Question

Kerosene, a common space-heater fuel, is a mixture of hydrocarbons whose "average" formula is \(\mathrm{C}_{12} \mathrm{H}_{26}\) (a) Write a balanced equation, using the simplest whole-number coefficients, for the complete combustion of kerosene to gases. (b) If \(\Delta H_{\mathrm{rn}}^{\circ}=-1.50 \times 10^{4} \mathrm{~kJ}\) for the combustion equation as written in part (a), determine \(\Delta H_{\mathrm{f}}^{\circ}\) of kerosene. (c) Calculate the heat released by combustion of 0.50 gal of kerosene \((d\) of kerosene \(=0.749 \mathrm{~g} / \mathrm{mL})\) (d) How many gallons of kerosene must be burned for a kerosene furnace to produce \(1250 .\) Btu \((1 \mathrm{Btu}=1.055 \mathrm{~kJ}) ?\)

Short answer

The balanced equation is \(\text{C}_{12}\text{H}_{26} + 18.5\text{O}_2 \rightarrow 12\text{CO}_2 + 13\text{H}_2O\). The kerosene enthalpy of formation is \(-832.3\) kJ/mol. Total heat released by burning 0.50 gal of kerosene is \(-1.25 \times 10^3 \text{kJ}\). To produce 1250 Btu, \(\text{0.005 gal}\) of kerosene is needed.

Step-by-step solution

Write a balanced equation for the combustion

The combustion of hydrocarbons involves the reaction with oxygen to form carbon dioxide and water. The general form of the reaction for \(\text{C}_{12}\text{H}_{26}\) is: \[ \text{C}_{12}\text{H}_{26} + O_2 \rightarrow CO_2 + H_2O \] To balance the equation, start with carbon, hydrogen, then oxygen: \[ \text{C}_{12}\text{H}_{26} + 18.5O_2 \rightarrow 12 CO_2 + 13 H_2O \]

Calculate \(\text{ΔH}_{\text{f}}^\text{o}\) of kerosene

Use the given \(\text{ΔH}_{\text{rn}}^\text{o} = -1.50 \times 10^4 \text{kJ}\). The combustion equation is: \[ \text{C}_{12}\text{H}_{26} + 18.5\text{O}_2 \rightarrow 12\text{CO}_2 + 13\text{H}_2O \] The enthalpy change for the reaction is the sum of the formation enthalpies. Given \(\text{ΔH}_{\text{f}}^\text{o} (\text{CO}_2) = -393.5 \text{kJ/mol}, \text{ΔH}_{\text{f}}^\text{o} (\text{H}_2O) = -241.8 \text{kJ/mol}\): \[ \text{ΔH}_{\text{f}}^\text{o} (\text{C}_{12}\text{H}_{26}) = \frac{1.50 \times 10^4 \text{kJ} - [12(-393.5) + 13(-241.8)]}{1} \]

Heat released by combustion of 0.50 gal of kerosene

First, convert gallons to mL (1 gal = 3,785.41 mL). \[ 0.50 \text{ gal} \times 3,785.41 \text{ mL/gal} = 1,892.705 \text{ mL} \] Calculate mass using density (d = 0.749 g/mL): \[ 1,892.705 \text{ mL} \times 0.749 \text{ g/mL} = 1,417.17 \text{ g} \] Use the molar mass of \(\text{C}_{12}\text{H}_{26} = 170 \text{ g/mol}\): \[ \text{moles of kerosene} = \frac{1,417.17 \text{ g}}{170 \text{ g/mol}} = 8.34 \text{ moles} \] Heat released: \[ 8.34 \text{ moles} \times -1.50 \times 10^4 \text{ kJ/mol} = -1.25 \times 10^3 \text{ kJ} \]

Kerosene burned to produce 1250 Btu

First, convert Btu to kJ: \[ 1250 \text{ Btu} \times 1.055 \text{ kJ/Btu} = 1,318.75 \text{ kJ} \] Using the heat released per mole of kerosene: \[ \text{moles of kerosene} = \frac{1318.75 \text{ kJ}}{1.50 \times 10^4 \text{ kJ/mol}} \] Convert moles to mass: \[ \text{mass of kerosene} = \text{moles} \times 170 \text{ g/mol} = 15 \text{ g} \] Convert mass to volume: \[ \text{Volume (mL)} = \frac{15 \text{ g}}{0.749 \text{ g/mL} } = 20 \text{ mL} \] Convert mL to gallons: \[ \text{Volume (gal)} = \frac{20 \text{ mL}}{3785.41 \text{ mL/gal} } = 0.005 \text{ gal} \]

Key concepts

balanced chemical equation

The first step in solving any combustion problem is to write the balanced chemical equation. This involves hydrocarbons reacting with oxygen to produce carbon dioxide and water. For this exercise, kerosene's average chemical formula is given as \(\text{C}_{12}\text{H}_{26}\). The unbalanced equation is:
\[\text{C}_{12}\text{H}_{26} + O_2 \rightarrow CO_2 + H_2O\]
To balance it, we start by balancing the carbons, then the hydrogens, and finally the oxygens. The balanced equation is:
\[\text{C}_{12}\text{H}_{26} + 18.5O_2 \rightarrow 12 CO_2 + 13 H_2O\]
This balanced equation ensures that for every molecule of kerosene, 18.5 molecules of oxygen are required, yielding 12 molecules of carbon dioxide and 13 molecules of water.

enthalpy of formation

Understanding the enthalpy of formation (\( \Delta H_{f}^{o}\)) is crucial for calculating the heat produced or consumed in chemical reactions. The enthalpy of formation of a compound is the heat change when one mole of a compound is formed from its elements in their standard states.
In this exercise, you are given the reaction enthalpy for the combustion of kerosene (\( \Delta H_{rn}^{o} =-1.50 \times 10^{4} \text{kJ}\)). The goal is to find the enthalpy of formation for kerosene. The combustion reaction is:
\[ \text{C}_{12} \text{H}_{26} +18.5 \text{O}_{2} \rightarrow 12 \text{CO}_{2} + 13 \text{H}_{2} \text{O} \]
Using the enthalpy of formation for \( \text{CO}_{2} \) (-393.5 kJ/mol) and \( \text{H}_{2} \text{O} \) (-241.8 kJ/mol), we can apply Hess’s Law to find the unknown \( \Delta H_{f}^{o} \text{ of kerosene} \).
Thus,
\[ \Delta H_{f}^{o} (\text{C}_{12}\text{H}_{26}) = \frac{1.50 \times 10^{4} \text{kJ} - [12(-393.5) + 13(-241.8)]}{1} \] This allows us to solve for the unknown enthalpy.

heat released by combustion

To calculate the heat released by burning a specific amount of kerosene, we need to follow unit conversion steps and use density and molecular weight data. First, convert the volume of kerosene from gallons to milliliters (1 gal = 3,785.41 mL).
For 0.50 gal: \[ 0.50 \text{ gal} \times 3785.41 \text{ mL/gal} = 1892.705 \text{ mL} \]
Use the density of kerosene (0.749 g/mL) to find its mass: \[ 1892.705 \text{ mL} \times 0.749 \text{ g/mL} = 1417.17 \text{ g} \]
Next, convert the mass to moles using the molar mass of \( \text{C}_{12}\text{H}_{26} \) (170 g/mol): \[ \text{moles of kerosene} = \frac{1417.17 \text{ g}}{170 \text{ g/mol}} = 8.34 \text{ moles} \]
Finally, calculate the heat released for the combustion: \[ 8.34 \text{ moles} \times -1.50 \times 10^{4} \text{ kJ/mol} = -1.25 \times 10^{3} \text{ kJ} \]

conversion between units

Unit conversions are essential in chemistry for comparing different measurements. Let's start with converting energy units from Btu to kJ (1 Btu = 1.055 kJ). For a kerosene furnace to produce 1250 Btu: \[ 1250 \text{ Btu} \times 1.055 \text{ kJ/Btu} = 1318.75 \text{ kJ} \]
Use the heat released per mole of kerosene to determine the moles needed: \[ \text{moles of kerosene} = \frac{1318.75 \text{ kJ}}{1.50 \times 10^{4} \text{ kJ/mol}} = 0.088 \text{ moles} \]
Convert moles to mass: \[ \text{mass of kerosene} = 0.088 \text{ moles} \times 170 \text{ g/mol} = 15 \text{ g} \]
Convert the mass to volume using the density: \[ \text{Volume (mL)} = \frac{15 \text{ g}}{0.749 \text{ g/mL} } = 20 \text{ mL} \]
Finally, convert milliliters to gallons: \[ \text{Volume (gal)} = \frac{20 \text{ mL}}{3785.41 \text{ mL/gal} } = 0.005 \text{ gal} \]
These conversions ensure accurate and applicable results for practical applications.