Answer:
The answer is explained below:
Explanation:
Given the chemical equation:
N2H4(g)+H2(g)→2NH3(g)
The standard enthalpy of formation is given by the following formula:
ΔH^0 rxn = ∑ B reactants - ∑ product
-187.78 kJ/mol = [( 1x B N-N) + (1 x B N-H ) + (1 x b H-H)] - [ 6 x B N-H]
-187.78 kJ/mol = [( 1x B N-N) +(4 x 391 kJ/mol) + (1 x 436 kJ/mol)] - [ 6x 391 kJ/mol ]
-187.78 kJ/mol = B N-N + (1564 + 436 - 2346) kJ/mol
B = 158.22 kJ/mol
So, in this case the enthalpy of N-N bond is 158 kJ/mol
The enthalpy of the nitrogen-nitrogen bond in N2H4 should be considered as the 158 kJ/mol.
The following formula should be used
ΔH^0 rxn = ∑ B reactants - ∑ product
-187.78 kJ/mol = [( 1x B N-N) + (1 x B N-H ) + (1 x b H-H)] - [ 6 x B N-H]
-187.78 kJ/mol = [( 1x B N-N) +(4 x 391 kJ/mol) + (1 x 436 kJ/mol)] - [ 6x 391 kJ/mol ]
-187.78 kJ/mol = B N-N + (1564 + 436 - 2346) kJ/mol
B = 158.22 kJ/mol
So, the enthalpy of N-N bond is 158 kJ/mol.
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