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A certain reaction has a \(K_{\text {eq }}\) value of \(1.5 \times 10^{-6}\). (a) Would this be a practical reaction from which to isolate pure product? Explain your answer.

Short Answer

Expert verified
No, this reaction would not be practical for isolating pure product because the small \(K_{\text{eq}}\) value of \(1.5 \times 10^{-6}\) indicates that the reaction favors the reactants, meaning there would be a significantly higher concentration of reactants compared to products at equilibrium, resulting in a low yield of the desired product.

Step by step solution

01

Analyze the given equilibrium constant

The given reaction has an equilibrium constant, \(K_{\text{eq}}\), of \(1.5 \times 10^{-6}\). This is a small value, which implies that the reaction favors the reactants much more than the products.
02

Determine if the reaction is practical for isolating product

Since the \(K_{\text{eq}}\) value is small, it indicates that at the equilibrium position, there will be a significantly higher concentration of reactants compared to products. Therefore, this reaction would not be considered practical for isolating pure product, as it would not provide a high yield of the desired product.

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Key Concepts

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

Chemical Equilibrium
Chemical equilibrium refers to a state in a reversible chemical reaction where the concentrations of both reactants and products remain constant over time. This happens when the rate of the forward reaction equals the rate of the reverse reaction.
Understanding chemical equilibrium is crucial in predicting the extent of a reaction under a given set of conditions. It helps in determining how much product can be formed when reactants are mixed. A key part of studying chemical equilibrium is the equilibrium constant, denoted as \(K_{eq}\). This constant provides insight into the relative amounts of products and reactants at equilibrium.
  • If \(K_{eq}\) is much greater than 1, the reaction strongly favors the formation of products.
  • If \(K_{eq}\) is much less than 1, the reaction tends to favor the reactants, making it difficult to obtain a large quantity of the product.
  • A \(K_{eq}\) close to 1 suggests a roughly equal amount of reactants and products at equilibrium.
Knowing the value of \(K_{eq}\) helps chemists evaluate whether a particular reaction is practical for producing a desired chemical output.
Reaction Yield
Reaction yield refers to the amount of product formed in a chemical reaction, often expressed as a percentage of the theoretical maximum. It's important for chemists to assess the efficiency and feasibility of a reaction.
The yield is highly influenced by the equilibrium constant \(K_{eq}\). In our exercise, a small \(K_{eq}\) of \(1.5 \times 10^{-6}\) suggests that reaching equilibrium results in a small amount of product. This low yield situation poses challenges when isolating enough product for practical uses.
  • A high reaction yield is favorable for industrial and laboratory processes because it minimizes waste and maximizes the output of desired substances.
  • A low yield may not only reduce efficiency but also increase costs and resource consumption.
Therefore, evaluating the reaction yield alongside the equilibrium constant is vital in deciding a reaction's practical applicability.
Le Chatelier's Principle
Le Chatelier's Principle is a guideline used to predict how a change in conditions affects a chemical equilibrium. It states that if a dynamic equilibrium is disturbed by changing the conditions, the system responds by shifting the equilibrium position to counteract the change, aiming to re-establish equilibrium.
Understanding this principle is crucial when dealing with reactions with low \(K_{eq}\) values, as it allows for manipulation to potentially increase product yield. For example:
  • Increasing the concentration of reactants can drive the equilibrium towards product formation.
  • Adjusting temperature and pressure may also favor either the forward or reverse reaction, depending on the reaction specifics.
This principle is a powerful tool for chemists aiming to optimize reaction conditions for better yields, especially in cases where the natural equilibrium position does not provide a substantial amount of product.

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Most popular questions from this chapter

Will \(K_{\text {eq }}\) for an exothermic reaction increase or decrease when the reaction mixture is (a) heated and (b) cooled? Explain your answer.

Suppose the reaction \(\mathrm{A}_{2}+\mathrm{B}_{2} \rightleftarrows 2 \mathrm{AB}\) proceeds via a one-step mechanism involving a collision between one \(\mathrm{A}_{2}\) molecule and one \(\mathrm{B}_{2}\) molecule. Suppose also that this reaction is reversible, and that the forward reaction is inherently much faster than the reverse reaction. (a) Does the equilibrium lie to the left or to the right? Explain your choice in terms of the reactant and product concentrations necessary to establish equal forward and reverse rates. (b) Does an analysis in terms of the relationship \(K_{\mathrm{eq}}=k_{\mathrm{f}} / k_{\mathrm{r}}\) yield the same answer as in (a)? Explain.

For the reaction \(2 \mathrm{NO}_{2}(g) \rightleftarrows \mathrm{N}_{2} \mathrm{O}_{4}(g)\) \(K_{\text {eq }}=0.500\). What is the equilibrium molar concentration of \(\mathrm{NO}_{2}(g)\) if \(\left[\mathrm{N}_{2} \mathrm{O}_{4}\right]=0.248 \mathrm{M}\) ?

Sparingly soluble aluminum hydroxide dissolves in water to yield an equilibrium hydroxide ion concentration of \(8.58 \times 10^{-9} \mathrm{M}\). (a) Write the balanced equilibrium equation for aluminum hydroxide dissolving in water. (b) Write the \(K_{\mathrm{sp}}\) expression for aluminum hydroxide. (c) What is the equilibrium concentration of aluminum ion? (d) Calculate the value of \(K_{\mathrm{sp}}\) for aluminum hydroxide (show your calculation).

How would the value of the equilibrium constant for a one-step reaction calculated as \(k_{\mathrm{f}} / k_{\mathrm{r}}\) compare with the value calculated from the concentrations of all substances present at equilibrium?

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