Question

The enthalpy change for the following reaction is 95.4 kJ. Using bond energies, estimate the N-H bond energy in N2H4(g).

N2(g) + 2H2(g) -----> N2H4 (g)

I need help finding the N-H bond energy, I dont understand how to get the correct answer.

Answer 1

The reaction for the formation of n_2 H_4 (g) is as follows: N_2 (g) = 2 h_2 (g) rightarrow N_2 H_4 (g) The bonds in the reaction and products are follows N equiv N h - H rightarrow The bond enthapies (BF) are as follows: N = n 945 kJ/mol N-N 163 kJ/mol H-H 436 kJ/mol N-H ? calculate the bond energy of N-H as follows: Delta h = sigma BE_bondsformed - sigma BE_bondsformed = 2BE_H-H + BE_N = N - $BE_N - H - BE_N - N Rearrange the above to get BE_N - H BE_N-H = Delta H - 2BE_H-H -BE_N = N + BE_N_N/4 = 95.4 KJ/mol -(2 mol times 436 kJ/mol)-945 kJ/mol + 163 kJ/mol/4 mol = 389.7 kJ/mol Therefore the bond energy of N-H bond is 389.7 kJ/mol