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Q.

AB, A2 and B2 are diatomic molecules, if the bond enthalpies of A2, AB and B2 are in the ratio of 1 : 1 : 0.5 and the enthalpy of formation of AB from A2 and B2 is -100 kJ mol-1, combustion of A2 gives AO and ∆HC=-1200 kJ mol-1. Bond energy of O=O bond is 500 kJ mol-1. What is the bond enthalpy of (A - O) bond ?

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a

400 kJ mol-1

b

1650 kJ mol-1

c

1200 kJ mol-1

d

200 kJ mol-1

answer is B.

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Detailed Solution

Bond enthalpy (BE) of A2 = x,  AB = x and B2 = x212A2+12B2 → AB        -100 kJ/mol-1                 .....(i) 12A2+12O2 → AO        -1200 kJ/mol (given) A2+O2 → 2AO               -2400 kJ/mol           ....(ii) O=O→ 500 kJ/mol From eq. i 12 B.EA2 + 12 B.EB2- B.EAB=-100 12x + x4-x=-100 -x4=-100 x= 400 kJ/mol Now, from eq. ii B.EA2 +  B.EO2-2 B.EAO=-2400 x++500-2 B.EAO=-2400 400 + 500 - 2 B.EAO=-2400 - B.EA-O=-3300  B.EA-O=1650 kJ/mol.
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