How to calculate the enthalpy of formation of hydrazine from tabulated values? 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}) &amp;= 436~\mathrm{kJ/mol}\\
\Delta{}H_\mathrm{B}(\ce{N2}) &amp;= 944~\mathrm{kJ/mol}\\
\Delta{}H_\mathrm{B}(\ce{N–N}) &amp;= 163~\mathrm{kJ/mol}\\
\Delta{}H_\mathrm{B}(\ce{N–H}) &amp;= 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?
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.