The standard enthalpy of formation of NH3 is -46.0 kJ mol-1. If the enthalpy of formation of H2 from its atoms is -436 kJ mol-1 and that of N2 is -712 kJ mol-1, the average bond enthalpy of N-H bond in NH3 is

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-964 kJ mol-1

+352 kJ mol-1

+1056 kJ mol-1

-1102 kJ mol-1

answer is B.

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Detailed Solution

12N2 +32H2 ⇌NH3 ∆Hr =∆HNH3 -12∆HN≡N +32∆HH-H ∆Hr =46--712x12 +32-436 ∆HNH3 =356+654+46 3∆HN-H =1056 ∆HN-H =10563=352

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