Chapter 4, Problem 4E.5E

(a)

Interpretation Introduction

Interpretation:

The reaction enthalpy for the given reaction has to be calculated.

The given reaction is,

  ${\text{3C}}_{\text{2}}{\text{H}}_{\text{2}}\left(\text{g}\right)\stackrel{}{\to }{\text{C}}_{\text{6}}{\text{H}}_{\text{6}}\left(\text{g}\right)$

Answer to Problem 4E.5E

The reaction enthalpy for the given reaction is $\text{-597 kJ}\cdot {\text{mol}}^{\text{-1}}\text{.}$

Explanation of Solution

The given reaction is,

  ${\text{3C}}_{\text{2}}{\text{H}}_{\text{2}}\left(\text{g}\right)\stackrel{}{\to }{\text{C}}_{\text{6}}{\text{H}}_{\text{6}}\left(\text{g}\right)$

In the reactant side, the bond enthalpy of ${\text{C}}_{\text{2}}{\text{H}}_{\text{2}}$ is calculated as,

The bond enthalpy of $\text{C-H}$ bond is $\text{412}\text{kJ}\cdot {\text{mol}}^{\text{-1}}.$

  $2\times 412\text{kJ}\cdot {\text{mol}}^{\text{-1}}=\text{824kJ}\cdot {\text{mol}}^{\text{-1}}$

The bond enthalpy of $\text{C}\equiv \text{C}$ is $\text{837}\text{kJ}\cdot {\text{mol}}^{\text{-1}}.$

  ${\text{1×837 kJ×mol}}^{\text{-1}}\text{=837 kJ}\cdot {\text{mol}}^{\text{-1}}$

The total bond enthalpy is $3\left(837\text{kJ}\cdot {\text{mol}}^{\text{-1}}+824\text{kJ}\cdot {\text{mol}}^{\text{-1}}\right)=4983\text{kJ}\cdot {\text{mol}}^{\text{-1}}$

The total bond enthalpy in reactant side is $4983\text{kJ}\cdot {\text{mol}}^{\text{-1}}.$

In the product side, the bond enthalpy of ${\text{C}}_{\text{6}}{\text{H}}_{\text{6}}$ is calculated as,

The bond enthalpy of $\text{C=C}$ bond is $\text{518}\text{kJ}\cdot {\text{mol}}^{\text{-1}}\text{.}$

  $\text{6×518}\text{kJ}\cdot {\text{mol}}^{\text{-1}}=\text{3108}\text{kJ}\cdot {\text{mol}}^{\text{-1}}$

The bond enthalpy of $\text{C-H}$ bond is $\text{412}\text{kJ}\cdot {\text{mol}}^{\text{-1}}.$

  $\text{6×412}\text{kJ}\cdot {\text{mol}}^{\text{-1}}\text{=2472kJ}\cdot {\text{mol}}^{\text{-1}}$

The total bond enthalpy is $\left(\text{3108 kJ}\cdot {\text{mol}}^{\text{-1}}\text{+2472}\text{kJ}\cdot {\text{mol}}^{\text{-1}}\right)=5580\text{kJ}\cdot {\text{mol}}^{\text{-1}}$

The total bond enthalpy in product side is $\text{-5580}\text{kJ}\cdot {\text{mol}}^{\text{-1}}\text{.}$

Negative sign indicates the bonds are broken.

The reactant enthalpy is,

  $\begin{array}{l}{\text{ΔH}}^{\text{o}}\text{=4983}\text{kJ}\cdot {\text{mol}}^{\text{-1}}\text{+}\left(\text{-5580}\text{kJ}\cdot {\text{mol}}^{\text{-1}}\right)\\ {\text{ΔH}}^{\text{o}}\text{=-597}\text{kJ}\cdot {\text{mol}}^{\text{-1}}\end{array}$

(b)

Interpretation Introduction

Interpretation:

The reaction enthalpy for the given reaction has to be calculated.

The given reaction is,

  ${\text{CH}}_{\text{4}}\left(\text{g}\right){\text{+4Cl}}_{\text{2}}\left(\text{g}\right)\stackrel{}{\to }{\text{CCl}}_{\text{4}}\left(\text{g}\right)\text{+4HCl}\left(\text{g}\right)$

Answer to Problem 4E.5E

The reaction enthalpy for the given reaction is $\text{-460}\text{kJ}\cdot {\text{mol}}^{\text{-1}}\text{.}$

Explanation of Solution

The given reaction is,

  ${\text{CH}}_{\text{4}}\left(\text{g}\right){\text{+4Cl}}_{\text{2}}\left(\text{g}\right)\stackrel{}{\to }{\text{CCl}}_{\text{4}}\left(\text{g}\right)\text{+4HCl}\left(\text{g}\right)$

In the reactant side,

The bond enthalpy of $\text{C-H}$ bond is $\text{412}\text{kJ}\cdot {\text{mol}}^{\text{-1}}.$

  $4\times 412\text{kJ}\cdot {\text{mol}}^{\text{-1}}=\text{1648 kJ}\cdot {\text{mol}}^{\text{-1}}$

The bond enthalpy of ${\text{Cl}}_{\text{2}}$ bond is $\text{242}\text{kJ}\cdot {\text{mol}}^{\text{-1}}.$

  $4\times 242\text{kJ}\cdot {\text{mol}}^{\text{-1}}=96\text{8 kJ}\cdot {\text{mol}}^{\text{-1}}$

The total bond enthalpy is $\left(\text{1648}\text{kJ}\cdot {\text{mol}}^{\text{-1}}\text{+968}\text{kJ}\cdot {\text{mol}}^{\text{-1}}\right)=2616\text{kJ}\cdot {\text{mol}}^{\text{-1}}$

The total bond enthalpy in the reactant side is $\text{2616}\text{kJ}\cdot {\text{mol}}^{\text{-1}}.$

In the product side,

The bond enthalpy of $\text{C-Cl}$ bond is $\text{338}\text{kJ}\cdot {\text{mol}}^{\text{-1}}.$

  $4\times 338\text{kJ}\cdot {\text{mol}}^{\text{-1}}=\text{1352 kJ}\cdot {\text{mol}}^{\text{-1}}$

The bond enthalpy of $\text{HCl}$ bond is $\text{431}\text{kJ}\cdot {\text{mol}}^{\text{-1}}.$

  $4\times 431\text{kJ}\cdot {\text{mol}}^{\text{-1}}=1724\text{kJ}\cdot {\text{mol}}^{\text{-1}}$

The total bond enthalpy is $\left(\text{1352}\text{kJ}\cdot {\text{mol}}^{\text{-1}}\text{+1724}\text{kJ}\cdot {\text{mol}}^{\text{-1}}\right)=3076\text{kJ}\cdot {\text{mol}}^{\text{-1}}$

The total bond enthalpy in the product side is $\text{-3076}\text{kJ}\cdot {\text{mol}}^{\text{-1}}\text{.}$

Negative sign indicates the bonds are broken.

The reactant enthalpy is,

  $\begin{array}{l}{\text{ΔH}}^{\text{o}}\text{=2616}\text{kJ}\cdot {\text{mol}}^{\text{-1}}\text{+}\left(\text{-3076}\text{kJ}\cdot {\text{mol}}^{\text{-1}}\right)\\ {\text{ΔH}}^{\text{o}}\text{=-460}\text{kJ}\cdot {\text{mol}}^{\text{-1}}\end{array}$

The reaction enthalpy for the given reaction is $\text{-460}\text{kJ}\cdot {\text{mol}}^{\text{-1}}\text{.}$

(c)

Interpretation Introduction

Interpretation:

The reaction enthalpy for the given reaction has to be calculated.

The given reaction is,

  ${\text{CH}}_{\text{4}}\left(\text{g}\right){\text{+CCl}}_{\text{4}}\left(\text{g}\right)\stackrel{}{\to }{\text{CHCl}}_{\text{3}}\left(\text{g}\right){\text{+CH}}_{\text{3}}\text{Cl}\left(\text{g}\right)$

Answer to Problem 4E.5E

The reaction enthalpy for the given reaction is $\text{0}\text{kJ}\cdot {\text{mol}}^{\text{-1}}\text{.}$

Explanation of Solution

The given reaction is,

  ${\text{CH}}_{\text{4}}\left(\text{g}\right){\text{+CCl}}_{\text{4}}\left(\text{g}\right)\stackrel{}{\to }{\text{CHCl}}_{\text{3}}\left(\text{g}\right){\text{+CH}}_{\text{3}}\text{Cl}\left(\text{g}\right)$

In the reactant side,

The bond enthalpy of ${\text{CH}}_{\text{4}}$ is calculated as,

The bond enthalpy of $\text{C-H}$ bond is $\text{412}\text{kJ}\cdot {\text{mol}}^{\text{-1}}.$

  $4\times 412\text{kJ}\cdot {\text{mol}}^{\text{-1}}=\text{1648 kJ}\cdot {\text{mol}}^{\text{-1}}$

The bond enthalpy of ${\text{CCl}}_{\text{4}}$ is calculated as,

The bond enthalpy of $\text{C-Cl}$ bond is $\text{338}\text{kJ}\cdot {\text{mol}}^{\text{-1}}.$

  $4\times 338\text{kJ}\cdot {\text{mol}}^{\text{-1}}=1352\text{ kJ}\cdot {\text{mol}}^{\text{-1}}$

The total bond enthalpy is $\left(\text{1648}\text{kJ}\cdot {\text{mol}}^{\text{-1}}\text{+1352}\text{kJ}\cdot {\text{mol}}^{\text{-1}}\right)=3000\text{kJ}\cdot {\text{mol}}^{\text{-1}}$

The total bond enthalpy in the reactant side is $\text{3000}\text{kJ}\cdot {\text{mol}}^{\text{-1}}.$

In the product side,

The bond enthalpy of ${\text{CHCl}}_{\text{3}}$ is calculated as,

The bond enthalpy of $\text{C-H}$ bond is $\text{412}\text{kJ}\cdot {\text{mol}}^{\text{-1}}.$

  $1\times 412\text{kJ}\cdot {\text{mol}}^{\text{-1}}=\text{412 kJ}\cdot {\text{mol}}^{\text{-1}}$

The bond enthalpy of $\text{C-Cl}$ bond is $\text{338}\text{kJ}\cdot {\text{mol}}^{\text{-1}}.$

  $3\times 338\text{kJ}\cdot {\text{mol}}^{\text{-1}}=\text{1014 kJ}\cdot {\text{mol}}^{\text{-1}}$

The bond enthalpy of ${\text{CH}}_{\text{3}}\text{Cl}$ is calculated as,

The bond enthalpy of $\text{C-H}$ bond is $\text{412}\text{kJ}\cdot {\text{mol}}^{\text{-1}}.$

  $3\times 412\text{kJ}\cdot {\text{mol}}^{\text{-1}}=1236\text{ kJ}\cdot {\text{mol}}^{\text{-1}}$

The bond enthalpy of $\text{C-Cl}$ bond is $\text{338}\text{kJ}\cdot {\text{mol}}^{\text{-1}}.$

  $1\times 338\text{kJ}\cdot {\text{mol}}^{\text{-1}}=338\text{ kJ}\cdot {\text{mol}}^{\text{-1}}$

The total bond enthalpy is

$\left(\text{1014}\text{kJ}\cdot {\text{mol}}^{\text{-1}}\text{+412}\text{kJ}\cdot {\text{mol}}^{\text{-1}}+1236\text{kJ}\cdot {\text{mol}}^{\text{-1}}+338\text{kJ}\cdot {\text{mol}}^{\text{-1}}\right)=3000\text{kJ}\cdot {\text{mol}}^{\text{-1}}$

The total bond enthalpy in the product side is $\text{-3000}\text{kJ}\cdot {\text{mol}}^{\text{-1}}.$

Negative sign indicates the bonds are broken.

The reactant enthalpy is,

  $\begin{array}{l}{\text{ΔH}}^{\text{o}}\text{=300}\text{kJ}\cdot {\text{mol}}^{\text{-1}}\text{+}\left(\text{-3000}\text{kJ}\cdot {\text{mol}}^{\text{-1}}\right)\\ {\text{ΔH}}^{\text{o}}\text{=0}\text{kJ}\cdot {\text{mol}}^{\text{-1}}\end{array}$

The reaction enthalpy for the given reaction is $\text{0}\text{kJ}\cdot {\text{mol}}^{\text{-1}}\text{.}$