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Chapter 14: Problem 131

In theory, all reactions are reversible, but in practice, some are not. Explain why.

Short Answer

Expert verified
In theory, all reactions are reversible; however, in practice, some reactions are irreversible due to factors such as kinetic factors, thermodynamic factors, side reactions, and product removal. Kinetic factors can make the reverse reaction too slow to be observed, while thermodynamic factors can drive the reaction to completion by releasing a large amount of energy. Side reactions can cause products to undergo additional reactions, making it difficult to revert to the original reactants, and continuous removal of products can prevent the reverse reaction from occurring. Examples of practically irreversible reactions include combustion and precipitation reactions.

Step by step solution

01

Understand reversible and irreversible reactions

In reversible reactions, reactants form products, and products can react with each other to form reactants again. These reactions reach a state of equilibrium, where the rates of the forward reaction (conversion of reactants to products) and the reverse reaction (conversion of products to reactants) become equal. On the other hand, in irreversible reactions, the forward reaction occurs, and the products formed do not revert to the original reactants.
02

Factors contributing to irreversible reactions

There are several factors that lead to some reactions being irreversible, including: 1. Kinetic factors: The reaction rate may be too slow for the reverse reaction to be observed or measured. This makes the reaction seem irreversible. 2. Thermodynamic factors: Some reactions produce a large amount of energy, which drives the reaction to completion, making the reverse reaction less favorable. 3. Side reactions: Some products can undergo side reactions, which makes it difficult for the reaction to go back to the original reactants. 4. Removal of products: If products are removed continuously from the reaction mixture, the reverse reaction is unlikely to occur, and the reaction becomes irreversible.
03

Examples of irreversible reactions

Here are some examples of reactions that are considered irreversible in practice due to the factors mentioned above: 1. Combustion: Burning a fuel, such as methane, in the presence of oxygen is highly exothermic and thermodynamically favored, making the reverse reaction highly unlikely. 2. Precipitation reactions: In reactions where a solid is formed, such as the formation of a salt from a reaction between a metal ion and an anion, the solid formed can be removed, making the reverse reaction practically impossible. In conclusion, while theoretically, all reactions are reversible, several factors make some reactions practically irreversible. Examples include combustion reactions and precipitation reactions.

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Key Concepts

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

Reaction Kinetics
Reaction kinetics is the study of the rates at which chemical reactions occur and the factors that affect these rates. It's crucial for understanding why some reactions seem irreversible. The speed of a reaction depends on several factors such as temperature, concentration, and the presence of catalysts.
  • Rate of Reaction: If the reverse reaction occurs too slowly, it might not be noticeable, giving an illusion of irreversibility.
  • Catalysts: These substances speed up reactions without being consumed. They can make reactions reversible by accelerating both forward and reverse reactions.
Reaction kinetics helps us understand when a reaction has effectively reached a point of no return due to the speed at which reactants are converted to products.
Chemical Equilibrium
Chemical equilibrium is a key concept in reversible reactions. It occurs when the forward and reverse reactions happen at the same rate, resulting in no net change in the concentration of products and reactants.
  • Dynamic Process: Even at equilibrium, reactions continue to occur, but their effects cancel each other out.
  • Le Chatelier鈥檚 Principle: If a system at equilibrium is disturbed, it will adjust to minimize the disturbance, often re-establishing equilibrium conditions.
Understanding equilibrium helps explain why some reactions appear irreversible if one side of the reaction is heavily favored.
Thermodynamics
Thermodynamics explores the energy changes in reactions. It explains why some reactions are irreversible by focusing on the energetics involved.
  • Exothermic Reactions: These reactions release energy, often driving reactions to completion.
  • Gibbs Free Energy: Spontaneity of reactions is determined by changes in Gibbs free energy (\( \Delta G \)). For forward reactions with large negative \( \Delta G \), the reverse reaction is less likely.
Thermodynamics helps us understand that when a reaction releases a large amount of energy, the products are less likely to revert to reactants.
Combustion Reactions
Combustion reactions are classic examples of irreversible reactions. They involve burning substances in oxygen, producing energy rapidly.
  • Highly Exothermic: The release of energy drives the reaction forward.
  • Complete Conversion: Organic compounds oxidize to form carbon dioxide and water, making reversibility impractical.
Combustion reactions illustrate how energy release and product stability favor only the forward reaction.
Precipitation Reactions
Precipitation reactions occur when two solutions react to form an insoluble solid known as a precipitate.
  • Solid Formation: The precipitate can be removed from the mixture, preventing the reverse reaction.
  • Solubility Product: The equilibrium position is far to the right, meaning almost complete conversion to products.
These reactions showcase how the removal of products from the reaction mixture can make a reversible reaction practically irreversible.

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Most popular questions from this chapter

What is "dynamic" about the equilibrium that is established when a sparingly soluble salt is added to water?

The process of photosynthesis in plants converts carbon dioxide and water to glucose and oxygen: \(6 \mathrm{CO}_{2}(g)+6 \mathrm{H}_{2} \mathrm{O}(l) \rightleftarrows \mathrm{C}_{6} \mathrm{H}_{12} \mathrm{O}_{6}(s)+6 \mathrm{O}_{2}(g) \quad \Delta E_{\mathrm{rxn}}=2801 \mathrm{~kJ}\) (a) Write the equilibrium constant expression for this conversion. (b) How would the equilibrium be affected if \(\mathrm{CO}_{2}(g)\) were added? (c) How would the equilibrium be affected if \(\mathrm{H}_{2} \mathrm{O}(l)\) were added? (d) How would the equilibrium be affected if the reaction vessel were warmed? (e) How would the equilibrium be affected if a catalyst were added?

For the reaction \(\mathrm{CH}_{4}(g)+2 \mathrm{H}_{2} \mathrm{~S}(g) \rightleftarrows \mathrm{CS}_{2}(g)+4 \mathrm{H}_{2}(g)\) \(K_{\text {eq }}=3.59\) at \(900^{\circ} \mathrm{C}\). After the reaction has run for 10 min at \(900^{\circ} \mathrm{C}\), the concentrations are \(\left[\mathrm{CH}_{4}\right]=\) \(1.15 \mathrm{M} ;\left[\mathrm{H}_{2} \mathrm{~S}\right]=1.20 \mathrm{M} ;\left[\mathrm{CS}_{2}\right]=1.51 \mathrm{M} ;\left[\mathrm{H}_{2}\right]=\) \(1.08 \mathrm{M}\). Is this reaction at equilibrium?

At \(25^{\circ} \mathrm{C}\), the solubility of iron(III) hydroxide in water is \(4.49 \times 10^{-10} \mathrm{M}\). (a) What is the solubility in grams per liter? (b) What is the molar equilibrium concentration of each ion? (c) How many grams of iron(III) hydroxide could you dissolve in a 20,000-gallon swimming pool?

Consider a saturated aqueous solution of \(\mathrm{AgCl}\), a salt that is only sparingly soluble in water. What happens to this solution if a saturated solution of NaCl (a water-soluble salt) is added to it? (Hint: If \(\left[\mathrm{Ag}^{+}(a q)\right] \times\left[\mathrm{Cl}^{-}(a q)\right]>K_{\mathrm{sp}^{\prime}}\), precipitation will occur.)
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