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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
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
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 : 1xcapital 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 xx = -200 KJ/0.25 = -800 KJBond dissociation energy for X2 = -800 KJ
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 + 0.5Y2 → XYΔH(reaction) = ΣΔH(reactantbonds) − ΣΔH(productbonds)ΔH(reaction) = 0.5ΔH(X-X) + 0.5ΔH(Y-Y) − ΔH(X-Y)−200 = 0.5(a) + 0.5(0.5a) −(a)−200 = −0.25(a)a = 800The bond dissociation energy of X2 will be 800kJ/molThanks and Regards
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