Consider the reaction between hydrazine and hydrogen to produce ammonia, n2h4(g)+h2(g)→2nh3(g). use enthalpies of formation and bond enthalpies to estimate the enthalpy of the nitrogen-nitrogen bond in n2h4.

The enthalpy of the nitrogen-nitrogen bond in N2H4 is 162.6 kJ

Explanation:

For the reaction: N2H4(g)+H2(g)→2NH3(g), the enthalpy change of reaction is

ΔH rxn = 2 ΔHºf NH3 - ΔHºf N2H4

but we also know that the ΔH rxn is calculated by accounting   the sum of number of bonds formed and bonds broken as follows:

ΔH rxn = 6H (N-H) + 4 (N-H) + 2H (H-H)  

where H is the bond enthalpy .When bonds are broken H is positive, and negative when formed,  in the product there are 6 N-H bonds , and in the reactants 4 N-H and 1 H-H bonds).

Consulting an appropiate reference handbook or table the following values are used:

ΔHºf (NH3) = -46 kJ/mol

ΔHºf (N2H4) = 95.94 kJ/mol

(The enthalpy of fomation of hydrogen in its standard state is zero)

H (N-H) = 391 kJ

H (H-H) = 432 kJ

H (N-N) = ?

So plugging our values:

ΔH rxn =  2mol ( -46.0 kJ/mol) - 1mol(95.4 kJ/mol) = -187.40 kJ

-187.40 kJ = 6(-391 kJ) + 4 (391 kJ) + 432 +  H(N-N)

-187.40 kJ = -350 kJ + H(N-N)

H(N-N) = 162.6 kJ

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explanation:

they are all solids.