I've been given a problem asking:

Estimate the standard enthalpy of formation of hydrazine, $\ce{N2H4(g)}$, from the following data. \begin{align} \Delta{}H_\mathrm{B}(\ce{H2}) &= 436~\mathrm{kJ/mol}\\ \Delta{}H_\mathrm{B}(\ce{N2}) &= 944~\mathrm{kJ/mol}\\ \Delta{}H_\mathrm{B}(\ce{N–N}) &= 163~\mathrm{kJ/mol}\\ \Delta{}H_\mathrm{B}(\ce{N–H}) &= 388~\mathrm{kJ/mol}\\ \end{align}

While I know that $\Delta{}H_\mathrm{B}(\ce{N–N})$ refers to the energy released from breaking that bond, I'm not sure what $\Delta{}H_\mathrm{B}(\ce{H2})$ or $\Delta{}H_\mathrm{B}(\ce{N2})$ refers to. Could someone please help me understand?

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    $\begingroup$ The latter are the heats of formation of hydrogen and nitrogen. Do you know how to use them together with the other data? $\endgroup$ – Klaus-Dieter Warzecha Mar 17 '15 at 5:10

In order to "make" hydrazine, $\ce{H2N-NH2}$ from hydrogen and nitrogen atoms, you have to

  • break $\ce{N2}$ once, giving 2 nitrogen atoms
  • break $\ce{H2}$ twice, giving 4 hydrogen atoms
  • form one $\ce{N-N}$ bond
  • form four $\ce{N-H}$ bonds

You have all the enthalpies that you need. Just pay attention to the coefficients and the signs.

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