/*! This file is auto-generated */ .wp-block-button__link{color:#fff;background-color:#32373c;border-radius:9999px;box-shadow:none;text-decoration:none;padding:calc(.667em + 2px) calc(1.333em + 2px);font-size:1.125em}.wp-block-file__button{background:#32373c;color:#fff;text-decoration:none} Problem 83 A mixture of carbon monoxide, hy... [FREE SOLUTION] | 91Ó°ÊÓ

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A mixture of carbon monoxide, hydrogen, and methanol, \(\mathrm{CH}_{3} \mathrm{OH}\), is at equilibrium according to the equation $$ \mathrm{CO}(g)+2 \mathrm{H}_{2}(g) \rightleftharpoons \mathrm{CH}_{3} \mathrm{OH}(g) $$ At \(250^{\circ} \mathrm{C}\), the mixture is \(0.096 M \mathrm{CO}, 0.191 M \mathrm{H}_{2}\), and \(0.015 M\) \(\mathrm{CH}_{3} \mathrm{OH}\). What is \(K_{c}\) for this reaction at \(250^{\circ} \mathrm{C}\) ?

Short Answer

Expert verified
The equilibrium constant \( K_c \) at \(250^{\circ} \text{C}\) is 4.28.

Step by step solution

01

Understanding the Reaction and Equilibrium

We are given a reaction: \(\text{CO}(g) + 2 \text{H}_{2}(g) \rightleftharpoons \text{CH}_{3}\text{OH}(g)\). The concentrations at equilibrium are \( \text{CO} = 0.096~M \), \( \text{H}_{2} = 0.191~M \), and \( \text{CH}_{3}\text{OH} = 0.015~M \). We need to find the equilibrium constant, \( K_c \), for this reaction at \(250^{\circ} \text{C}\).
02

Writing the Expression for Kc

The equilibrium constant expression for the reaction is given by:\[ K_c = \frac{[\text{CH}_{3}\text{OH}]}{[\text{CO}][\text{H}_{2}]^2} \]Here, the concentrations of the reactants and products are used to calculate \( K_c \).
03

Substituting In the Given Concentrations

Substitute the given concentrations into the \( K_c \) expression:\[ K_c = \frac{0.015}{0.096 \times (0.191)^2} \]
04

Calculating Kc Value

Perform the calculations:Calculate \( (0.191)^2 = 0.036481 \).Then, calculate \( 0.096 \times 0.036481 = 0.003502176 \).Finally,\[ K_c = \frac{0.015}{0.003502176} = 4.284 \]
05

Conclusion

The equilibrium constant \( K_c \) for the reaction at \(250^{\circ} \text{C}\) is found to be approximately 4.28.

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Key Concepts

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

Equilibrium Constant
The equilibrium constant, denoted as \( K_c \), is a crucial component in understanding how chemical reactions behave when they are in a state of balance, or equilibrium. The equilibrium constant provides a numerical value that shows the ratio of the concentration of products to reactants at equilibrium for a given reaction. This concept is foundational in chemistry because it helps predict how the concentrations of different substances in a reaction mixture will change over time. The value of \( K_c \) is different for each reaction and depends only on the temperature at which the reaction occurs. For example, if \( K_c \) is large, it indicates that the products are favored at equilibrium, whereas a small \( K_c \) suggests that the reactants are more prevalent. This insight helps chemists understand the position of equilibrium and guides decisions in industrial applications where maximizing product yield is essential.
Reaction Quotient
The reaction quotient, \( Q \), is similar to the equilibrium constant, but with a key difference: it represents the ratio of products to reactants at any point in time, not just at equilibrium. By calculating \( Q \) and comparing it with the equilibrium constant \( K_c \), one can determine which direction a reaction needs to progress to reach equilibrium. If \( Q < K_c \), the reaction favors the formation of products to reach equilibrium. Conversely, if \( Q > K_c \), the reaction shifts towards the reactants. When \( Q = K_c \), the system is already at equilibrium. This comparison is an essential tool for predicting the behavior of reactions under different conditions and adjusting processes accordingly. Understanding the reaction quotient allows chemists to manipulate reaction conditions, such as temperature or concentration, to steer reactions in the desired direction.
Equilibrium Expression
An equilibrium expression is a mathematical formula we use to describe the relationship between the concentrations of reactants and products in a chemical reaction at equilibrium. It forms the basis for calculating the equilibrium constant \( K_c \). The expression for a generic reaction, \( aA + bB \rightleftharpoons cC + dD \), can be written as:\[ K_c = \frac{[C]^c [D]^d}{[A]^a [B]^b} \]In this formula, \([A], [B], [C], \text{and} [D]\) represent the equilibrium concentrations of the chemicals, and \(a, b, c, \text{and} d\) are their respective stoichiometric coefficients. This equation highlights how the nature and quantities of the substances involved in a reaction affect the equilibrium state. Recognizing how to construct and use equilibrium expressions is key to solving problems and understanding dynamic chemical systems.
Gaseous Reactions
Gaseous reactions are a common type of chemical reaction where the reactants and products are in the gas phase. They are especially interesting in the context of equilibrium because changes in pressure, volume, and temperature can significantly affect the position of equilibrium. For gaseous reactions, the equilibrium constant is often expressed in terms of partial pressures, rather than concentrations, although both can be related through the ideal gas law. This is crucial since gases behave differently from liquids and solids, responding more noticeably to changes in these conditions. In practice, when dealing with gaseous reactions, adjustments to the reaction conditions can help in achieving a more favorable equilibrium state. For example, increasing the pressure of a system can drive the reaction towards the side with fewer gas molecules, based on Le Chatelier’s principle. Understanding these concepts helps in optimizing processes such as chemical synthesis, where controlling the environment is necessary for desired outcomes.

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Most popular questions from this chapter

Explain why pure liquids and solids can be ignored when writing the equilibrium-constant expression.

When a continuous stream of hydrogen gas, \(\mathrm{H}_{2}\), passes over hot magnetic iron oxide, \(\mathrm{Fe}_{3} \mathrm{O}_{4}\), metallic iron and water vapor form. When a continuous stream of water vapor passes over hot metallic iron, the oxide \(\mathrm{Fe}_{3} \mathrm{O}_{4}\) and \(\mathrm{H}_{2}\) form. Explain why the reaction goes in one direction in one case but in the reverse direction in the other.

Sulfur trioxide, used to manufacture sulfuric acid, is obtained commercially from sulfur dioxide. $$ 2 \mathrm{SO}_{2}(g)+\mathrm{O}_{2}(g) \rightleftharpoons 2 \mathrm{SO}_{3}(g) $$ The equilibrium constant \(K_{c}\) for this reaction is \(4.17 \times 10^{-2}\) at \(727^{\circ} \mathrm{C}\). What is the direction of reaction when a mixture that is \(0.20 \mathrm{M} \mathrm{SO}_{2}, 0.10 \mathrm{M} \mathrm{O}_{2}\), and \(0.40 \mathrm{M} \mathrm{SO}_{3}\) approaches equilibrium?

A 2.00-L vessel contains \(1.00 \mathrm{~mol} \mathrm{~N}_{2}, 1.00 \mathrm{~mol} \mathrm{H}_{2}\), and \(2.00 \mathrm{~mol} \mathrm{NH}_{3} .\) What is the direction of reaction (forward or reverse) needed to attain equilibrium at \(400^{\circ} \mathrm{C}\) ? The equilibrium constant \(K_{c}\) for the reaction $$ \mathrm{N}_{2}(g)+3 \mathrm{H}_{2}(g) \rightleftharpoons 2 \mathrm{NH}_{3}(g) $$ is \(0.51\) at \(400^{\circ} \mathrm{C}\).

Hydrogen fluoride decomposes according to the following equation: $$ 2 \mathrm{HF}(g) \rightleftharpoons \mathrm{H}_{2}(g)+\mathrm{F}_{2}(g) $$ The value of \(K_{c}\) at room temperature is \(1.0 \times 10^{-95} .\) From the magnitude of \(K_{c}\), do you think the decomposition occurs to any great extent at room temperature? If an equilibrium mixture in a 1.0-L vessel contains \(1.0 \mathrm{~mol} \mathrm{HF}\), what is the amount of \(\mathrm{H}_{2}\) formed? Does this result agree with what you expect from the magnitude of \(K_{c}\) ?

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