How to calculate the enthalpy of formation of hydrazine from tabulated values?


> 
> 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?




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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Standard Enthalpies of Reaction


The magnitude of $\Delta H\_\mathrm f^\circ$ is the sum of the standard enthalpies of formation of the products, each multiplied by its appropriate coefficient, **minus** the sum of the standard enthalpies of formation of the reactants, also multiplied by their coefficients:


Here is the table of enthalpies required to break bounds for the different molecular assemblies.


\begin{array}{lcccc}\hline
\textbf{Bounds} & \ce{H-H} &\ce{N-H}&\ce{N-N} &\ce{N#N} \\ \hline
 \Delta H\_\mathrm B=-\Delta H\_\mathrm f^\circ & 436 & 388 & 163 & 944 \\\hline
\end{array}


This table gives the amount of energy in $\pu{kJ mol-1}$ required for *breaking chemical bounds*. Thus it is called $-\Delta H\_\mathrm B^\circ$ and means the opposite of *forming chemical bounds* $\Delta H\_\mathrm f^\circ$.


The sign convention for $\Delta H\_\mathrm f^\circ$ is the same as for any enthalpy change: $\Delta H\_\mathrm f^\circ\lt0$ if heat is *released* when elements combine to form a compound and $\Delta H\_\mathrm f^\circ\gt0$ if heat is *absorbed* according to the usual conventions.


* Producing $\ce{N2(g)}$ requires that: given $\ce{2N}$, produce $\ce{N2}$, 1 time, so it means


$$\Delta H\_\mathrm f^\circ[\ce{N2}]=\pu{-944 kJ mol-1}$$


* Producing two $\ce{H2(g)}$ requires that: given $\ce{4 H}$, produce $\ce{H2}$, 2 times, so it means


$$2\Delta H\_\mathrm f^\circ[\ce{H2}]=\pu{-872 kJ mol-1}$$


* Producing $\ce{N2H4(g)}$ from $\ce{N2H4(l)}$ requires absorbing $$\Delta\_\mathrm{vap} H^\circ=\pu{+44.7 kJ mol-1}$$
* Reacting $\ce{N2H4(g)}$ requires that: to produce $\ce{4 H}$ and $\ce{2 N}$, the experimenter has to destroy 4 bounds $\ce{N-H}$ and he has to destroy 1 bound $\ce{N-N}$, so it means


$$\Delta H\_\mathrm f^\circ[\ce{N2H4}]=\pu{-1715 kJ mol-1}$$


Finally, by making the sum of the energy needed to produce hydrogen and nitrogen, **minus** the energy got from the reaction of the hydrazine molecule, we get a standard enthalpy of reaction:


$$\Delta\_\mathrm rH^\circ=\pu{-56.3 kJ mol-1}$$


proving that the decomposition of hydrazine is exothermic.


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