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Once a reaction begins, the rate of the reverse reaction gradually speeds up. Explain why this is so.

Short Answer

Expert verified
In a reversible chemical reaction, the system eventually reaches a state of equilibrium, where the rate of the forward and reverse reactions is equal. Initially, there are more reactants than products, so the forward reaction is more prevalent. As reactants are converted into products, the concentration of products increases, leading to an increased likelihood of the reverse reaction occurring. As a result, the rate of the reverse reaction gradually speeds up until it becomes equal to the rate of the forward reaction, re-establishing equilibrium.

Step by step solution

01

Introduction to Chemical Equilibria and Reaction Rates

In a reversible chemical reaction, reactants are converted into products, and products are converted back into reactants. As these forward and reverse reactions occur simultaneously, the concentration of reactants and products change over time. Eventually, the system reaches a state of equilibrium, where the rate of the forward and reverse reactions is equal, and the concentration of the reactants and products no longer change.
02

Le Chatelier's Principle

Le Chatelier's Principle states that if a system in equilibrium is subjected to a change in concentration of the reactants or products, the system will adjust the equilibrium to counteract the effect of the change. This happens in such a way as to minimize the change, and restore the equilibrium condition.
03

Initial Reaction Rates

At the beginning of a reaction, there are usually more reactants than products. Consequently, the forward reaction is initially more prevalent than the reverse reaction, as there is a higher concentration of reactants. As the forward reaction progresses, some of the reactants will be converted into products, which leads to an increase in the concentration of the products.
04

Increasing Rate of Reverse Reaction

As the products' concentration increases, the likelihood of the reverse reaction occurring also increases. This is because there are now more product molecules available to collide with each other and convert back into reactants. As a result, the rate of the reverse reaction gradually speeds up.
05

Re-establishing Equilibrium

As the reverse reaction speeds up, the rate of the reverse reaction will eventually become equal to the rate of the forward reaction. When this occurs, the system reaches a new equilibrium state. This is why the rate of the reverse reaction gradually increases once a chemical reaction begins.

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Key Concepts

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

Reaction Rates
When discussing chemical reactions, reaction rates are fundamental to understanding how quickly reactants convert into products. In a reversible reaction, two opposing reactions happen: the forward reaction, where reactants turn into products, and the reverse reaction, where products convert back into reactants.
Initially, the forward reaction is often faster because there are more reactants available to interact. As the reaction proceeds, the concentration of reactants decreases, making the forward reaction slower. Simultaneously, the concentration of products increases, leading to a quicker reverse reaction.
This interplay continues until both the forward and reverse reactions happen at the same rate, reaching chemical equilibrium. The concept of reaction rates helps predict how long it will take for a system to reach this equilibrium.
Le Chatelier's Principle
Le Chatelier's Principle is a key concept in chemical equilibrium, describing how a system adjusts to changes. If a change is applied to a system at equilibrium, such as altering the concentration of reactants or products, the system responds by shifting the equilibrium position to reduce that change.
For example, adding more reactants will push the equilibrium towards forming more products to reduce the reactants' concentration. Conversely, removing some products would shift the equilibrium to produce more products to re-establish balance.
  • This principle helps in predicting the direction that a reaction will take in response to various changes.
  • It is widely used in industrial applications to optimize conditions for desired chemical products.
By understanding this principle, chemists can better control and manipulate reactions to achieve desired outcomes.
Reversible Reactions
Reversible reactions are those in which the reactants can form products, and the products can revert back to reactants.
This dual-pathway nature means these reactions don't proceed to completion but instead reach a state of equilibrium. Unlike irreversible reactions, reversible reactions are dynamic, with ongoing forward and reverse reactions even at equilibrium.
  • At the start, the forward reaction dominates due to a larger pool of reactants.
  • As products accumulate, the reverse reaction speeds up.
  • Eventually, both reaction rates balance out, establishing equilibrium.
This characteristic of reversible reactions is crucial in many natural and industrial processes, such as the Haber process for ammonia synthesis or the functioning of biological systems. Understanding reversible reactions allows for better control of chemical processes by manipulating conditions to favor either the forward or reverse reaction.

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Most popular questions from this chapter

On the basis of \(K_{\text {eq }}\) values, which reaction goes essentially to completion? How would you describe the other reaction? (a) \(2 \mathrm{H}_{2}(g)+\mathrm{O}_{2}(g) \leftrightarrows 2 \mathrm{H}_{2} \mathrm{O}(g)\); \(K_{\mathrm{eq}}=3 \times 10^{81}\) (b) \(2 \mathrm{HF}(g) \rightleftarrows \mathrm{H}_{2}(g)+\mathrm{F}_{2}(g)\) \(K_{\mathrm{eq}}=1 \times 10^{-95}\)

\(K_{\text {eq }}=3.9 \times 10^{-11}\) for the dissolution of calcium fluoride in water: \(\mathrm{CaF}_{2}(s) \rightleftarrows \mathrm{Ca}^{2+}(a q)+2 \mathrm{~F}^{-}(a q)\) (a) What is another name for \(K_{\mathrm{eq}}\) for this reaction? (b) If the equilibrium calcium ion concentration in a saturated aqueous solution of calcium fluoride is \(3.3 \times 10^{-4} \mathrm{M}\), what is the equilibrium fluoride ion concentration? (c) Which is larger, the rate constant for the forward reaction or the rate constant for the reverse reaction? (d) Which is larger, \(E_{a}\) for the forward reaction or \(E_{\mathrm{a}}\) for the reverse reaction? (e) Which is larger, the rate of the forward reaction or the rate of the reverse reaction? (f) For lithium carbonate, \(K_{\mathrm{sp}}=0.0011\). Write the balanced chemical equation and the equilibrium expression for the dissolution of \(\mathrm{Li}_{2} \mathrm{CO}_{3}\) in water. (g) Which is more soluble in water, calcium fluoride or lithium carbonate?

Would the solubility of \(\mathrm{PbI}_{2}(s)\) be greater in water or in an aqueous solution of NaI? Explain your answer. (Hint: If \(\left[\mathrm{Pb}^{2+}\right] \times\left[\mathrm{I}^{-}\right]^{2}>K_{\mathrm{sp}}\) ' precipitation will occur.)

State Le Châtelier's principle using the words undo and partially.

Consider the gas-phase reaction \(3 \mathrm{O}_{2}(g) \rightleftarrows 2 \mathrm{O}_{3}(g) .\) Suppose \(K_{\mathrm{eq}}\) for this reaction is \(\sim 1\) (it is not, but assume it is for this problem). Suppose you want pure ozone \(\left(\mathrm{O}_{3}\right)\) that is uncontaminated with oxygen \(\left(\mathrm{O}_{2}\right)\). (a) Why can't you simply remove the oxygen from the reaction vessel once the reaction has come to equilibrium to obtain pure ozone? (b) In fact, \(K_{\text {eq }}\) for this reaction at room temperature is \(2.5 \times 10^{-29}\). Knowing this, how important would you say Le Châtelier's principle is for this reaction when it comes to influencing the amount of ozone present at equilibrium? Explain.

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