Question

033. For the reaction
N2H4(9)-->N_H2(g) + H2(g);
AH° = 109 kJ/mole
Calculate the bond enthalpy of N = N.
Given : B.E. of (N-N) = 163 kJ/mol
B.E. of (N - H) = 391 kJ/mol
B.E. of (H - H) = 436 kJ/mol
(1) 182 kJ/mol
(2400 kJ/mol
(3) 300 kJ/mol
(4) 218 kJ/mol​

Answer 1

Answer:

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