Chapter 6, Problem 6.99P

(a)

Interpretation Introduction

Interpretation:

The balanced equation for the combustion of kerosene is to be written.

Concept introduction:

A combustion reaction is the reaction in which reactant is reacted with molecular oxygen to form the product. Heat is released and the energy is produced in the reaction. Molecular oxygen is employed as an oxidizing agent in these reactions.

Answer to Problem 6.99P

The balanced equation for the combustion of kerosene is,

${\text{2C}}_{\text{12}}{\text{H}}_{26}\left(l\right)+37{\text{O}}_2\left(g\right)\to 24{\text{CO}}_2\left(g\right)+26{\text{H}}_{\text{2}}\text{O}\left(\text{g}\right)$

Explanation of Solution

Kerosene $\left({\text{C}}_{\text{12}}{\text{H}}_{26}\right)$ reacts with molecular oxygen to form carbon dioxide and water molecule. The molecular equation foe the combustion of Kerosene $\left({\text{C}}_{\text{12}}{\text{H}}_{26}\right)$ is,

${\text{2C}}_{\text{12}}{\text{H}}_{26}\left(l\right)+37{\text{O}}_2\left(g\right)\to 24{\text{CO}}_2\left(g\right)+26{\text{H}}_{\text{2}}\text{O}\left(\text{g}\right)$

Conclusion

A combustion reaction is the reaction in which reactant is reacted with molecular oxygen to form carbon dioxide and water molecule.

(b)

Interpretation Introduction

Interpretation:

$\Delta H_{\text{f}}^{°}$ of kerosene when $\Delta H_{\text{rxn}}^{°}$ is $-1.5\times {10}^4\text{kJ}$ is to be calculated.

Concept introduction:

The standard enthalpy of reaction is calculated by the summation of standard enthalpy of formation of the product minus the summation of standard enthalpy of formation of product at the standard conditions. The formula to calculate the standard enthalpy of reaction $\left(\Delta H_{\text{rxn}}^{°}\right)$ is as follows:

$\Delta H_{\text{rxn}}^{°}=\sum m\Delta H_{\text{f (products)}}^{°}-\sum m\Delta H_{\text{f (reactants)}}^{°}$

Here, m and n are the stoichiometric coefficients of reactants and product in the balanced chemical equation.

Answer to Problem 6.99P

$\Delta H_{\text{f}}^{°}$ of kerosene when $\Delta H_{\text{rxn}}^{°}$ is $-1.5\times {10}^4\text{kJ}$ is $-3.66\times {10}^2\text{kJ/mol}$.

Explanation of Solution

The balanced chemical equation for the combustion of kerosene $\left({\text{C}}_{\text{12}}{\text{H}}_{26}\right)$ is as follows:

${\text{2C}}_{\text{12}}{\text{H}}_{26}\left(l\right)+37{\text{O}}_2\left(g\right)\to 24{\text{CO}}_2\left(g\right)+26{\text{H}}_{\text{2}}\text{O}\left(\text{g}\right)$

The standard state of oxygen is ${\text{O}}_2\left(g\right)$ so the standard enthalpy of formation for ${\text{O}}_2$ is zero.

The formula to calculate the standard enthalpy of a given reaction $\left(\Delta H_{\text{rxn}}^{°}\right)$ is as follows:

$\Delta H_{\text{rxn}}^{°}=\left[\begin{array}{l}\left\{24\Delta H_{\text{f}}^{°}\left[{\text{CO}}_2\left(g\right)\right]+26\Delta H_{\text{f}}^{°}\left[{\text{H}}_2\text{O}\left(g\right)\right]\right\}\\ -\left\{2\Delta H_{\text{f}}^{°}\left[{\text{C}}_{\text{12}}{\text{H}}_{26}\left(l\right)\right]+32\Delta H_{\text{f}}^{°}\left[{\text{O}}_2\left(g\right)\right]\right\}\end{array}\right]$ (1)

Rearrange equation (1) to calculate $2\Delta H_{\text{f}}^{°}$ of ${\text{C}}_2{\text{H}}_2\left(g\right)$ is as follows:

$2\Delta H_{\text{f}}^{°}\left[{\text{C}}_{\text{12}}{\text{H}}_{26}\left(l\right)\right]=\left[\begin{array}{l}\left(24\Delta H_{\text{f}}^{°}\left[{\text{CO}}_2\left(g\right)\right]+26\Delta H_{\text{f}}^{°}\left[{\text{H}}_2\text{O}\left(g\right)\right]\right)-\\ \left(\Delta H_{\text{rxn}}^{°}+32\Delta H_{\text{f}}^{°}\left[{\text{O}}_2\left(g\right)\right]\right)\end{array}\right]$ (2)

Substitute $-393.5\text{kJ/mol}$ for $\Delta H_{\text{f}}^{°}\left[{\text{CO}}_2\left(g\right)\right]$, $-241.826\text{kJ/mol}$ for $\Delta H_{\text{f}}^{°}\left[{\text{H}}_2\text{O}\left(g\right)\right]$ and $-1.5\times {10}^4\text{kJ}$ for $\Delta H_{\text{rxn}}^{°}$ and 0 for $\Delta H_{\text{f}}^{°}\left[{\text{O}}_2\left(g\right)\right]$ in the equation (2).

$\begin{array}{c}\left(2\text{mol}\right)\Delta H_{\text{f}}^{°}\left[{\text{C}}_{\text{12}}{\text{H}}_{26}\left(l\right)\right]=\left[\begin{array}{l}\left(\text{24 mol}\right)\left(-393.5\text{kJ/mol}\right)+24\left(-241.826\text{kJ/mol}\right)-\\ \left(\left(-1.5\times {10}^4\text{kJ}\right)+37\left(0\right)\right)\end{array}\right]\\ \Delta H_{\text{f}}^{°}\left[{\text{C}}_{\text{12}}{\text{H}}_{26}\left(l\right)\right]=\frac{-\text{731}\text{.476}\text{kJ}}{\text{2}\text{mol}}\\ =-\text{365}\text{.738}\text{kJ/mol}\\ \approx -3.66\times {10}^2\text{kJ/mol}\end{array}$

Conclusion

$\Delta H_{\text{f}}^{°}$ of kerosene when $\Delta H_{\text{rxn}}^{°}$ is $-1.5\times {10}^4\text{kJ}$ is $-3.66\times {10}^2\text{kJ/mol}$.

(c)

Interpretation Introduction

Interpretation:

The heat released by the combustion of $0.50\text{ gal}$ of kerosene $\left({\text{C}}_{\text{12}}{\text{H}}_{26}\right)$ is to be determined.

Concept introduction:

Heat $\left(q\right)$ is the energy transferred as a consequence of the difference of the temperature between the system and surrounding.

The heat released by the system is negative and the heat taken by the system is positive.

The amount of heat released by the system is equal to the amount of heat absorbed by the surrounding.

The conversion factor to convert $\text{gal}$ into $\text{qt}$ is:

$\text{1}\text{gal}=4\text{qt}$

The conversion factor to convert $\text{L}$ into $\text{qt}$ is:

$\text{1}\text{L}=1.057\text{qt}$

Answer to Problem 6.99P

The heat released by the combustion of $0.50\text{ gal}$ of kerosene $\left({\text{C}}_{\text{12}}{\text{H}}_{26}\right)$ is $-6.2\times {10}^4\text{kJ}$.

Explanation of Solution

The formula to calculate moles of kerosene $\left({\text{C}}_{\text{12}}{\text{H}}_{26}\right)$ is as follows:

${\text{Moles of C}}_{\text{12}}{\text{H}}_{26}=\left(\frac{{\text{volume of C}}_{\text{12}}{\text{H}}_{26}}{{\text{molar mass of C}}_{\text{12}}{\text{H}}_{26}}\right)\left({\text{density of C}}_{\text{12}}{\text{H}}_{26}\right)$ (3)

Substitute $0.50\text{ gal}$ for volume of ${\text{C}}_{\text{12}}{\text{H}}_{26}$, $0.749\text{g/mL}$ for density of ${\text{C}}_{\text{12}}{\text{H}}_{26}$ and $170.33\text{g/mol}$ for molar mass of ${\text{C}}_{\text{12}}{\text{H}}_{26}$ in the equation (1).

$\begin{array}{c}{\text{Moles of C}}_8{\text{H}}_{18}=\left(\frac{0.50\text{ gal}}{170.33\text{g/mol}}\right)\left(\frac{\text{4}\text{qt}}{\text{1}\text{gal}}\right)\left(\frac{1\text{L}}{1.057\text{qt}}\right)\left(\frac{1000\text{mL}}{1\text{L}}\right)\left(0.749\text{g/mL}\right)\\ =8.3204282\text{mol}\end{array}$

The heat released by $2\text{mol}$ of ${\text{C}}_{\text{12}}{\text{H}}_{26}$ is $-1.5\times {10}^4\text{kJ}$.

Therefore, the heat released by $8.3204282\text{mol}$ of ${\text{C}}_{\text{12}}{\text{H}}_{26}$ is as follows:

$\begin{array}{c}\text{Heat}\text{released}=\left(8.3204282\text{mol}\right)\left(\frac{-1.5\times {10}^4\text{kJ}}{2\text{mol}}\right)\\ =-6.2403\times {10}^4\text{kJ}\\ \approx -6.2\times {10}^4\text{kJ}\end{array}$

Conclusion

The heat released by the combustion of $0.50\text{ gal}$ of kerosene $\left({\text{C}}_{\text{12}}{\text{H}}_{26}\right)$ is $-6.2\times {10}^4\text{kJ}$.

(d)

Interpretation Introduction

Interpretation:

The volume of kerosene in $\text{gal}$ that must be burned for a kerosene furnace to produce $1250\text{Btu}$ is to be calculated.

Concept introduction:

Heat $\left(q\right)$ is the energy transferred as a consequence of the difference of the temperature between the system and surrounding.

The heat released by the system is negative and the heat taken by the system is positive.

The amount of heat released by the system is equal to the amount of heat absorbed by the surrounding.

The conversion factor to convert $\text{Btu}$ into $\text{kJ}$$\text{qt}$ is:

$\text{1}\text{Btu}=1.055\text{kJ}$

Answer to Problem 6.99P

The volume of kerosene in $\text{gal}$ is $1.1\times {10}^{-2}\text{gal}$.

Explanation of Solution

The heat released by the combustion of $0.50\text{ gal}$ of kerosene $\left({\text{C}}_{\text{12}}{\text{H}}_{26}\right)$ is $6.2\times {10}^4\text{kJ}$.

The formula to calculate the volume is as follows:

$\text{Volume}=\left(\text{energy of kerotene}\right)\left(\frac{0.50\text{ gal}}{6.2\times {10}^4\text{kJ}}\right)$ (4)

Substitute $1250\text{Btu}$ for energy of kerotene in the equation (4).

$\begin{array}{c}\text{Volume}=\left(1250\text{Btu}\right)\left(\frac{1.055\text{kJ}}{1\text{Btu}}\right)\left(\frac{0.50\text{ gal}}{6.2\times {10}^4\text{kJ}}\right)\\ \approx 1.1\times {10}^{-2}\text{gal}\end{array}$

Conclusion

The volume of kerosene in $\text{gal}$ is $1.1\times {10}^{-2}\text{gal}$.