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Enthalpy of formation of A2B is +1000 KJ. Ratio of bond enthalpies of A2,B2 and A2B is 1 : 2 : 1. Bond enthalpy of B2would be

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answer is 2.

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Detailed Solution

Ratio of bond enthalpies of A2:B2:A2B is 1 : 2 : 1B.E of A2=xB.E of B2=2xB.E of 12B2=xB.E of A2B=x∴A2x+12B2x→ A2Bx;ΔH=+1000kJΔHB.D.E=HR−HP+1000kJ =2x -xx=1000kJB.D.E of B2 = 2x =+2000kJ

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