Chapter 5, Problem 5.43AP

Interpretation Introduction

(a)

Interpretation:

The $\Delta H°$ values for the halogenation of methane in the gas phase is an industrial method for the preparation of certain alkyl halides takes readily when $\text{X}=\text{Br}$ or $\text{X}=\text{Cl}$ but not when $\text{X}=\text{I}$ is to be calculated.

Concept introduction:

Halogenation is an electrophilic substitution reaction in which the halogen groups are introduced in the reacting compound.

The minimum amount of energy needed to dissociate a bond is called the bond dissociation energy. The term bond dissociation energy is only used for diatomic species. The covalent bonds have low value of bond dissociation energy.

Answer to Problem 5.43AP

The $\Delta H°$ values for the halogenation of methane in the gas phase is $-41\text{kJ}{\text{mol}}^{\text{-1}}$

Explanation of Solution

The reaction required in the given problem is written below.

${\text{X}}_{\text{2}}+{\text{CH}}_{\text{4}}\to {\text{H}}_{\text{3}}\text{C}-\text{X}+\text{H}-\text{X}$

When $\text{X}=\text{Br}$, one $\text{Br-Br}$ bond and one $\text{C}-\text{H}$ bond dissociates and a new $\text{C}-\text{Br}$ bond and one $\text{H}-\text{Br}$ is formed

The expression for the term $\Delta H°$ (enthalpy change) is given below.

$\Delta H°=\left(\Delta H_{\text{Br-Br}}^{\text{ο}}+\Delta H_{\text{C}-\text{H}}^{\text{ο}}\right)-\left(\Delta H_{\text{H}-\text{Br}}^{\text{ο}}+\Delta H_{\text{C}-\text{Br}}^{\text{ο}}\right)$…(1)

Where,

• $\Delta H_{\text{Br-Br}}^{\text{ο}}$ is the bond dissociation energy of bromine-bromine bonds.

• $\Delta H_{\text{C}-\text{H}}^{\text{ο}}$ is the bond dissociation energy of carbon-bromine bonds

• $\Delta H_{\text{H}-\text{Br}}^{\text{ο}}$ is the bond dissociation energy of hydrogen-bromine bonds

• $\Delta H_{\text{C}-\text{Br}}^{\text{ο}}$ is the bond dissociation energy of carbon-bromine bonds

The value of $\Delta H^o{}_{\text{Br-Br}}$ is $190\text{kJ}{\text{mol}}^{-1}$.

The value of $\Delta H^o{}_{\text{C}-\text{H}}$ is $439\text{kJ}{\text{mol}}^{-1}$.

The value of $\Delta H^o{}_{\text{C}-\text{Br}}$ is $302\text{kJ}{\text{mol}}^{-1}$.

The value of $\Delta H^o{}_{\text{H}-\text{Br}}$ is $368\text{kJ}{\text{mol}}^{-1}$.

Substitute the value of $\Delta H^o{}_{\text{Br-Br}}$, $\Delta H^o{}_{\text{C}-\text{Br}}$, $\Delta H^o{}_{\text{H}-\text{Br}}$ and $\Delta H^o{}_{\text{C}-\text{H}}$ in the equation (1).

$\begin{array}{rll}\Delta H^o& =& \left(\Delta H^o{}_{\text{Br-Br}}+\Delta H^o{}_{\text{C}-\text{H}}\right)-\left(\Delta H^o{}_{\text{H}-\text{Br}}+\Delta H^o{}_{\text{C}-\text{Br}}\right)\\ & =& \left(190+439\right)-\left(368+302\right)\text{kJ}{\text{mol}}^{\text{-1}}\\ & =& -41\text{kJ}{\text{mol}}^{\text{-1}}\end{array}$

The value obtained for the term $\Delta H^o$ is $-41\text{kJ}{\text{mol}}^{\text{-1}}$ and thus the negative value signifies that the reaction is exothermic as well as favorable.

Now, when $\text{X}=\text{Cl}$, one $\text{Cl}-\text{Cl}$ bond and one $\text{C}-\text{H}$ bond dissociates and a new $\text{C}-\text{Cl}$ bond and one $\text{H}-\text{Cl}$ is formed.

The expression for the term $\Delta H^o$ (enthalpy change) is given below.

$\Delta H°=\left(\Delta H_{\text{Cl}-\text{Cl}}^{\text{ο}}+\Delta H_{\text{C}-\text{H}}^{\text{ο}}\right)-\left(\Delta H_{\text{H}-\text{Cl}}^{\text{ο}}+\Delta H_{\text{C}-\text{Cl}}^{\text{ο}}\right)$…(2)

Where,

• $\Delta H_{\text{Cl}-\text{Cl}}^{\text{ο}}$ is the bond dissociation energy of chlorine-chlorine bonds.

• $\Delta H_{\text{C}-\text{H}}^{\text{ο}}$ is the bond dissociation energy of carbon-hydrogen bonds

• $\Delta H_{\text{H}-\text{Cl}}^{\text{ο}}$ is the bond dissociation energy of hydrogen-chlorine bonds

• $\Delta H_{\text{C}-\text{Cl}}^{\text{ο}}$ is the bond dissociation energy of carbon-chlorine bonds

The value of $\Delta H_{\text{Cl}-\text{Cl}}^{\text{ο}}$ is $239\text{kJ}{\text{mol}}^{-1}$.

The value of $\Delta H_{\text{C}-\text{H}}^{\text{ο}}$ is $439\text{kJ}{\text{mol}}^{-1}$.

The value of $\Delta H_{\text{C}-\text{Cl}}^{\text{ο}}$ is $350\text{kJ}{\text{mol}}^{-1}$.

The value of $\Delta H_{\text{H}-\text{Cl}}^{\text{ο}}$ is $431\text{kJ}{\text{mol}}^{-1}$.

Substitute the value of $\Delta H^o{}_{\text{Cl}-\text{Cl}}$, $\Delta H^o{}_{\text{C}-\text{Cl}}$, $\Delta H^o{}_{\text{H}-\text{Cl}}$ and $\Delta H^o{}_{\text{C}-\text{H}}$ in the equation (1).

$\begin{array}{rll}\Delta H^o& =& \left(\Delta H^o{}_{\text{Cl}-\text{Cl}}+\Delta H^o{}_{\text{C}-\text{H}}\right)-\left(\Delta H^o{}_{\text{H}-\text{Cl}}+\Delta H^o{}_{\text{C}-\text{Cl}}\right)\\ & =& \left(239+439\right)-\left(431+350\right)\text{kJ}{\text{mol}}^{\text{-1}}\\ & =& -103\text{kJ}{\text{mol}}^{\text{-1}}\end{array}$

The value obtained for the term $\Delta H^o$ when $\text{X}=\text{Cl}$ is $-103\text{kJ}{\text{mol}}^{-1}$ and thus the negative value signifies that the reaction highly exothermic as well as favorable.

The expression for the term $\Delta H^o$ (enthalpy change) is given below.

$\Delta H°=\left(\Delta H_{\text{I}-\text{I}}^{\text{ο}}+\Delta H_{\text{C}-\text{H}}^{\text{ο}}\right)-\left(\Delta H_{\text{H}-\text{I}}^{\text{ο}}+\Delta H_{\text{C}-\text{I}}^{\text{ο}}\right)$…(3)

When $\text{X}=\text{I}$, one $\text{I}-\text{I}$ bond and one $\text{C}-\text{H}$ bond dissociates and a new $\text{C}-\text{I}$ bond and one $\text{H}-\text{I}$ is formed.

Where,

• $\Delta H_{\text{I}-\text{I}}^{\text{ο}}$ is the bond dissociation energy of iodine-iodine bonds.

• $\Delta H_{\text{C}-\text{H}}^{\text{ο}}$ is the bond dissociation energy of carbon-hydrogen bonds

• $\Delta H_{\text{H}-\text{I}}^{\text{ο}}$ is the bond dissociation energy of hydrogen-iodine bonds

• $\Delta H_{\text{C}-\text{I}}^{\text{ο}}$ is the bond dissociation energy of carbon-iodine bonds

The value of $\Delta H^o{}_{\text{H}-\text{I}}$ is $297\text{kJ}{\text{mol}}^{-1}$.

The value of $\Delta H^o{}_{\text{C}-\text{H}}$ is $439\text{kJ}{\text{mol}}^{-1}$.

The value of $\Delta H^o{}_{\text{I}-\text{I}}$ is $149\text{kJ}{\text{mol}}^{-1}$.

The value of $\Delta H^o{}_{\text{C}-\text{I}}$ is $241\text{kJ}{\text{mol}}^{-1}$.

Substituting the value of $\Delta H^o{}_{\text{Br-Br}}$, $\Delta H^o{}_{\text{C}-\text{H}}$, $\Delta H^o{}_{\text{I}-\text{I}}$ and $\Delta H^o{}_{\text{C}-\text{I}}$ in the equation (2) and solving for $\Delta H^o$.

$\begin{array}{rll}\Delta H°& =& \left(\Delta H_{\text{I}-\text{I}}^{\text{ο}}+\Delta H_{\text{C}-\text{H}}^{\text{ο}}\right)-\left(\Delta H_{\text{H}-\text{I}}^{\text{ο}}+\Delta H_{\text{C}-\text{I}}^{\text{ο}}\right)\\ & =& \left(149+439\right)-\left(297+241\right)\text{kJ}{\text{mol}}^{-1}\\ & =& 50\text{kJ}{\text{mol}}^{-1}\end{array}$

The value obtained for the term $\Delta H°$ when $\text{X}=\text{I}$ is $50\text{kJ}{\text{mol}}^{-1}$ and thus, the positive value signifies that the reaction is endothermic and so it is unfavorable.

Conclusion

The value obtained for the term $\Delta H°$ when $\text{X}=\text{Br}$, $\text{X}=\text{Cl}$ and $\text{X}=\text{I}$ are $-40\text{kJ}{\text{mol}}^{\text{-1}}$, $-103\text{kJ}{\text{mol}}^{-1}$ and $50\text{kJ}{\text{mol}}^{-1}$ respectively.

Interpretation Introduction

(b)

Interpretation:

An explanation as to why samples of methyl iodide $\left({\text{H}}_{\text{3}}\text{C-I}\right)$ that are contaminated with traces of $\text{HI}$ darken with the color of iodine on standing a long time is to be stated.

Concept introduction:

Halogenation is an electrophilic substitution reaction in which the halogen groups are introduced in the reacting compound.

The minimum amount of energy needed to dissociate a bond is called the bond dissociation energy. The term bond dissociation energy is only used for diatomic species. The covalent bonds have low value of bond dissociation energy.

Answer to Problem 5.43AP

The value obtained for the term $\Delta H^o$ is $-50\text{kJ}{\text{mol}}^{\text{-1}}$ and thus the negative value signifies that the reaction is exothermic as well as favorable. Therefore, the samples of methyl iodide $\left({\text{H}}_{\text{3}}\text{C-I}\right)$ that are contaminated with traces of $\text{HI}$ darken with the color of iodine on standing a long time.

Explanation of Solution

The reaction required in the given problem is written below.

${\text{X}}_{\text{2}}+{\text{CH}}_{\text{4}}\to {\text{H}}_{\text{3}}\text{C}-\text{X}+\text{H}-\text{X}$

The expression for the term $\Delta H^o$ (enthalpy change) is given below.

$\Delta H°=\left(\Delta H_{\text{H}-\text{I}}^{\text{ο}}+\Delta H_{\text{C}-\text{I}}^{\text{ο}}\right)-\left(\Delta H_{\text{I}-\text{I}}^{\text{ο}}+\Delta H_{\text{C}-\text{H}}^{\text{ο}}\right)$…(3)

When $\text{X}=\text{I}$, one $\text{I}-\text{I}$ bond and one $\text{C}-\text{H}$ bond dissociates and a new $\text{C}-\text{I}$ bond and one $\text{H}-\text{I}$ is formed.

Where,

• $\Delta H_{\text{I}-\text{I}}^{\text{ο}}$ is the bond dissociation energy of iodine-iodine bonds.

• $\Delta H_{\text{C}-\text{H}}^{\text{ο}}$ is the bond dissociation energy of carbon-hydrogen bonds

• $\Delta H_{\text{H}-\text{I}}^{\text{ο}}$ is the bond dissociation energy of hydrogen-iodine bonds

• $\Delta H_{\text{C}-\text{I}}^{\text{ο}}$ is the bond dissociation energy of carbon-iodine bonds

The value of $\Delta H_{\text{H}-\text{I}}^{\text{ο}}$ is $297\text{kJ}{\text{mol}}^{-1}$.

The value of $\Delta H_{\text{C}-\text{H}}^{\text{ο}}$ is $439\text{kJ}{\text{mol}}^{-1}$.

The value of $\Delta H_{\text{I}-\text{I}}^{\text{ο}}$ is $149\text{kJ}{\text{mol}}^{-1}$.

The value of $\Delta H_{\text{C}-\text{I}}^{\text{ο}}$ is $241\text{kJ}{\text{mol}}^{-1}$.

Substitute the value of $\Delta H^o{}_{\text{I-I}}$, $\Delta H^o{}_{\text{C}-\text{H}}$, $\Delta H^o{}_{\text{I}-\text{I}}$ and $\Delta H^o{}_{\text{C}-\text{I}}$ in the equation (3) and solve for $\Delta H^o$.

$\begin{array}{rll}\Delta H°& =& \left(\Delta H_{\text{H}-\text{I}}^{\text{ο}}+\Delta H_{\text{C}-\text{I}}^{\text{ο}}\right)-\left(\Delta H_{\text{I}-\text{I}}^{\text{ο}}+\Delta H_{\text{C}-\text{H}}^{\text{ο}}\right)\\ & =& \left(297+241\right)-\left(149+439\right)\text{kJ}{\text{mol}}^{-1}\\ & =& -50\text{kJ}{\text{mol}}^{-1}\end{array}$

The value obtained for the term $\Delta H°$ is $-50\text{kJ}{\text{mol}}^{-1}$ and thus the negative value signifies that the reaction exothermic as well as favorable. Therefore, the samples of methyl iodide $\left({\text{H}}_{\text{3}}\text{C}-\text{I}\right)$ that are contaminated with traces of $\text{HI}$ darken with the color of iodine on standing a long time.

Conclusion

The samples of methyl iodide $\left({\text{H}}_{\text{3}}\text{C}-\text{I}\right)$ that are contaminated with traces of $\text{HI}$ darken with the color of iodine on standing a long time because the value of $\Delta H°$ is negative and thus the reaction is highly exothermic as well as favorable.