Question

Assuming gasoline is pure \(\mathrm{C}_{8} \mathrm{H}_{1 \mathrm{~s}}(l)\), predict the signs of \(q\) and \(w\) for the process of combusting gasoline into \(\mathrm{CO}_{2}(g)\) and \(\mathrm{H}_{2} \mathrm{O}(\mathrm{g}) .\)

Short answer

In the combustion of gasoline (\(\mathrm{C}_{8} \mathrm{H}_{18}(l)\)) into \(\mathrm{CO}_{2}(g)\) and \(\mathrm{H}_{2} \mathrm{O}(g)\), the signs of \(q\) and \(w\) are both negative. This is because it is an exothermic reaction, releasing heat to the surroundings (q < 0), and the system does work on the surroundings by expanding the gaseous products against the pressure exerted by the surroundings (w < 0).

Step-by-step solution

Combustion Reaction

Combustion is an exothermic reaction, meaning it releases heat to the surroundings. The balanced chemical equation for the combustion of gasoline is: \( C_{8}H_{18}(l) + \frac{25}{2} O_{2}(g) \rightarrow 8 CO_{2}(g) + 9 H_{2}O(g) \)

Determine the Sign of \(q\) (Heat)

Since combustion is an exothermic reaction, heat is released by the system (reaction) to the surroundings. Therefore, the sign of \(q\) is negative. q < 0

Determine the Sign of \(w\) (Work)

In the combustion reaction, the system (reaction) does work on the surroundings by expanding the products (8 moles of CO₂ and 9 moles of H₂O) against the pressure exerted by the surroundings (since more moles of gaseous products are formed than the moles of gaseous reactants). This means work is done by the system on the surroundings. Therefore, the sign of \(w\) is negative. w < 0 In conclusion, both \(q\) and \(w\) are negative for the process of combusting gasoline into \(\mathrm{CO}_{2}\) and \(\mathrm{H}_{2} \mathrm{O}\).

Key concepts

Exothermic Reaction

Exothermic reactions are chemical processes that release energy in the form of heat. This means that during an exothermic reaction, the system (the set of reactants undergoing the change) transfers heat to its surroundings. A classic example of an exothermic process is a combustion reaction, like the burning of gasoline.

When gasoline combusts, it combines with oxygen to release energy, resulting in the production of carbon dioxide and water vapor. The chemical equation shows this release in mathematical terms:

• \( C_{8}H_{18}(l) + \frac{25}{2} O_{2}(g) \rightarrow 8 CO_{2}(g) + 9 H_{2}O(g) \)

This equation tells us that every mole of octane (gasoline) reacts with the specified amount of oxygen to produce energy. The negative sign of \(q\), the heat exchanged, signifies energy is leaving the system. Understanding that exothermic reactions always have negative \(q\) values is crucial when studying thermodynamic properties.

Thermodynamics

Thermodynamics deals with the principles governing energy exchanges in physical and chemical processes. It provides us with the tools to predict the signs of energy transfers, such as heat (\(q\)) and work (\(w\)). In our combustion scenario, thermodynamics tells us what happens to the energy in the system.

When gasoline burns, it expels heat to its surroundings, a key point in thermodynamics. This is why \(q\) is negative, indicating an outward transfer of energy. Additionally, work is associated with changes in volume, particularly in gas-producing reactions like combustion. As burning gasoline generates more moles of gas than it consumes, it expands, effectively pushing against atmospheric pressure, doing work on its surroundings. Consequently, \(w\) also carries a negative sign, denoting work done by the system.

By applying thermodynamic concepts, students can understand how different energy transfers occur and are represented in chemical equations.

Chemical Equation Balancing

Balancing chemical equations is fundamental to understanding chemical reactions. It ensures that the same number of each type of atom is present in both the reactants and products. For the combustion of gasoline, the initial equation must be carefully balanced to reflect the conservation of mass.

The balanced chemical equation is:

• \(C_{8}H_{18}(l) + \frac{25}{2} O_{2}(g) \rightarrow 8 CO_{2}(g) + 9 H_{2}O(g)\)

In this equation, we begin by balancing carbon and hydrogen from the octane, followed by the oxygen molecules. It's crucial to balance the equation properly to accurately depict the stoichiometry, representing how moles of reactants convert into moles of products.

By mastering the skill of balancing equations, students can ensure that their chemical reactions account for all atoms involved and predict the quantities of products formed.