Dissociation enthalpy of diatomic molecules `XY, X_(2) and Y_(2)` are too in the ratio of `1:1:0.5` the value. Enthalpy of formation of `XY. Delta_(f) H= - 200` kJ mol`""^(-1)`. Final the dissociation enthalpy of `X_(2)` ?

Options

A. 200 kJ mol`""^(-1)`
B. 300 kJ mol`""^(-1)`
C. 400 kJ mol`""^(-1)`
D. 800 kJ mol`""^(-1)`

Answer

D

Explanation

Reaction : `(1)/(2) X_(2) + (1)/(2) Y_(2) to XY, Delta_(f) H= - 200 "kJ mol"^(-1)`
Dissociation enthalpy `= 1: 0.5 : 1[X_(2) : Y_(2) :XY]`
`= 1x" " 0.5x" " 1x`
`DeltaH_(f) (XY) = [(1)/(2) ("Dissociation enthalpy of"x_(2) ) + (1)/(2) ("Dissociation enthalpy of"y)]-("Dissociation enthalpy"XY)`
`therefore - 200 = [ ( (1)/(2) (x) + (1)/(2) (0.5x)- x)]`
`therefore - 200 = (x)/(2) = (x)/(4) - x = - (1)/(4) x`
`therefore x= -2 00 (-(4)/(1) ) = + 800 "kJ"`
`therefore` Dissociation enthalpy of `X_(2) = 800 "kJ" = x`