Question

Calculate the enthalpy change (Δ_rH) for the reaction below,

        N₂(g) + 3 F₂(g) → 2 NF₃(g)

given the bond enthalpies of the reactants and products.

	Bond	Bond Enthalpy (kJ/mol×rxn)
	N–N	163
	N=N	418
	N≡N	945
	F–F	155
	N–F	283

Answer 1

In order to use bond energies, all species must be in the gaseous state. That is the case. The enthalpy of a reaction is the sum of the bond energies of the reactants and products. The bond energies of the reactants are positive since energy must be added to break bonds, and the bond energies of the products are negative since energy is released as bonds form.

N2(g) + 3F2(g) --> 2NF3(g)
ΔH = ΣBE
ΔH = N≡N + 3(F−F) + 6(N−F)
ΔH = +945 kJ + 3(+155 kJ) + 6(-283 kJ)
ΔH = -288 kJ