/*! 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 59 Explain, using Le Ch芒telier's p... [FREE SOLUTION] | 91影视

91影视

Log out

Learning Materials

Features

Discover

Chapter 18: Problem 59

Explain, using Le Ch芒telier's principle, why the equilibrium constant for the formation of NO from \(\mathrm{N}_{2}\) and \(\mathrm{O}_{2}\) increases with increasing temperature, whereas the equilibrium constant for the formation of \(\mathrm{NO}_{2}\) from NO and \(\mathrm{O}_{2}\) decreases with increasing temperature.

Short Answer

Expert verified
In conclusion, based on Le Ch芒telier's principle, the equilibrium constant for the formation of NO from \( N_{2} \) and \( O_{2} \) increases with increasing temperature because the reaction is endothermic, so the system favors the products when the temperature increases. On the other hand, the equilibrium constant for the formation of \( NO_{2} \) from NO and \( O_{2} \) decreases with increasing temperature because the reaction is exothermic, so the system favors the reactants when the temperature increases.

Step by step solution

01

Le Ch芒telier's principle and temperature

Le Ch芒telier's principle states that if a system at equilibrium is subjected to a change in temperature, pressure, or concentration of a component, the system will adjust its position to counteract the change. In the case of temperature, if a reaction is exothermic (releases heat), an increase in temperature will shift the equilibrium to favor the reactants, thus decreasing the equilibrium constant K. If a reaction is endothermic (absorbs heat), an increase in temperature will shift the equilibrium to favor the products, thus increasing the equilibrium constant K.
02

Reaction 1: Formation of NO from N2 and O2

The reaction for the formation of NO can be written as: \( N_{2}(g) + O_{2}(g) \rightleftharpoons 2NO(g) \) This reaction is endothermic, meaning it absorbs heat. Therefore, when the temperature is increased, the equilibrium will shift to favor the products (NO).
03

Effect of temperature on Reaction 1

As the reaction is endothermic, an increase in temperature will favor the formation of more NO molecules. This means that the equilibrium constant K for this reaction will increase as the temperature increases.
04

Reaction 2: Formation of NO2 from NO and O2

The reaction for the formation of NO2 can be written as: \( 2NO(g) + O_{2}(g) \rightleftharpoons 2NO_{2}(g) \) This reaction is exothermic, meaning it releases heat. Therefore, when the temperature is increased, the equilibrium will shift to favor the reactants (NO and O2).
05

Effect of temperature on Reaction 2

As the reaction is exothermic, an increase in temperature will favor the consumption of NO2 molecules and increase the formation of NO and O2. This means that the equilibrium constant K for this reaction will decrease as the temperature increases. In conclusion, using Le Ch芒telier's principle, we can explain that the equilibrium constant for the formation of NO from N2 and O2 increases with increasing temperature since it is an endothermic reaction, whereas the equilibrium constant for the formation of NO2 from NO and O2 decreases with increasing temperature since it is an exothermic reaction.

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.

Equilibrium Constant
The equilibrium constant, known as \( K \), quantifies the ratio of the concentrations of products to reactants at equilibrium for a particular reaction. It reflects the state of balance between the forward and reverse reactions. When discussing chemical reactions, knowing the value of \( K \) is crucial because:
  • A large \( K \) (>1) indicates a reaction that heavily favors the formation of products.
  • A small \( K \) (<1) indicates a reaction that favors reactants.
Changes in temperature can affect the equilibrium constant, especially when considering endothermic or exothermic reactions, which we'll discuss next. Understanding \( K \) provides insights into how reactions shift under different conditions.
Endothermic Reactions
Endothermic reactions are those that absorb energy, usually in the form of heat, from their surroundings. A common way to identify these reactions is:
  • They often feel cold to the touch as heat is absorbed from the environment.
  • The products have higher energy than the reactants.
Increasing the temperature in an endothermic reaction can shift the equilibrium towards the formation of products. This happens because the system attempts to counteract the increase in temperature by absorbing more heat. Consequently, the equilibrium constant \( K \) typically increases for an endothermic reaction as the temperature rises.
Exothermic Reactions
Exothermic reactions release energy to their surroundings, usually as heat. They can be recognized by:
  • Reactions feeling hot as heat is being released.
  • The products having lower energy than the reactants.
In exothermic reactions, an increase in temperature tends to shift the equilibrium towards the reactants. This is because the system reduces the excess heat by favoring the reaction that releases less heat (the reverse reaction). As a result, the equilibrium constant \( K \) decreases with a rise in temperature for exothermic reactions.
Temperature Effect on Equilibrium
The effect of temperature on equilibrium is well explained by Le Ch芒telier's Principle. It states that a system at equilibrium will adjust to counteract the effects of changes imposed on it, such as temperature changes.
In an **endothermic reaction**, increasing the temperature leads to more products being formed since the reaction absorbs heat. Therefore, \( K \) increases.
In contrast, in an **exothermic reaction**, increasing the temperature causes the formation of more reactants, as the system tries to absorb the extra heat by favoring the formation of less-energetic reactants. This results in a decreased \( K \).
Understanding how temperature influences the equilibrium allows chemists to manipulate conditions to favor the production of desired substances, making it a crucial aspect of chemical reaction engineering.

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) Write a chemical equation that describes the attack of acid rain on limestone, \(\mathrm{CaCO}_{3}\). (b) If a limestone sculpture were treated to form a surface layer of calcium sulfate, would this help to slow down the effects of acid rain? Explain.

The Baeyer-Villiger reaction is a classic organic oxidation reaction for converting ketones to lactones, as in this reaction: The reaction is used in the manufacture of plastics and pharmaceuticals. The reactant 3 -chloroperbenzoic acid is somewhat shock sensitive, however, and prone to explode Also, 3 -chlorobenzoic acid is a waste product. An alternative process being developed uses hydrogen peroxide and a catalyst consisting of tin deposited within a solid support. The catalyst is readily recovered from the reaction mixture. (a) What would you expect to be the other product of oxidation of the ketone to lactone by hydrogen peroxide? (b) What principles of green chemistry are addressed by use of the proposed process?

The degradation of \(\mathrm{CF}_{3} \mathrm{CH}_{2} \mathrm{~F}\) (an HFC) by OH radicals in the troposphere is first order in each reactant and has a rate constant of \(k=1.6 \times 10^{8} \mathrm{M}^{-1} \mathrm{~s}^{-1}\) at \(4^{\circ} \mathrm{C}\). If the tropospheric concentrations of \(\mathrm{OH}\) and \(\mathrm{CF}_{3} \mathrm{CH}_{2} \mathrm{~F}\) are \(8.1 \times 10^{5}\) and \(63 \times 10^{8}\) molecules \(\mathrm{cm}^{-3}\), respectively, what is the rate of reaction at this temperature in \(M / \mathrm{s} ?\)

In 1986 an electrical power plant in Taylorsville, Georgia, burned \(8,376,726\) tons of coal, a national record at that time. (a) Assuming that the coal was \(83 \%\) carbon and \(25 \%\) sulfur and that combustion was complete, calculate the number of tons of carbon dioxide and sulfur dioxide produced by the plant during the year. (b) If \(55 \%\) of the \(\mathrm{SO}_{2}\) could be removed by reaction with powdered \(\mathrm{CaO}\) to form \(\mathrm{CaSO}_{3}\), how many tons of \(\mathrm{CaSO}_{3}\) would be produced?

The estimated average concentration of \(\mathrm{NO}_{2}\) in air in the United States in 2006 was \(0.016\) ppm. (a) Calculate the partial pressure of the \(\mathrm{NO}_{2}\) in a sample of this air when the atmospheric pressure is 755 torr \((99.1 \mathrm{kPa}) .\) (b) How many molecules of \(\mathrm{NO}_{2}\) are present under these conditions at \(20^{\circ} \mathrm{C}\) in a room that measures \(15 \times 14 \times 8 \mathrm{ft}\) ?
See all solutions

Recommended explanations on Chemistry Textbooks

Organic Chemistry

Read Explanation

Physical Chemistry

Read Explanation

Kinetics

Read Explanation

The Earths Atmosphere

Read Explanation

Chemical Analysis

Read Explanation

Nuclear 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.