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)383 kJ/mol
B)475 kJ/mol
C)360 kJ/mol
D)391 kJ/mol
E)459 kJ/mol

Question

Quizplus · Accepted Answer

The enthalpy of formation of ammonia can be written as:

NH3(g) → N(g) + 3/2 H2(g)

ΔHf° = -46.3 kJ/mol

The bond enthalpy of N-H can be calculated by using the bond enthalpies of N≡N, H-H, and N≡H:

N≡N(g) + 3H-H(g) → 2N≡H(g)

ΔH = 2BE(N≡H) - BE(N≡N) - 3BE(H-H)

ΔH = 2(941.4 kJ/mol) - 418.4 kJ/mol - 3(436.4 kJ/mol)

ΔH = 1882.8 kJ/mol - 1252.2 kJ/mol

ΔH = 630.6 kJ/mol

The bond enthalpy of N-H is half of this value, since two N≡H bonds are broken and two N-H bonds are formed. Therefore:

BE(N-H) = 1/2 ΔH

BE(N-H) = 1/2 (630.6 kJ/mol)

BE(N-H) = 315.3 kJ/mol

Therefore, the best choice is D.

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

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