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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 100 kJ mol1. 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 mol1
Solution:- (D) 400KJ/mol
Given:-
B.E.A2:B.E.AB:B.E.B2=2:2:1
Let bond enthalpy of B2 be xKJ/mol
Therefore,
B.E.A2=B.E.AB=2x
Given that ΔHAB(formation)=100KJ/mol
12A2+12B2AB;ΔH=100KJ/mol
A2+B22AB;ΔH=200KJ/mol
As 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=3x4x
x=200
Therefore,
B.E.A2=2x=2(200)=400KJ/mol
Hence the bond energy of A2 is 400KJ/mol.

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