Chapter 3, Problem 152CP

Three processes that have been used for the industrial manufacture of acrylonitrile (CH₂CHCN), an important chemical used in the manufacture of plastics, synthetic rubber, and fibers, are shown below. Use bond energy values (Table 3-3) to estimate ∆E for each of the reactions.

a.

b.

The nitrogen-oxygen bond energy in nitric oxide (NO) is 630. kJ/mol.

c.

Interpretation Introduction

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

Concept introduction: In a chemical reaction, energy is 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 152CP

The required energy change is $\underset{\_}{-43kJ}$ and $\underset{\_}{37kJ}$.

Explanation of Solution

Given

The chemical reaction involved is,

Figure 1

Formula

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

In the first reaction,

Figure 2

For first reactant,

$4\text{C}-\text{H}\text{=}\frac{413\text{kJ}}{1\text{mol}}\times 4\text{mol=1652}\text{kJ}$

$2\text{C}-\text{O}\text{=}\frac{358\text{kJ}}{1\text{mol}}\times 2\text{mol=716}\text{kJ}$

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

Hence, the total energy required $=\left(1652+716+34\right)\text{kJ}=2715\text{kJ}$

For $\text{HC}\equiv \text{N}$,

$1\text{C}\equiv \text{N}\text{=}\frac{891\text{kJ}}{1\text{mol}}\times 1\text{mol=891}\text{kJ}$

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

Hence, the total energy required $=\left(891+431\right)\text{kJ}=1304\text{kJ}$ (1)

Now, the total energy required for the reactants combined $=\left(2715+1304\right)\text{kJ}\text{=}4019\text{kJ}$.

Product bonds,

$4\text{C}-\text{H}\text{=}\frac{413\text{kJ}}{1\text{mol}}\times 4\text{mol=1652}\text{kJ}$

$1\text{C}-\text{O}\text{=}\frac{358\text{kJ}}{1\text{mol}}\times 1\text{mol=358}\text{kJ}$

$2\text{C}-\text{C}\text{=}\frac{347\text{kJ}}{1\text{mol}}\times 2\text{mol=694}\text{kJ}$

$1\text{C}\equiv \text{N}\text{=}891\text{kJ}$

$1\text{O}-\text{H}\text{=}467\text{kJ}$

Hence,

The total energy released when the product is formed $=\left(1652+358+694+891+467\right)\text{kJ}=4062\text{kJ}$ (2)

So the change in energy for the first reaction is,

$\Delta {\rm H}=\left(4019-4062\right)kJ=\underset{\_}{-43kJ}$ (Using equation (1) and (2)) (3)

In the second reaction,

Figure 3

For the reactant,

$4\text{C}-\text{H}\text{=}\frac{413\text{kJ}}{1\text{mol}}\times 4\text{mol=1652}\text{kJ}$

$1\text{C}-\text{O}\text{=}\frac{358\text{kJ}}{1\text{mol}}\times 1\text{mol=358}\text{kJ}$

$2\text{C}-\text{C}\text{=}\frac{347\text{kJ}}{1\text{mol}}\times 2\text{mol=694}\text{kJ}$

$1\text{C}\equiv \text{N}\text{=}891\text{kJ}$

$1\text{O}-\text{H}\text{=}467\text{kJ}$

Hence,

The total energy released when the product is formed $=\left(1652+358+694+891+467\right)\text{kJ}=4062\text{kJ}$ (4)

For product,

$3\text{C}-\text{H}\text{=}\frac{413\text{kJ}}{1\text{mol}}\times 3\text{mol=1239}\text{kJ}$

$1\text{C}=\text{C}\text{=}\frac{614\text{kJ}}{1\text{mol}}\times 1\text{mol=614}\text{kJ}$

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

$1\text{C}\equiv \text{N}\text{=}891\text{kJ}$

Hence,

The total energy released when the product is formed $=\left(1239+614+891+347\right)\text{kJ}\text{=}\text{3091}\text{kJ}$ (5)

For water,

$2\text{O}-\text{H}\text{=}\text{2}\text{mol}\times \text{467}\text{kJ/mol}\text{=}\text{934}\text{kJ}$

So, the total energy of products $=\left(3091+934\right)\text{kJ}\text{=}\text{4025}\text{kJ}$ (6)

So the change in energy for the second reaction is,

$\Delta {\rm H}=\left(4062-4025\right)kJ=\underset{\_}{37kJ}$ (Using equation (4) and (6))

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 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 152CP

The required energy change is $\underset{\_}{-850kJ}$.

Explanation of Solution

Given

The given reaction is,

Figure 4

For the reactant side,

For ${\text{H}}_2\text{C}={\text{CHCH}}_3$,

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

$1\text{C}=\text{C}\text{=}\frac{614\text{kJ}}{1\text{mol}}\times 1\text{mol=614}\text{kJ}$

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

The energy required $=4\times \left(2478+614+347\right)\text{kJ}\text{=}13756\text{kJ}$ (since $4$ molecules are present) (1)

For $\text{NO}$,

$6\text{NO}\text{=}\left(\text{6}\times \text{630}\right)\text{kJ}\text{=}\text{3780}\text{kJ}$ (2)

Total reactant energy $=13756+3780\text{kJ}=17536\text{kJ}$ (using equation (1) and (2)) (3)

For products,

For ${\text{H}}_2\text{C}=\text{CHCN}$,

$3\text{C}-\text{H}\text{=}\frac{413\text{kJ}}{1\text{mol}}\times 3\text{mol=1239}\text{kJ}$

$1\text{C}=\text{C}\text{=}\frac{614\text{kJ}}{1\text{mol}}\times 1\text{mol=614}\text{kJ}$

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

$1\text{C}\equiv \text{N}\text{=}891\text{kJ}$

The total energy is $=4\times \left(1239+614+891+347\right)\text{kJ=12364}\text{kJ}$ (since, $4$ molecules are formed) (4)

For ${\text{H}}_2\text{O}$,

$2\text{O}-\text{H}\text{=}\frac{467\text{kJ}}{1\text{mol}}\times 2\text{mol=934}\text{kJ}$

Since $6$ molecules are formed, energy $=\left(6\times 934\right)\text{kJ=5604}\text{kJ}$ (5)

For ${\text{N}}_2$,

$1\text{N}-\text{N}\text{=}\frac{418\text{kJ}}{1\text{mol}}\times 1\text{mol=418}\text{kJ}$ (6)

The total energy for products is $=\left(12364+5604+418\right)\text{kJ}=18286\text{kJ}$ (7)

So the change in energy for the second reaction is,

$\Delta {\rm H}=\left(17536-18386\right)kJ=\underset{\_}{-850kJ}$ (Using equation (3) and (7))

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 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 152CP

The required energy change is $\underset{\_}{-1077kJ}$.

Explanation of Solution

Given

For the given reaction,

Figure 5

Energy for reactants,

For ${\text{H}}_2\text{C}={\text{CHCH}}_3$,

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

$1\text{C}=\text{C}\text{=}\frac{614\text{kJ}}{1\text{mol}}\times 1\text{mol=614}\text{kJ}$

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

Total energy $=2\times \left(2478+614+347\right)\text{kJ}=6878\text{kJ}$ (1)

(since $2$ molecules are formed)

For ${\text{NH}}_3$,

$3\text{N}-\text{H}\text{=}\frac{391\text{kJ}}{1\text{mol}}\times 3\text{mol=1173}\text{kJ}$

Total energy $=2\times 1173\text{kJ}=2346\text{kJ}$ (2)

(since $2$ molecules are formed)

For ${\text{O}}_2$,

$1\text{O}=\text{O}\text{=}\frac{495\text{kJ}}{1\text{mol}}\times 1\text{mol=495}\text{kJ}$ (3)

The total energy of reactants $=\left(6878+2346+495\right)\text{kJ}=10709\text{kJ}$ (4)

Energy for products,

For ${\text{H}}_2\text{C}=\text{CHCN}$,

$3\text{C}-\text{H}\text{=}\frac{413\text{kJ}}{1\text{mol}}\times 3\text{mol=1239}\text{kJ}$

$1\text{C}=\text{C}\text{=}\frac{614\text{kJ}}{1\text{mol}}\times 1\text{mol=614}\text{kJ}$

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

$1\text{C}\equiv \text{N}\text{=}891\text{kJ}$

The total energy $=2\times \left(1239+614+891+347\right)\text{kJ}=6182\text{kJ}$ (5)

(since $2$ molecules are formed)

For ${\text{H}}_2\text{O}$,

$2\text{O}-\text{H}\text{=}\frac{467\text{kJ}}{1\text{mol}}\times 2\text{mol=934}\text{kJ}$

Since $6$ molecules are formed, energy $=\left(6\times 934\right)\text{kJ=5604}\text{kJ}$ (6)

The total energy for products $=\left(6182+5604\right)\text{kJ}\text{=}\text{11786}\text{kJ}$ (7)

So the change in energy for the second reaction is,

$\Delta {\rm H}=\left(10709-11786\right)kJ=\underset{\_}{-1077kJ}$ (Using equation (4) and (7))

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.