Question

if the bond dissociation energies of XY,X2,Y2 are in the ratio 1:1:0.5 and change in enthalpy of formation of XY is 200kJ per mol. then find the bond dissociation enthalpy of X2.

Answer 1

hey!! Mate actually may e there's a mistake in the question you have here ...the bond formation enthalpy of XY is actually ""-200KJ"" but since you have given here it as ""+200KJ"" hence the answer is.....if you would have given it as -200KJ then the answer would be ""+800KJ""

Answer 2

Answer:The answer is 800kj/mol

Explanation:

X2 +Y2 --->2XY

Delta H=(B.E of X-X) +(B.E of Y-Y) - (2*B.E of X-X)

If BE of X-Y =a,

Then BE of X-X=a and

BE of Y-Y=a/2

Delta Hf(X-Y) = - 200

So, - 400 for 2 moles of X-Y

-400=a+a/2-2a

-400=-a/2

Therefore a=+800 kj/mol