Chapter U5.11, Problem 7E

(a)

Interpretation Introduction

Interpretation: The net energy exchange for the given combustion reaction of methanol and butanol needs to be determined.

Concept Introduction: In a combustion reaction, hydrocarbon reacts with oxygen to form carbon dioxide and water.

Explanation of Solution

The given reactions are as follows:

  ${\text{Methanol: 2CH}}_{\text{4}}\text{O}\left(l\right)+3{\text{O}}_{\text{2}}\left(g\right)\to 2{\text{CO}}_{\text{2}}\left(g\right)+4{\text{H}}_{\text{2}}\text{O}\left(l\right)$

And,

  ${\text{Butanol: C}}_{\text{4}}{\text{H}}_{\text{10}}\text{O}\left(l\right)+6{\text{O}}_{\text{2}}\left(g\right)\to 4{\text{CO}}_{\text{2}}\left(g\right)+5{\text{H}}_{\text{2}}\text{O}\left(l\right)$

The net energy change can be calculated using the following formula:

  $\Delta H={\displaystyle \sum B.E\left(\text{reactant}\right)}-{\displaystyle \sum B.E\left(\text{product}\right)}$

For methanol:

  ${\text{2CH}}_{\text{4}}\text{O}\left(l\right)+3{\text{O}}_{\text{2}}\left(g\right)\to 2{\text{CO}}_{\text{2}}\left(g\right)+4{\text{H}}_{\text{2}}\text{O}\left(l\right)$

There are 3 C-H bonds, 1 C-O bond, and 1 O-H bond in CH₃OH and 1 O=O bond in O₂ thus the enthalpy of the bonds broken can be calculated as follows:

  $\begin{array}{l}{\displaystyle \sum B.E\left(\text{reactant}\right)=}2\left[3\times C-H+C-O+H-O\right]+\left(3\times O=O\right)\\ =2\left[3\times 412+360+463\right]+\left(3\times 496\right)=4118+1488\\ =5\text{606 kJ}\end{array}$

Now, there are 2 C=O bonds in CO₂ and 2 H-O bonds in H₂O thus the enthalpy of formation bonds is:

  $\begin{array}{l}{\displaystyle \sum B.E\left(\text{product}\right)=}=2\left[2\times C=O\right]+4\left(2\times H-O\right)\\ =2\left[2\times 743\right]+4\left(2\times 463\right)\\ =2972+3704=6676\text{ kJ}\end{array}$

Now, the enthalpy of reaction can be calculated as follows:

  $\Delta H={\displaystyle \sum B.E\left(\text{reactant}\right)}-{\displaystyle \sum B.E\left(\text{product}\right)}$

Or,

  $\begin{array}{c} \Delta H=5606\text{ kJ }-6676\text{ kJ}\\ =-1070\text{ kJ}\end{array}$

Thus, the enthalpy of combustion of 1 mol methanol will be:

  $\begin{array}{c}\Delta H= \frac{-1070\text{ kJ}}{2\text{ mol}}\\ =-535\text{ kJ/mol}\end{array}$

Similarly,

For butanol:

  ${\text{C}}_{\text{4}}{\text{H}}_{\text{10}}\text{O}\left(l\right)+6{\text{O}}_{\text{2}}\left(g\right)\to 4{\text{CO}}_{\text{2}}\left(g\right)+5{\text{H}}_{\text{2}}\text{O}\left(l\right)$

There are 3 C-C bonds, 9 C-H bonds, 1 C-O bond, and 1 O-H bond in ${\text{C}}_{\text{4}}{\text{H}}_{\text{10}}\text{O}$ and 1O=O bond in O₂ thus the enthalpy of the bonds broken can be calculated as follows:

  $\begin{array}{l}{\displaystyle \sum B.E\left(\text{reactant}\right)=}\left[3\times C-C+9\times C-H+C-O+H-O\right]+\left(6\times O=O\right)\\ =\left[3\times 346+9\times 412+360+463\right]+\left(6\times 496\right)\\ =\left[1038+3708+360+463\right]+\left(2976\right)\\ =\left(5569\right)+2976\\ =\text{ 8545 kJ}\end{array}$

Now, there are 2 C=O bonds in CO₂ and 2 H-O bonds in H₂O thus the enthalpy of formation bonds is:

  $\begin{array}{l}{\displaystyle \sum B.E\left(\text{product}\right)=}\text{ 4}\left[2\times C=O\right]+5\left(2\times H-O\right)\\ =4\left[2\times 743\right]+5\left(2\times 463\right)\\ =\text{ 5944 }+\text{ 4630}=10574\text{ kJ}\end{array}$

Now, the enthalpy of reaction can be calculated as follows:

  $\Delta H={\displaystyle \sum B.E\left(\text{reactant}\right)}-{\displaystyle \sum B.E\left(\text{product}\right)}$

Or,

  $\begin{array}{c} \Delta H=8545\text{ kJ }-10574\text{ kJ}\\ =\text{ -2029 kJ}\end{array}$

Thus, the enthalpy of combustion of 1 mol butanol is -2029 kJ/mol.

(b)

Interpretation Introduction

Interpretation: The net energy exchange for each reaction needs to be compared with the known heat of combustion values.

Concept Introduction: The net energy change can be calculated using the following formula:

  $\Delta H={\displaystyle \sum B.E\left(\text{reactant}\right)}-{\displaystyle \sum B.E\left(\text{product}\right)}$

Explanation of Solution

From the bond enthalpy, the net energy exchange for the combustion of methanol and butanol is calculated as -535 kJ/mol and -2029 kJ/mol respectively.

The known heat of combustion values for methanol and butanol are -726 kJ/mol and -2671 kJ/mol. Thus, the calculated values are slightly less than the known values for the heat of combustion.