,

\(\begin{array}{l}{\rm{2Hg}}\left( l \right) + {\rm{C}}{{\rm{l}}_{\rm{2}}}\left( {\rm{g}} \right) \to {\rm{H}}{{\rm{g}}_{\rm{2}}}{\rm{C}}{{\rm{l}}_{\rm{2}}}\left( {\rm{s}} \right)..........(3){\rm{ \Delta H}} = \left( { - {\rm{224}} - {\rm{41}}{\rm{.2}}} \right){\rm{kJ}}\\{\rm{ = }} - 265.2{\rm{ kJ}}\\{\rm{Now we will multiply the whole eqaution by }} - 1,{\rm{(equation as well as the change in enthalpy calculated)}}\\{\rm{we will get, }}\\{\rm{H}}{{\rm{g}}_{\rm{2}}}{\rm{C}}{{\rm{l}}_{\rm{2}}}\left( {\rm{s}} \right) \to {\rm{2Hg}}\left( l \right) + {\rm{C}}{{\rm{l}}_{\rm{2}}}\left( {\rm{g}} \right)...........(4){\rm{ \Delta H}} = 265.2{\rm{ kJ}}\end{array}\)

Hence, the change in enthalpy of the given equation will be \({\rm{\Delta H}} = 265.2{\rm{ kJ}}\).