Question

Propane \(\left(\mathrm{C}_{3} \mathrm{H}_{8}\right)\) gas \(\left(d=1.83 \mathrm{kg} / \mathrm{m}^{3}\right)\) is used in most gas grills. What volume (in liters) of propane is needed to generate \(273.8 \mathrm{kJ}\) of heat? $$\begin{array}{r} \mathrm{C}_{3} \mathrm{H}_{8}(\mathrm{g})+5 \mathrm{O}_{2}(\mathrm{g}) \longrightarrow 3 \mathrm{CO}_{2}(\mathrm{g})+4 \mathrm{H}_{2} \mathrm{O}(\mathrm{l}) \\ \Delta H^{\circ}=-2219.9 \mathrm{kJ} \end{array}$$

Short answer

The volume of propane needed to generate 273.8 kJ of heat is \[\frac{\left(\frac{273.8}{2219.9} \times 44.09\right)}{1.83} \, \mathrm{L}\]

Step-by-step solution

Calculate the Moles of Propane Needed

First, identify the amount of heat needed, which is given as \(273.8 \, \mathrm{kJ}\). We also have the heat of combustion of propane \(\left(\Delta H^{\circ}\right) = -2219.9 \, \mathrm{kJ}\). The number of moles of propane (\(\mathrm{C}_{3} \mathrm{H}_{8}\)) required can be calculated using the equation: \[\mathrm{n} = \frac{\mathrm{Heat} \, \mathrm{needed}}{\mathrm{Heat} \, \mathrm{generated} \, \mathrm{by} \, \mathrm{one} \, \mathrm{mole} \, \mathrm{of} \, \mathrm{Propane}}\] Substituting the given values gives: \[\mathrm{n} = \frac{273.8}{2219.9} \, \mathrm{mol}\]

Calculate the Mass of Propane Needed

The mass of propane needed (\(\mathrm{m}\)) can be calculated from the number of moles and the molecular weight of propane. The molecular weight of propane (\(\mathrm{C}_{3} \mathrm{H}_{8}\)) is \(3(12.01) + 8(1.01) = 44.09 \, \mathrm{g/mol}\). Hence the mass needed can be calculated using the equation: \[\mathrm{m} = \mathrm{n} \times \mathrm{molar} \, \mathrm{mass}\] Substituting the values gives: \[\mathrm{m} = \frac{273.8}{2219.9} \times 44.09\]

Calculate the Volume of Propane Needed

The volume of propane needed (\(\mathrm{V}\)) can be calculated from the mass and the given density using the equation: \[\mathrm{V} = \frac{\mathrm{m}}{\mathrm{Density}}\] Substituting the values gives: \[\mathrm{V} = \frac{\left(\frac{273.8}{2219.9} \times 44.09\right)}{1.83} \, \mathrm{L}\]

Key concepts

Propane Combustion

Propane combustion is a chemical process in which propane (\(\text{C}_3\text{H}_8\)) reacts with oxygen to produce carbon dioxide and water. This type of combustion releases a significant amount of energy in the form of heat, which makes propane a popular fuel choice for heating and cooking.

• The balanced chemical equation for propane combustion is:
• \(\text{C}_3\text{H}_8(\text{g}) + 5\text{O}_2(\text{g}) \rightarrow 3\text{CO}_2(\text{g}) + 4\text{H}_2\text{O}(\text{l})\)

This reaction is highly exothermic because it releases more energy than it consumes, as evidenced by the negative enthalpy change \(\Delta H^\circ = -2219.9\, \text{kJ/mol}\). Negative \(\Delta H^\circ\) indicates heat release, highlighting the energy efficiency in burning propane for practical energy use.

Thermochemistry

Thermochemistry is the branch of chemistry that studies the energy and heat involved in chemical reactions. It allows us to understand how much energy is released or absorbed during a chemical transformation.

• In our example, thermochemistry is used to determine the amount of heat that propane combustion can produce.

When propane is burned, it releases 2219.9 kJ of energy per mole of propane. This is known as the enthalpy of combustion. If one wants to generate a specific amount of heat, it is possible to calculate the requisite amount of propane using \[\text{\(n\)} = \frac{\text{Heat needed }}{\text{Heat generated by one mole of propane}} \] For \(273.8\, \text{kJ}\), this can be solved to find the moles of propane needed.

Gas Density

Gas density is an important physical property that helps relate the mass of a gas to its volume. It is generally expressed in terms of kg/m³. Understanding gas density allows us to convert between the mass of a gas and the volume it occupies.

• Propane has a density of \(1.83\,\text{kg/m}^3\).

This means that in each cubic meter of the gas, you have \(1.83\,\text{kg}\) of propane. Once the mass required is known, we can use the formula \[\mathrm{Volume} = \frac{\mathrm{Mass}}{\mathrm{Density}}\] to calculate the volume of propane necessary for generating a certain amount of heat, like \(273.8\,\text{kJ}\), by applying the density value.

Stoichiometry

Stoichiometry is the branch of chemistry that deals with the quantitative relationships between the reactants and products in a chemical reaction. It involves using balanced chemical equations to calculate the amount of reactants required or products formed.

• In combustion reactions, stoichiometry helps in determining the amount of oxygen needed to fully combust propane.

For the equation \(\text{C}_3\text{H}_8 + 5\text{O}_2 \rightarrow 3\text{CO}_2 + 4\text{H}_2\text{O}\), stoichiometry indicates that one mole of propane reacts with five moles of oxygen. By knowing the moles of propane needed (calculated from the thermochemistry section), we can figure out how much volume of propane gas is needed to produce a set amount of heat using the stoichiometric ratios derived from the balanced equation.