Problem 61 - General Chemistry

Question

Propane, \(\mathrm{C}_{3} \mathrm{H}_{8}\), is a common fuel gas. Use the following to calculate the grams of propane you would need to provide \(369 \mathrm{~kJ}\) of heat. $$ \begin{aligned} \mathrm{C}_{3} \mathrm{H}_{8}(g)+5 \mathrm{O}_{2}(g) & \longrightarrow 3 \mathrm{CO}_{2}(g)+4 \mathrm{H}_{2} \mathrm{O}(g) \\ \Delta H=-2043 \mathrm{~kJ} \end{aligned} $$

Step-by-Step Solution

Answer

You need approximately 7.97 grams of propane.

Step 1: Understand the Reaction

The given chemical equation is of propane combusting with oxygen to produce carbon dioxide and water. The enthalpy change \( \Delta H \) is \(-2043\, \text{kJ}\), which means that this amount of heat is released when 1 mole of propane combusts.

Step 2: Determine Heat Requirement per Mole

We need to determine the grams of propane required to produce \(369\, \text{kJ}\) of heat. Recognize that \(-2043\, \text{kJ}\) is released per mole of propane from the chemical reaction.

Step 3: Calculate Moles of Propane Needed

Calculate the moles of propane needed using the proportion: \(\text{Moles of } \mathrm{C}_3\mathrm{H}_8 = \frac{369\, \text{kJ}}{2043\, \text{kJ/mol}}\).If you solve this, you will find:\(\text{Moles of } \mathrm{C}_3\mathrm{H}_8 \approx 0.1807\, \text{mol}\).

Step 4: Convert Moles to Grams

The molar mass of propane \( \mathrm{C}_3\mathrm{H}_8 \) is calculated as follows:- Carbon: \(12.01\, \text{g/mol} \times 3 = 36.03\, \text{g/mol}\)- Hydrogen: \(1.008\, \text{g/mol} \times 8 = 8.064\, \text{g/mol}\)Total molar mass of propane = \(44.094\, \text{g/mol}\).Now, multiply the moles by the molar mass to find the grams: \( 0.1807\, \text{mol} \times 44.094\, \text{g/mol} \approx 7.97\, \text{g}\).

Key Concepts

Combustion Reaction

A combustion reaction occurs when a substance combines with oxygen, releasing energy in the form of heat and light. In the case of propane (\(\mathrm{C}_{3} \mathrm{H}_{8}\)), this gas combines with oxygen (\(\mathrm{O}_{2}\)) to produce carbon dioxide (\(\mathrm{CO}_{2}\)) and water (\(\mathrm{H}_{2} \mathrm{O}\)). This process is exothermic, meaning it releases energy. The balanced chemical reaction for the combustion of propane is:\( \mathrm{C}_{3} \mathrm{H}_{8}(g) + 5 \mathrm{O}_{2}(g) \longrightarrow 3 \mathrm{CO}_{2}(g) + 4 \mathrm{H}_{2} \mathrm{O}(g) \).
This equation indicates that one mole of propane requires five moles of oxygen to completely combust and produce three moles of carbon dioxide and four moles of water. Such reactions are essential in everyday life as they power vehicles, heat our homes, and much more. Understanding combustion reactions helps us efficiently use fuels and minimize pollution, as efficient combustion reduces the amount of harmful byproducts.

Molar Mass

Molar mass is the weight of one mole of a given substance. It's crucial in converting moles into grams, which is a practical measurement unit in chemistry. The molar mass of a molecule is calculated by adding up the atomic masses of its constituent elements. This is derived from the periodic table. Let's take propane (\( \mathrm{C}_{3} \mathrm{H}_{8} \)), for example:
- Carbon has an atomic mass of approximately \(12.01 \, \text{g/mol}\). With three carbon atoms, the total mass contributed by carbon is \( 3 \times 12.01 = 36.03 \, \text{g/mol}\).
- Hydrogen has an atomic mass of about \(1.008 \, \text{g/mol}\). With eight hydrogen atoms, the total mass from hydrogen is \( 8 \times 1.008 = 8.064 \, \text{g/mol}\).
Combining these, the total molar mass of propane is calculated as \(36.03 \, \text{g/mol} + 8.064 \, \text{g/mol} = 44.094 \, \text{g/mol}\).
Knowing the molar mass allows chemists to convert between the mass of substances and the amount in moles, which is fundamental for measuring chemical reactions precisely.

Energy Conversion

Energy conversion in chemical reactions is the transformation of potential chemical energy into other forms of energy, most often heat. For instance, when propane combusts, its stored chemical potential energy is irradiated as heat, which we measure in kilojoules (kJ). In our combustion example, the energy conversion can be seen in the reaction's enthalpy change (\( \Delta H = -2043 \, \text{kJ} \)).
- \(\Delta H\) denotes the amount of heat energy released or absorbed during a reaction.
- A negative \(\Delta H\) means the reaction releases energy (exothermic).
In this instance, when one mole of propane combusts, \(2043 \, \text{kJ}\) is released. Importantly, understanding energy conversion assists us in calculating the amount of fuel needed for specific energy productions, as in determining how many grams of propane we need to produce \(369 \, \text{kJ}\) of energy. This is vital for applications in energy management and harnessing energy effectively.

Chemical Thermodynamics

Chemical thermodynamics involves studying the relationship between heat and other forms of energy during chemical processes. It helps us predict whether reactions can occur spontaneously, determine energy changes, and understand how energy is distributed.
Some key concepts include:

• **Enthalpy (\(\Delta H\))**: Represents heat changes at constant pressure. For our reaction, \(\Delta H = -2043 \, \text{kJ} \) indicates that the reaction is exothermic.
• **Entropy (\(\Delta S\))**: Reflects disorder or randomness in a system. Generally, gaseous products have higher entropy than solids or liquids.
• **Gibbs Free Energy (\(\Delta G\))**: Combines enthalpy and entropy to determine reaction spontaneity. A negative \(\Delta G\) suggests a spontaneous reaction.

By integrating these principles, chemical thermodynamics allows us to anticipate and control chemical processes. This understanding not only aids in solving textbook problems like those involving combustion but also applies to industries where efficient energy use is critical, such as manufacturing and environmental science.