Chapter 3, Problem 153CP

(a) (i)

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 153CP

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))

(ii)

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 153CP

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))

(iii)

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 153CP

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.

(b)

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 153CP

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 $\text{CL}-20$.