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JEE Main 2010 Chemistry question

ChemistryThermodynamicsJEE Main 2010

The standard enthalpy of formation of is −46.0 kJ . If the enthalpy of formation of from its atoms is −436 kJ and that of is −712 kJ , the average bond enthalpy of N − H bond in is

Answer: (2)

Solution

: \frac\frac3E_{N-H}3E_{N-H}$ = 356 + 654 + 46

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Question ID VK-039233