Chapter 6, Problem 6.74P

Kerosene, a common space-heater fuel, is a mixture of hydrocarbons whose “average” formula is ${\text{C}}_{12}{\text{H}}_{26}.$

1. Write a balanced equation, using the simples whole-number coefficients, for the complete combustion of kerosene to gases.
2. If $\text{Δ}{\text{H}}_{rxn}^o=-1.50\times {10}^4\text{ kJ}$ for the combustion equation as written in part (a), determine $\text{Δ}{\text{H}}_f^o$ of kerosene.
3. Calculate the heat released by combustion of 0.50 gal of kerosene $(d\text{of kerosene = 0}\text{.749 g/mL)}\text{.}$ How many gallons of kerosene must be burned for a kerosene furnace to produce 1250. Btu $(1\text{Btu = 1}\text{.055 kJ)?}$

Interpretation Introduction

Interpretation:The balanced equation for the complete combustion of kerosene to gases needs to be determined.

Concept Introduction:In a balanced chemical equation, the number of atoms of individual element as reactant is equal to the number of atoms of that element as a product.

Answer to Problem 6.74P

The balanced equation is,

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

Explanation of Solution

Kerosene reacts with oxygen to give carbon dioxide and water.

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

The above equation is balanced chemical equation as the number of atoms of each element is equal in both the sides of the reaction, that is, reactant side and product side.

Interpretation Introduction

Interpretation:The value of $\Delta H^0{}_f$ of kerosene needs to be determined.

Concept Introduction : The amount of energy released on the combustion of an organic compound is known as enthalpy of combustion. The formula of determination of $\Delta H_{\text{rxn}}$ is as follows:

  $\Delta H^0{}_{\text{rxn}}={\displaystyle \sum _{}^{}m\Delta H^0{}_{\text{f}}(\text{products})-}{\displaystyle \sum n\Delta H^0{}_{\text{f}}}(\text{reactants})$

Here, ${\displaystyle \sum }$ represents the sum of terms, m and n are the number of moles of products and reactants respectively. The term $H^0{}_{\text{f}}$ is standard heat of formation

Answer to Problem 6.74P

  ${\text{ΔH}}^{\text{0}}{}_{\text{f}}\text{of}\left[{\text{C}}_{\text{12}}{\text{H}}_{\text{26}}\text{(l)}\right]\text{=}-\text{3}\text{.66}\text{×}{\text{10}}^{\text{2}}\text{kJ}$

Explanation of Solution

The given values are as follows:

  $\begin{array}{l}{\text{ΔH}}^{\text{0}}{}_{\text{rxn}}\text{=}-\text{1}\text{.50}\text{×}{\text{10}}^{\text{4}}\text{kJ,}\\ {\text{ΔH}}^{\text{0}}{}_{\text{f}}\text{for}{\text{CO}}_{\text{2}}\text{=}-\text{393}\text{.5}\text{kJ/mol,}\\ {\text{ΔH}}^{\text{0}}{}_{\text{f}}\text{of}{\text{H}}_{\text{2}}\text{O}\text{(g)=}-\text{241}\text{.826}\text{kJ/mol,}\\ {\text{ΔH}}^{\text{0}}{}_{\text{f}}\text{of}{\text{O}}_{\text{2}}\text{(g)=}\text{0}\text{kJ/mol}\end{array}$

The heat of formation of kerosene is calculated by the formula:

  ${\text{ΔH}}^{\text{0}}\text{rxn=}\{\text{24}{\text{ΔH}}^{\text{0}}{}_{\text{f}}\left[{\text{CO}}_{\text{2}}\text{(g)}\right]\text{+}\text{26}{\text{ΔH}}^{\text{0}}{}_{\text{f}}\left[{\text{H}}_{\text{2}}\text{O}\text{(g)}\right]\}\text{-}\left\{\text{2}{\text{ΔH}}^{\text{0}}{}_{\text{f}}\left[{\text{C}}_{\text{12}}{\text{H}}_{\text{26}}\text{(l)}\right]\text{+}\text{37}{\text{ΔH}}^{\text{0}}{}_{\text{f}}\left[{\text{O}}_{\text{2}}\text{(g)}\right]\right\}$

Substitute all the above values.

  $\begin{array}{c}\left[{\text{CO}}_{\text{2}}\text{(g)}\right]\text{+}\text{26}{\text{ΔH}}^{\text{0}}{}_{\text{f}}\left[{\text{H}}_{\text{2}}\text{O}\text{(g)}\right]\}-\{\begin{array}{l}\text{2}{\text{ΔH}}^{\text{0}}{}_{\text{f}}\left[{\text{C}}_{\text{12}}{\text{H}}_{\text{26}}\text{(l)}\right]\\ \text{+}\text{37}{\text{ΔH}}^{\text{0}}{}_{\text{f}}\left[{\text{O}}_{\text{2}}\text{(g)}\right]\end{array}\}\\ -\text{1}\text{.50}\text{×}{\text{10}}^{\text{4}}\text{kJ}\text{=}\{\text{24}\text{×}\left(-\text{393}\text{.5}\text{kJ/mol}\right)\text{+}\text{26}\text{×}\left(-\text{241}\text{.826}\text{kJ/mol}\right)\}-\{\begin{array}{l}\text{2}{\text{ΔH}}^{\text{0}}{}_{\text{f}}\left[{\text{C}}_{\text{12}}{\text{H}}_{\text{26}}\text{(l)}\right]\\ \text{+}\text{37}\text{×}\left(\text{0}\text{kJ/mol}\right)\end{array}\}\end{array}$

  $\begin{array}{c}-\text{1}\text{.50}\text{×}{\text{10}}^{\text{4}}\text{kJ}\text{=}-\text{9444}\text{kJ}-\text{6287}\text{.476}\text{kJ}-\text{2}{\text{ΔH}}^{\text{0}}{}_{\text{f}}\left[{\text{C}}_{\text{12}}{\text{H}}_{\text{26}}\text{(l)}\right]\text{-}\text{0}\text{kJ}\\ \text{2}{\text{ΔH}}^{\text{0}}{}_{\text{f}}\left[{\text{C}}_{\text{12}}{\text{H}}_{\text{26}}\text{(l)}\right]\text{=}-\text{731}\text{.476}\text{kJ}\\ {\text{ΔH}}^{\text{0}}{}_{\text{f}}\left[{\text{C}}_{\text{12}}{\text{H}}_{\text{26}}\text{(l)}\right]\text{=}-\text{3}\text{.66}\text{×}{\text{10}}^{\text{2}}\text{kJ}\end{array}$.

Interpretation Introduction

Interpretation:The heat produced by combustion of 0.50 gal of kerosene needs to be determined.

Concept Introduction : Combustion of Kerosene gives carbon dioxide and water.

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

Answer to Problem 6.74P

Heat produced by the combustion of kerosne is $-\text{6}\text{.3}\text{×}{\text{10}}^{\text{4}}\text{kJ}$

Explanation of Solution

Amount of given kerosene =0.50 gal

Density of kerosene = 0.749 g/mL

  $\begin{array}{l}\text{Now,}\text{volume}\text{=}\text{0}\text{.50}\text{gal}\text{×}\frac{\text{3}\text{.785}\text{L}}{\text{1}\text{gal}}\text{=}\text{1}\text{.89}\text{L}\\ \text{And}\text{Mass}\text{=}\text{1}\text{.89}\text{L}\text{×}\frac{\text{0}\text{.749}\text{g}}{\text{mL}}\text{×}\frac{\text{1000}\text{mL}}{\text{1}\text{L}}\text{=}\text{1}\text{.42}\text{×}{\text{10}}^{\text{3}}\text{g}\end{array}$

Calculate the moles using mass and molar mass.

  $\text{1}\text{.42}\text{×}{\text{10}}^{\text{3}}\text{g}\text{×}\frac{\text{1}\text{mol}{\text{C}}_{\text{12}}{\text{H}}_{\text{26}}}{\text{170}\text{.34}\text{g}}\text{=}\text{8}\text{.34}\text{mol}{\text{C}}_{\text{12}}{\text{H}}_{\text{26}}$

  $\text{8}\text{.34}\text{mol}{\text{C}}_{\text{12}}{\text{H}}_{\text{26}}\text{×}\frac{\text{-1}\text{.50}\text{×}{\text{10}}^{\text{4}}\text{kJ}}{\text{2}\text{mol}{\text{C}}_{\text{12}}{\text{H}}_{\text{26}}}\text{=}\text{-6}\text{.3}\text{×}{\text{10}}^{\text{4}}\text{kJ}$

  $\begin{array}{l}\\ \text{Therefore}\text{the}\text{heat}\text{produced=}\text{-6}\text{.3}\text{×}{\text{10}}^{\text{4}}\text{kJ}\end{array}$

Interpretation Introduction

Interpretation:The amount of kerosene that must be burned for a kerosene furnace to produce 1250.Btu needs to be determined.

Concept Introduction: Heating values for combustion of kerosene is 43.1 MJ/kg and 46.2 MJ/kg for lower and higher heating values respectively. Combustion is the reaction of a compound with oxygen. It generates energy.

Answer to Problem 6.74P

The volume of kerosene required is $\text{1}\text{.0}\text{×}{\text{10}}^{\text{-2}}\text{gal}$

Explanation of Solution

Amount of kerosene required is $\text{1}\text{.0}\text{×}{\text{10}}^{\text{-2}}\text{gal}$

  $\text{1250}\text{.}\text{Btu}\text{×}\frac{\text{1}\text{.055}\text{kJ}}{\text{1}\text{Btu}}\text{=}\text{1}\text{.32}\text{×}{\text{10}}^{\text{3}}\text{kJ}$

Amount of heat produced by 0.50 gal is $\text{-}\text{6}\text{.3}\text{×}{\text{10}}^{\text{4}}\text{kJ}$ .

  $\text{1}\text{.32}\text{×}{\text{10}}^{\text{3}}\text{kJ}\text{×}\frac{\text{0}\text{.50}\text{gal}}{\text{6}\text{.3}\text{×}{\text{10}}^{\text{4}}\text{kJ}}\text{=}\text{1}\text{.0}\text{×}{\text{10}}^{\text{-2}}\text{gal}$