JEE MAIN - Chemistry (2005 - No. 4)
If the bond dissociation energies of XY, X2 and Y2 (all diatomic molecules) are in the ratio of 1:1:0.5 and $$\Delta H_f$$ for the formation of XY is -200 kJ mole-1. The bond
dissociation energy of X2 will be :
100 kJ mol-1
200 kJ mol-1
300 kJ mol-1
800 kJ mol-1
Explanation
$${X_2} + {Y_2} \to 2XY,$$ $$\Delta H = 2\left( { - 200} \right).$$
Let $$x$$ be the bond dissociation energy of $${X_2}.$$
Then $$\Delta H = - 400$$
$$ = {\xi _{x - x}} + {\xi _{y - y}} - 2{\xi _{x - y}}$$
$$ = x + 0.5x - 2x$$
$$ = - 0.5x$$
or $$\,\,\,x = {{400} \over {0.5}} = 800\,kJ\,mo{l^{ - 1}}$$
Let $$x$$ be the bond dissociation energy of $${X_2}.$$
Then $$\Delta H = - 400$$
$$ = {\xi _{x - x}} + {\xi _{y - y}} - 2{\xi _{x - y}}$$
$$ = x + 0.5x - 2x$$
$$ = - 0.5x$$
or $$\,\,\,x = {{400} \over {0.5}} = 800\,kJ\,mo{l^{ - 1}}$$
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