The standard enthalpy of formation of ammonia at 25^$\circ$C is -46.3 kJ/mol. Estimate the N-H bond enthalpy at this temperature.(Given: BE(N$\equiv$N) = 941.4 kJ/mol, BE(H-H) = 436.4 kJ/mol) A) 360 kJ/mol B) 383 kJ/mol C) 391 kJ/mol D) 459 kJ/mol E) 475 kJ/mol

Question

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The N-H bond enthalpy can be estimated using the following equation:N-H bond enthalpy = (BE(N≡N) + 3BE(H-H) - ΔH°f(NH3)) / 6Substituting the given values into the equation, we get:N-H bond enthalpy = (941.4 kJ/mol + 3(436.4 kJ/mol) - (-46.3 kJ/mol)) / 6 = 391 kJ/molTherefore, the N-H bond enthalpy at 25°C is 391 kJ/mol.

The Standard Enthalpy of Formation of Ammonia at 25 ∘\circ∘ C Is -46

The Standard Enthalpy of Formation of Ammonia at 25^$\circ$C Is -46