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Chapter 14: Problem 5

Why might oxygen contained in an iron tank be treated as inert in a thermodynamic analysis even though iron oxidizes in the presence of oxygen?

Short Answer

Expert verified
Oxygen is treated as inert in thermodynamic analysis because its reaction rate with iron is very slow, making its impact negligible.

Step by step solution

01

Identify the context

Understand that the question is about the behavior of oxygen and iron in a thermodynamic analysis. It asks why oxygen is considered inert in this scenario.
02

Understand oxidation process

Iron oxidizes when it reacts with oxygen, forming iron oxide. This reaction is generally slow at room temperature.
03

Consider conditions of thermodynamic analysis

In a thermodynamic analysis, processes are assumed to be in equilibrium or moving towards equilibrium in a controlled way. This usually ignores slow chemical reactions unless they are of primary interest.
04

Determine the inert definition

In thermodynamics, an inert substance is one that does not participate in the reactions being considered. This simplifies the calculations and focuses only on relevant interactions.
05

Evaluate the role of oxygen and iron

Even though iron can oxidize, the rate of oxidation is very slow at normal temperatures. During the analysis period, changes in oxygen levels due to oxidation are negligible.
06

Conclude why oxygen is considered inert

Given the slow rate of oxidation, oxygen's impact on the thermodynamic properties within the timeframe of analysis is minimal. Therefore, it can be treated as inert.

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Key Concepts

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

Oxidation Process
Oxidation is a chemical reaction where a substance loses electrons, often involving oxygen. When iron oxidizes, it reacts with oxygen to form iron oxide (rust). This process is generally slow at room temperature but can be accelerated by factors like moisture or higher temperatures. Understanding the oxidation process helps us grasp why oxygen is treated as inert in some thermodynamic analyses. Even though iron rusts over time, the rate is so slow under standard conditions that the change in oxygen levels is negligible within a typical analysis period. This slow reaction means it doesn't significantly impact the thermodynamic properties we're studying.
Thermodynamic Equilibrium
In thermodynamics, equilibrium refers to a state where all processes occur at steady rates, resulting in no net change in the system. This concept is crucial because it allows us to simplify calculations by assuming that reactions proceed to completion or not at all. In the context of oxygen in an iron tank, we're often interested in the system's behavior over shorter timescales. Since the oxidation process is slow, the system can be modeled as if oxygen does not react with iron, simplifying our analysis. At equilibrium, only the primary reactions of interest are considered, while slow or negligible reactions, like iron oxidation at room temperature, are ignored.
Reaction Rates
Reaction rates describe how quickly a chemical reaction occurs. They are influenced by factors like temperature, concentration, and presence of catalysts. In the case of iron and oxygen, the rate of oxidation is slow at room temperature. This slow reaction rate means that within a short timescale, the amount of oxygen reacting with iron is insignificant. By treating oxygen as inert, we focus on other more relevant interactions, making our thermodynamic analysis more manageable. Understanding reaction rates is essential for determining which substances can be considered inactive (or inert) in a given analysis and helps simplify complex systems into more understandable models.

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Most popular questions from this chapter

If the ionization-equilibrium constants for \(\mathrm{Cs} \rightleftarrows \mathrm{Cs}^{+}+\mathrm{e}^{-}\) at 1600 and \(2000 \mathrm{~K}\) are \(K=0.78\) and \(K=15.63\), respectively, estimate the enthalpy of ionization, in \(\mathrm{kJ} / \mathrm{kmol}\), at \(1800 \mathrm{~K}\) using the van't Hoff equation.

The following exercises involve oxides of nitrogen: (a) One \(\mathrm{kmol}\) of \(\mathrm{N}_{2} \mathrm{O}_{4}\) dissociates at \(25^{\circ} \mathrm{C}, 1 \mathrm{~atm}\) to form an equilibrium ideal gas mixture of \(\mathrm{N}_{2} \mathrm{O}_{4}\) and \(\mathrm{NO}_{2}\) in which the amount of \(\mathrm{N}_{2} \mathrm{O}_{4}\) present is \(0.8154 \mathrm{kmol}\). Determine the amount of \(\mathrm{N}_{2} \mathrm{O}_{4}\) that would be present in an equilibrium mixture at \(25^{\circ} \mathrm{C}, 0.5 \mathrm{~atm}\). (b) A gaseous mixture consisting of \(1 \mathrm{kmol}\) of \(\mathrm{NO}, 10 \mathrm{kmol}\) of \(\mathrm{O}_{2}\), and \(40 \mathrm{kmol}\) of \(\mathrm{N}_{2}\) reacts to form an equilibrium ideal gas mixture of \(\mathrm{NO}_{2}, \mathrm{NO}\), and \(\mathrm{O}_{2}\) at \(500 \mathrm{~K}, 0.1 \mathrm{~atm}\). Determine the composition of the equilibrium mixture. For \(\mathrm{NO}+\frac{1}{2} \mathrm{O}_{2} \rightleftarrows \mathrm{NO}_{2}, K=120\) at \(500 \mathrm{~K}\) (c) An equimolar mixture of \(\mathrm{O}_{2}\) and \(\mathrm{N}_{2}\) reacts to form an equilibrium ideal gas mixture of \(\mathrm{O}_{2}, \mathrm{~N}_{2}\), and NO. Plot the mole fraction of \(\mathrm{NO}\) in the equilibrium mixture versus equilibrium temperature ranging from 1200 to \(2000 \mathrm{~K}\). Why are oxides of nitrogen of concern?

An equimolar mixture of \(\mathrm{CO}\) and \(\mathrm{O}_{2}\) reacts to form an equilibrium mixture of \(\mathrm{CO}_{2}, \mathrm{CO}\), and \(\mathrm{O}_{2}\) at \(3000 \mathrm{~K}\). Determine the effect of pressure on the composition of the equilibrium mixture. Will lowering the pressure while keeping the temperature fixed increase or decrease the amount of \(\mathrm{CO}_{2}\) present? Explain.

Carbon at \(25^{\circ} \mathrm{C}, 1\) atm enters a reactor operating at steady state and burns with oxygen entering at \(127^{\circ} \mathrm{C}, 1 \mathrm{~atm}\). The entering streams have equal molar flow rates. An equilibrium mixture of \(\mathrm{CO}_{2}, \mathrm{CO}\), and \(\mathrm{O}_{2}\) exits at \(2727^{\circ} \mathrm{C}, 1\) atm. Determine, per kmol of carbon, (a) the composition of the exiting mixture. (b) the heat transfer between the reactor and its surroundings, in \(\mathrm{kJ}\). Neglect kinetic and potential energy effects.

Gaseous propane \(\left(\mathrm{C}_{3} \mathrm{H}_{8}\right.\) ) at \(25^{\circ} \mathrm{C}, 1\) atm enters a reactor operating at steady state and burns with \(80 \%\) of theoretical air entering separately at \(25^{\circ} \mathrm{C}, 1 \mathrm{~atm}\). An equilibrium mixture of \(\mathrm{CO}_{2}, \mathrm{CO}, \mathrm{H}_{2} \mathrm{O}(\mathrm{g}), \mathrm{H}_{2}\), and \(\mathrm{N}_{2}\) exits at \(1227^{\circ} \mathrm{C}, 1 \mathrm{~atm}\). Determine the heat transfer between the reactor and its surroundings, in kJ per kmol of propane entering. Neglect kinetic and potential energy effects.
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