ThermochemistryMedium
Question
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
Options
A.- 964 kJ mol-1
B.+352 kJ mol-1
C.+ 1056 kJ mol-1
D.- 1102 kJ mol-1
Solution
Enthalpy of formation of NH3 = - 46 kJ/mole
∴ N2 + 3H2 → 2NH3 ᐃ Hf = - 2 × 46 kJ mol
Bond breaking is endothermic and Bond formation is exothermic
Assuming ′x′ is the bond energy of N - H bond (kJ mol-1)
∴ 712 + (3 × 436)- 6x = - 46 x 2
∴ x = 352 kJ/mol
∴ N2 + 3H2 → 2NH3 ᐃ Hf = - 2 × 46 kJ mol
Bond breaking is endothermic and Bond formation is exothermic
Assuming ′x′ is the bond energy of N - H bond (kJ mol-1)
∴ 712 + (3 × 436)- 6x = - 46 x 2
∴ x = 352 kJ/mol
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