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5 January, 02:45

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.

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  1. 5 January, 02:50
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    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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