Question

Calculate the avg. N-H bond energy in NH₃(g). Delta Hf for NH₃(g) = -46.11kJ/mol, Delta Hf for N(g) = 472.7kJ/mol, Delta Hf for H(g) = 218kJ/mol, Delta Hf for N₂(g) = 0kj/mol, Delta Hf for H₂(g) = 0kJ/mol.

a) -46.11kJ
b) -15.4kJ
c) -390.9kJ
d) 264.5kJ
e) 1173kJ

Answer 1

To find the average N-H bond energy, the enthalpy change for the formation of 2 moles of NH3(g) is divided by 6, yielding 15.37 kJ/mol. This indicates that each N-H bond has an average bond energy of -15.4 kJ/mol, aligning with option (b).

Step-by-step explanation:

To calculate the average N-H bond energy in NH3(g), we need to consider the standard enthalpy of formation ΔHf for the compounds and elements involved in creating ammonia. The reaction to form ammonia from its elements in their standard states is N2(g) + 3H2(g) → 2NH3(g). ΔHf for NH3(g) is given as -46.11 kJ/mol, and ΔHf for the elements in their standard states, N2(g) and H2(g), is 0 kJ/mol. We use the bond energy concept which states that the sum of all bond energies in a molecule equals the enthalpy change for breaking all the bonds.

To find the average N-H bond energy, we first calculate the enthalpy change of the reaction:

• ΔHrxn = [2 mol NH3 × ΔHf (NH3)] – [ΔHf (N2) + 3 × ΔHf (H2)]
• Substituting the given values: ΔHrxn = [2 mol × (-46.11 kJ/mol)] – [0 kJ + 3 × 0 kJ] = -92.22 kJ

Since 6 N-H bonds are formed in producing 2 moles of NH3, the average N-H bond energy is the enthalpy change of the reaction divided by 6.

1. Average N-H bond energy = -ΔHrxn / number of N-H bonds formed
2. Substituting the values: Average N-H bond energy = -(-92.22 kJ) / 6 = 15.37 kJ/mol

Therefore, the average N-H bond energy in NH3 is 15.37 kJ/mol, which closely matches to option (b) -15.4 kJ/mol.