A2 + B2  –> 2AB.    ΔH0f = –200kJmol–1
AB, A2 and B2 are diatomic molecules. If the bond enthalpies of A2, B2 and AB are in the ratio 1:0.5:1, then the bond enthalpy of A2 is _______ kJmol–1 . (Nearest integer)

Answer: 800
Solution: A2 + B2  –> 2AB
Enthalpy change of formation of 1 mol of AB = –200 kJmol–1
Enthalpy change of formation of 2 mol of AB = –400 kJmol–1
Let Bond Enthalpy of A2 = x
Bond Enthalpy of B2 = 0.5x
Bond Enthalpy of AB= x

The standard enthalpy change for the reaction can be calculated using bond enthalpies. In this case, it is equal to the sum of the bond enthalpies broken(Reactants) minus the sum of the bond enthalpies formed(Product).

ΔH°f = (Bond enthalpy of A2 + Bond enthalpy of B2) – (2 * Bond enthalpy of AB)
– 400 = (x + 0.5x) – 2x – 400 = – 0.5x
=> x = 800 kJmol–1

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