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13\. Why is combustion inherently an irreversible process?

Short Answer

Expert verified
Combustion is inherently irreversible due to energy dispersal, increased entropy, and practical constraints on reversing the reaction.

Step by step solution

01

Define Combustion

Combustion is a chemical process in which a substance reacts with oxygen to give off heat. It typically produces carbon dioxide, water, and other gases.
02

Understand Irreversible Process

An irreversible process is one that cannot be undone by reversing the change. This means that once the process has occurred, the system cannot be returned to its original state without external intervention.
03

Analyze the Energy Conversion

In combustion, chemical energy in the fuel is converted to heat and light. This energy transformation is highly dispersive and results in increased entropy, making it difficult to reverse.
04

Consider Entropy Changes

Combustion significantly increases entropy due to the production of gases that disperse into the surroundings. This disorder cannot be easily reduced, contributing to the irreversibility.
05

Examine Practical Constraints

In real-world scenarios, recovering the original fuel and oxygen from the combustion products would require substantial external work and energy input, often making it practically unfeasible.
06

Conclusion

Sum up the reasons why combustion cannot be reversed due to the energy dispersion, increase in entropy, and practical constraints on restoring the original substances.

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Key Concepts

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

Combustion Process
Combustion is an essential chemical reaction that involves a substance, generally a fuel, reacting with oxygen to produce heat, light, and other byproducts. This process is commonly seen in everyday activities, such as burning wood in a fireplace or gasoline in a car engine. The typical products of combustion are carbon dioxide (CO2), water (H2O), and other gases. This process is exothermic, meaning it releases energy, primarily in the form of heat and light. During combustion, the chemical bonds of the fuel are broken, and new bonds with oxygen are formed, leading to the release of energy.
Irreversible Process
An irreversible process is a process that cannot return to its initial state without external intervention. In other words, once the process occurs, you cannot simply reverse it by undoing the steps. The combustion process is an excellent example of an irreversible process. Consider a piece of wood that has been burnt. To revert the ashes, gases, and heat back to the original piece of wood and oxygen, a significant amount of external energy would be needed. This transformation involves such a high level of energy dispersion and entropy increase that reversing it becomes practically impossible.
Entropy in Thermodynamics
Entropy is a fundamental concept in thermodynamics that measures the level of disorder or randomness in a system. During the combustion process, entropy increases significantly because the solid fuel is broken down into gaseous products that spread out into the environment. The second law of thermodynamics states that the total entropy of an isolated system can never decrease over time. In the case of combustion, the increase in entropy manifests as the spread and mixing of molecules in the gases produced. This substantial increase in entropy underscores why combustion is considered an irreversible process.
Energy Conversion
Energy conversion during combustion involves transforming the chemical energy stored in the fuel into heat and light energy. This conversion is inherently dispersive, meaning that the energy from the combustion spreads out and becomes less concentrated. Chemical energy is stored in the bonds of the fuel molecules, and during combustion, these bonds are broken and new bonds are formed with oxygen. This process releases energy, which disperses into the surroundings as heat and light. The dispersive nature of this energy transformation makes it challenging, if not impossible, to recapture the energy and revert to the original fuel and oxygen.
Practical Constraints in Thermodynamics
Practical constraints in thermodynamics further explain why the combustion process is irreversible. In theory, it might be possible to recreate the original substances from the combustion products, but in reality, this would require an enormous amount of energy and work. Extracting the combustion products like CO2 and H2O and converting them back into fuel and oxygen would not only be energy-intensive but also technically very challenging. These practical constraints highlight the infeasibility of reversing combustion on a macroscopic scale. It also emphasizes why combustion is a one-way process leading to permanent energy conversion and increased entropy.

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Most popular questions from this chapter

13.43 For a natural gas with a molar analysis of \(86.5 \% \mathrm{CH}_{4}, 8 \%\) \(\mathrm{C}_{2} \mathrm{H}_{6}, 2 \% \mathrm{C}_{3} \mathrm{H}_{8}, 3.5 \% \mathrm{~N}_{2}\), determine the lower heating value, in \(\mathrm{kJ}\) per \(\mathrm{kmol}\) of fuel and in \(\mathrm{kJ}\) per \(\mathrm{kg}\) of fuel, at \(25^{\circ} \mathrm{C}, 1 \mathrm{~atm}\).

13.8D Fuel or chemical leaks and spills can have catastrophic ramifications; thus the hazards associated with such events must be well understood. Prepare a memorandum for one of the following: (a) Experience with interstate pipelines shows that propane leaks are usually much more hazardous than leaks of natural gas or liquids such as gasoline. Why is this so? (b) The most important parameter in determining the accidental rate of release from a fuel or chemical storage vessel is generally the size of the opening. Roughly how much faster would such a substance be released from a \(1-\mathrm{cm}\) hole than. from a 1 -mm hole? What are the implications of this?

13.60 Streams of hydrogen \(\left(\mathrm{H}_{2}\right)\) and oxygen \(\left(\mathrm{O}_{2}\right)\), each at \(1 \mathrm{~atm}\), enter a fuel cell operating at steady state and water vapor exits at \(1 \mathrm{~atm}\). If the cell operates isothermally at (a) \(300 \mathrm{~K}\), (b) \(400 \mathrm{~K}\), and (c) \(500 \mathrm{~K}\), determine the maximum theoretical work that can be developed by the cell in each case, in kJ per kmol of hydrogen flowing, and comment. Heat transfer with the surroundings takes place at the cell temperature, and kinetic and potential energy effects can be ignored.

13.52 Carbon monoxide (CO) at \(25^{\circ} \mathrm{C}, 1\) atm enters an insulated reactor operating at steady state and reacts completely with the theoretical amount of air entering in a separate stream at \(25^{\circ} \mathrm{C}, 1 \mathrm{~atm}\). The products of combustion exit as a mixture at \(1 \mathrm{~atm}\). For the reactor, determine the rate of entropy production, in \(\mathrm{kJ} / \mathrm{K}\) per kmol of CO entering. Neglect kinetic and potential energy effects.

\(13.49\) Liquid methanol \(\left(\mathrm{CH}_{3} \mathrm{OH}\right)\) at \(25^{\circ} \mathrm{C}, 1\) atm enters an insulated reactor operating at steady state and burns completely with air entering at \(100^{\circ} \mathrm{C}, 1 \mathrm{~atm}\). If the combustion products exit at \(1256^{\circ} \mathrm{C}\), determine the percent excess air used. Neglect kinetic and potential energy effects.

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