When writing an equation, how is a reversible reaction distinguished from a nonreversible reaction?

Short Answer

Expert verified
A reversible reaction is distinguished by a double-headed arrow (\( \rightleftharpoons \)), whereas a nonreversible reaction is represented by a single-headed arrow (\( \rightarrow \)).

Step by step solution

01

Understanding Reversible Reactions

A reversible reaction is a chemical reaction where the reactants form products, which then react together to give the reactants back. These reactions reach an equilibrium state where the reactants and products are present at constant concentrations.
02

Representing Reversible Reactions in Equations

Reversible reactions are represented by a double-headed arrow (\( \rightleftharpoons \)). This symbol indicates that the reaction can proceed in both the forward and reverse directions.
03

Distinguishing Nonreversible Reactions in Equations

Nonreversible reactions, on the other hand, are typically represented by a single-headed arrow (\( \rightarrow \)). This signifies that the reaction proceeds in only one direction, from the reactants to the products, with no significant reverse reaction occurring.

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Key Concepts

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

Chemical Equilibrium
Imagine a tug of war where both teams are equally strong; the rope might wiggle back and forth, but it stays basically in the same place. Something similar happens in chemical equilibrium, a state in chemical reactions where the rate of the forward reaction (formation of products from reactants) equals the rate of the backward reaction (reformation of reactants from the products). As a result, the amounts of reactants and products remain constant over time, just like the teams in tug of war.

This doesn't mean the reactions have stopped. On the contrary, both reactions are happening continuously and at the same speed, ensuring the overall concentrations of reactants and products stay the same. It's like having a constant exchange of players between the teams but always maintaining the same team sizes.

Reaching chemical equilibrium is a dynamic process. It suggests that a system is stable and not having any net change, which is critical in processes ranging from industrial chemical synthesis to the functioning of biological systems.
Chemical Reaction Representation
How we represent a chemical reaction is vital to understanding the process it illustrates. Think of chemical reaction equations as recipes: they tell us what we need (reactants) and what we'll get (products). For a reversible reaction, we write the reactants and products with a double-headed arrow (\( \rightleftharpoons \) between them. This arrow speaks of the capability of the reaction to go in both directions - forward to create products and backward to revert to reactants.

In texts, diagrams, and even lab reports, the precise use of symbols can convey a significant amount of information. Consider the difference between using a double-headed arrow versus a single-headed one; it's like the difference between a two-way street and a one-way street. Accurate representation matters in the accuracy and the understanding of chemical behavior. Through proper representation in equations, scientists can quickly ascertain the nature of the reaction and predict its behavior under different conditions.
Nonreversible Reactions
Now, let's consider nonreversible reactions. These are the chemical equivalent of a one-way trip – once you start, there's no going back. Such reactions are typified by their finality; once the reactants turn into products, that's it. They are often represented by a single-headed arrow (\( \rightarrow \) that points from the reactants to the products.

Nonreversible reactions are common in instances where energy is released in forms such as light or heat, often making it impossible for the reaction to go in reverse. These reactions are fundamental to many industrial processes, such as combustion in engines or the production of materials like cement. Understanding that a reaction is nonreversible is crucial because it tells chemists the process is complete and that the products won't spontaneously revert to reactants. This concept is essential when planning chemical syntheses, as it assures the desired product won't disappear over time.

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Most popular questions from this chapter

How will an increase in temperature affect each of the following equilibria? How will a decrease in the volume of the reaction vessel affect each? (a) \(2 \mathrm{NH}_{3}(g) \rightleftharpoons \mathrm{N}_{2}(g)+3 \mathrm{H}_{2}(g) \quad \Delta H=92 \mathrm{kJ}\) (b) \(\mathrm{N}_{2}(g)+\mathrm{O}_{2}(g) \rightleftharpoons 2 \mathrm{NO}(g) \quad \Delta H=181 \mathrm{kJ}\) (c) \(2 \mathrm{O}_{3}(g) \rightleftharpoons 3 \mathrm{O}_{2}(g) \quad \Delta H=-285 \mathrm{kJ}\) (d) \(\mathrm{CaO}(s)+\mathrm{CO}_{2}(g) \rightleftharpoons \mathrm{CaCO}_{3}(s) \quad \Delta H=-176 \mathrm{kJ}\)

Suggest four ways in which the concentration of hydrazine, \(\mathrm{N}_{2} \mathrm{H}_{4}\), could be increased in an equilibrium described by the following equation: \(\mathrm{N}_{2}(g)+2 \mathrm{H}_{2}(g) \rightleftharpoons \mathrm{N}_{2} \mathrm{H}_{4}(g) \quad \Delta H=95 \mathrm{kJ}\)

Among the solubility rules previously discussed is the statement: Carbonates, phosphates, borates, and arsenates - except those of the ammonium ion and the alkali metals-are insoluble. (a) Write the expression for the equilibrium constant for the reaction represented by the equation \(\mathrm{CaCO}_{3}(s) \rightleftharpoons \mathrm{Ca}^{2+}(a q)+\mathrm{CO}_{3}^{2-}(a q) .\) Is \(K_{c} > 1, < 1,\) or \(\approx 1 ?\) Explain your answer. (b) Write the expression for the equilibrium constant for the reaction represented by the equation \(3 \mathrm{Ba}^{2+}(a q)+2 \mathrm{PO}_{4}^{3-}(a q) \rightleftharpoons \mathrm{Ba}_{3}\left(\mathrm{PO}_{4}\right)_{2}(s) .\) Is \(K_{c} > 1, < 1,\) or \(\approx 1 ?\) Explain your answer.

Nitrogen and oxygen react at high temperatures. (a) Write the expression for the equilibrium constant \(\left(K_{c}\right)\) for the reversible reaction \(\mathrm{N}_{2}(g)+\mathrm{O}_{2}(g) \rightleftharpoons 2 \mathrm{NO}(g) \quad \Delta H=181 \mathrm{kJ}\) (b) What will happen to the concentrations of \(\mathrm{N}_{2}, \mathrm{O}_{2},\) and \(\mathrm{NO}\) at equilibrium if more \(\mathrm{O}_{2}\) is added? (c) What will happen to the concentrations of \(\mathrm{N}_{2}, \mathrm{O}_{2},\) and \(\mathrm{NO}\) at equilibrium if \(\mathrm{N}_{2}\) is removed? (d) What will happen to the concentrations of \(\mathrm{N}_{2}, \mathrm{O}_{2},\) and NO at equilibrium if NO is added? (e) What will happen to the concentrations of \(\mathrm{N}_{2}, \mathrm{O}_{2},\) and \(\mathrm{NO}\) at equilibrium if the volume of the reaction vessel is decreased? (f) What will happen to the concentrations of \(\mathrm{N}_{2}, \mathrm{O}_{2},\) and \(\mathrm{NO}\) at equilibrium if the temperature of the system is increased?

How can the pressure of water vapor be increased in the following equilibrium? \(\mathrm{H}_{2} \mathrm{O}(l) \rightleftharpoons \mathrm{H}_{2} \mathrm{O}(g) \quad \Delta H=41 \mathrm{kJ}\)

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