Warning: foreach() argument must be of type array|object, bool given in /var/www/html/web/app/themes/studypress-core-theme/template-parts/header/mobile-offcanvas.php on line 20

Assuming gasoline is pure \(\mathrm{C}_{8} \mathrm{H}_{18}(l),\) predict the signs of \(q\) and \(w\) for the process of combusting gasoline into \(\mathrm{CO}_{2}(g)\) and \(\mathrm{H}_{2} \mathrm{O}(g)\)

Short Answer

Expert verified
During the combustion of gasoline (C₈H₁₈) into CO₂(g) and H₂O(g), both q and w are negative. This means that heat is released to the surroundings (exothermic process) and the system does work against the surroundings due to the expansion of gases. Therefore, \(q < 0\) and \(w < 0\).

Step by step solution

01

1. Write the balanced chemical equation for the combustion of gasoline

For the combustion of C₈H₁₈, the balanced chemical equation is: \[C_{8}H_{18}(l) + 12.5O_{2}(g) \rightarrow 8CO_{2}(g) + 9H_{2}O(g)\]
02

2. Determine the signs of q

Combustion reactions are exothermic, meaning they release heat. As a result, the system loses heat to the surroundings, making the value of q negative. So, the sign of q is negative: \[q < 0\]
03

3. Determine the signs of w

To find the sign of the work (w) done during the combustion process, we need to consider the expansion of gases. In the balanced chemical equation, we see that 12.5 moles of O₂(g) are consumed and 17 moles of products (8 moles of CO₂ and 9 moles of H₂O) are formed, which means more gas molecules are produced. This results in an expansion of the gases, as more particles are created and occupy a larger volume. In an expansion, the system does work against the surroundings; thus, the sign of work is negative. So, the sign of w is negative: \[w < 0\]
04

4. Conclusion

During the combustion of gasoline (C₈H₁₈) into CO₂(g) and H₂O(g), both q and w are negative. This means that heat is released to the surroundings and the system does work against the surroundings in the expansion of gases. q is negative: \(q < 0\) w is negative: \(w < 0\)

Unlock Step-by-Step Solutions & Ace Your Exams!

  • Full Textbook Solutions

    Get detailed explanations and key concepts

  • Unlimited Al creation

    Al flashcards, explanations, exams and more...

  • Ads-free access

    To over 500 millions flashcards

  • Money-back guarantee

    We refund you if you fail your exam.

Over 30 million students worldwide already upgrade their learning with Vaia!

Key Concepts

These are the key concepts you need to understand to accurately answer the question.

Exothermic Reactions
Exothermic reactions are a key concept in understanding combustion. In an exothermic reaction, energy is released as heat during the transformation from reactants to products. This release of energy occurs because the energy required to break the bonds in the reactants is less than the energy released when new bonds form in the products.
For instance, when gasoline (\(\mathrm{C}_{8} \mathrm{H}_{18}(l)\) ) burns, it undergoes an exothermic reaction, releasing heat into the surroundings. This heat release is why the sign of \(q\) (heat) is negative during combustion, indicating that the system loses energy to its surroundings.
Exothermic reactions are prevalent in everyday life, not just in combustion. They include common processes like freezing water and rusting iron. Each time an exothermic reaction occurs, energy flows outward, which is fundamental to the reactions we observe, such as lighting a car engine with gasoline.
Chemical Equations
Chemical equations are a symbolic representation of a chemical reaction, illustrating how reactants transform into products. They are vital in predicting the outcomes of chemical processes.
In the combustion of gasoline, the balanced chemical equation is:\[C_{8}H_{18}(l) + 12.5O_{2}(g) \rightarrow 8CO_{2}(g) + 9H_{2}O(g)\]This equation shows us that one molecule of gasoline (\(\mathrm{C}_{8} \mathrm{H}_{18}\) ) reacts with 12.5 molecules of oxygen (\(\mathrm{O}_{2}\) ) to produce carbon dioxide (\(\mathrm{CO}_{2}\) ) and water vapor (\(\mathrm{H}_{2} \mathrm{O}\) ).
Writing balanced chemical equations is crucial as it ensures the conservation of mass and atoms, adhering to the law of conservation of matter. Understanding these equations enables us to analyze chemical reactions, predict products, and measure energy changes, such as heat released or absorbed.
Gas Expansion
Gas expansion occurs when the number of gas molecules increases during a reaction, often leading to a larger volume. In the context of combustion, understanding gas expansion helps explain why energy is expended by the system when gases evolve.
In the balanced equation for gasoline combustion, 12.5 moles of oxygen gas become 17 moles of total gas products. This net increase in the number of molecules means that the gases occupy more space, causing them to do work on their surroundings as they expand. This is why the sign of \(w\) (work) is negative in such reactions.
During gas expansion, the system performs work against external pressure. This work can be visualized as a balloon inflating, where the expanding gas inside pushes against the exterior pressure to expand. As gases spread out, they require energy to overcome surrounding forces, which is reflected in the negative sign of work during expansion.
Thermodynamics
Thermodynamics is the study of energy, heat, and their transformations. It provides a framework for understanding the energy exchanges in chemical reactions, including combustion.
In thermodynamics, we analyze two main factors in a reaction: heat transfer (\(q\)) and work done (\(w\)). Both are crucial in determining the energy change of a system. In the combustion of gasoline, as gasoline burns, it releases energy as heat (\(q < 0\)) and performs work through gas expansion (\(w < 0\)).
Thermodynamics involves understanding the laws governing energy changes:
  • The First Law, which states that energy is conserved, it cannot be created or destroyed.
  • The Second Law, which states that energy transfer will always result in increased entropy or disorder.
In a combustion reaction, the released energy primarily manifests as heat transferred to the surroundings, while work is done to accommodate gas expansion, both obeying these fundamental laws.

One App. One Place for Learning.

All the tools & learning materials you need for study success - in one app.

Get started for free

Most popular questions from this chapter

A system absorbs 35 \(\mathrm{J}\) of heat and has 25 \(\mathrm{J}\) of work performed on it. The system then returns to its initial state by a second step. If 5 \(\mathrm{J}\) of heat are given off in the second step, how much work is done by the system in the second step?

Calculate \(w\) and \(\Delta E\) when 1 mole of a liquid is vaporized at its boiling point \(\left(80 .^{\circ} \mathrm{C}\right)\) and 1.00 atm pressure. \(\Delta H_{\text { vap }}\) for the liquid is 30.7 \(\mathrm{kJ} / \mathrm{mol}\) at \(80 .^{\circ} \mathrm{C} .\)

The combustion of 0.1584 g benzoic acid increases the temperature of a bomb calorimeter by \(2.54^{\circ} \mathrm{C}\) . Calculate the heat capacity of this calorimeter. (The energy released by combustion of benzoic acid is 26.42 \(\mathrm{kJ} / \mathrm{g} .\) ) A \(0.2130-\mathrm{g}\) sample of vanillin \(\left(\mathrm{C}_{8} \mathrm{H}_{8} \mathrm{O}_{3}\right)\) is then burned in the same calorimeter, and the temperature increases by \(3.25^{\circ} \mathrm{C}\) . What is the energy of combustion per gram of vanillin? Per mole of vanillin?

A gas absorbs 45 kJ of heat and does 29 kJ of work. Calculate \(\Delta E .\)

A sample of nickel is heated to \(99.8^{\circ} \mathrm{C}\) and placed in a coffeecup calorimeter containing 150.0 \(\mathrm{g}\) water at \(23.5^{\circ} \mathrm{C}\) . After the metal cools, the final temperature of metal and water mixture is \(25.0^{\circ} \mathrm{C}\) . If the specific heat capacity of nickel is 0.444 \(\mathrm{J} /^{\prime} \mathrm{C} \cdot \mathrm{g}\) what mass of nickel was originally heated? Assume no heat loss to the surroundings.

See all solutions

Recommended explanations on Chemistry Textbooks

View all explanations

What do you think about this solution?

We value your feedback to improve our textbook solutions.

Study anywhere. Anytime. Across all devices.

Sign-up for free