**A _{2} + B_{2} –> 2AB. ΔH^{0}_{f} = –200kJmol^{–1}**

AB, A_{2} and B_{2} are diatomic molecules. If the bond enthalpies of A_{2}, B_{2} and AB are in the ratio 1:0.5:1, then the bond enthalpy of A_{2} is _______ kJmol^{–1} . (Nearest integer)

**Answer: 800**

**Solution:**A

_{2}+ B

_{2}–> 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 A

_{2 }= x

Bond Enthalpy of B

_{2 }= 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**

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