Chapter 14, Problem 14.79P

Interpretation Introduction

Interpretation:

The value of standard enthalpy $\left(\Delta H{°}_{\text{rxn}}\right)$ of combustion per gram of ${\text{CH}}_{\text{3}}{\text{CH}}_{\text{2}}\text{OH}\left(l\right)$ and that of ${\text{C}}_8{\text{H}}_{18}\left(l\right)$ has to be determined. Also, substance with more energy per gram has to be determined.

Concept Introduction:

Thermodynamics is a study of energy transfers that can be done by either heat or work. The energy transferred through work involves force. When work is positive then system gains energy while when work is negative then system loses energy. Since energy as heat is not state function thus it order to find change in energy produced or evolved, change in enthalpy of reaction $\Delta H{°}_{\text{rxn}}$ as a state function has been introduced. The expression to calculate $\Delta H{°}_{\text{rxn}}$ is as follows:

  $\Delta H_{\text{rxn}}=H_{\text{prod}}-H_{\text{react}}$

The useful property of $\Delta H{°}_{\text{rxn}}$ is that it is an additive property.

Answer to Problem 14.79P

The value of standard enthalpy $\left(\Delta H{°}_{\text{rxn}}\right)$ of combustion per gram for ${\text{CH}}_{\text{3}}{\text{CH}}_{\text{2}}\text{OH}\left(l\right)$ and ${\text{C}}_8{\text{H}}_{18}\left(l\right)$ are $-29.67\text{ kJ/g}$ and $-47.89\text{ kJ/g}$ respectively.

Explanation of Solution

The chemical equation for combustion reaction of ${\text{CH}}_{\text{3}}{\text{CH}}_{\text{2}}\text{OH}\left(l\right)$ is as follows:

  ${\text{CH}}_{\text{3}}{\text{CH}}_{\text{2}}\text{OH}\left(l\right)+3{\text{O}}_2\left(g\right)\to 2{\text{CO}}_2\left(g\right)+3{\text{H}}_{\text{2}}\text{O}\left(l\right)$        (1)

The expression of $\Delta H{°}_{\text{rxn}}$ for chemical equation (1) is as follows:

  $\Delta H{°}_{\text{rxn}}=\left[\begin{array}{l}\left(2\right)\Delta H{°}_{\text{f}}\left({\text{CO}}_2\right)\\ +\left(3\right)\Delta H{°}_{\text{f}}\left({\text{H}}_{\text{2}}\text{O}\right)\end{array}\right]-\left[\begin{array}{l}\left(1\right)\Delta H{°}_{\text{f}}\left({\text{CH}}_{\text{3}}{\text{CH}}_{\text{2}}\text{OH}\right)\\ +\left(3\right)\Delta H{°}_{\text{f}}\left({\text{O}}_2\right)\end{array}\right]$        (2)

Substitute $-393.5{\text{ kJ·mol}}^{\text{–1}}$ for $\Delta H{°}_{\text{f}}\left({\text{CO}}_2\right)$, $-285.8{\text{ kJ·mol}}^{\text{–1}}$ for $\Delta H{°}_{\text{f}}\left({\text{H}}_{\text{2}}\text{O}\right)$, $-277.6{\text{ kJ·mol}}^{\text{–1}}$ for $\Delta H{°}_{\text{f}}\left({\text{CH}}_{\text{3}}{\text{CH}}_{\text{2}}\text{OH}\right)$, and $0{\text{ kJ·mol}}^{\text{–1}}$ for $\Delta H{°}_{\text{f}}\left({\text{O}}_2\right)$, and  in equation (2).

  $\begin{array}{c}\Delta H{°}_{\text{rxn}}=\left[\begin{array}{l}\left(2\right)\left(-393.5{\text{ kJ·mol}}^{\text{–1}}\right)\\ +\left(3\right)\left(-285.8{\text{ kJ·mol}}^{\text{–1}}\right)\end{array}\right]-\left[\begin{array}{l}\left(1\right)\left(-277.6{\text{ kJ·mol}}^{\text{–1}}\right)\\ +\left(3\right)\left(0{\text{ kJ·mol}}^{\text{–1}}\right)\end{array}\right]\\ =-1366.8{\text{ kJ·mol}}^{\text{–1}}\end{array}$

The expression to calculate standard enthalpy of combustion per gram of the fuels ${\text{CH}}_{\text{3}}{\text{CH}}_{\text{2}}\text{OH}\left(l\right)$ is as follows:

  $\Delta H{°}_{\text{rxn}}\left(\text{kJ/g}\right)=\frac{\Delta H{°}_{\text{rxn}}\left(\text{kJ/mol}\right)}{\text{Molar mass}\left(\text{g/mol}\right)}$        (3)

Substitute $-1366.8{\text{ kJ·mol}}^{\text{–1}}$ for $\Delta H{°}_{\text{rxn}}\left(\text{kJ/mol}\right)$ and $\text{46}\text{.07 g/mol}$ for molar mass in equation (3).

  $\begin{array}{c}\Delta H{°}_{\text{rxn}}\left(\text{kJ/g}\right)=\frac{-1366.8{\text{ kJ·mol}}^{\text{–1}}}{\text{46}\text{.07 g/mol}}\\ =-29.67\text{ kJ/g}\end{array}$

The chemical equation for combustion reaction of ${\text{C}}_{\text{8}}{\text{H}}_{\text{18}}\left(g\right)$ is as follows:

  ${\text{C}}_{\text{8}}{\text{H}}_{\text{18}}\left(g\right)+\frac{25}{2}{\text{O}}_2\left(g\right)\to 8{\text{CO}}_2\left(g\right)+9{\text{H}}_{\text{2}}\text{O}\left(l\right)$        (4)

The expression of $\Delta H{°}_{\text{rxn}}$ for chemical equation (4) is as follows:

  $\Delta H{°}_{\text{rxn}}=\left[\begin{array}{l}\left(8\right)\Delta H{°}_{\text{f}}\left({\text{CO}}_2\right)\\ +\left(9\right)\Delta H{°}_{\text{f}}\left({\text{H}}_{\text{2}}\text{O}\right)\end{array}\right]-\left[\begin{array}{l}\left(1\right)\Delta H{°}_{\text{f}}\left({\text{C}}_{\text{8}}{\text{H}}_{\text{18}}\right)\\ +\left(\frac{25}{2}\right)\Delta H{°}_{\text{f}}\left({\text{O}}_2\right)\end{array}\right]$        (5)

Substitute $-393.5{\text{ kJ·mol}}^{\text{–1}}$ for $\Delta H{°}_{\text{f}}\left({\text{CO}}_2\right)$, $-285.8{\text{ kJ·mol}}^{\text{–1}}$ for $\Delta H{°}_{\text{f}}\left({\text{H}}_{\text{2}}\text{O}\right)$, $-250.1{\text{ kJ·mol}}^{\text{–1}}$ for $\Delta H{°}_{\text{f}}\left({\text{C}}_{\text{8}}{\text{H}}_{\text{18}}\right)$, and $0{\text{ kJ·mol}}^{\text{–1}}$ for $\Delta H{°}_{\text{f}}\left({\text{O}}_2\right)$ in equation (5).

  $\begin{array}{c}\Delta H{°}_{\text{rxn}}=\left[\begin{array}{l}\left(8\right)\left(-393.5{\text{ kJ·mol}}^{\text{–1}}\right)\\ +\left(9\right)\left(-285.8{\text{ kJ·mol}}^{\text{–1}}\right)\end{array}\right]-\left[\begin{array}{l}\left(1\right)\left(-250.1{\text{ kJ·mol}}^{\text{–1}}\right)\\ +\left(\frac{25}{2}\right)\left(0{\text{ kJ·mol}}^{\text{–1}}\right)\end{array}\right]\\ =-5470.1{\text{ kJ·mol}}^{\text{–1}}\end{array}$

The expression to calculate standard enthalpy of combustion for unitgram of the fuels ${\text{C}}_{\text{8}}{\text{H}}_{\text{18}}\left(g\right)$ is as follows:

  $\Delta H{°}_{\text{rxn}}\left(\text{kJ/g}\right)=\frac{\Delta H{°}_{\text{rxn}}\left(\text{kJ/mol}\right)}{\text{Molar mass}\left(\text{g/mol}\right)}$        (6)

Substitute $-5470.1{\text{ kJ·mol}}^{\text{–1}}$ for $\Delta H{°}_{\text{rxn}}\left(\text{kJ/mol}\right)$ and $114.23\text{ g/mol}$ for molar mass in equation (6).

  $\begin{array}{c}\Delta H{°}_{\text{rxn}}\left(\text{kJ/g}\right)=\frac{-5470.1{\text{ kJ·mol}}^{\text{–1}}}{114.23\text{ g/mol}}\\ =-47.89\text{ kJ/g}\end{array}$

The substance which has more energy in unit gram of the fuel is octane.