Chapter 17, Problem 17.ACP

(a)

Interpretation Introduction

Interpretation:

The combustion reaction for one mole of octane has to be stated.

Concept Introduction:

The chemical reactions in which an organic compound reacts with oxygen and produces carbon dioxide and water are referred to as combustion reactions.  These types of chemical reactions are exothermic in nature as large amount of heat is also produced as by product of the reaction.

Answer to Problem 17.ACP

The chemical equation for complete combustion reaction of octane is shown below.

  ${\text{C}}_8{\text{H}}_{18}+\frac{25}{2}{\text{O}}_2\to 8{\text{CO}}_2+9{\text{H}}_2\text{O}$

Explanation of Solution

The main products produced by the complete combustion of octane are carbon dioxide and water.  The chemical equation representing the combustion reaction of octane is shown below.

  ${\text{C}}_8{\text{H}}_{18}+\frac{25}{2}{\text{O}}_2\to 8{\text{CO}}_2+9{\text{H}}_2\text{O}$

The chemical reaction shown above illustrates the burning of one mole of the octane in the presence of the oxygen to produce eight moles of carbon dioxide and nine moles of water.

(b)

Interpretation Introduction

Interpretation:

The chemical reaction for the hydrogenation process has to be stated.  Also the amount of hydrogen gas produced by the combustion of one mole of octane has to be determined.

Concept Introduction:

The chemical reaction which takes place between molecular hydrogen and another compound, in the presence of a catalyst such as nickel, palladium or platinum is termed as hydrogenation reaction.  This process is commonly used to reduce the unsaturated compounds.

Answer to Problem 17.ACP

The chemical equation which represents the first step is shown below.

  ${\text{C}}_8{\text{H}}_{18}+4{\text{O}}_2\to 8\text{CO}+9{\text{H}}_2$

The chemical equation which represents the second step is shown below.

  $8\text{CO}+8{\text{H}}_2\text{O}\to 8{\text{CO}}_2+8{\text{H}}_2$

Seventeen moles of ${\text{H}}_2$ are produced by the combustion of one mole of octane.

Explanation of Solution

The steps which include the hydrogenation process are shown below.

The first step is the partial oxidation of the octane which leads to the formation of carbon monoxide and hydrogen.

The chemical equation which represents the first step is shown below.

  ${\text{C}}_8{\text{H}}_{18}+4{\text{O}}_2\to 8\text{CO}+9{\text{H}}_2$

The second step leads to the production of carbon dioxide and hydrogen gas when carbon monoxide combines with water.  The mixture of gas obtained by the combination of carbon dioxide and hydrogen gas is referred to as the shift gas.

The chemical equation which represents the second step is shown below.

  $8\text{CO}+8{\text{H}}_2\text{O}\to 8{\text{CO}}_2+8{\text{H}}_2$

Hence the overall reaction will get on addition of chemical equation for first and second step which is shown below.

  $\begin{array}{l}\underset{\_}{\begin{array}{c}{\text{C}}_8{\text{H}}_{18}+4{\text{O}}_2\to 8\text{CO}+9{\text{H}}_2\\ 8\text{CO}+8{\text{H}}_2\text{O}\to 8{\text{CO}}_2+8{\text{H}}_2\end{array}}\\ {\text{C}}_8{\text{H}}_{18}+4{\text{O}}_2+8{\text{H}}_2\text{O}\to 8{\text{CO}}_2+17{\text{H}}_2\end{array}$

Thus, the overall reaction so obtained is shown below.

  ${\text{C}}_8{\text{H}}_{18}+4{\text{O}}_2+8{\text{H}}_2\text{O}\to 8{\text{CO}}_2+17{\text{H}}_2$

According to the above chemical equation, seventeen moles of ${\text{H}}_2$ are produced by the combustion of one mole of octane.

(c)

Interpretation Introduction

Interpretation:

The chemical equations are to be combined and have to be shown that the overall reaction is same as that of the combustion of octane.

Concept Introduction:

The zero-emission fuel burned with oxygen is referred to as hydrogen fuel.  There are many applications of it as it is used in fuel cells or internal combustion engines.  The hydrogen reacts with oxygen and leads to the formation of water and also releases energy as the by product.

Answer to Problem 17.ACP

The result is same as that for the combustion reaction of octane.

Explanation of Solution

The chemical equation representing the reaction takes place for the hydrogen gas as a fuel is shown below.

  $17{\text{H}}_2+\frac{17}{2}{\text{O}}_2\to 17{\text{H}}_2\text{O}$

The overall chemical equation so obtained in part (b) is shown below.

  ${\text{C}}_8{\text{H}}_{18}+4{\text{O}}_2+8{\text{H}}_2\text{O}\to 8{\text{CO}}_2+17{\text{H}}_2$

The chemical equation for the combustion reaction of octane is obtained by addition of reactions mentioned above.  Thus, the result obtained is shown below.

  $\begin{array}{l}\underset{\_}{\begin{array}{c}17{\text{H}}_2+\frac{17}{2}{\text{O}}_2\to 17{\text{H}}_2\text{O}\\ {\text{C}}_8{\text{H}}_{18}+4{\text{O}}_2+8{\text{H}}_2\text{O}\to 8{\text{CO}}_2+17{\text{H}}_2\end{array}}\\ {\text{C}}_8{\text{H}}_{18}+\frac{25}{2}{\text{O}}_2\to 8{\text{CO}}_2+9{\text{H}}_2\text{O}\end{array}$

Hence, the result so obtained by the addition of above two reactions is the same as the combustion reaction of octane.

(d)

Interpretation Introduction

Interpretation:

The Gibbs energy produced in thehydrogen fuel gas reaction and that of the combustion of the octane has to be compared.

Concept Introduction:

The general expression for the calculation of Gibbs free energy for a reaction is shown below.

  $\begin{array}{l}\Delta G_f^{\circ }={\displaystyle \sum \left\{\left(\text{Moles}\text{of}\text{product}\right)\times \Delta G_f^{\circ }\left(\text{product}\right)\right\}}\\ -{\displaystyle \sum \left\{\left(\text{Moles}\text{of}\text{reactant}\right)\times \Delta G_f^{\circ }\left(\text{reactant}\right)\right\}}\end{array}$

Answer to Problem 17.ACP

The Gibbs energy produced in thehydrogen fuel gas reaction is less negative and has smaller magnitude as compared to the combustion of the octane.

Explanation of Solution

The chemical equation representing the combustion reaction of octane is shown below.

  ${\text{C}}_8{\text{H}}_{18}+\frac{25}{2}{\text{O}}_2\to 8{\text{CO}}_2+9{\text{H}}_2\text{O}$

The elements which are present in their elemental states have $\Delta G_f^{\circ }=0$.  Thus, the standard Gibbs free energy for the formation of oxygen, $\Delta G_f^{\circ }.{\text{O}}_2\left(\text{g}\right)$ is $0$.

The standard Gibbs free energy for the formation of water, $\Delta G_f^{\circ }.{\text{H}}_2\text{O}\left(\text{g}\right)$, octane, $\Delta G_f^{\circ }.{\text{C}}_8{\text{H}}_{18}\left(\text{g}\right)$ and carbon dioxide, $\Delta G_f^{\circ }.{\text{CO}}_2\left(\text{g}\right)$ is $-228.572\text{kJ}/\text{mol}$, $6.707\text{kJ}/\text{mol}$ and $-394.359\text{kJ}/\text{mol}$ respectively.

The expression for the calculation of Gibbs free energy for the combustion reaction of octane is shown below.

  $\Delta G_f^{\circ }=\left(\begin{array}{l}\left(\left(8\text{mol}\right)\Delta G_f^{\circ }\left({\text{CO}}_2\right)+\left(9\text{mol}\right)\Delta G_f^{\circ }\left({\text{H}}_2\text{O}\right)\right)-\\ \left(\left(\frac{25}{2}\text{mol}\right)\Delta G_f^{\circ }\left({\text{O}}_2\right)+\left(1\text{mol}\right)\Delta G_f^{\circ }\left({\text{C}}_8{\text{H}}_{18}\right)\right)\end{array}\right)$        (1)

The value of $\Delta G_f^{\circ }\left({\text{CO}}_2\right)$ is $-394.359\text{kJ}/\text{mol}$.

The value of $\Delta G_f^{\circ }\left({\text{H}}_2\text{O}\right)$ is $-228.572\text{kJ}/\text{mol}$.

The value of $\Delta G_f^{\circ }\left({\text{O}}_2\right)$ is $0$.

The value of $\Delta G_f^{\circ }\left({\text{C}}_8{\text{H}}_{18}\right)$ is $6.707\text{kJ}/\text{mol}$.

Substitute the value of $\Delta G_f^{\circ }\left({\text{CO}}_2\right)$, $\Delta G_f^{\circ }\left({\text{H}}_2\text{O}\right)$, $\Delta G_f^{\circ }\left({\text{O}}_2\right)$ and $\Delta G_f^{\circ }\left({\text{C}}_8{\text{H}}_{18}\right)$ in equation (1).

  $\begin{array}{c}\Delta G_f^{\circ }=\left(\begin{array}{l}\left(\left(8\text{mol}\right)\left(-394.359\text{kJ}/\text{mol}\right)+\left(9\text{mol}\right)\left(-228.572\text{kJ}/\text{mol}\right)\right)-\\ \left(\begin{array}{l}\left(\frac{25}{2}\text{mol}\right)\left(0\right)\\ +\left(1\text{mol}\right)\left(6.707\text{kJ}/\text{mol}\right)\end{array}\right)\end{array}\right)\\ =-3154.872\text{kJ}-2057.148\text{kJ}-6.707\text{kJ}\\ =-5218.727\text{kJ}\end{array}$

Thus, the energy produced during the combustion of one mole of octane is $-5218.727\text{kJ}$.

The chemical equation for the hydrogen gas shift reaction is shown below.

  $17{\text{H}}_2+\frac{17}{2}{\text{O}}_2\to 17{\text{H}}_2\text{O}$

The elements which are present in their elemental states have $\Delta G_f^{\circ }=0$.  Thus, the standard Gibbs free energy for the formation of both oxygen, $\Delta G_f^{\circ }.{\text{O}}_2\left(\text{g}\right)$ and hydrogen, $\Delta G_f^{\circ }.{\text{H}}_2\left(\text{g}\right)$ is $0$.

The standard Gibbs free energy for the formation of water, $\Delta G_f^{\circ }.{\text{H}}_2\text{O}\left(\text{g}\right)$ is $-228.572\text{kJ}/\text{mol}$.

The expression for the calculation of Gibbs free energy for the hydrogen gas shift reaction is shown below.

  $\Delta G_f^{\circ }=\left(\left(17\text{mol}\right)\Delta G_f^{\circ }\left({\text{H}}_2\text{O}\right)-\left(\begin{array}{l}\left(\frac{17}{2}\text{mol}\right)\Delta G_f^{\circ }\left({\text{O}}_2\right)\\ +\left(17\text{mol}\right)\Delta G_f^{\circ }\left({\text{H}}_2\right)\end{array}\right)\right)$        (2)

The value of $\Delta G_f^{\circ }\left({\text{H}}_2\text{O}\right)$ is $-228.572\text{kJ}/\text{mol}$.

The value of $\Delta G_f^{\circ }\left({\text{O}}_2\right)$ is $0$.

The value of $\Delta G_f^{\circ }\left({\text{H}}_2\right)$ is $0$.

Substitute the value of $\Delta G_f^{\circ }\left({\text{H}}_2\text{O}\right)$, $\Delta G_f^{\circ }\left({\text{O}}_2\right)$ and $\Delta G_f^{\circ }\left({\text{H}}_2\right)$ in equation (2).

  $\begin{array}{c}\Delta G_f^{\circ }=\left\{\left(17\text{mol}\right)\left(-228.572\text{kJ}/\text{mol}\right)\right\}-\left\{\left(\frac{17}{2}\text{mol}\right)\left(0\right)+\left(17\text{mol}\right)\left(0\right)\right\}\\ =-3885.724\text{kJ}\end{array}$

Thus, the energy produced during the hydrogen gas shift reaction is $-3885.724\text{kJ}$.

Therefore, the amount of the Gibbs energy produced in the hydrogen fuel gas reaction is less negative and has smaller magnitude as compared to that of the combustion of the octane.