What symbol and name are used to replace the ratio \(k_{\mathrm{f}} / k_{\mathrm{r}}\) for a reaction?

Short Answer

Expert verified
The symbol for the ratio \(k_{\mathrm{f}} / k_{\mathrm{r}}\) is either \(K_\mathrm{c}\) or \(K_\mathrm{eq}\), and the name for this ratio is the equilibrium constant.

Step by step solution

01

Identify the ratio

The given ratio is \(k_{\mathrm{f}} / k_{\mathrm{r}}\), where \(k_{\mathrm{f}}\) is the forward reaction rate constant, and \(k_{\mathrm{r}}\) is the reverse reaction rate constant.
02

Symbol and name for the ratio

The symbol used to represent this ratio is \(K_\mathrm{c}\) or \(K_\mathrm{eq}\), and the name for this ratio is the equilibrium constant.

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Key Concepts

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

Reaction Rate Constant
In chemical kinetics, the reaction rate constant is a crucial factor that quantifies the speed at which a chemical reaction proceeds. For any given reaction, whether forward or reverse, we denote the rate constant with the letter 'k'. The forward reaction rate constant, represented by \( k_{\text{f}} \), is associated with the conversion of reactants to products. Conversely, the reverse reaction rate constant, denoted \( k_{\text{r}} \), is linked to the transformation of products back into reactants.

Generally, the value of the rate constant varies with temperature and the presence of a catalyst, following the principles outlined by the Arrhenius equation. It's important to understand that the magnitude of these constants provides insight into the reaction's tendency to proceed in a particular direction under given conditions. A larger \( k_{\text{f}} \) suggests a reaction that favors the formation of products, while a larger \( k_{\text{r}} \) indicates a reaction inclined to revert to its reactants.
Forward Reaction
The term 'forward reaction' refers to the process in which reactants are turned into products. It is the initial direction of a chemical reaction before reaching a state of equilibrium. The speed at which this occurs is quantified by the forward reaction rate constant \( k_{\text{f}} \).

A rate equation, which involves the rate constant and the concentrations of the reactants, usually describes the reaction. For example, in a simple reaction where A and B react to form C, the rate of the forward reaction could be represented by \( k_{\text{f}} [A][B] \), emphasizing that the rate is directly proportional to the concentration of the reactants and the constant \( k_{\text{f}} \).

Understanding the dynamics of the forward reaction is essential in predicting how a reaction proceeds over time and, fundamentally, in designing processes and conditions favorable for desired product yields.
Reverse Reaction
In contrast to the forward reaction, the reverse reaction is the process where products can revert back to form the original reactants. It counteracts the forward reaction and is a natural part of the dynamic equilibrium in a closed system. The rate at which the reverse reaction occurs is expressed by the reverse reaction rate constant \( k_{\text{r}} \).

This reaction rate is similarly governed by a rate equation, and for the reverse process of our earlier example, it could be represented as \( k_{\text{r}} [C] \). This indicates that the rate of the reverse reaction is directly proportional to the concentration of the product and the constant \( k_{\text{r}} \).

The reverse reaction is an integral concept in understanding chemical equilibrium, as it directly impacts the final composition of reactants and products in a mixture once equilibrium is established.
Chemical Equilibrium
Chemical equilibrium occurs in a closed system when the rates of the forward and reverse reactions are equal, leading to no net change in the concentrations of reactants and products over time. At this point, the reaction has not stopped; rather, it is dynamically balanced, with the forward and reverse reactions continuously occurring at the same rate.

The relationship between the rate constants for the forward and reverse reactions is captured by the equilibrium constant, represented by \( K_{\text{c}} \) or \( K_{\text{eq}} \). It is defined as the ratio \( K_{\text{eq}} = \frac{k_{\text{f}}}{k_{\text{r}}} \), signifying the ratio of the reaction rate constants. The numerical value of the equilibrium constant gives insight into the product/reactant ratio at equilibrium and, thus, the position of equilibrium. A high \( K_{\text{eq}} \) value means that the equilibrium heavily favors the products, while a low value suggests predominant reactants.

Understanding equilibrium constants is invaluable for predicting the extent of a reaction and determining the conditions required to optimize the yield of specific products. It is a cornerstone concept in both academic studies and practical applications such as industrial chemical synthesis.

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Most popular questions from this chapter

Sparingly soluble magnesium hydroxide dissolves in water to yield an equilibrium magnesium ion concentration of \(1.44 \times 10^{-4} \mathrm{M}\). (a) Write the balanced equilibrium equation for magnesium hydroxide dissolving in water. (b) Write the \(K_{\mathrm{sp}}\) expression for magnesium hydroxide. (c) What is the equilibrium concentration of hydroxide ion? (d) Calculate the value of \(K_{\mathrm{sp}}\) for magnesium hydroxide (show your calculation).

At \(25^{\circ} \mathrm{C}\), the solubility in water of the moderately soluble salt silver acetate, \(\mathrm{AgC}_{2} \mathrm{H}_{3} \mathrm{O}_{2}\) is \(10.6 \mathrm{~g} / \mathrm{L}\). (a) Write the chemical equation for the dissolving of silver acetate in water. (b) Write the \(K_{\mathrm{sp}}\) expression for silver acetate. (c) Calculate the value of \(K_{\mathrm{sp}}\) (show your work).

Suppose you have an endothermic reaction with \(K_{\text {eq }}\) approximately equal to \(1 \times 10^{3}\). How could you adjust the temperature of this reaction to drive it to completion? Explain your answer.

An 8.00-L reaction vessel at \(491^{\circ} \mathrm{C}\) contains \(0.650\) mole of \(\mathrm{H}_{2}, 0.275\) mole of \(\mathrm{I}_{2}\), and \(3.00\) moles of HI. Assuming that the reaction is at equilibrium, determine the value of \(K_{\text {eq }}\) and comment on where the equilibrium lies. The reaction is: \(\mathrm{H}_{2}(g)+\mathrm{I}_{2}(g) \rightleftarrows 2 \mathrm{HI}(g)\)

What does a value of \(K_{\text {eq }}\) greater than \(10^{3}\) imply? Prove that your answer is correct by using the general expression \(K_{\mathrm{eq}}=[\) Products \(] /[\) Reactants \(]\).

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