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Chapter 14: Problem 7

What qualitative information can you get from the magnitude of the equilibrium constant?

Short Answer

Expert verified
The magnitude of the equilibrium constant indicates whether a reaction favors products (large \( K \)), reactants (small \( K \)), or is balanced (\( K \approx 1 \)).

Step by step solution

01

Understanding the Equilibrium Constant

The equilibrium constant, often denoted as \( K \), represents the ratio of the concentrations of products to reactants at equilibrium for a reversible chemical reaction at a given temperature. Mathematically, for the reaction \( aA + bB ightleftharpoons cC + dD \), it is expressed as \( K = \frac{[C]^c [D]^d}{[A]^a [B]^b} \).
02

Interpreting Large Equilibrium Constants

If \( K \) is much greater than 1, it indicates that, at equilibrium, the concentration of products is much greater than the concentration of reactants. This suggests that the reaction heavily favors the formation of products. Practically, this means the reaction proceeds almost to completion.
03

Interpreting Small Equilibrium Constants

Conversely, if \( K \) is much less than 1, it implies that the concentration of reactants is much greater than that of the products at equilibrium. This signifies that the reaction does not proceed far and favors the reactants.
04

Equilibrium Constant Close to One

When \( K \) is close to 1, this indicates that neither reactants nor products are particularly favored, and significant amounts of both are present at equilibrium. It suggests that the reaction proceeds to a significant extent but does not favor one side.

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Key Concepts

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

Chemical Equilibrium
In a chemical reaction, equilibrium signifies a balance between the forward and reverse reactions. Imagine a seesaw perfectly balanced in the center. At this point, the rate at which reactants transform into products equals the rate at which products revert back to reactants.
This creates a dynamic state where both processes occur simultaneously but the concentrations of reactants and products remain constant over time.
This state, known as chemical equilibrium, is crucial in understanding how reactions behave under various conditions.
  • In a reversible reaction, equilibrium indicates no further net change in concentrations.
  • Even though reactants and products are continuously interconverting, their ratio remains stable.
  • The presence of chemical equilibrium does not mean the reactants and products are equal, but rather their rates of change are equal.
Reaction Completion
The concept of reaction completion centers around how far a reaction proceeds toward the formation of products once equilibrium is achieved. Using the equilibrium constant, denoted as \( K \), we can infer a lot about this aspect.
  • If \( K \) is significantly larger than 1, product formation is heavily favored, suggesting the reaction almost proceeds to completion.
  • Reactions with large \( K \) values are often considered to have reached completion because almost all reactants have transformed into products.
  • Conversely, if \( K \) is small, the reaction barely proceeds towards product formation, showing that reactants are favored and completion is less likely.

In practice, this means knowing \( K \) helps predict how much of each substance will be present when the reaction is at rest, whether most, some, or very little reactants become products.
Concentration Ratio
The equilibrium constant is fundamentally about the concentration ratio of products to reactants at equilibrium. This ratio helps chemists understand the depth or extent of a reaction.
When a reaction reaches equilibrium, this doesn't imply equal concentrations of reactants and products; rather, it refers to a specific ratio as dictated by \( K \).
  • The concentration ratio is pivotal in calculating and predicting the position of equilibrium in chemical reactions.
  • This ratio, represented by \( K = \frac{[C]^c [D]^d}{[A]^a [B]^b} \), indicates which side of the reaction is favored.
  • The value of \( K \) provides a snapshot of the relative amounts of substances in a reaction at any given moment.

By examining this ratio, chemists can deduce whether a reaction is likely to yield more products or retain more reactants under given conditions. This insight is crucial for optimizing chemical processes and predicting outcomes.

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Most popular questions from this chapter

Suppose liquid water and water vapor exist in equilibrium in a closed container. If you add a small amount of liquid water to the container, how does this affect the amount of water vapor in the container? If, instead, you add a small amount of water vapor to the container, how does this affect the amount of liquid water in the container?

An experimenter places the following concentrations of gases in a closed container: \([\mathrm{NOBr}]=7.13 \times 10^{-2} M,[\mathrm{NO}]=\) \(1.58 \times 10^{-2} M,\left[\mathrm{Br}_{2}\right]=1.29 \times 10^{-2} M .\) These gases then react: $$ 2 \mathrm{NOBr}(g) \rightleftharpoons 2 \mathrm{NO}(g)+\mathrm{Br}_{2}(g) $$ At the temperature of the reaction, the equilibrium constant \(K_{c}\) is \(3.07 \times 10^{-4}\). Calculate the reaction quotient, \(Q_{c}\), from the initial concentrations and determine whether the concentration of NOBr increases or decreases as the reaction approaches equilibrium. a. \(Q_{c}=6.33 \times 10^{-4} ;\) the concentration of \(\mathrm{NOBr}\) decreases b. \(Q_{c}=6.33 \times 10^{-4}\); the concentration of NOBr increases c. \(Q_{c}=1.58 \times 10^{4}\); the concentration of NOBr increases d. \(Q_{c}=4.65 \times 10^{-4}\); the concentration of NOBr decreases e. \(Q_{c}=4.65 \times 10^{-4} ;\) the concentration of NOBr increases

The equilibrium-constant expression for a gas reaction is $$ K_{c}=\frac{\left[\mathrm{H}_{2} \mathrm{O}\right]^{2}\left[\mathrm{SO}_{2}\right]^{2}}{\left[\mathrm{H}_{2} \mathrm{~S}\right]^{2}\left[\mathrm{O}_{2}\right]^{3}} $$ Write the balanced chemical equation corresponding to this expression.

Consider the reaction \(\mathrm{N}_{2} \mathrm{O}_{4}(g) \rightleftharpoons 2 \mathrm{NO}_{2}(g) .\) Draw a graph illustrating the changes of concentrations of \(\mathrm{N}_{2} \mathrm{O}_{4}\) and \(\mathrm{NO}_{2}\) as equilibrium is approached. Describe how the rates of the forward and reverse reactions change as the mixture approaches dynamic equilibrium. Why is this called a dynamic equilibrium?

Ammonium hydrogen sulfide, \(\mathrm{NH}_{4} \mathrm{HS}\), is unstable at room temperature and decomposes: $$ \mathrm{NH}_{4} \mathrm{HS}(s) \rightleftharpoons \mathrm{NH}_{3}(g)+\mathrm{H}_{2} \mathrm{~S}(g) $$ You have placed some solid ammonium hydrogen sulfide in a closed flask. Which of the following would produce less hydrogen sulfide, \(\mathrm{H}_{2} \mathrm{~S}\), which is a poisonous gas? a. Removing some \(\mathrm{NH}_{3}\) from the flask b. Adding some \(\mathrm{NH}_{3}\) to the flask c. Removing some of the \(\mathrm{NH}_{4} \mathrm{HS}\) d. Increasing the pressure in the flask by adding helium gas Explain each of your answers.
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