Chapter 9, Problem 9.101PAE

Interpretation Introduction

Interpretation:

The quantity of energy released by 25.2 billion barrels of oil should be calculated.

Concept Introduction:

Oil can be converted to gasoline. Pure gasoline contains octane. Combustion of octane produces carbon dioxide and water vapor. Energy is also released during the combustion of octane.

Answer to Problem 9.101PAE

Solution:

$\text{6}{\text{.89×10}}^{\text{16}}\text{kJ}$ of energy would be released.

Explanation of Solution

1 barrel of oil can produce 19.5 gallons = 73815.5 mL of gasoline

We know, 1 gallon = 3785.41 mL

Therefore, 19.5 gallons $\text{=}\frac{\left(\text{3785}\text{.41 mL}\right)\text{×}\left(\text{19}\text{.5 gallon}\right)}{\left(\text{1 gallon}\right)}\text{=73815}\text{.5 mL}$

25.2 billion or 25,200,000,000 barrels of oil can produce

[Since, 1 billion = 1,000,000,000]

$\text{=}\frac{\left(\text{73815}\text{.5 mL gasoline}\right)\text{×}\left(\text{25,200,000,000 barrels of oil}\right)}{\left(\text{1 barrel of oil}\right)}\text{=1}{\text{.86×10}}^{\text{15}}\text{ mL}$ of gasoline.

Gasoline contains pure octane with a density of 0.692 g/mL.

Mass of octane in $\text{1}{\text{.86×10}}^{\text{15}}\text{mL}$ of gasoline

$\begin{array}{l}\text{=}\left(\text{1}{\text{.86×10}}^{\text{15}}\text{ mL×0}\text{.692 g/mL}\right)\\ \text{=1}{\text{.287×10}}^{\text{15}}\text{ g}\end{array}$

Moles of octane in $\text{1}{\text{.86×10}}^{\text{15}}\text{ mL}$ of gasoline

$\begin{array}{l}\text{=}\frac{\text{1}{\text{.287×10}}^{\text{15}}\text{ g}}{\text{114}\text{.23 g/mol}}\\ \text{=1}{\text{.126×10}}^{\text{13}}\text{ moles}\end{array}$

The balanced chemical equation for the combustion of octane is as follows:

$\begin{array}{l}{\text{2C}}_{\text{8}}{\text{H}}_{\text{18}}{\text{(l)+25O}}_{\text{2}}\text{(g)}\to {\text{16CO}}_{\text{2}}{\text{(g)+18H}}_{\text{2}}\text{O(g)}\\ {\text{C}}_{\text{8}}{\text{H}}_{\text{18}}\text{(l)+12}{\text{.5O}}_{\text{2}}\text{(g)}\to {\text{8CO}}_{\text{2}}{\text{(g)+9H}}_{\text{2}}\text{O(g)}\end{array}$

From Hess’s law, we know,

Change in enthalpy in a chemical reaction $$\left({\text{ΔH}}_{\text{rxn}}\right)\text{=}{\displaystyle \sum {\text{ΔH}}_{\text{f}}^{\text{o}}{}_{\text{(products)}}}\text{-}{\displaystyle \sum {\text{ΔH}}_{\text{f(reactants)}}^{\text{o}}}$$

Change in enthalpy for the combustion of one-mole octane

$\begin{array}{l}\text{=}\left({\text{8ΔH}}_{{\text{f(CO}}_{\text{2}}\text{)}}^{\text{o}}{\text{+9ΔH}}_{{\text{f(H}}_{\text{2}}\text{O)}}^{\text{o}}\right)\text{-}\left({\text{ΔH}}_{{\text{f(C}}_{\text{8}}{\text{H}}_{\text{18}}}^{\text{o}}\text{+12}{\text{.5ΔH}}_{{\text{f(O}}_{\text{2}}\text{)}}^{\text{o}}\right)\\ \text{=}\left[\text{8}\left(\text{-393}\text{.5}\right)\right]\text{+9}\left(\text{-285}\text{.8}\right)\text{]-}\left[\left(\text{-250}\text{.3}\right)\text{+12}\text{.5}\left(\text{0}\right)\right]\text{ kJ}\\ \text{=}\left[\left(\text{-3148-2572}\text{.2}\right)\text{+250}\text{.3}\right]\text{ kJ}\\ \text{=}\left(\text{-5720}\text{.2+250}\text{.3}\right)\text{ kJ}\\ \text{=-5469}\text{.9 kJ}\end{array}$

Change in enthalpy during combustion of $\text{1}{\text{.26×10}}^{\text{13}}\text{ moles}$ octane

$\begin{array}{l}\text{=}\left(\text{-5469}\text{.9×1}{\text{.26×10}}^{\text{13}}\right)\text{ kJ}\\ \text{=-6}{\text{.89×10}}^{\text{16}}\text{ kJ}\end{array}$

Therefore, $\text{6}{\text{.89×10}}^{\text{16}}\text{kJ}$ of energy would be released as heat.

Conclusion

The quantity of heat energy released by 25.2 billion barrels of oil has been calculated.