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if the bond dissociation energies of xy,x2,y2(all diatomic molecules) are in the ratio of 1:1:0.5 &delta H for the formation of xy is -200kj/mol, the bond dissociation energy of x2will be (a) 400kj/mol (b)300kj/mol (c)200kj/mol (d)800kj/mol please explain me

if the bond dissociation energies of xy,x2,y2(all diatomic molecules) are in the ratio of 1:1:0.5 &delta H for the formation of xy is -200kj/mol, the bond dissociation energy of x2will be
(a) 400kj/mol             (b)300kj/mol
(c)200kj/mol               (d)800kj/mol
please explain me

Grade:12th pass

3 Answers

Aashna
79 Points
5 years ago
Answer is 400KJRatios are given for bde now lets convert ratio into no.i.e 1:1:0.5 to x x anf 0.5x for X2 XY and Y2 respectivelyH of form = heatof reactant -heat of product-200=1.5x-2x (x is bde of X2)x=400KJ
Amritpal Singh
11 Points
5 years ago
The reaction component can be written as:X2 →2X.............1Y2 →2Y.................2XY →X + Y..............3X + Y →XY.................4to obtain capital delta H4 = capital deltaH1/2 + capital delta H2/2 - capital deltaH3...................5ratio between capital deltaH1,capital deltaH2 andcapital deltaH3 is 1 : 0.5 : 1 = 1x : 0.5x : 1x​capital delta H 4 = 200 KJPutting all the values in Eq 5 we get200 KJ = 1/2 x + 1/4 x - 1x.............................note ; 0.5 x/2 = 1/4 x​x = -200 KJ/0.25 = -800 KJBond dissociation energy for X2 = -800 KJ
Rishi Sharma
askIITians Faculty 646 Points
3 years ago
Dear Student,
Please find below the solution to your problem.

ΔH for the formation of XY is −200kJmol−1.
The bond dissociation energies of X2​,Y2​ and XY are in the ratio of 1:0.5:1.
Let the bond dissociation energy of X2​ will be a kJ/mol.
The bond dissociation energy of Y2​ will be 0.5a kJ/mol.
The bond dissociation energy of XY will be a kJ/mol.
0.5 ​X2 ​+ 0.5​Y2 ​→ XY
ΔH(reaction) = ΣΔH(reactantbonds) − ΣΔH(productbonds)
ΔH(reaction) = 0.5​ΔH(X-X) + 0.5​ΔH(Y-Y) − ΔH(X-Y)
−200 = 0.5(a) + 0.5(0.5a) −(a)
−200 = −0.25(a)
a = 800
The bond dissociation energy of X2​ will be 800kJ/mol

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