Question

For the reaction N₂(g)+3H₂(g)→2NH₃(g),with ∆H₂₉₈= -92.2kj/mol, calculate the average bond enthalpy of N-N.

a) -92.2 kJ/mol
b) -184.4 kJ/mol
c) -46.1 kJ/mol
d) -23.05 kJ/mol

Answer 1

The provided information is insufficient to calculate the average bond enthalpy of the N-N bond since bond enthalpies for H₂ and NH₃ are also needed.

Step-by-step explanation:

The question involves calculating the average bond enthalpy of the N-N bond in the reaction N₂(g) + 3H₂(g) → 2NH₃(g). Given that the enthalpy change (ΔH) for the formation of ammonia (NH₃) from nitrogen (N₂) and hydrogen (H₂) is -92.2 kJ/mol, we can deduce the average bond enthalpy of the N-N bond. However, the provided information is insufficient to directly calculate the bond enthalpy of N-N as it requires the bond enthalpies of all reactants and products. Typically, to find the average bond enthalpy, one would use the bond enthalpies for each type of bond broken and formed throughout the reaction.

Since the question instructs to ignore typos or irrelevant parts, and the provided statements do not give enough data to perform the calculation, we cannot provide an exact answer from the given choices without additional bond enthalpy values for the H₂ molecules and the N-H bonds in NH₃. The student would need additional data regarding bond enthalpies from a comprehensive data source to complete this calculation.