Log out Go to App
Go to App

Learning Materials

Features

Discover

Chapter 13: Problem 23

Consider the reaction $$\mathrm{CO}(g)+\mathrm{H}_{2}(g) \leftrightharpoons \mathrm{CH}_{3} \mathrm{OH}(g)$$ Suppose the system is at equilibrium, and then an additional mole of \(\mathrm{N}_{2} \mathrm{O}(g)\) is injected into the system at constant temperature. Once the reaction reestablishes equilibrium, has the amount of \(\mathrm{N}_{2} \mathrm{O}\) increased or decreased from its original equilibrium amount? Explain. What happens to the value of the equilibrium constant with this change?

Short Answer

Expert verified
After adding N2O to the system, the equilibrium composition and the amount of N2O remain the same, since the total pressure change does not affect the given reaction's equilibrium. The value of the equilibrium constant, K, remains unchanged as well since the temperature is constant and the reaction's equilibrium is unaffected.

Step by step solution

01

Identify the reaction and given information

The given reaction is: \[\mathrm{CO}(g) + \mathrm{H}_2(g) \leftrightharpoons \mathrm{CH}_3\mathrm{OH}(g)\] We know that the system is initially at equilibrium, and then an additional mole of N2O is injected into the system at constant temperature.
02

Consider Le Châtelier's principle

Le Châtelier's principle states that if a change is made to a system at equilibrium, the system will adjust to oppose the change and restore the equilibrium. In this case, since N2O is not part of the reaction, adding N2O will not directly affect the reaction. However, the addition of N2O will increase the total pressure by increasing the total number of moles of gas.
03

Analyze the pressure change

The addition of N2O increases the total pressure inside the container. According to Le Châtelier's principle, the reaction will attempt to counteract this change by favoring the reaction direction that reduces pressure. In this case, the reaction will favor the direction where there are fewer moles of gas produced. Counting the moles of gas in the reaction, we see that 1 mole of CO and 1 mole of H2 produce 1 mole of CH3OH. There is no net change in the total moles of gas in the reaction. As a result, the reaction will not shift to counteract the pressure change, and the equilibrium will remain the same.
04

Determine the effect on the amount of N2O and the equilibrium constant

Since the total pressure change does not affect the given reaction's equilibrium, the amount of N2O will remain the same after equilibrium is reestablished. It neither increases nor decreases from its original amount. Furthermore, since the temperature is constant and the equilibrium of the reaction remains unchanged, the value of the equilibrium constant, K, will also remain the same. In summary, after adding N2O to the system, the equilibrium composition remains the same, and the amount of N2O stays the same. The value of the equilibrium constant remains unchanged as well.

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.

Start your free trial

Over 30 million students worldwide already upgrade their learning with Vaia!

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

At a particular temperature, 12.0 moles of \(\mathrm{SO}_{3}\) is placed into a 3.0 -L rigid container, and the \(\mathrm{SO}_{3}\) dissociates by the reaction $$2 \mathrm{SO}_{3}(g) \rightleftharpoons 2 \mathrm{SO}_{2}(g)+\mathrm{O}_{2}(g)$$At equilibrium, 3.0 moles of \(\mathrm{SO}_{2}\) is present. Calculate \(K\) for this reaction.

For the reaction $$\mathrm{PCl}_{5}(g) \rightleftharpoons \mathrm{PCl}_{3}(g)+\mathrm{Cl}_{2}(g)$$ at \(600 . \mathrm{K}\) , the equilibrium constant, \(K_{\mathrm{p}},\) is \(11.5 .\) Suppose that 2.450 \(\mathrm{g} \mathrm{PCl}_{5}\) is placed in an evacuated 500 -mL bulb, which is then heated to \(600 . \mathrm{K}\) . a. What would be the pressure of \(\mathrm{PCl}_{5}\) if it did not dissociate? b. What is the partial pressure of \(\mathrm{PCl}_{5}\) at equilibrium? c. What is the total pressure in the bulb at equilibrium? d. What is the percent dissociation of PCl_ at equilibrium?

The creation of shells by mollusk species is a fascinating process. By utilizing the \(\mathrm{Ca}^{2+}\) in their food and aqueous environment, as well as some complex equilibrium processes, a hard calcium carbonate shell can be produced. One important equilibrium reaction in this complex process is $$\mathrm{HCO}_{3}^{-}(a q) \leftrightharpoons \mathrm{H}^{+}(a q)+\mathrm{CO}_{3}^{2-}(a q) K=5.6 \times 10^{-11}$$ If 0.16 mole of \(\mathrm{HCO}_{3}^{-}\) is placed into 1.00 \(\mathrm{L}\) of solution, what will be the equilibrium concentration of \(\mathrm{CO}_{3}^{2-2}\) ?

At a particular temperature, \(K=2.0 \times 10^{-6}\) for the reaction $$2 \mathrm{CO}_{2}(g) \rightleftharpoons 2 \mathrm{CO}(g)+\mathrm{O}_{2}(g)$$ If 2.0 moles of \(\mathrm{CO}_{2}\) is initially placed into a 5.0 -L vessel, calculate the equilibrium concentrations of all species.

In a study of the reaction $$3 \mathrm{Fe}(s)+4 \mathrm{H}_{2} \mathrm{O}(g) \rightleftharpoons \mathrm{Fe}_{3} \mathrm{O}_{4}(s)+4 \mathrm{H}_{2}(g)$$ at 1200 \(\mathrm{K}\) it was observed that when the equilibrium partial pressure of water vapor is 15.0 torr, the total pressure at equilibrium is 36.3 torr. Calculate the value of \(K_{\mathrm{p}}\) for this reaction at 1200 \(\mathrm{K}\) . Hint: Apply Dalton's law of partial pressures.)
See all solutions

Recommended explanations on Chemistry Textbooks

Inorganic Chemistry

Read Explanation

Ionic and Molecular Compounds

Read Explanation

Making Measurements

Read Explanation

Nuclear Chemistry

Read Explanation

Chemistry Branches

Read Explanation

Kinetics

Read Explanation
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.

Sign-up for free