Question

# AB, A2, and B2 are diatomic molecules. If the bond enthalpies of A2, AB, and B2 are in the ratio 2:2:1 and ethalpy of formation AB from A2 and B2 is âˆ’100 kJ molâˆ’1. What is the bond energy of A2?

A
200 kJ mol1
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B
100 kJ mol1
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C
300 kJ mol1
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D
400 kJ mol1
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Solution

## The correct option is D 400 kJ mol−1Solution:- (D) 400KJ/molGiven:-B.E.A2:B.E.AB:B.E.B2=2:2:1Let bond enthalpy of B2 be xKJ/molTherefore,B.E.A2=B.E.AB=2xGiven that ΔHAB(formation)=−100KJ/mol 12A2+12B2⟶AB;ΔH=−100KJ/molA2+B2⟶2AB;ΔH=−200KJ/molAs we know that,ΔH= Bond enthalpy of reactant − Bond enthalpy of product∴ΔH=(B.E.A2+B.E.B2)−(2×B.E.AB)−200=(2x+x)−(2×2x)−200=3x−4x⇒x=200Therefore,B.E.A2=2x=2(200)=400KJ/molHence the bond energy of A2 is 400KJ/mol.

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