Chapter 6, Problem 98QRT

(a)

Interpretation Introduction

Interpretation:

The average $\text{C-H}$ bond energy in methane has to be calculated.

Concept Introduction:

Bond energy means the amount of energy need to break a mole of molecules into its constituent atoms.

The expression for the enthalpy change is:

  ${\text{Δ}}_{\text{r}}{\text{H}}^{\text{°}}\text{= }\sum {\text{nΔ}}_{\text{f}}{\text{H}}^{\text{°}}\left(\text{products}\right)\text{-}\sum {\text{nΔ}}_{\text{f}}{\text{H}}^{\text{°}}\left(\text{reactants}\right)$

Explanation of Solution

Given,

    $\begin{array}{l}{\text{C(g)=Δ}}_{\text{f}}{\text{H}}^{\text{o}}\text{=716}\text{.7kJ/mol}\\ \\ {\text{H(g)=Δ}}_{\text{f}}{\text{H}}^{\text{o}}\text{=218}\text{kJ/mol}\end{array}$

The average $\text{C-H}$ bond energy in methane can be calculated as

  $\begin{array}{l}{\text{Δ}}_{\text{r}}{\text{H}}^{\text{o}}{\text{=Δ}}_{\text{f}}{\text{H}}^{\text{o}}{\text{{C(g)}+4Δ}}_{\text{f}}{\text{H}}^{\text{o}}{\text{{H(g)}-Δ}}_{\text{f}}{\text{H}}^{\text{o}}{\text{{CH}}_{\text{4}}\text{(g)}}\\ \\ \text{=(716}\text{.7}\text{kJ/mol)+}\text{4(218}\text{.0kJ/mol)-4(-74}\text{.81kJ/mol)=1663}\text{.5kJ/mol}{\text{CH}}_{\text{4}}\end{array}$

Therefore, the average bond energy is

  $\left(\frac{\text{1663}\text{.5}\text{kJ}}{\text{1}\text{mol}{\text{CH}}_{\text{4}}}\right)\text{×}\left(\frac{\text{1}\text{mol}{\text{CH}}_{\text{4}}}{\text{4}\text{mol}\text{C-H}\text{bonds}}\right)\text{=415}\text{.88kJ/mol}\text{bonds}$

(b)

Interpretation Introduction

Interpretation:

${\text{Δ}}_{\text{r}}{\text{H}}^{\text{o}}$ for the given reaction has to be estimated.

  ${\text{CH}}_{\text{4}}\text{(g)}\to \text{C(g)}\text{+}{\text{2H}}_{\text{2}}\text{(g)}$

Explanation of Solution

There are four $\text{C–H}$ bonds in methane broken, and two $\text{H–H}$ bonds formed.

  $\begin{array}{l}{\text{Δ}}_{\text{r}}{\text{H° = 4 D}}_{\text{C–H}}{\text{– 4 D}}_{\text{H–H}}\text{.}\hfill \\ \hfill \\ \text{ΔH° = 4 }\left(\text{416 kJ/mol}\right)\text{ – 2 }\left(\text{436 kJ/mol}\right)\text{ = 792 kJ/mol}\hfill \end{array}$

(c)

Interpretation Introduction

Interpretation:

The average $\text{C-H}$ bond energy in ${\text{CH}}_{\text{3}}\text{,}{\text{CH}}_{\text{2}}\text{,}\text{and}\text{CH}$ has to be calculated.

Explanation of Solution

To break the $\text{C-H}$ bond and form $\text{H-H}$ bonds.

Figure 1

Average Bond Energy in gas-phase ${\text{CH}}_{\text{3}}$

    ${\text{CH}}_3\text{(g)}\to \text{C(g)}\text{+}3\text{H(g)}$

Given,

$\begin{array}{l}{\text{C(g)=Δ}}_{\text{f}}{\text{H}}^{\text{o}}\text{=716}\text{.7kJ/mol}\\ \\ {\text{H(g)=Δ}}_{\text{f}}{\text{H}}^{\text{o}}\text{=218}\text{kJ/mol}\end{array}$

  $\begin{array}{l}{\text{Δ}}_{\text{r}}{\text{H}}^{\text{o}}{\text{=Δ}}_{\text{f}}{\text{H}}^{\text{o}}{\text{{C(g)}+3Δ}}_{\text{f}}{\text{H}}^{\text{o}}{\text{{H(g)}-Δ}}_{\text{f}}{\text{H}}^{\text{o}}{\text{{CH}}_3\text{(g)}}\\ \\ \text{=(716}\text{.7}\text{kJ/mol)+}3\text{(218}\text{.0kJ/mol)}\text{-}\text{(146 kJ/mol)=1224}\text{.7}\text{kJ/mol}{\text{CH}}_3\end{array}$

Therefore, the average bond energy is

  $\left(\frac{\text{1224}\text{.7}\text{kJ}}{\text{1}\text{mol}{\text{CH}}_3}\right)\text{×}\left(\frac{\text{1}\text{mol}{\text{CH}}_3}{3\text{C-H}\text{bonds}}\right)\text{=}\text{408}\text{.23}\text{kJ/mol}\text{bonds}$

Average Bond Energy in gas-phase ${\text{CH}}_2$

    ${\text{CH}}_2\text{(g)}\to \text{C(g)}\text{+}2\text{H(g)}$

Given,

    $\begin{array}{l}{\text{C(g)=Δ}}_{\text{f}}{\text{H}}^{\text{o}}\text{=716}\text{.7kJ/mol}\\ \\ {\text{H(g)=Δ}}_{\text{f}}{\text{H}}^{\text{o}}\text{=218}\text{kJ/mol}\end{array}$

  $\begin{array}{l}{\text{Δ}}_{\text{r}}{\text{H}}^{\text{o}}{\text{=Δ}}_{\text{f}}{\text{H}}^{\text{o}}{\text{{C(g)}+2Δ}}_{\text{f}}{\text{H}}^{\text{o}}{\text{{H(g)}-Δ}}_{\text{f}}{\text{H}}^{\text{o}}{\text{{CH}}_2\text{(g)}}\\ \\ \text{=(716}\text{.7}\text{kJ/mol)+}2\text{(218}\text{.0kJ/mol)}\text{-}\text{(392}\text{.5 kJ/mol)=760}\text{.2}\text{kJ/mol}{\text{CH}}_2\end{array}$

Therefore, the average bond energy is

  $\left(\frac{\text{760}\text{.2}\text{kJ}}{\text{1}\text{mol}{\text{CH}}_{\text{2}}}\right)\text{×}\left(\frac{\text{1}\text{mol}{\text{CH}}_{\text{2}}}{\text{2}\text{mol}\text{bonds}}\right)\text{=}\text{380}\text{.1}\text{kJ/mol}\text{bonds}$

Average Bond Energy in gas-phase $\text{CH}$

    $\text{CH(g)}\to \text{C(g)}\text{+}\text{H(g)}$

Given,

    $\begin{array}{l}{\text{C(g)=Δ}}_{\text{f}}{\text{H}}^{\text{o}}\text{=716}\text{.7kJ/mol}\\ \\ {\text{H(g)=Δ}}_{\text{f}}{\text{H}}^{\text{o}}\text{=218}\text{kJ/mol}\end{array}$

  $\begin{array}{l}{\text{Δ}}_{\text{r}}{\text{H}}^{\text{o}}{\text{=Δ}}_{\text{f}}{\text{H}}^{\text{o}}{\text{{C(g)}+Δ}}_{\text{f}}{\text{H}}^{\text{o}}{\text{{H(g)}-Δ}}_{\text{f}}{\text{H}}^{\text{o}}\text{{CH(g)}}\\ \\ \text{=(716}\text{.7}\text{kJ/mol)+}\text{(218}\text{.0kJ/mol)}\text{-}\text{(596}\text{.3 kJ/mol)=338}\text{.4}\text{kJ/mol}\text{CH}\end{array}$

Therefore, the average bond energy is

  $\left(\frac{338.4\text{kJ}}{\text{1}\text{mol}\text{CH}}\right)\text{×}\left(\frac{\text{1}\text{mol}\text{CH}}{1\text{mol}\text{bonds}}\right)\text{=}\text{338}\text{.4}\text{kJ/mol}\text{bonds}$

It is easier to break a bond in an unstable molecule than in a stable molecule.