Question

2 NH₃(g) → 3 H₂(g) + N₂(g) ΔH°298 = 92 kJ/molrxn According to the information above, what is the standard enthalpy of formation, ΔH°f , for NH₃(g) at 298 K ?

Answer 1

The standard enthalpy of formation for NH₃(g) at 298 K is -46.1 kJ/mol, determined by reversing the given decomposition reaction and dividing the enthalpy change by two.

Step-by-step explanation:

The standard enthalpy of formation, ΔH°f, can be determined for NH₃(g) using the provided reaction and its enthalpy change. The provided reaction is the decomposition of ammonia into hydrogen and nitrogen:

2 NH₃(g) → 3 H₂(g) + N₂(g) with ΔH°₂₉₈ = 92 kJ/mol

To find the enthalpy of formation of NH₃(g), we consider the reverse reaction, which is the formation of NH₃(g) from its elements in their standard states:

N₂(g) + 3 H₂(g) → 2 NH₃(g) with ΔH° = −92.2 kJ/mol

Since the formation of 2 moles of NH₃(g) releases 92.2 kJ of energy, the formation of 1 mole of NH₃(g) would release half of this energy:

ΔH°f for NH₃(g) = -92.2 kJ / 2 = -46.1 kJ/mol

Answer 2

The standard enthalpy of formation, ΔH°f, for NH₃(g) at 298 K can be calculated using Hess's law and the given reaction: N₂(g) + 3H₂(g) → 2NH₃(g). The ΔH°f for NH₃(g) is -92.2 kJ/mol.

Step-by-step explanation:

The standard enthalpy of formation, ΔH°f, for NH₃(g) at 298 K can be determined by using Hess's law and the given reaction:N₂(g) + 3H₂(g) → 2NH₃(g)

Since the standard enthalpy change, ΔH°, for this reaction is -92.2 kJ, the standard enthalpy of formation, ΔH°f, for NH₃(g) can be calculated by multiplying ΔH° by the stoichiometric coefficient of NH₃:

ΔH°f for NH₃(g) = (2 mol NH₃ × -92.2 kJ/mol) / 2 = -92.2 kJ/mol