/*! 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 73 Consider the equilibrium $$ ... [FREE SOLUTION] | 91Ó°ÊÓ

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Consider the equilibrium $$ \mathrm{FeO}(s)+\mathrm{CO}(g) \rightleftharpoons \mathrm{Fe}(s)+\mathrm{CO}_{2}(g) $$ When carbon dioxide is removed from the equilibrium mixture (say, by passing the gases through water to absorb \(\mathrm{CO}_{2}\) ), what is the direction of net reaction as the new equilibrium is achieved?

Short Answer

Expert verified
The reaction will shift to the right to produce more \( \mathrm{CO}_{2} \).

Step by step solution

01

Identify the Change in the System

Removing carbon dioxide (\( \mathrm{CO}_{2} \)) from the equilibrium mixture constitutes a disturbance to the established equilibrium. According to Le Chatelier's Principle, the system will adjust itself to minimize the effect of this disturbance.
02

Apply Le Chatelier's Principle

Since \( \mathrm{CO}_{2} \) is being removed from the right side of the equilibrium reaction, the system will respond by favoring the forward reaction to produce more \( \mathrm{CO}_{2} \). In this way, the system works to counteract the change and re-establish equilibrium.
03

Determine the Direction of the Reaction

The removal of \( \mathrm{CO}_{2} \) shifts the equilibrium to the right, thereby favoring the forward reaction where \( \mathrm{FeO} \) and \( \mathrm{CO} \) are converted to \( \mathrm{Fe} \) and additional \( \mathrm{CO}_{2} \) to replace the removed \( \mathrm{CO}_{2} \).

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Key Concepts

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

Le Chatelier's Principle
Le Chatelier's Principle is a fundamental concept in understanding chemical equilibrium. It states that if a dynamic equilibrium is disturbed by changing the conditions, the system will adjust itself to counteract the disturbance and a new equilibrium is established. In simpler terms, the principle predicts the direction of shift in equilibrium to minimize the effect of the change.
For example, when a component of a chemical reaction at equilibrium is removed or altered, as in the case of removing carbon dioxide (\( \mathrm{CO}_{2} \)) from our reaction:\[\mathrm{FeO}(s)+\mathrm{CO}(g) \rightleftharpoons\mathrm{Fe}(s)+\mathrm{CO}_{2}(g)\]The system reacts by shifting the equilibrium to oppose the change.
  • This means producing more of the removed component.
  • In our reaction, the forward reaction is favored to produce more \( \mathrm{CO}_{2} \).
This shift is how the system attempts to restore balance and reach a new equilibrium point.
Equilibrium Shift
An equilibrium shift refers to the movement of the equilibrium position in response to an external change, such as concentration, pressure, or temperature. These shifts are crucial to understand because they tell us which direction a reaction will favor after a disturbance. In our chemical reaction, removing \( \mathrm{CO}_{2} \) leads to an equilibrium shift to the right.
When you encounter a change in the equilibrium:
  • Identify which component is altered or removed.
  • Determine how the system will shift to compensate the change.
  • Check if the reaction shifts towards the products (right) or reactants (left).
In this scenario, since \( \mathrm{CO}_{2} \) is removed, the equilibrium shifts right to produce more \( \mathrm{CO}_{2} \) to replace it, thus enhancing the forward reaction.
Reaction Direction
Understanding the reaction direction is key for predicting the outcome of a disturbance in equilibrium. The direction can either be forward, favoring product formation, or reverse, favoring reactant generation. For the reaction at hand, when \( \mathrm{CO}_{2} \) is removed from:\[\mathrm{FeO}(s)+\mathrm{CO}(g) \rightleftharpoons\mathrm{Fe}(s)+\mathrm{CO}_{2}(g)\]The reaction direction moves forward. Why does this happen?
  • Removing a product like \( \mathrm{CO}_{2} \) from the right prompts the creation of more products.
  • The system "wants" to replace lost products to regain equilibrium, hence favoring the forward direction.
By predicting the new direction, you can foresee potential increases in product concentration or even changes in the speed of both forward and reverse reactions.

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Most popular questions from this chapter

The amount of nitrogen dioxide formed by dissociation of dinitrogen tetroxide, $$ \mathrm{N}_{2} \mathrm{O}_{4}(g) \rightleftharpoons 2 \mathrm{NO}_{2}(g) $$ increases as the temperature rises. Is the dissociation of \(\mathrm{N}_{2} \mathrm{O}_{4}\) endothermic or exothermic?

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 bromide dissociates when heated according to the equation $$ 2 \mathrm{HBr}(g) \rightleftharpoons \mathrm{H}_{2}(g)+\mathrm{Br}_{2}(g) $$ The equilibrium constant \(K_{c}\) equals \(1.6 \times 10^{-2}\) at \(200^{\circ} \mathrm{C}\). What are the moles of substances in the equilibrium mixture at \(200^{\circ} \mathrm{C}\) if we start with \(0.010 \mathrm{~mol} \mathrm{HBr}\) in a \(1.0\) - \(\mathrm{L}\) vessel?

Which of the following reactions involve homogeneous equilibria and which involve heterogeneous equilibria? Explain the difference. a. \(2 \mathrm{NO}(g)+\mathrm{O}_{2}(g) \rightleftharpoons 2 \mathrm{NO}_{2}(g)\) b. \(2 \mathrm{Cu}\left(\mathrm{NO}_{3}\right)_{2}(s) \rightleftharpoons 2 \mathrm{CuO}(s)+4 \mathrm{NO}_{2}(g)+\mathrm{O}_{2}(g)\) c. \(2 \mathrm{~N}_{2} \mathrm{O}(g) \rightleftharpoons 2 \mathrm{~N}_{2}(g)+\mathrm{O}_{2}(g)\) d. \(2 \mathrm{NH}_{3}(g)+3 \mathrm{CuO}(s) \rightleftharpoons 3 \mathrm{H}_{2} \mathrm{O}(g)+\mathrm{N}_{2}(g)+3 \mathrm{Cu}(s)\)

Explain why the equilibrium constant for a gaseous reaction can be written in terms of partial pressures instead of concentrations.

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