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Chapter 14: Problem 89

A gaseous mixture containing \(1.00 \mathrm{~mol}\) each of \(\mathrm{CO}\), \(\mathrm{H}_{2} \mathrm{O}, \mathrm{CO}_{2},\) and \(\mathrm{H}_{2}\) is exposed to a zinc oxide-copper oxide catalyst at \(1000^{\circ} \mathrm{C}\). The reaction is $$ \mathrm{CO}(g)+\mathrm{H}_{2} \mathrm{O}(g) \rightleftharpoons \mathrm{CO}_{2}(g)+\mathrm{H}_{2}(g) $$ and the equilibrium constant \(K_{c}\) is 0.58 at \(1000^{\circ} \mathrm{C}\). What is the direction of reaction (forward or reverse) as the mixture attains equilibrium?

Short Answer

Expert verified
The reaction proceeds in the reverse direction as it approaches equilibrium.

Step by step solution

01

Identify Initial Concentrations

The initial concentrations for each component of the mixture are given as 1.00 mol each. Hence, the initial concentrations are:\[ [CO] = [H_2O] = [CO_2] = [H_2] = 1.00 \] mol/L since the volume of the vessel containing the mixture is constant.
02

Expression for Reaction Quotient \(Q_c\)

The reaction quotient \(Q_c\), which helps determine the direction of the reaction, is given by:\[ Q_c = \frac{[CO_2][H_2]}{[CO][H_2O]} \]Initially, the concentrations of all species are equal, so:\[ Q_c = \frac{(1.00)(1.00)}{(1.00)(1.00)} = 1.00 \]
03

Compare \(Q_c\) to \(K_c\)

Compare the calculated \(Q_c\) with the given equilibrium constant \(K_c = 0.58\).Since \(Q_c (1.00) > K_c (0.58)\), the reaction will shift in the direction that reduces \(Q_c\), which means the reaction will proceed in the reverse direction to reach equilibrium.

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Key Concepts

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

Reaction Quotient
The reaction quotient, often represented as \( Q_c \), is a key concept in understanding chemical equilibrium. It is used to predict the direction in which a chemical reaction will proceed.
  • \( Q_c \) is calculated similarly to the equilibrium constant \( K_c \), but it applies to any point in the reaction before equilibrium is reached.
  • It is expressed as a ratio of the concentrations of products to reactants: \[ Q_c = \frac{[CO_2][H_2]}{[CO][H_2O]} \]
  • In this formula, each concentration is raised to the power of its coefficient in the balanced chemical equation.
By determining \( Q_c \), we can tell whether the system is already at equilibrium or needs to shift in one direction to reach it. If \( Q_c = K_c \), the system is at equilibrium.
Equilibrium Constant
The equilibrium constant, denoted as \( K_c \), is a fundamental concept in chemical equilibria, providing a measure of the relative concentrations of products and reactants at equilibrium.
  • It is specific to a particular reaction at a given temperature.
  • For a reaction: \[ \text{CO}(g) + \text{H}_2\text{O}(g) \rightleftharpoons \text{CO}_2(g) + \text{H}_2(g) \]
  • The equilibrium constant expression is: \[ K_c = \frac{[CO_2][H_2]}{[CO][H_2O]} \]
In this scenario, \( K_c \) is 0.58 at 1000掳C. This value gives us insight into the system's position at equilibrium. A \( K_c \) different from 1 suggests unequal concentrations of reactants and products at equilibrium.
Direction of Reaction
The direction of reaction is determined by comparing the reaction quotient \( Q_c \) with the equilibrium constant \( K_c \).
  • If \( Q_c < K_c \), the reaction will proceed in the forward direction, producing more products.
  • If \( Q_c > K_c \), the reaction will shift reverse to form more reactants.
  • If \( Q_c = K_c \), then the system is at equilibrium, and no net change occurs.
In our example, since \( Q_c = 1.00 \) and \( K_c = 0.58 \), \( Q_c \) is greater than \( K_c \). Therefore, the reaction will move in the reverse direction to reach equilibrium by consuming the products \( \text{CO}_2 \) and \( \text{H}_2 \), and forming more \( \text{CO} \) and \( \text{H}_2\text{O} \).
Catalyst Effect
A catalyst is a substance that increases the rate of a chemical reaction without being consumed in the process. It does not affect the position of equilibrium or the equilibrium constant \( K_c \), but it helps the system reach equilibrium faster.
  • Catalysts speed up both the forward and reverse reactions equally, ensuring that equilibrium is achieved more quickly.
  • They lower the activation energy needed for the reaction to proceed.
  • In this exercise, a zinc oxide-copper oxide catalyst is used. It accelerates the reaction but does not alter the equilibrium position or the comparison between \( Q_c \) and \( K_c \).
So, while the catalyst helps the components react faster, it does not change whether the reaction shifts forward or reverse to reach equilibrium.

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Most popular questions from this chapter

Phosgene, \(\mathrm{COCl}_{2}\), is a toxic gas used in the manufacture of urethane plastics. The gas dissociates at high temperature. $$ \mathrm{COCl}_{2}(g) \rightleftharpoons \mathrm{CO}(g)+\mathrm{Cl}_{2}(g) $$ At \(720^{\circ} \mathrm{C}\), the equilibrium constant \(K_{c}\) is \(3.63 \times 10^{-3}\). Find the percentage of phosgene that dissociates at this temperature when \(8.60 \mathrm{~mol}\) of phosgene is placed in a \(27.1-\mathrm{L}\) vessel.

List four ways in which the yield of ammonia in the reaction $$ \mathrm{N}_{2}(g)+3 \mathrm{H}_{2}(g) \rightleftharpoons 2 \mathrm{NH}_{3}(g) ; \Delta H^{\circ}<0 $$ can be improved for a given amount of \(\mathrm{H}_{2}\). Explain the principle behind each way.

Methanol, \(\mathrm{CH}_{3} \mathrm{OH},\) formerly known as wood alcohol, is manufactured commercially by the following reaction: $$ \mathrm{CO}(g)+2 \mathrm{H}_{2}(g) \rightleftharpoons \mathrm{CH}_{3} \mathrm{OH}(g) $$ A 1.500-L vessel was filled with \(0.1500 \mathrm{~mol} \mathrm{CO}\) and 0.3000 mol \(\mathrm{H}_{2}\). When this mixture came to equilibrium at \(500 \mathrm{~K}\), the vessel contained \(0.1187 \mathrm{~mol} \mathrm{CO}\). How many moles of each substance were in the vessel?

A vessel originally contained \(0.0200 \mathrm{~mol}\) iodine monobromide (IBr), \(0.050 \mathrm{~mol} \mathrm{I}_{2}\), and \(0.050 \mathrm{~mol} \mathrm{Br}_{2}\). The equilibrium constant \(K_{c}\) for the reaction $$ \mathrm{I}_{2}(g)+\mathrm{Br}_{2}(g) \rightleftharpoons 2 \mathrm{IBr}(g) $$ is \(1.2 \times 10^{2}\) at \(150^{\circ} \mathrm{C}\). What is the direction (forward or reverse) needed to attain equilibrium at \(150^{\circ} \mathrm{C} ?\)

A 2.50-L vessel contains \(1.75 \mathrm{~mol} \mathrm{~N}_{2}, 1.75 \mathrm{~mol} \mathrm{H}_{2}\), and \(0.346 \mathrm{~mol} \mathrm{NH}_{3}\). What is the direction of reaction (forward or reverse) needed to attain equilibrium at \(401^{\circ} \mathrm{C} ?\) The equilibrium constant \(K_{c}\) for the reaction $$ \mathrm{N}_{2}(g)+3 \mathrm{H}_{2}(g) \rightleftharpoons 2 \mathrm{NH}_{3}(g) $$ is 0.50 at \(401^{\circ} \mathrm{C}\)
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