Question

The compound hexaazaisowurtzitane is one of the highest-energy explosives known. The compound, also known as $\mathbf{CL}\mathbf{-}\mathbf{20}$, was first synthesized in 1987. The method of synthesis and detailed performance data are still classified information because of $\mathbf{CL}\mathbf{-}\mathbf{20}\mathbf{\text{'}}\mathbf{s}$potential military application in rocket boosters and in warheads of “smart” weapons. The structure of $\mathbf{CL}\mathbf{-}\mathbf{20}$, is

In such shorthand structures, each point where lines meet represents a carbon atom. In addition, the hydrogens attached to the carbon atoms are omitted. Each of the six carbon atoms has one hydrogen atom attached. Three possible reactions for the explosive decomposition of$\mathbf{CL}\mathbf{-}\mathbf{20}$, are

1. ${\mathbf{C}}_{\mathbf{6}}{\mathbf{H}}_{\mathbf{6}}{\mathbf{N}}_{\mathbf{12}}{\mathbf{O}}_{\mathbf{12}}\mathbf{\left(}\mathbf{s}\mathbf{\right)}\mathbf{\to }\mathbf{6}\mathbf{CO}\mathbf{\left(}\mathbf{g}\mathbf{\right)}\mathbf{+}\mathbf{3}{\mathbf{H}}_{\mathbf{2}}\mathbf{O}\mathbf{\left(}\mathbf{g}\mathbf{\right)}\mathbf{+}\frac{\mathbf{3}}{\mathbf{2}}{\mathbf{O}}_{\mathbf{2}}\mathbf{\left(}\mathbf{g}\mathbf{\right)}$
2. ${\mathbf{C}}_{\mathbf{6}}{\mathbf{H}}_{\mathbf{6}}{\mathbf{N}}_{\mathbf{12}}{\mathbf{O}}_{\mathbf{12}}\mathbf{\left(}\mathbf{s}\mathbf{\right)}\mathbf{\to }\mathbf{3}\mathbf{CO}\mathbf{\left(}\mathbf{g}\mathbf{\right)}\mathbf{+}\mathbf{3}\mathbf{CO}\mathbf{\left(}\mathbf{g}\mathbf{\right)}\mathbf{+}\mathbf{6}{\mathbf{N}}_{\mathbf{2}}\mathbf{\left(}\mathbf{g}\mathbf{\right)}\mathbf{+}\mathbf{3}{\mathbf{H}}_{\mathbf{2}}\mathbf{O}\mathbf{\left(}\mathbf{g}\mathbf{\right)}$
   
3. ${\mathbf{C}}_{\mathbf{6}}{\mathbf{H}}_{\mathbf{6}}{\mathbf{N}}_{\mathbf{12}}{\mathbf{O}}_{\mathbf{12}}\mathbf{\left(}\mathbf{s}\mathbf{\right)}\mathbf{\to }\mathbf{6}{\mathbf{CO}}_{\mathbf{2}}\mathbf{\left(}\mathbf{g}\mathbf{\right)}\mathbf{+}\mathbf{6}{\mathbf{N}}_{\mathbf{2}}\mathbf{\left(}\mathbf{g}\mathbf{\right)}\mathbf{+}\mathbf{3}{\mathbf{H}}_{\mathbf{2}}\mathbf{\left(}\mathbf{g}\mathbf{\right)}$

   
   

a. Use bond energies to estimate $∆H$for these three reactions.

b. Which of the three reactions releases the largest amount of energy per kilogram of $\mathbf{CL}\mathbf{-}\mathbf{20}\mathbf{,}$?

Short answer

a. $∆\mathrm{H}$for the given three reactions

1. - 2635.5 kJ
2. - 3471 kJ
3. - 3543 kJ

b. The reaction (iii) releases the largest amount of energy per kilogram of CL- 20 that is -8085kJ/ kg.

Step-by-step solution

Estimate ∆H  for the reaction given using bond energies.

Determining $∆$H for given three reactants:

localid="1663770557362" $$\begin{gathered} 3\mathrm{C}-\mathrm{C}\to 3\times 347=1041\mathrm{kJ} \\ 6\mathrm{C}-\mathrm{H}\to 6\times 313=2478\mathrm{kJ} \\ 12\mathrm{C}-\mathrm{N}\to 12\times 305=3660\mathrm{kJ} \\ 6\mathrm{N}-\mathrm{N}\to 6\times 160=960\mathrm{kJ} \\ 6\mathrm{N}-\mathrm{O}\to 6\times 201=1206\mathrm{kJ} \\ 6\mathrm{N}=\mathrm{O}\to 6\times 607=3642\mathrm{kJ} \\ \mathrm{So},∆{\mathrm{H}}_{\mathrm{reactants}}=12,987\mathrm{kJ} \end{gathered}$$

Estimate ∆H  for the reaction (i) given using bond energies.

1. $C_6{H}_6{N}_{12}{O}_{12}\left(s\right)\to 6CO\left(g\right)+3H_{2}O\left(g\right)+\frac{3}{2}{O}_2\left(g\right)$

Determining $∆H$for products of above reaction

$$\begin{gathered} 6C\equiv 0\to 6\times 1072kJ=6432kJ \\ 6N\equiv N\to 6\times 941kJ=5646kJ \\ 6O‐H\to 6\times 467kJ=2802kJ \\ \frac{3}{2}O=O\to \frac{3}{2}\times 495kJ=742.5kJ \end{gathered}$$

So, $∆H_{products}=15,622.5kJ$

Now, $∆H_{reaction}=12,987-15,622.5=-2635.5kJ$

Estimate ∆H  for the reaction (ii) given using bond energies.

$$\begin{gathered} \mathrm{ii}.{\mathrm{C}}_6{\mathrm{H}}_6{\mathrm{N}}_{12}{\mathrm{O}}_{12}\left(\mathrm{S}\right)\to 3\mathrm{CO}\left(\mathrm{g}\right)+3\mathrm{CO}\left(\mathrm{g}\right)+6{\mathrm{N}}_2\left(\mathrm{g}\right)+3{\mathrm{H}}_2\mathrm{O}\left(\mathrm{g}\right) \\ \mathrm{Determining}∆H\mathrm{for}\mathrm{products}\mathrm{of}\mathrm{above}\mathrm{reaction} \\ 3\mathrm{C}\equiv \mathrm{O}\to 3\times 1072\mathrm{kJ}=3216\mathrm{kJ} \\ 6\mathrm{N}\equiv \mathrm{N}\to 6\times 941\mathrm{kJ}=5646\mathrm{kJ} \\ 6\mathrm{O}-\mathrm{H}\to 6\times 467\mathrm{kJ}=2802\mathrm{kJ} \\ 6\mathrm{C}=\mathrm{O}\to 6\times 799\mathrm{kJ}=4794\mathrm{kJ} \\ \mathrm{So},∆{\mathrm{H}}_{\mathrm{products}}=16,458\mathrm{kJ} \\ \mathrm{Now},∆{\mathrm{H}}_{\mathrm{reaction}}=12,987-16,458=-3471\mathrm{kJ} \end{gathered}$$

Estimate ∆H for the reaction (iii) given using bond energies.

$$\begin{gathered} \mathrm{iii}.{\mathrm{C}}_6{\mathrm{H}}_6{\mathrm{N}}_{12}{\mathrm{O}}_{12}\left(\mathrm{S}\right)\to 6{\mathrm{CO}}_2\left(\mathrm{g}\right)+6{\mathrm{N}}_2\left(\mathrm{g}\right)+3{\mathrm{H}}_2\left(\mathrm{g}\right) \\ \mathrm{Determining}∆H\mathrm{for}\mathrm{products}\mathrm{of}\mathrm{above}\mathrm{reaction} \\ 6\mathrm{N}\equiv \mathrm{N}\to 6\times 941\mathrm{kJ}=5646\mathrm{kJ} \\ 3\mathrm{H}-\mathrm{H}\to 3\times 432\mathrm{kJ}=1296\mathrm{kJ} \\ 12\mathrm{C}=\mathrm{O}\to 12\times 799\mathrm{kJ}=9588\mathrm{kJ} \\ \mathrm{So},∆{\mathrm{H}}_{\mathrm{products}}=16,530\mathrm{kJ} \\ \mathrm{Now},∆{\mathrm{H}}_{\mathrm{reaction}}=12,987-16,530=-3543\mathrm{kJ} \end{gathered}$$

Determine which of the three reactions releases the largest amount of energy per kilogram of CL-20.

From the above step we can see that reaction

$C_6{\mathrm{H}}_6{\mathrm{N}}_{12}{\mathrm{O}}_{12}\left(\mathrm{S}\right)\to 6{\mathrm{CO}}_2\left(\mathrm{g}\right)+6{\mathrm{N}}_2\left(\mathrm{g}\right)+3{\mathrm{H}}_2\left(\mathrm{g}\right)$

releases the largest amount of energy per kilogram of CL-20 .

So, the energy released per kilogram of CL-20 is:

$\frac{-3543\mathrm{kJ}}{438.23\mathrm{g}}\times \frac{1\mathrm{mol}}{1\mathrm{mol}}\times \frac{1000\mathrm{g}}{1\mathrm{kg}}=-8085\mathrm{kJ}/\mathrm{kg}$