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) ?

Dissociation enthalpy of diatomic molecules XY, X_(2) and Y_(2) are to

1.	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) ?
Answer» 200 kJ mol`""^(-1)` 300 kJ mol`""^(-1)` 400 kJ mol`""^(-1)` 800 kJ mol`""^(-1)` SOLUTION :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`