\( \mathrm{~A} 3.00-\mathrm{L}\) reaction vessel is filled with \(0.150 \mathrm{~mol}\) \(\mathrm{CO}, 0.0900 \mathrm{~mol} \mathrm{H}_{2}\), and \(0.180 \mathrm{~mol} \mathrm{CH}_{3} \mathrm{OH}\). Equilibrium is reached in the presence of a zinc oxidechromium(III) oxide catalyst; and at \(300^{\circ} \mathrm{C}, K_{c}=\) \(1.1 \times 10^{-2}\) for the reaction, \(\mathrm{CO}(\mathrm{g})+2 \mathrm{H}_{2}(\mathrm{~g}) \rightleftharpoons\) \(\mathrm{CH}_{3} \mathrm{OH}(\mathrm{g})\). (a) As the reaction approaches equilibrium, will the molar concentration of \(\mathrm{CH}_{3} \mathrm{OH}\) increase, decrease, or remain unchanged? (b) What is the equilibrium composition of the mixture?

Short Answer

Expert verified
As the reaction approaches equilibrium, the molar concentration of \( \mathrm{CH}_3\mathrm{OH} \) will increase if \( Q < K_c \. The equilibrium composition requires solving for \( x \) using an ICE table and the equilibrium constant expression.

Step by step solution

01

Write the Balanced Equation

Write down the balanced chemical equation for the reaction: \[ \mathrm{CO(g)} + 2\mathrm{H}_2(\mathrm{g}) \rightleftharpoons \mathrm{CH}_3\mathrm{OH}(\mathrm{g}) \]
02

Determine the Initial Concentrations

Determine the initial molar concentrations of reactants and products. The concentration is given by the number of moles divided by the volume, thus for \( \mathrm{CO} \): \( \frac{0.150\, \mathrm{mol}}{3.00\, \mathrm{L}} = 0.0500 \, \mathrm{M} \), for \( \mathrm{H}_2 \): \( \frac{0.0900\, \mathrm{mol}}{3.00\, \mathrm{L}} = 0.0300 \, \mathrm{M} \), and for \( \mathrm{CH}_3\mathrm{OH} \): \( \frac{0.180\, \mathrm{mol}}{3.00\, \mathrm{L}} = 0.0600 \, \mathrm{M} \).
03

Calculate Q to Predict Direction of Reaction

The reaction quotient \( Q \) is calculated using the initial concentrations. \[ Q = \frac{[\mathrm{CH}_3\mathrm{OH}]_{\text{initial}}}{[\mathrm{CO}]_{\text{initial}}[\mathrm{H}_2]^2_{\text{initial}}} = \frac{0.0600}{0.0500 \times (0.0300)^2} \] This value can be compared to the equilibrium constant \( K_c \) to predict the direction the reaction will proceed to reach equilibrium.
04

Predict the Change in Concentration

If \( Q < K_c \), the reaction will proceed to the right, making more products. If \( Q > K_c \), the reaction will proceed to the left, making more reactants. If \( Q = K_c \), the reaction is already at equilibrium.
05

Determine the Direction to Equilibrium

Calculate the value of \( Q \) and compare it to \( K_c \). If \( Q \) is less than \( K_c = 1.1 \times 10^{-2} \), \( \mathrm{CH}_3\mathrm{OH} \) will increase as the system moves towards equilibrium.
06

Write the Equilibrium Expression

Write the expression for the equilibrium constant \( K_c \): \[ K_c = \frac{[\mathrm{CH}_3\mathrm{OH}]_{\text{eq}}}{[\mathrm{CO}]_{\text{eq}}[\mathrm{H}_2]^2_{\text{eq}}} \]
07

Set Up the ICE Table

An ICE (Initial, Change, Equilibrium) table can be set up to find the equilibrium concentrations of the reactants and products. Let the change for \( \mathrm{CO} \) be \( -x \), for \( \mathrm{H}_2 \) be \( -2x \), and for \( \mathrm{CH}_3\mathrm{OH} \) be \( +x \).
08

Solve for x

Substitute the equilibrium concentrations into the equilibrium expression in terms of \( x \) and solve for \( x \) to find the changes in concentration.
09

Calculate Equilibrium Concentrations

Use the value of \( x \) calculated to find equilibrium concentrations of each species.
10

Verify with Kc

Verify that the calculated equilibrium concentrations satisfy the equilibrium constant \( K_c \) equation.

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Key Concepts

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

Equilibrium Constant
Understanding chemical reactions often involves appreciating how they balance out over time. That's where the equilibrium constant, typically expressed as \( K_c \), comes into play. It quantifies the relationship between the concentrations of products and reactants when a reaction has reached a state where the rates of the forward and reverse reactions are equal, known as equilibrium. The constant is calculated using a specific formula, which for the provided chemical reaction \( \mathrm{CO(g)} + 2 \mathrm{H}_2(\mathrm{g}) \rightleftharpoons \mathrm{CH}_3\mathrm{OH}(\mathrm{g}) \) would be:
\[ K_c = \frac{[\mathrm{CH}_3\mathrm{OH}]_{\text{eq}}}{[\mathrm{CO}]_{\text{eq}}[\mathrm{H}_2]^2_{\text{eq}}} \]This equation demonstrates the balance that the system tries to maintain despite any changes in conditions. Its precision allows us to predict how the system will respond to those changes and aids in understanding chemical behavior at a molecular level.
Reaction Quotient
The reaction quotient, denoted as \( Q \), is a valuable tool in assessing the current state of a reaction. It closely resembles the equilibrium constant formula but uses the initial concentrations instead of the equilibrium concentrations. By comparing \( Q \) to \( K_c \), we can predict the direction the reaction is likely to take to reach equilibrium. If \( Q < K_c \), the reaction will proceed forward to generate more products. Conversely, if \( Q > K_c \), the reaction will shift to produce more reactants. This concept not only helps in predicting the changes but also plays a fundamental role in the calculation steps of finding the equilibrium composition of a reaction.
ICE Table
An ICE table, which stands for Initial, Change, and Equilibrium, is a strategic format to lay out the concentrations of reactants and products during a reaction. The table aids in simplifying the process of determining how concentrations evolve from initial states to equilibrium. An example of an ICE table for the given reaction can be visualized as columns for each substance involved and rows for their initial concentration, change in concentration, and equilibrium concentration respectively. To solve for unknowns, such as the change in concentration needed to reach equilibrium, the ICE table is invaluable as it organizes and clarifies the process, making the mathematical journey of finding equilibrium compositions much less daunting.
Le Chatelier's Principle
Le Chatelier's Principle provides a qualitative understanding of how a system at equilibrium responds to external changes. It states that if an external stress (such as a change in concentration, pressure, or temperature) is applied to a system at equilibrium, the system will adjust its equilibrium position to counteract that stress. This principle underlies much of chemical intuition and equilibrium analysis. In practice, it may lead to increasing the concentration of reactants or products, changing the pressure by modifying volume, or altering the temperature, each having the potential to shift the equilibrium position. Recognizing and applying Le Chatelier's Principle is crucial when examining the impact of any change on a chemical system's equilibrium.

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Most popular questions from this chapter

The value of the equilibrium constant \(K\) for the reaction \(2 \mathrm{SO}_{2}(\mathrm{~g})+\mathrm{O}_{2}(\mathrm{~g}) \rightleftharpoons 2 \mathrm{SO}_{3}(\mathrm{~g})\) is \(3.0 \times 10^{4}\) at \(700 \mathrm{~K}\). Determine the value of \(K_{c}\) for the reactions (a) \(2 \mathrm{SO}_{2}(\mathrm{~g})+\mathrm{O}_{2}(\mathrm{~g}) \rightleftharpoons 2 \mathrm{SO}_{3}(\mathrm{~g})\) and \((\mathrm{b}) \mathrm{SO}_{3}(\mathrm{~g}) \rightleftharpoons\) \(\mathrm{SO}_{2}(\mathrm{~g})+\frac{1}{2} \mathrm{O}_{2}(\mathrm{~g})\)

State whether reactants or products will be favored by compression in each of the following equilibria. If no change occurs, explain why that is so. (a) \(2 \mathrm{O}_{3}(\mathrm{~g})=3 \mathrm{O}_{2}(\mathrm{~g})\) (b) \(\mathrm{H}_{2} \mathrm{O}(\mathrm{g})+\mathrm{C}(\mathrm{s}) \rightleftharpoons \mathrm{H}_{2}(\mathrm{~g})+\mathrm{CO}(\mathrm{g})\) (c) \(4 \mathrm{NH}_{3}(\mathrm{~g})+5 \mathrm{O}_{2}(\mathrm{~g})=4 \mathrm{NO}(\mathrm{g})+6 \mathrm{H}_{2} \mathrm{O}(\mathrm{g})\) (d) \(2 \mathrm{HD}(\mathrm{g}) \rightleftharpoons \mathrm{H}_{2}(\mathrm{~g})+\mathrm{D}_{2}(\mathrm{~g})\) (e) \(\mathrm{Cl}_{2}(\mathrm{~g})=2 \mathrm{Cl}(\mathrm{g})\)

Use Le Chatelier's principle to predict the effect that the change given in the first column of the following table has on the quantity in the second column for the following equilibrium system: $$ \begin{gathered} 5 \mathrm{CO}(\mathrm{g})+\mathrm{I}_{2} \mathrm{O}_{5}(\mathrm{~s}) \rightleftharpoons \mathrm{I}_{2}(\mathrm{~g})+5 \mathrm{CO}_{2}(\mathrm{~g}) \\\ \Delta H^{\circ}=-1175 \mathrm{~kJ} \end{gathered} $$

Write the equilibrium expressions \(K_{c}\) for the following reactions. (a) \(\mathrm{CO}(\mathrm{g})+\mathrm{Cl}_{2}(\mathrm{~g}) \rightleftharpoons \mathrm{COCl}(\mathrm{g})+\mathrm{Cl}(\mathrm{g})\) (b) \(\mathrm{H}_{2}(\mathrm{~g})+\mathrm{Br}_{2}(\mathrm{~g}) \rightleftharpoons 2 \mathrm{HBr}(\mathrm{g})\) (c) \(2 \mathrm{H}_{2} \mathrm{~S}(\mathrm{~g})+3 \mathrm{O}_{2}(\mathrm{~g})=2 \mathrm{SO}_{2}(\mathrm{~g})+2 \mathrm{H}_{2} \mathrm{O}(\mathrm{g})\)

A mixture consisting of \(1.1 \mathrm{mmol} \mathrm{SO}_{2}\) and \(2.2 \mathrm{mmol} \mathrm{O}_{2}\) in a \(250-\mathrm{mL}\) container was heated to \(500 \mathrm{~K}\) and allowed to reach equilibrium. Will more sulfur trioxide be formed if that equilibrium mixture is cooled to \(25^{\circ} \mathrm{C}\) ? For the reaction \(2 \mathrm{SO}_{2}(\mathrm{~g})+\) \(\mathrm{O}_{2}(\mathrm{~g})=2 \mathrm{SO}_{3}(\mathrm{~g}), \mathrm{K}=2.5 \times 10^{10}\) at \(500 \mathrm{~K}\) and \(4.0 \times 10^{24}\) at \(25^{\circ} \mathrm{C}\).

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