Chapter 8, Problem 155CP

The compound hexaazaisowurtzitane is one of the highest-energy explosives known (C & E News, Jan. 17, 1994, p. 26). The compound, also known as CL-20, was first synthesized in 1987. The method of synthesis and detailed performance data are still classified because of CL-20’s potential military application in rocket boosters and in warheads of “smart” weapons. The structure of CL-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. Finally, assume that the two O atoms in the NO₂ groups are attached to N with one single bond and one double bond.

Three possible reactions for the explosive decomposition ofCL-20 are

i. ${\text{C}}_{\text{6}}{\text{H}}_{\text{6}}{\text{N}}_{\text{12}}{\text{O}}_{\text{12}}\left(s\right)\to \text{6CO}\left(g\right)+6{\text{N}}_{\text{2}}\left(g\right)+{\text{3H}}_{\text{2}}\text{O}\left(g\right)+\frac{3}{2}{\text{O}}_2\left(g\right)$

ii. C₆H₆N₁₂O₁₂(s) → 3CO(g) + 3CO₂(g) + 6N₂(g) + 3H₂O(g)

iii. C₆H₆N₁₂O₁₂(s) → 6CO₂(g) + 6N₂(g) + 3H₂ (g)

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

b. Which of the above reactions releases the largest amount of energy per kilogram of CL-20?

Interpretation Introduction

Interpretation: The change in energy for the given chemical reactions has to be calculated.

Concept introduction: In a chemical reaction, energy is either gained, endothermic reactions, or released, exothermic reactions. The change in energy can be stated as the difference between the energy required to break the bonds in case of reactants and the energy released on the formation of the products.

To determine: The change in energy for the stated reactions.

Answer to Problem 155CP

The change in energy $=\underset{\_}{-2635.5kJ}$

Explanation of Solution

Given

The chemical reaction involved is,

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

Formula

$\text{The change in energy }=\left(\begin{array}{l}\text{Energy required to break}\\ \text{the bonds in reactants}\end{array}\right)–\left(\begin{array}{l}\text{Energy released when}\\ \text{products}\text{are}\text{formed}\end{array}\right)$

Energy for reactants,

$6\text{C}-\text{H}\text{=}\frac{413\text{kJ}}{1\text{mol}}\times 6\text{mol=2478}\text{kJ}$

$6\text{N}=\text{O}\text{=}\frac{607\text{kJ}}{1\text{mol}}\times 6\text{mol=3642}\text{kJ}$

$3\text{C}-\text{C}\text{=}\frac{347\text{kJ}}{1\text{mol}}\times 3\text{mol=1041}\text{kJ}$

$12\text{C}-\text{N}\text{=}\frac{305\text{kJ}}{1\text{mol}}\times 12\text{mol}=3660\text{kJ}$

$6\text{N}-\text{N}\text{=}\frac{160\text{kJ}}{1\text{mol}}\times 6\text{mol=960}\text{kJ}$

The total energy $=\left(3660+1041+960+3642+2478\right)\text{kJ}\text{=}\text{12987}\text{kJ}$ (1)

For products,

$6\text{C}\equiv \text{O}\text{=}\frac{1072\text{kJ}}{1\text{mol}}\times 6\text{mol=6432}\text{kJ}$

$6\text{N}\equiv \text{N}\text{=}\frac{941\text{kJ}}{1\text{mol}}\times 6\text{mol=5646}\text{kJ}$

$6\text{O}-\text{H}\text{=}\frac{467\text{kJ}}{1\text{mol}}\times 6\text{mol=2802}\text{kJ}$

$\frac{3}{2}\text{O}=\text{O}\text{=}\frac{495\text{kJ}}{1\text{mol}}\times \frac{3}{2}\text{mol}=742.5\text{kJ}$

The total energy $=\left(6432+5646+2802+742.5\right)\text{kJ}\text{=}\text{15622}\text{.5}\text{kJ}$ (2)

The change in energy $=\left(12987-15622.5\right)kJ=\underset{\_}{-2635.5kJ}$ (from equation (1) and (2))

Interpretation Introduction

Interpretation: The change in energy for the given chemical reactions has to be calculated.

Concept introduction: In a chemical reaction, energy is either gained, endothermic reactions, or released, exothermic reactions. The change in energy can be stated as the difference between the energy required to break the bonds in case of reactants and the energy released on the formation of the products.

To determine: The change in energy for the stated reactions.

Answer to Problem 155CP

The change in energy $=\underset{\_}{-3147kJ}$

Explanation of Solution

Given

The chemical reaction involved is,

${\text{C}}_6{\text{H}}_6{\text{N}}_{12}{\text{O}}_{12}\left(\text{s}\right)\to 3\text{CO}\left(\text{g}\right)+3{\text{CO}}_2\left(\text{g}\right)+6{\text{N}}_2\left(\text{g}\right)+3{\text{H}}_2\text{O}\left(\text{g}\right)$

Formula

$\text{The change in energy }=\left(\begin{array}{l}\text{Energy required to break}\\ \text{the bonds in reactants}\end{array}\right)–\left(\begin{array}{l}\text{Energy released when}\\ \text{products}\text{are}\text{formed}\end{array}\right)$

Energy for reactants,

$6\text{C}-\text{H}\text{=}\frac{413\text{kJ}}{1\text{mol}}\times 6\text{mol=2478}\text{kJ}$

$6\text{N}=\text{O}\text{=}\frac{607\text{kJ}}{1\text{mol}}\times 6\text{mol=3642}\text{kJ}$

$3\text{C}-\text{C}\text{=}\frac{347\text{kJ}}{1\text{mol}}\times 3\text{mol=1041}\text{kJ}$

$12\text{C}-\text{N}\text{=}\frac{305\text{kJ}}{1\text{mol}}\times 12\text{mol}=3660\text{kJ}$

$6\text{N}-\text{N}\text{=}\frac{160\text{kJ}}{1\text{mol}}\times 6\text{mol=960}\text{kJ}$

The total energy $=\left(3660+1041+960+3642+2478\right)\text{kJ}\text{=}\text{12987}\text{kJ}$ (1)

For products,

$3\text{C}\equiv \text{O}\text{=}\frac{1072\text{kJ}}{1\text{mol}}\times 3\text{mol=3216}\text{kJ}$

$6\text{C}=\text{O}\text{=}\frac{745\text{kJ}}{1\text{mol}}\times 6\text{mol=4470}\text{kJ}$

$6\text{N}-\text{N}\text{=}\frac{941\text{kJ}}{1\text{mol}}\times 6\text{mol=5646}\text{kJ}$

$6\text{O}-\text{H}\text{=}\frac{467\text{kJ}}{1\text{mol}}\times 6\text{mol}=2802\text{kJ}$

The total energy $=\left(3216+4470+5646+2802\right)\text{kJ}\text{=}\text{16134}\text{kJ}$ (2)

The change in energy $=\left(12987-16134\right)kJ=\underset{\_}{-3147kJ}$ (from equation (1) and (2))

Interpretation Introduction

Interpretation: The change in energy for the given chemical reactions has to be calculated.

Concept introduction: In a chemical reaction, energy is either gained, endothermic reactions, or released, exothermic reactions. The change in energy can be stated as the difference between the energy required to break the bonds in case of reactants and the energy released on the formation of the products.

To determine: The change in energy for the stated reactions.

Answer to Problem 155CP

The change in energy $=\underset{\_}{-4191kJ}$

Explanation of Solution

Given

The chemical reaction involved is,

${\text{C}}_6{\text{H}}_6{\text{N}}_{12}{\text{O}}_{12}\left(\text{s}\right)\to 6{\text{CO}}_2\left(\text{g}\right)+6{\text{N}}_2\left(\text{g}\right)+6{\text{H}}_2\left(\text{g}\right)$

Formula

$\text{The change in energy }=\left(\begin{array}{l}\text{Energy required to break}\\ \text{the bonds in reactants}\end{array}\right)–\left(\begin{array}{l}\text{Energy released when}\\ \text{products}\text{are}\text{formed}\end{array}\right)$

Energy for reactants,

$6\text{C}-\text{H}\text{=}\frac{413\text{kJ}}{1\text{mol}}\times 6\text{mol=2478}\text{kJ}$

$6\text{N}=\text{O}\text{=}\frac{607\text{kJ}}{1\text{mol}}\times 6\text{mol=3642}\text{kJ}$

$3\text{C}-\text{C}\text{=}\frac{347\text{kJ}}{1\text{mol}}\times 3\text{mol=1041}\text{kJ}$

$12\text{C}-\text{N}\text{=}\frac{305\text{kJ}}{1\text{mol}}\times 12\text{mol}=3660\text{kJ}$

$6\text{N}-\text{N}\text{=}\frac{160\text{kJ}}{1\text{mol}}\times 6\text{mol=960}\text{kJ}$

The total energy $=\left(3660+1041+960+3642+2478\right)\text{kJ}\text{=}\text{12987}\text{kJ}$ (1)

For products,

$12\text{C}=\text{O}\text{=}\frac{745\text{kJ}}{1\text{mol}}\times 12\text{mol=8940}\text{kJ}$

$6\text{N}-\text{N}\text{=}\frac{941\text{kJ}}{1\text{mol}}\times 6\text{mol=5646}\text{kJ}$

$6\text{H}-\text{H}\text{=}\frac{432\text{kJ}}{1\text{mol}}\times 6\text{mol}=2592\text{kJ}$

The total energy $=\left(8940+5646+2592\right)\text{kJ}\text{=}\text{17178}\text{kJ}$ (2)

The change in energy $=\left(12987-17178\right)kJ=\underset{\_}{-4191kJ}$ (from equation (1) and (2))

Conclusion

The change in energy can be stated as the difference between the energy required to break the bonds in case of reactants and the energy released on the formation of the products.

Interpretation Introduction

Interpretation: The change in energy for the given chemical reactions has to be calculated.

Concept introduction: In a chemical reaction, energy is either gained, endothermic reactions, or released, exothermic reactions. The change in energy can be stated as the difference between the energy required to break the bonds in case of reactants and the energy released on the formation of the products.

To determine: The reaction that releases the larger amount of energy per kilogram of $\text{CL}-20$.

Answer to Problem 155CP

The reaction (iii) releases the largest amount of energy per kilogram of $\text{CL}-20$.

Explanation of Solution

One mole of ${\text{C}}_6{\text{H}}_6{\text{N}}_{12}{\text{O}}_{12}$ gives $438\text{g}$.

In case of the (i) reaction,

$438\text{g}$ of the reactant gives energy $=-2635.5\text{kJ}$

Hence, $1\text{kg}$ of the reactant gives energy $=\frac{-2635.5}{438}\times 1000\text{kJ}\text{=}\text{-6017}\text{.12}\text{kJ}$

In case of the (ii) reaction,

$438\text{g}$ of the reactant gives energy $=-3147\text{kJ}$

Hence, $1\text{kg}$ of the reactant gives energy $=\frac{-3147}{438}\times 1000\text{kJ}\text{=}\text{-7184}\text{.9}\text{kJ}$

In case of the (iii) reaction,

$438\text{g}$ of the reactant gives energy $=-4191\text{kJ}$

Hence, $1\text{kg}$ of the reactant gives energy $=\frac{-4191}{438}\times 1000\text{kJ}\text{=}\underset{\_}{-9568.49kJ}$

The reaction (iii) releases the largest amount of energy per kilogram of $\text{CL}-20$.

Conclusion

The third stated reaction releases the largest amount of energy per kilogram of $\text{CL}-20$.