The standard enthalpy of formation of ammonia at 25°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

Quizplus · Accepted Answer

The enthalpy change for the formation of ammonia can be expressed as:

N2(g) + 3H2(g) -> 2NH3(g) ΔHf° = -46.3 kJ/mol

To calculate the N-H bond enthalpy, we can use the following equation:

ΔHf°(NH3) = 3ΔHf°(H-H) + ΔHf°(N≡N) - 2ΔHf°(N-H)

Substituting the given values:

-46.3 kJ/mol = 3(436.4 kJ/mol) + 941.4 kJ/mol - 2ΔHf°(N-H)
ΔHf°(N-H) = (3(436.4 kJ/mol) + 941.4 kJ/mol + 46.3 kJ/mol)/2 
ΔHf°(N-H) = 391 kJ/mol

Therefore, the N-H bond enthalpy at 25°C is estimated to be 391 kJ/mol. The correct choice is C.

The Standard Enthalpy of Formation of Ammonia at 25°C Is ≡\equiv≡

The Standard Enthalpy of Formation of Ammonia at 25°C Is $\equiv$