/*! 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 87 A 2.00-L vessel contains \(1.00 ... [FREE SOLUTION] | 91影视

91影视

Log out

Learning Materials

Features

Discover

Chapter 14: Problem 87

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 \(400^{\circ} \mathrm{C}\).

Short Answer

Expert verified
The reaction will shift in the reverse direction to reach equilibrium.

Step by step solution

01

Write out the equilibrium expression

The reaction is \( \mathrm{N}_{2}(g)+3 \mathrm{H}_{2}(g) \rightleftharpoons 2 \mathrm{NH}_{3}(g) \). The equilibrium constant expression \( K_c \) for this reaction is given by \( K_c = \frac{[\mathrm{NH}_{3}]^2}{[\mathrm{N}_{2}][\mathrm{H}_{2}]^3} \).
02

Calculate the initial concentrations

Find the initial concentrations of all the gases. Using the formula \( \text{Concentration} = \frac{{\text{moles}}}{{\text{volume}}} \), for \( \mathrm{N}_{2} \), \( [\mathrm{N}_{2}] = \frac{1.00}{2.00} = 0.50 \ \mathrm{mol/L} \). For \( \mathrm{H}_{2} \), \( [\mathrm{H}_{2}] = \frac{1.00}{2.00} = 0.50 \ \mathrm{mol/L} \). For \( \mathrm{NH}_{3} \), \( [\mathrm{NH}_{3}] = \frac{2.00}{2.00} = 1.00 \ \mathrm{mol/L} \).
03

Determine the reaction quotient Q

Calculate \( Q_c \) using the initial concentrations: \( Q_c = \frac{[\mathrm{NH}_{3}]^2}{[\mathrm{N}_{2}][\mathrm{H}_{2}]^3} = \frac{(1.00)^2}{(0.50)(0.50)^3} = \frac{1.00}{0.125} = 8.00 \).
04

Compare Q and K to determine the direction of reaction

Since \( Q_c > K_c \) (8.00 > 0.51), the reaction quotient is greater than the equilibrium constant, indicating that the reaction will shift in the reverse direction to reach equilibrium.

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 91影视!

Key Concepts

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

Reaction Quotient
The reaction quotient, often represented as \( Q \), is a snapshot of a chemical reaction's status at any given point in time. It is calculated in a similar way to the equilibrium constant, \( K \), but unlike \( K \), \( Q \) can be computed using the initial concentrations of reactants and products. To find \( Q \), you simply plug the concentrations of reactants and products into the equilibrium expression.
For the nitrogen, hydrogen, and ammonia reaction presented, the reaction quotient \( Q_c \) is determined by the formula:
  • \( Q_c = \frac{[\mathrm{NH}_3]^2}{[\mathrm{N}_2][\mathrm{H}_2]^3} \)
  • Using the initial concentrations \([\mathrm{N}_2] = 0.50 \ \mathrm{mol/L}\), \([\mathrm{H}_2] = 0.50 \ \mathrm{mol/L}\), and \([\mathrm{NH}_3] = 1.00 \ \mathrm{mol/L}\).
  • The calculated \( Q_c \) is 8.00.
By assessing \( Q \), you can determine whether a reaction is at equilibrium or in which direction it needs to proceed to achieve equilibrium.
Equilibrium Constant
The equilibrium constant \( K \) helps us understand the ratio of concentrations of products to reactants at equilibrium. For a specific temperature, \( K \) remains constant, reflecting the position of equilibrium. This is unique because it offers insight into the extent of a reaction. For the considered reaction, at \(400^{\circ}\mathrm{C}\), \( K_c = 0.51\).
In the context of the given equation:
  • \( \mathrm{N}_2(g) + 3 \mathrm{H}_2(g) \rightleftharpoons 2 \mathrm{NH}_3(g) \).
  • The equilibrium expression for this reaction is \( K_c = \frac{[\mathrm{NH}_3]^2}{[\mathrm{N}_2][\mathrm{H}_2]^3} \).
  • It shows that at equilibrium, the ratio of squared ammonia concentration to nitrogen and cubed hydrogen must match \( K_c \).
Understanding \( K \) allows predicting reaction behavior, crucial for chemical processes and balancing chemical reactions.
Le Chatelier's Principle
Le Chatelier's Principle is an insightful guideline that predicts how a system at equilibrium reacts to changes鈥攕uch as concentration, temperature, or pressure. When a system is disturbed, it tends to compensate by shifting the equilibrium position. In this problem, the principle helps to foresee how the reaction will shift upon comparing \( Q \) and \( K \):
  • Since \( Q_c = 8.00 \), which is greater than \( K_c = 0.51 \).
  • According to Le Chatelier's Principle, the reaction will shift in the reverse direction.
  • This shift aims to reduce the concentration of excess products \( (\mathrm{NH}_3) \) and increase the amount of reactants \( (\mathrm{N}_2 \text{ and } \mathrm{H}_2) \).
Ultimately, Le Chatelier's Principle gives valuable predictions about the direction of response needed to restore equilibrium whenever there is a disturbance.

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

A 6.00-L reaction vessel at \(491^{\circ} \mathrm{C}\) contained \(0.488 \mathrm{~mol}\) \(\mathrm{H}_{2}, 0.206 \mathrm{~mol} \mathrm{I}_{2}\), and \(2.250 \mathrm{~mol}\) HI. Assuming that the substances are at equilibrium, find the value of \(K_{c}\) at \(491^{\circ} \mathrm{C}\) for the reaction of hydrogen and iodine to give hydrogen iodide. The equation is $$ \mathrm{H}_{2}(g)+\mathrm{I}_{2}(g) \rightleftharpoons 2 \mathrm{HI}(g) $$

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 \(400^{\circ} \mathrm{C}\), the equilibrium constant \(K_{c}\) is \(8.05 \times 10^{-4}\). Find the percentage of phosgene that dissociates at this temperature when \(1.00\) mol of phosgene is placed in a 25.0-L vessel.

An equilibrium mixture of \(\mathrm{SO}_{3}, \mathrm{SO}_{2}\), and \(\mathrm{O}_{2}\) at \(727^{\circ} \mathrm{C}\) is \(0.0160 \mathrm{MSO}_{3}, 0.0056 \mathrm{MSO}_{2}\), and \(0.0021 \mathrm{M} \mathrm{O}_{2}\). What is the value of \(K_{c}\) for the following reaction? $$ \mathrm{SO}_{2}(g)+\frac{1}{2} \mathrm{O}_{2}(g) \rightleftharpoons \mathrm{SO}_{3}(g) $$

Iodine, \(\mathrm{I}_{2}\), is a blue-black solid, but it easily vaporizes to give a violet vapor. At high temperatures, this molecular substance dissociates to atoms: $$ \mathrm{I}_{2}(g) \rightleftharpoons 2 \mathrm{I}(g) $$ An absent-minded professor measured equilibrium constants for this dissociation at two temperatures, \(700^{\circ} \mathrm{C}\) and \(800^{\circ} \mathrm{C}\). He obtained the \(K_{p}\) values \(0.01106\) and \(0.001745\), but he couldn't remember which value went with what temperature. Can you please help him assign temperatures to the equilibrium constants? State your argument carefully.

Phosgene, \(\mathrm{COCl}_{2}\), used in the manufacture of polyurethane plastics, is prepared from \(\mathrm{CO}\) and \(\mathrm{Cl}_{2}\). $$ \mathrm{CO}(g)+\mathrm{Cl}_{2}(g) \rightleftharpoons \mathrm{COCl}_{2}(g) $$ An equilibrium mixture at \(395^{\circ} \mathrm{C}\) contains \(0.012 \mathrm{~mol} \mathrm{CO}\) and \(0.025 \mathrm{~mol} \mathrm{Cl}_{2}\) per liter, as well as \(\mathrm{COCl}_{2}\). If \(K_{c}\) at \(395^{\circ} \mathrm{C}\) is \(1.23 \times 10^{3}\), what is the concentration of \(\mathrm{COCl}_{2}\) ?
See all solutions

Recommended explanations on Chemistry Textbooks

Chemistry Branches

Read Explanation

Organic Chemistry

Read Explanation

Inorganic Chemistry

Read Explanation

Kinetics

Read Explanation

Ionic and Molecular Compounds

Read Explanation

Physical Chemistry

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.