Estimate the enthalpy of formation for the following reaction using bond dissociation energies. Indicate if the reaction is endothermic or exothermic (circle answer) ? N2 (g) + H2 (g) à N2H4 (g) BE (N N) = 946 kJ/mol BE (H - H) = 436 kJ/mol BE (N - H) = 389 kJ/mol BE (N - N) = 163 kJ/mol

337 kJ

It's an endothermic reaction.

Explanation:

The bond energy is endothermic, so the break must be exothermic, then the enthalpy change can be calculated by the difference between the bond energy of the products and the reactants (the minus signal comes from that the bond will be a break in the reactants). Thus,

ΔH = ∑n*H products - ∑n*H reactants, where n is the number of bonds.

The reaction given is:

N₂ (g) + H₂ (g) → N₂H₄ (g)

In N₂ there is 1 N≡N bond; in H₂ there is 1 H-H bond, and in N₂H₄ there is 1 N-N bond and 4 N-H bonds.

ΔH = (163 + 4*389) - (946 + 436)

ΔH = 337 kJ

Because ΔH > 0, the reaction is absorbing heat, and it's endothermic.