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?

  • 1
    $\begingroup$ 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.


Standard Enthalpies of Reaction

The magnitude of $\Delta H^o_f$ 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.

Bounds $H-H$ $N-H$ $N-N$ $N \equiv N$
$\Delta H_B=-\Delta H^{\circ}_f$ 436 388 163 944

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

The sign convention for $\Delta H^{\circ}_f$ is the same as for any enthalpy change: $\Delta H^o_f < 0 $ if heat is released when elements combine to form a compound and $\Delta H^{\circ}_f > 0$ if heat is absorbed according to the usual conventions.

  • Producing $N_2(g)$ requires that: given 2 $N$, produce $N_2$, 1 time, so it means $$\Delta H^{\circ}_{f}[N_2]=-944\,kJ.mol^{-1}$$

  • Producing two $H_2(g)$ requires that: given 4 $H$, produce $H_2$, 2 times, so it means $$2\Delta H^{\circ}_{f}[H_2]=-872\,kJ.mol^{-1}$$

  • Producing $N_2H_4(g)$ from $N_2H_4(l)$ requires absorbing $\Delta_{vap} H^{\circ} =\, + \, 44.7\, kJ.mol^{-1}$

  • Reacting $N_2H_4(g)$ requires that: to produce 4 $H$ and 2 $N$, the experimenter has to destroy 4 bounds $N-H$ and he has to destroy 1 bound $N-N$, so it means $$\Delta H^{\circ}_{f}[H_4N_2]=-1715\,kJ.mol^{-1}$$

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: $$\Delta_r H^{\circ} = -56.3\,kJ.mol^{-1}$$ proving that the decomposition of hydrazine is exothermic.


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