Chapter 14, Problem 14.82P

Interpretation Introduction

Interpretation:

The value of standard enthalpy $\left(\Delta H{°}_{\text{rxn}}\right)$ of combustion per gram of ${\text{C}}_{\text{2}}{\text{H}}_{\text{2}}\left(g\right)$, ${\text{C}}_{\text{2}}{\text{H}}_{\text{4}}\left(g\right)$, ${\text{C}}_{\text{2}}{\text{H}}_{\text{6}}\left(g\right)$ 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 14.82P

The value of $\Delta H{°}_{\text{rxn}}\left(\text{kJ/g}\right)$ for ${\text{C}}_{\text{2}}{\text{H}}_{\text{2}}\left(g\right)$, ${\text{C}}_{\text{2}}{\text{H}}_{\text{4}}\left(g\right)$, and ${\text{C}}_{\text{2}}{\text{H}}_{\text{6}}\left(g\right)$ are $-54.08\text{ kJ/g}$, $-50.3\text{ kJ/g}$, and $-51.9\text{ kJ/g}$ respectively.

Explanation of Solution

The chemical equation for combustion reaction of ${\text{C}}_{\text{2}}{\text{H}}_{\text{2}}\left(g\right)$ is as follows:

  ${\text{C}}_{\text{2}}{\text{H}}_{\text{2}}\left(g\right)+\frac{5}{2}{\text{O}}_2\left(g\right)\to 2{\text{CO}}_2\left(g\right)+{\text{H}}_{\text{2}}\text{O}\left(l\right)$        (1)

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

  $\Delta H{°}_{\text{rxn}}=\left[\begin{array}{l}\left(2\right)\Delta H{°}_{\text{f}}\left({\text{CO}}_2\right)\\ +\left(1\right)\Delta H{°}_{\text{f}}\left({\text{H}}_{\text{2}}\text{O}\right)\end{array}\right]-\left[\begin{array}{l}\left(1\right)\Delta H{°}_{\text{f}}\left({\text{C}}_{\text{2}}{\text{H}}_{\text{2}}\right)\\ +\left(\frac{5}{2}\right)\Delta H{°}_{\text{f}}\left({\text{O}}_2\right)\end{array}\right]$        (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 $\Delta H{°}_{\text{f}}\left({\text{H}}_{\text{2}}\text{O}\right)$, $+227.4{\text{ kJ·mol}}^{\text{–1}}$ for $\Delta H{°}_{\text{f}}\left({\text{C}}_{\text{2}}{\text{H}}_{\text{2}}\right)$, and $0{\text{ kJ·mol}}^{\text{–1}}$ for $\Delta H{°}_{\text{f}}\left({\text{O}}_2\right)$ in equation (2).

  $\begin{array}{c}\Delta H{°}_{\text{rxn}}=\left[\begin{array}{l}\left(2\right)\left(-393.5{\text{ kJ·mol}}^{\text{–1}}\right)\\ +\left(1\right)\left(-285.8{\text{ kJ·mol}}^{\text{–1}}\right)\end{array}\right]-\left[\begin{array}{l}\left(1\right)\left(+227.4{\text{ kJ·mol}}^{\text{–1}}\right)\\ +\left(\frac{5}{2}\right)\left(0{\text{ kJ·mol}}^{\text{–1}}\right)\end{array}\right]\\ =-1300.2{\text{ kJ·mol}}^{\text{–1}}\end{array}$

The expression to calculate standard enthalpy of combustion per gram of the fuels ${\text{C}}_{\text{2}}{\text{H}}_{\text{2}}\left(g\right)$ is as follows:

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

Substitute $-1300.2{\text{ kJ·mol}}^{\text{–1}}$ for $\Delta H{°}_{\text{rxn}}\left(\text{kJ/mol}\right)$ and $24.04\text{ g/mol}$ for molar mass in equation (3).

  $\begin{array}{c}\Delta H{°}_{\text{rxn}}\left(\text{kJ/g}\right)=\frac{-1300.2{\text{ kJ·mol}}^{\text{–1}}}{24.04\text{ g/mol}}\\ =-54.08\text{ kJ/g}\end{array}$

The chemical equation for combustion reaction of ${\text{C}}_{\text{2}}{\text{H}}_{\text{4}}\left(g\right)$ is as follows:

  ${\text{C}}_{\text{2}}{\text{H}}_{\text{4}}\left(g\right)+3{\text{O}}_2\left(g\right)\to 2{\text{CO}}_2\left(g\right)+2{\text{H}}_{\text{2}}\text{O}\left(l\right)$        (4)

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

  $\Delta H{°}_{\text{rxn}}=\left[\begin{array}{l}\left(2\right)\Delta H{°}_{\text{f}}\left({\text{CO}}_2\right)\\ +\left(2\right)\Delta H{°}_{\text{f}}\left({\text{H}}_{\text{2}}\text{O}\right)\end{array}\right]-\left[\begin{array}{l}\left(1\right)\Delta H{°}_{\text{f}}\left({\text{C}}_{\text{2}}{\text{H}}_{\text{4}}\right)\\ +\left(3\right)\Delta H{°}_{\text{f}}\left({\text{O}}_2\right)\end{array}\right]$        (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 $\Delta H{°}_{\text{f}}\left({\text{H}}_{\text{2}}\text{O}\right)$, $+52.4{\text{ kJ·mol}}^{\text{–1}}$ for $\Delta H{°}_{\text{f}}\left({\text{C}}_{\text{2}}{\text{H}}_{\text{4}}\right)$, and $0{\text{ kJ·mol}}^{\text{–1}}$ for $\Delta H{°}_{\text{f}}\left({\text{O}}_2\right)$, and  in equation (5).

  $\begin{array}{c}\Delta H{°}_{\text{rxn}}=\left[\begin{array}{l}\left(2\right)\left(-393.5{\text{ kJ·mol}}^{\text{–1}}\right)\\ +\left(2\right)\left(-285.8{\text{ kJ·mol}}^{\text{–1}}\right)\end{array}\right]-\left[\begin{array}{l}\left(1\right)\left(+52.4{\text{ kJ·mol}}^{\text{–1}}\right)\\ +\left(3\right)\left(0{\text{ kJ·mol}}^{\text{–1}}\right)\end{array}\right]\\ =-1411{\text{ kJ·mol}}^{\text{–1}}\end{array}$

The expression to calculate standard enthalpy of combustion per gram of the fuels ${\text{C}}_{\text{2}}{\text{H}}_{\text{4}}\left(g\right)$ 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 $-1411{\text{ kJ·mol}}^{\text{–1}}$ for $\Delta H{°}_{\text{rxn}}\left(\text{kJ/mol}\right)$ and $28.05\text{ g/mol}$ for molar mass in equation (6).

  $\begin{array}{c}\Delta H{°}_{\text{rxn}}\left(\text{kJ/g}\right)=\frac{-1411{\text{ kJ·mol}}^{\text{–1}}}{28.05\text{ g/mol}}\\ =-50.3\text{ kJ/g}\end{array}$

The chemical equation for combustion reaction of ${\text{C}}_{\text{2}}{\text{H}}_{\text{6}}\left(g\right)$ is as follows:

  ${\text{C}}_{\text{2}}{\text{H}}_{\text{6}}\left(g\right)+\frac{7}{2}{\text{O}}_2\left(g\right)\to 2{\text{CO}}_2\left(g\right)+3{\text{H}}_{\text{2}}\text{O}\left(l\right)$        (7)

The expression of $\Delta H{°}_{\text{rxn}}$ for chemical equation (7) is as follows:

  $\Delta H{°}_{\text{rxn}}=\left[\begin{array}{l}\left(2\right)\Delta H{°}_{\text{f}}\left({\text{CO}}_2\right)\\ +\left(3\right)\Delta H{°}_{\text{f}}\left({\text{H}}_{\text{2}}\text{O}\right)\end{array}\right]-\left[\begin{array}{l}\left(1\right)\Delta H{°}_{\text{f}}\left({\text{C}}_{\text{2}}{\text{H}}_{\text{6}}\right)\\ +\left(\frac{7}{2}\right)\Delta H{°}_{\text{f}}\left({\text{O}}_2\right)\end{array}\right]$        (8)

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 $\Delta H{°}_{\text{f}}\left({\text{H}}_{\text{2}}\text{O}\right)$, $-84.0{\text{ kJ·mol}}^{\text{–1}}$ for $\Delta H{°}_{\text{f}}\left({\text{C}}_{\text{2}}{\text{H}}_{\text{6}}\right)$, and $0{\text{ kJ·mol}}^{\text{–1}}$ for $\Delta H{°}_{\text{f}}\left({\text{O}}_2\right)$, and  in equation (8).

  $\begin{array}{c}\Delta H{°}_{\text{rxn}}=\left[\begin{array}{l}\left(2\right)\left(-393.5{\text{ kJ·mol}}^{\text{–1}}\right)\\ +\left(3\right)\left(-285.8{\text{ kJ·mol}}^{\text{–1}}\right)\end{array}\right]-\left[\begin{array}{l}\left(1\right)\left(-84.0{\text{ kJ·mol}}^{\text{–1}}\right)\\ +\left(\frac{7}{2}\right)\left(0{\text{ kJ·mol}}^{\text{–1}}\right)\end{array}\right]\\ =-1560.4{\text{ kJ·mol}}^{\text{–1}}\end{array}$

The expression to calculate standard enthalpy of combustion per gram of the fuels ${\text{C}}_{\text{2}}{\text{H}}_{\text{6}}\left(g\right)$ is as follows:

  $\Delta H{°}_{\text{rxn}}\left(\text{kJ/g}\right)=\frac{\Delta H{°}_{\text{rxn}}\left(\text{kJ/mol}\right)}{\text{Molar mass}\left(\text{g/mol}\right)}$        (9)

Substitute $-1560.4{\text{ kJ·mol}}^{\text{–1}}$ for $\Delta H{°}_{\text{rxn}}\left(\text{kJ/mol}\right)$ and $30.07\text{ g/mol}$ for molar mass in equation (9).

  $\begin{array}{c}\Delta H{°}_{\text{rxn}}\left(\text{kJ/g}\right)=\frac{-1560.4{\text{ kJ·mol}}^{\text{–1}}}{30.07\text{ g/mol}}\\ =-51.9\text{ kJ/g}\end{array}$

Since value of $\Delta H{°}_{\text{rxn}}\left(\text{kJ/g}\right)$ does not follow any trend as the mass of compound increases thus there is no trend.