1.

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 DeltaH_(f) for the formation of XY is -200 kJ mol^(-1). The bond dissociation energy of X_(2) will be

Answer»

400 kJ `MOL^(-1)`
300 kJ `mol^(-1)`
20 kJ `mol^(-1)`
none of these

Solution :Formation of XY is known as
`X_(2)+Y_(2) to 2XY`
`DeltaH=(BE)_(X-X)+(BE)_(Y-Y)-2(BE)_(X-Y)` ,brgt If `(BE)` of `X-Y=a`
then (B E) of (X-X)=a
and (B E) of (Y-Y)=`(a)/(2)`
`therefore DeltaH_(f)(X-Y)=-200KJ`
`therefore-400` (for 2 mol XY)`=a+(a)/(2)-2a`
`-400=-(a)/(2)impliesa=+800kJ`
The bond DISSOCIATION ENERGY of `X_(2)=800" kJ "mol^(-1)`


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