Question
If the bond dissociation energies of $$XY,{X_2}$$ and $${Y_2}$$ ( all
diatomic molecules ) are in the ratio of 1 : 1 : 0.5 and $$\Delta {H_f}$$ for the formation of $$XY$$ is $$ - 200\,kJ\,mo{l^{ - 1}}.$$ The bond dissociation energy of $${X_2}$$ will be
A.
$$400\,kJ\,mo{l^{ - 1}}$$
B.
$$800\,kJ\,mo{l^{ - 1}}$$
C.
$$200\,k{J^{ - 1}}$$
D.
$$100\,k{J^{ - 1}}$$
Answer :
Solution :
$$\eqalign{
& {X_2} + {Y_2} \to 2XY,\,\Delta H = 2\left( { - 200} \right) \cr
& {\text{Let}}\,x\,{\text{be}}\,{\text{the}}\,{\text{bond}}\,{\text{dissociation}}\,{\text{energy}}\,{\text{of}}\,{X_2}.\,{\text{Then}} \cr
& \Delta H = - 400 = {\xi _{x - x}} + {\xi _{y - y}} - 2{\xi _{x - y}} = x + 0.5x - 2x = - 0.5x \cr
& or\,x = \frac{{400}}{{0.5}} = 800\,kJ\,mo{l^{ - 1}} \cr} $$