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A reaction has a high rate constant but a small equilibrium constant. What does this mean in terms of producing an industrial product?

Short Answer

Expert verified
Fast reaction with low product yield at equilibrium; strategies needed to increase product yield.

Step by step solution

01

Understanding Rate Constant

The rate constant of a reaction provides information about the speed at which reactants are converted into products. A high rate constant indicates the reaction occurs quickly, with reactants rapidly forming products.
02

Understanding Equilibrium Constant

The equilibrium constant expresses the ratio of concentrations of products to reactants at equilibrium. A small equilibrium constant implies that, at equilibrium, the concentration of reactants is higher than that of products, suggesting products do not form in large amounts.
03

Interpreting Both Constants

Given the high rate constant and small equilibrium constant, this scenario suggests that reactants convert to products rapidly, but the reverse reaction is also favored, leading to a high concentration of reactants at equilibrium.
04

Implication for Industrial Production

In an industrial context, while the reaction can achieve product formation quickly due to the high rate, the low equilibrium constant means that achieving high yields of the desired product is challenging. Additional techniques may be necessary to shift equilibrium, such as removing products or changing reaction conditions.

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Key Concepts

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

Rate Constant
The rate constant is a value that helps us understand how fast a chemical reaction takes place. It is part of the rate equation: \[ \text{Rate} = k [A]^m [B]^n \]where \( k \) represents the rate constant, and \([A]\) and \([B]\) are the concentrations of reactants. The exponents \( m \) and \( n \) indicate how the rate changes as the concentration of each reactant changes. A high rate constant means that the reaction can occur very quickly, converting reactants into products at a fast pace. This is often desirable in industrial settings where speed is important for maximizing throughput and efficiency. By understanding and controlling the rate constant, chemists can optimize conditions to speed up production, reducing costs and increasing output.
Equilibrium Constant
The equilibrium constant, denoted as \( K_{eq} \), is crucial for understanding the balance between the reactants and products in a chemical reaction. It is calculated using the formula: \[ K_{eq} = \frac{[C]^c [D]^d}{[A]^a [B]^b} \]where \([C]\) and \([D]\) are the concentrations of products, and \([A]\) and \([B]\) are the concentrations of reactants at equilibrium. The exponents \( a, b, c, \) and \( d \) are their stoichiometric coefficients. When the equilibrium constant is small, it indicates that even though reactants may be quickly converted to products initially, the reverse reaction is also significant. Thus, the final concentrations at equilibrium favor the reactants over the products. This can limit the practicality of producing a specific product in large quantities without additional manipulation of the reaction conditions.
Industrial Chemistry
Industrial chemistry is the field where chemical processes and reactions are applied for large scale manufacturing. While chemical reactions in this setting need to be fast and efficient, using the rate constant to speed up reactions is just one part of the puzzle. In cases where the equilibrium constant is low, indicating that the reaction might not favor the desired product, industrial chemists need to take steps to shift the balance. Some strategies might include:
  • Altering temperature and pressure conditions to favor desired reaction direction.
  • Removing products as they are formed to prevent them from reversing back to reactants.
  • Introducing catalysts to enhance reaction rates without altering equilibrium positions.
By understanding both the rate and equilibrium constants, industries can better design processes to minimize waste and maximize production yield.
Reaction Yield
Reaction yield is a measure of how much product is obtained from a given amount of reactants. In the context of a reaction with a high rate constant but a low equilibrium constant, achieving a high reaction yield can be challenging. This is because, although the reaction is initially fast, the equilibrium state heavily favors the reactants, leading to low product formation at equilibrium. To improve yield in these situations, strategies such as shifting the equilibrium position to favor products are employed. This can be done by changing conditions—like increasing pressure for gaseous reactions or continuously removing one of the products. By adopting these strategies, even reactions not naturally favorable for high yield can be optimized for industrial processes, ensuring that businesses can maintain profitability while meeting production demands. Understanding these concepts allows chemists to effectively contribute to sustainable and successful industrial operations.

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Most popular questions from this chapter

Consider the reaction \(\mathrm{A} \longrightarrow \mathrm{B}\), for which you wish to determine the rate. You do not have any convenient method for determining the amount of B formed. You do, however, have a method for determining the amount of A left as the reaction proceeds. Does it make any difference if you determine the rate in terms of disappearance of A rather than appearance of B? Why or why not?

It can be shown that a mathematical relationship exists between rate constants and equilibrium constants. For example, consider the following generic reaction, where the rate constant \(k\) refers to the rate of the forward reaction and the rate constant \(k^{\prime}\) refers to the rate of the reverse reaction. $$A+B \frac{k}{\overline{k^{\prime}}} C+D$$ Verify that the equilibrium constant for a reaction is equal to the ratio of the rate constants for the forward and reverse reactions. $$K=\frac{k}{k^{\prime}}$$

What would happen to the reaction that produces ammonia if water is present in the reaction mixture? $$\mathrm{N}_{2}(g)+3 \mathrm{H}_{2}(g) \rightleftharpoons 2 \mathrm{NH}_{3}(g)$$ Hint: Ammonia is very soluble in water.

Consider the reaction of chloromethane with \(\mathrm{OH}^{-}\) in aqueous solution. $$\mathrm{CH}_{3} \mathrm{Cl}(g)+\mathrm{OH}^{-}(a q) \stackrel{k}{\rightleftharpoons} \mathrm{CH}_{3} \mathrm{OH}(a q)+\mathrm{Cl}^{-}(a q)$$ At room temperature, the rate constant for the forward reaction, \(k,\) is \(6 \times 10^{-6}\) and the equilibrium constant, \(K,\) is equal to \(1 \times 10^{16}\). Calculate the rate constant for the reverse reaction, \(k^{\prime}\) at room temperature.

Assume that two different reactions are taking place at the same temperature. In reaction A, two different spherical molecules collide to yield a product. In reaction \(\mathrm{B}\), the shape of the colliding molecules is rodlike. Each reaction has the same number of collisions per second and the same activation energy. Which reaction goes faster?

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