1.

The bond dissociation energies of X_(2), Y_(2) and XY are in the ratio of 1 : 0.5 : 1, DeltaH for the formation of XY is - 200 kJ mol^(-1). The bond dissociation energy of X_(2) will be

Answer»

200 KJ `mol^(-1)`
100 kJ `mol^(-1)`
800 kJ `mol^(-1)`
400 kJ `mol^(-1)`

Solution :Let B.E.Of`x_(2), y_(2)` & xy are x kJ `mol^(-1)`, `0.5 x kJ mol^(-1)` and x kJ `mol^(-1)`RESPECTIVELY.
`(1)/(2)x_(2)+(1)/(2)y_(2)rarrxy, DeltaH=-200 kJ mol^(-1)`
`DeltaH=-200=Sigma(B.E)_("REACTANT")-Sigma(B.E)_("PRODUCT")`
`=[(1)/(2)XX(x)+(1)/(2)xx(0.5x)]-[1xx(x)]`
`"B.E. of "X_(2)=x=800 kJ mol^(-1)`.


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