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Suppose a reaction has the equilibrium constant \(K=1.3 \times 10^{8}\). What does the magnitude of this constant tell you about the relative concentrations of products and reactants that will be present once equilibrium is reached? Is this reaction likely to be a good source of the products?

Short Answer

Expert verified
The given equilibrium constant, \(K = 1.3 \times 10^8\), is a large value, indicating that the reaction favors the formation of products. At equilibrium, the concentration of products will be much higher than that of reactants. Therefore, this reaction is likely to be a good source of the products.

Step by step solution

01

Understanding the equilibrium constant

The equilibrium constant (K) is a number that describes the ratio of the concentrations of products to reactants at equilibrium. A large K value indicates that the reaction favors the formation of products, while a small K value indicates that the reaction favors the formation of reactants.
02

Analyzing the given K value

In this problem, the equilibrium constant K is given as \(1.3 \times 10^{8}\). This is a large K value, which means that the reaction favors the formation of products.
03

Determining the relative concentrations of products and reactants at equilibrium

As the magnitude of the equilibrium constant is large, it indicates that the concentration of products in the reaction mixture will be much higher than that of the reactants when the equilibrium is reached. In other words, the products are being formed to a far greater extent compared to the amount of reactants remaining at equilibrium.
04

Evaluating if the reaction is a good source of products

Since the reaction favors the formation of products (due to large K value), it is likely to be a good source of products. In fact, with a value of \(1.3 \times 10^{8}\), the reaction is highly product-favorable, and most reactants will be converted into products at equilibrium. In conclusion, the magnitude of the equilibrium constant in this problem indicates that the reaction strongly favors the formation of products, and that the concentration of products will be much higher than that of reactants at equilibrium. This reaction is, therefore, likely to be a good source of the products.

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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 is the point where the rate of the forward reaction equals the rate of the reverse reaction. At this stage, the concentrations of reactants and products remain constant, though not necessarily equal. Achieving equilibrium doesn't mean that reactions stop. Instead, it's about a balance where changes continue in both directions at the same rate.
This balance is vital in predicting how a reaction behaves and is described by the equilibrium constant, denoted as \( K \). Understanding chemical equilibrium helps us anticipate how much product will form under certain conditions.
The equilibrium position is influenced by various factors such as temperature, pressure, and concentration. By adjusting these, we can shift the equilibrium to favor either reactants or products, all without changing \( K \).
  • Forward and reverse reaction rates are equal
  • Concentrations of products and reactants stay constant
  • \( K \) remains constant at a given temperature
Reaction Favorability
Reaction favorability refers to whether a reaction naturally prefers to form products or maintain reactants at equilibrium. The equilibrium constant \( K \) provides insight here. A large \( K \) value suggests product favorability, meaning the reaction tends to proceed towards forming more products.
Conversely, a small \( K \) indicates that reactants are favored, and the reaction doesn’t generate many products. Understanding favorability can help chemists decide whether a reaction is efficient or needs further adjustment.
Imagine a reaction where \( K \) is significantly large, like \( 1.3 \times 10^{8} \). In this case, the reaction is heavily slanted towards product formation. This insight allows us to predict outcomes more accurately.
  • Large \( K \): Product-favored
  • Small \( K \): Reactant-favored
  • \( K \) magnitude indicates reaction direction preference
Product Formation
When we talk about product formation in a reaction, it's crucial to consider how much of the reactants convert into products. A significant equilibrium constant hints at dominant product formation. With a large \( K \), nearly all reactants might transform into products at equilibrium.
This conversion efficiency is valuable in industries where maximizing product yield is essential, such as pharmaceuticals or chemical manufacturing.
In other words, the larger the \( K \), the greater the conversion of reactants into products, indicating a successful reaction in terms of output.
  • High \( K \) means significant product formation
  • Industries rely on efficient product conversion
  • Reaction success ties to product yield
Concentration of Products and Reactants
The equilibrium constant determines the ratio of the concentration of products to reactants at equilibrium. For a reaction with a large \( K \), the concentration of products will be much higher compared to the reactants.
This is because equilibrium shifts to favor product formation, using up most of the reactants. This scenario is ideal when producing large quantities of a desired product, because most of the reactants have been transformed.
Conversely, if \( K \) is small, reactants are prevalent, meaning limited product formation. This balance between product and reactant concentrations is crucial for understanding reaction dynamics.
  • Large \( K \): Higher product concentration
  • Small \( K \): Higher reactant concentration
  • Concentration ratios aid in predicting reaction outcomes

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Most popular questions from this chapter

A \(1.604-g\) sample of methane \(\left(\mathrm{CH}_{4}\right)\) gas and \(6.400 \mathrm{~g}\) oxygen gas are sealed into a 2.50-L vessel at \(411^{\circ} \mathrm{C}\) and are allowed to reach equilibrium. Methane can react with oxygen to form gaseous carbon dioxide and water vapor, or methane can react with oxygen to form gaseous carbon monoxide and water vapor. At equilibrium, the pressure of oxygen is \(0.326 \mathrm{~atm}\), and the pressure of water vapor is \(4.45\) atm. Calculate the pressures of carbon monoxide and carbon dioxide present at equilibrium.

For the reaction below, \(K_{\mathrm{p}}=1.16\) at \(800 .{ }^{\circ} \mathrm{C}\). $$ \mathrm{CaCO}_{3}(s) \rightleftharpoons \mathrm{CaO}(s)+\mathrm{CO}_{2}(g) $$ If a \(20.0-\mathrm{g}\) sample of \(\mathrm{CaCO}_{3}\) is put into a \(10.0\) - \(\mathrm{L}\) container and heated to \(800 .{ }^{\circ} \mathrm{C}\), what percentage by mass of the \(\mathrm{CaCO}_{3}\) will react to reach equilibrium?

A gaseous material \(\mathrm{XY}(g)\) dissociates to some extent to produce \(\mathrm{X}(g)\) and \(\mathrm{Y}(g)\) : $$ \mathrm{XY}(g) \rightleftharpoons \mathrm{X}(g)+\mathrm{Y}(g) $$ A \(2.00-\mathrm{g}\) sample of \(\mathrm{XY}\) (molar mass \(=165 \mathrm{~g} / \mathrm{mol}\) ) is placed in a container with a movable piston at \(25^{\circ} \mathrm{C}\). The pressure is held constant at \(0.967\) atm. As \(X Y\) begins to dissociate, the piston moves until \(35.0\) mole percent of the original XY has dissociated and then remains at a constant position. Assuming ideal behavior, calculate the density of the gas in the container after the piston has stopped moving, and determine the value of \(K\) for this reaction of \(25^{\circ} \mathrm{C}\).

At a particular temperature, \(K=2.0 \times 10^{-6}\) for the reaction $$ 2 \mathrm{CO}_{2}(g) \rightleftharpoons 2 \mathrm{CO}(g)+\mathrm{O}_{2}(g) $$ If \(2.0\) moles of \(\mathrm{CO}_{2}\) is initially placed into a \(5.0-\mathrm{L}\) vessel, calculate the equilibrium concentrations of all species.

At \(900^{\circ} \mathrm{C}, K_{\mathrm{p}}=1.04\) for the reaction $$ \mathrm{CaCO}_{3}(s) \rightleftharpoons \mathrm{CaO}(s)+\mathrm{CO}_{2}(g) $$ At a low temperature, dry ice (solid \(\mathrm{CO}_{2}\) ), calcium oxide, and calcium carbonate are introduced into a \(50.0-\mathrm{L}\) reaction chamber. The temperature is raised to \(900^{\circ} \mathrm{C}\), resulting in the dry ice converting to gaseous \(\mathrm{CO}_{2}\). For the following mixtures, will the initial amount of calcium oxide increase, decrease, or remain the same as the system moves toward equilibrium at \(900^{\circ} \mathrm{C} ?\) a. \(655 \mathrm{~g} \mathrm{CaCO}_{3}, 95.0 \mathrm{~g} \mathrm{CaO}, P_{\mathrm{CO}_{2}}=2.55 \mathrm{~atm}\) b. \(780 \mathrm{~g} \mathrm{CaCO}_{3}, 1.00 \mathrm{~g} \mathrm{CaO}, P_{\mathrm{CO}_{2}}=1.04 \mathrm{~atm}\) c. \(0.14 \mathrm{~g} \mathrm{CaCO}_{3}, 5000 \mathrm{~g} \mathrm{CaO}, P_{\mathrm{CO}_{2}}=1.04 \mathrm{~atm}\) d. \(715 \mathrm{~g} \mathrm{CaCO}_{3}, 813 \mathrm{~g} \mathrm{CaO}, P_{\mathrm{CO}_{2}}=0.211 \mathrm{~atm}\)

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