Chemistry heat question help? Estimate the enthalpy of formation of hydrazine N2H4 for the following reaction:N2 + 2H2 -----> N2H4Given the bond energies in kJ/mol are:Nitrogen to nitrogen triple bond = 941.4Nitrogen to nitrogen single bond = 393Nitrogen to hydrogen single bond = 393Hydrogen to hydrogen single bond = 436.4  

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The Heat of Formation is equal to the sum of the energy in the bonds of the products minus the sum of the energy in the bonds of the reactants, or

ΔH = Σ ΔHf products - Σ ΔHf reactants

In this case, ΔHf reactants is equal to the bond energies...

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The Heat of Formation is equal to the sum of the energy in the bonds of the products minus the sum of the energy in the bonds of the reactants, or

ΔH = Σ ΔHf products - Σ ΔHf reactants

In this case, ΔHf reactants is equal to the bond energies of the Nitrogen molecule and the two Hydrogen molecules.  The Nitrogen is triple bonded, so

ΔHf reactants = 9.41 kJ/mol + 2(4.36.4 kJ/mol) = 18.13 kj/mol

Similarly, the hydrazine molecule has 4 Nitrogen to Hydrogen bonds, and one Nitrogen to Nitrogen single bond, so

ΔHf products =  3.93 kJ/mol + 4(3.93 kJ/mol) = 19.65 kJ/mol

therefore

ΔH = 19.65 kJ/mol - 18.13 kJ/mol = 1.52 kJ/mol

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