/*! 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 68 Hydrogen for use in ammonia prod... [FREE SOLUTION] | 91Ó°ÊÓ

91Ó°ÊÓ

Hydrogen for use in ammonia production is produced by the reaction $$ \mathrm{CH}_{4}(g)+\mathrm{H}_{2} \mathrm{O}(g) \stackrel{\text { Ni catalyst }}{=\frac{750^{\circ} \mathrm{C}}{7}} \mathrm{CO}(g)+3 \mathrm{H}_{2}(g) $$ What will happen to a reaction mixture at equilibrium if a. \(\mathrm{H}_{2} \mathrm{O}(g)\) is removed? b. the temperature is increased (the reaction is endothermic)? c. an inert gas is added to a rigid reaction container? d. \(\mathrm{CO}(g)\) is removed? e. the volume of the container is tripled?

Short Answer

Expert verified
a. The reaction will shift to the left, producing more CHâ‚„ and less CO and Hâ‚‚. b. The reaction will shift to the right, producing more CO and Hâ‚‚. c. There will be no change in the equilibrium. d. The reaction will shift to the right, producing more CO and Hâ‚‚ while consuming CHâ‚„ and Hâ‚‚O. e. The reaction will shift to the right, producing more CO and Hâ‚‚, while consuming CHâ‚„ and Hâ‚‚O.

Step by step solution

01

Consider Le Chatelier's principle

Le Chatelier's principle states that if a stress is applied to a system at equilibrium, the system will adjust to reduce the stress, and the equilibrium will shift to either the left or right (favoring either reactants or products). Let's analyze the effects of the given changes one by one.
02

a. Remove Hâ‚‚O(g)

According to Le Chatelier's principle, if you remove one of the reactants (Hâ‚‚O in this case), the reaction will shift in the direction that consumes less of this reactant in order to re-establish equilibrium. This means that the reaction will shift to the left (favoring reactants), producing more CHâ‚„ and less CO and Hâ‚‚.
03

b. Increase the temperature

Given that the reaction is endothermic, increasing the temperature will cause the reaction to shift in the direction that absorbs the added heat, which is the direction of the reaction that requires heat (again, because it's endothermic). This means the reaction will shift to the right (favoring products), producing more CO and Hâ‚‚ in response to the increased temperature.
04

c. Add an inert gas to a rigid reaction container

Adding an inert gas to a rigid reaction container will increase the total pressure without affecting the partial pressures of the reactants and products since the inert gas does not participate in the reaction. Le Chatelier's principle predicts that there will be no change to the equilibrium since the partial pressures of the reactants and products remain unchanged.
05

d. Remove CO(g)

If CO(g) is removed from the reaction mixture, Le Chatelier's principle states that the reaction will adjust to restore equilibrium. The reaction will shift in the direction that produces more CO, which is the right side of the equation. As a result, more CO and Hâ‚‚ will be produced while consuming CHâ‚„ and Hâ‚‚O.
06

e. Triple the volume of the container

By tripling the volume of the container, the pressure decreases. According to Le Chatelier's principle, the equilibrium will shift in the direction that increases pressure, which is the side that produces more moles of gas. In this reaction, the right side produces more moles of gas due to 3 moles of Hâ‚‚ being produced. Therefore, the reaction will shift to the right (favoring products) and produce more CO and Hâ‚‚, while consuming CHâ‚„ and Hâ‚‚O.

Unlock Step-by-Step Solutions & Ace Your Exams!

  • Full Textbook Solutions

    Get detailed explanations and key concepts

  • Unlimited Al creation

    Al flashcards, explanations, exams and more...

  • Ads-free access

    To over 500 millions flashcards

  • Money-back guarantee

    We refund you if you fail your exam.

Over 30 million students worldwide already upgrade their learning with 91Ó°ÊÓ!

Key Concepts

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

Chemical Equilibrium
Imagine a busy city intersection with vehicles moving from all directions. Over time, the traffic flows achieve a balance, where the number of cars entering equals the number of cars leaving. This steady state is similar to what happens in a chemical equilibrium.

In a chemical reaction, reactants convert into products, and under certain conditions, the products can also revert into reactants. When the rates of the forward and reverse reactions are equal, the reaction has reached a state called chemical equilibrium. At this point, the concentration of reactants and products remains constant over time, not because the reactions have stopped, but because they are occurring at the same rate. It's important to note that equilibrium does not mean the reactants and products are present in equal quantities, but that their ratios are stable.

Understanding chemical equilibrium is crucial in predicting how a reaction will respond to changes in conditions, such as temperature, pressure, and concentration, which is guided by Le Chatelier's principle.
Endothermic Reactions
Reactions come in two main flavors: endothermic and exothermic. An endothermic reaction is akin to a heat-loving sponge, soaking up energy from its environment. During such a reaction, energy is absorbed from the surroundings, typically in the form of heat, which results in a temperature decrease around the reaction site. This is like how melting ice absorbs warmth from a drink, keeping it cool.

In the case of endothermic reactions, applying more heat — raising the temperature — encourages the reaction to proceed in the direction that absorbs heat, thus shifting equilibrium towards the products. This kind of response aligns with Le Chatelier's principle, reflecting how reactions adjust to changes in their environment. In the context of the provided exercise, increasing the temperature would favor the formation of more carbon monoxide (CO) and hydrogen gas (H₂) because we're feeding the reaction's need for energy.
Reaction Shifts
The concept of reaction shifts is central to understanding how chemical reactions adapt to changes, honoring Le Chatelier's principle. Picture a tug-of-war contest where the teams are perfectly matched, maintaining a fixed rope position. If one team suddenly loses a player (representing a change in conditions), the opposing team gains the advantage, and the rope moves in their direction.

In chemical terms, a 'shift' describes the direction the reaction moves to re-establish equilibrium when it is disturbed. Removing a reactant or a product, changing the temperature, or altering the pressure can all cause a shift. For instance, if we remove CO or Hâ‚‚O as seen in our exercise, the reaction will shift towards producing more of what was removed to balance the 'loss'. Similarly, adding an inert gas does not cause a shift if the volume is constant, because it doesn't affect reactant or product concentrations. By interpreting reaction shifts, chemists can influence reaction outcomes, control product formation, and optimize industrial processes.

One App. One Place for Learning.

All the tools & learning materials you need for study success - in one app.

Get started for free

Most popular questions from this chapter

Consider the reaction \(\mathrm{A}(g)+2 \mathrm{B}(g) \rightleftharpoons \mathrm{C}(g)+\mathrm{D}(g)\) in a 1.0-L rigid flask. Answer the following questions for each situation (a-d): i. Estimate a range (as small as possible) for the requested substance. For example, [A] could be between \(95 M\) and \(100 M\),ii. Explain how you decided on the limits for the estimated range. iii. Indicate what other information would enable you to narrow your estimated range. iv. Compare the estimated concentrations for a through d, and explain any differences. a. If at equilibrium \([\mathrm{A}]=1 M,\) and then 1 mole of \(\mathrm{C}\) is added, estimate the value for [A] once equilibrium is reestablished. b. If at equilibrium \([\mathrm{B}]=1 M,\) and then 1 mole of \(\mathrm{C}\) is added, estimate the value for \([\mathbf{B}]\) once equilibrium is reestablished. c. If at equilibrium \([\mathrm{C}]=1 M,\) and then 1 mole of \(\mathrm{C}\) is added, estimate the value for \([\mathrm{C}]\) once equilibrium is reestablished. d. If at equilibrium \([\mathrm{D}]=1 M,\) and then 1 mole of \(\mathrm{C}\) is added, estimate the value for [D] once equilibrium is reestablished.

Consider the reaction \(\mathrm{A}(g)+\mathrm{B}(g) \rightleftharpoons \mathrm{C}(g)+\mathrm{D}(g) .\) A friend asks the following: "I know we have been told that if a mixture of \(\mathrm{A}, \mathrm{B}, \mathrm{C},\) and \(\mathrm{D}\) is at equilibrium and more of \(\mathrm{A}\) is added, more \(\mathrm{C}\) and \(\mathrm{D}\) will form. But how can more \(\mathrm{C}\) and \(\mathrm{D}\) form if we do not add more \(\mathrm{B} ?^{\prime \prime}\) What do you tell your friend?

For the following endothermic reaction at equilibrium: $$2 \mathrm{SO}_{3}(g) \rightleftharpoons 2 \mathrm{SO}_{2}(g)+\mathrm{O}_{2}(g)$$ which of the following changes will increase the value of \(K ?\) a. increasing the temperature b. decreasing the temperature c. removing \(\mathrm{SO}_{3}(g)\) (constant \(T\) ) d. decreasing the volume (constant \(T\) ) e. adding Ne(g) (constant \(T\) ) f. adding \(\mathrm{SO}_{2}(g)\) (constant \(T\) ) g. adding a catalyst (constant \(T\) )

Predict the shift in the equilibrium position that will occur for each of the following reactions when the volume of the reaction container is increased. a. \(\mathrm{N}_{2}(g)+3 \mathrm{H}_{2}(g) \rightleftharpoons 2 \mathrm{NH}_{3}(g)\) b. \(\mathrm{PCl}_{5}(g) \rightleftharpoons \mathrm{PCl}_{3}(g)+\mathrm{Cl}_{2}(g)\) c. \(\mathrm{H}_{2}(g)+\mathrm{F}_{2}(g) \rightleftharpoons 2 \mathrm{HF}(g)\) d. \(\operatorname{COCl}_{2}(g) \rightleftharpoons \operatorname{CO}(g)+\mathrm{Cl}_{2}(g)\) e. \(\mathrm{CaCO}_{3}(s) \rightleftharpoons \mathrm{CaO}(s)+\mathrm{CO}_{2}(g)\)

A sample of \(\mathrm{N}_{2} \mathrm{O}_{4}(g)\) is placed in an empty cylinder at \(25^{\circ} \mathrm{C}\). After equilibrium is reached the total pressure is 1.5 atm and \(16 \%\) (by moles) of the original \(\mathrm{N}_{2} \mathrm{O}_{4}(g)\) has dissociated to \(\mathrm{NO}_{2}(g)\) a. Calculate the value of \(K_{\mathrm{p}}\) for this dissociation reaction at \(25^{\circ} \mathrm{C}\) b. If the volume of the cylinder is increased until the total pressure is 1.0 atm (the temperature of the system remains constant), calculate the equilibrium pressure of \(\mathrm{N}_{2} \mathrm{O}_{4}(g)\) and \(\mathrm{NO}_{2}(g)\) c. What percentage (by moles) of the original \(\mathrm{N}_{2} \mathrm{O}_{4}(g)\) is dissociated at the new equilibrium position (total pressure \(=\) \(1.00 \mathrm{atm}) ?\)

See all solutions

Recommended explanations on Chemistry Textbooks

View all explanations

What do you think about this solution?

We value your feedback to improve our textbook solutions.

Study anywhere. Anytime. Across all devices.