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Calculate enthalpy of formation of hydrazine, N₂H₄ (l) at 25°c and 1 atm pressure. 3N₂H₄ (l) N₂ (g) 4 NH₃ (g) ΔH°ᵣₓₙ = -336 kj/mol ΔH° NH₃ (g) = -46.3 kj/mol

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Final answer:

The enthalpy of the formation of hydrazine can be calculated by constructing a Hess's Law cycle using the given reaction enthalpies for hydrazine and ammonia.

Step-by-step explanation:

The enthalpy of the formation of hydrazine, N₂H₄(l), at 25°C and 1 atm pressure can be calculated using the given reaction:

3N₂H₄(l) → N₂(g) + 4NH₃(g)

The enthalpy change for this reaction is ΔH°ᵣₓₙ = -336 kJ/mol. It is important to note that the enthalpy of formation refers to the formation of 1 mole of the substance, so we need to divide the enthalpy change by the stoichiometric coefficient of N₂H₄ in the reaction, which is 3.

Therefore, the enthalpy of formation of hydrazine, N₂H₄(l), at 25°C and 1 atm pressure is:

ΔH°f = ΔH°ᵣₓₙ / 3 = -336 kJ/mol / 3 = -112 kJ/mol

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