The bond dissociation energies of `X_(2), Y_(2)`, and XY are in the ratio of 1 : 0.5 : 1. `Delta H` for the formation of XY is `-100 kJ mol^(-1).` The bond dissociation energy of `X_(2)` will be___ kJ` mol^(-1).`

Answer

400

Explanation

The ratio of the bond dissociation energies of `X _(2), Y_(2) and XY is 1 : 0.5 : 1`
` therefore ` Bond dissociation energies of `X _(2), Y_(2)and XY` are :
`Delta H (X - X) = x`
`DeltaH (Y-Y) = 0.5 x`
`DeltaH(X -Y) = x`
The chemical equation for formation of XY is :
`(1)/(2) X _(2) + (1)/(2) Y_(2) to XY`
`Delta H = sum m Delta H` (reactant bonds)
`- sum n Delta H`(product bonds)
` therefore -100`
`= [ (1)/(2) Delta H (X-X) + (1)/(2) Delta H (Y-Y)]`
`- Delta H ( X-Y)`
`-100 = [ (1)/(2) x + (1)/(2) xx 0.5 x ] - x`
`- 100 = [(x)/(2) + (x)/(4) ] - x`
`- 100 = ( 6x)/(8) - x`
`-100 = ( 6x -8x)/(8)`
`-100 = ( - 2x )/(8)`
`therefore x = (100 xx 8)/(2) = 400`
` therefore`Bond dissociation energy of
`X _(2) = 400 kJ mol ^(-1)`