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?


2 Answers 2


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.


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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