/*! 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 27 When air is heated at very high ... [FREE SOLUTION] | 91Ó°ÊÓ

91Ó°ÊÓ

When air is heated at very high temperatures in an automobile engine, the air pollutant nitric oxide is produced by the reaction $$ \mathrm{N}_{2}(g)+\mathrm{O}_{2}(g) \rightleftharpoons 2 \mathrm{NO}(g) \quad \Delta H^{\circ}=+182.6 \mathrm{~kJ} $$ How does the equilibrium amount of NO vary with an increase in temperature?

Short Answer

Expert verified
The equilibrium amount of NO increases with temperature.

Step by step solution

01

Identify the Reaction Type

This is an endothermic reaction because the enthalpy change (\(\Delta H^{\circ}\)) is positive, indicating that heat is absorbed.
02

Apply Le Chatelier's Principle

According to Le Chatelier’s principle, when the temperature of a system in dynamic equilibrium is increased, the equilibrium position will shift to counteract this change. For endothermic reactions, the equilibrium shifts to the right, leading to an increase in product formation.
03

Determine Effect on Product Formation

Since the equilibrium shifts to the right with increased temperature in this endothermic reaction, the concentration of \( \text{NO} \) will increase.

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.

Endothermic Reactions
Most chemical reactions are classified based on their energy changes as either endothermic or exothermic. Endothermic reactions absorb energy from their surroundings in the form of heat. In these reactions, the temperature of the surroundings drops as the reaction proceeds because the heat is taken in by the reactants to break bonds. A key identifier of an endothermic reaction is a positive enthalpy change (\( \Delta H^{\circ} \)), as seen in the chemical equation provided for the production of nitric oxide (NO), where \( \Delta H^{\circ} = +182.6 \mathrm{~kJ} \).
  • A positive \( \Delta H^{\circ} \) means energy is absorbed.
  • The system gains energy, while the surroundings lose heat.
  • Tends to occur spontaneously under high-temperature conditions.
Recognizing this type of reaction helps in predicting how the system will react to changes in temperature, as is further explained through Le Chatelier's Principle.
Le Chatelier's Principle
Le Chatelier's Principle is a fundamental concept used to predict the adjustment of a chemical equilibrium when it is subjected to a change in conditions. Think of it as a guide that tells us which way the balance will tip. When a system at equilibrium experiences a change in temperature, pressure, or concentration, the system adjusts to counteract that change and restore a new equilibrium. Temperature Change: For our endothermic reaction:
  • An increase in temperature adds more heat to the system.
  • According to Le Chatelier's Principle, the system will absorb this excess heat.
  • The equilibrium position shifts towards the side that consumes heat—in this case, towards the production of more NO.
By understanding and applying Le Chatelier's Principle, we can predict that increasing the temperature will result in a higher concentration of NO at equilibrium.
Enthalpy Change
Enthalpy change (\( \Delta H \)) is a measurement of the total energy change during a reaction. It tells us how much heat is absorbed or released as a reaction occurs.For endothermic reactions, like our nitric oxide production example:
  • The \( \Delta H^{\circ} \) is positive, indicating that the reaction absorbs heat.
  • It reflects the energy needed for breaking bonds in the reactant molecules, which is higher than the energy released when new bonds in the product are formed.
  • This positive enthalpy change is one reason why the reaction favors higher product formation upon heating.
Understanding the concept of enthalpy is crucial to grasping why reactions behave the way they do under different conditions. It tells us not just the direction of heat flow but also offers insight into the potential energy landscape during a chemical transformation.

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

Sulfur dioxide reacts with oxygen in a step in the production of sulfuric acid. \(2 \mathrm{SO}_{2}(g)+\mathrm{O}_{2}(g) \rightleftharpoons 2 \mathrm{SO}_{3}(g) \quad K_{c}=7.9 \times 10^{4}(800 \mathrm{~K})\) For an equilibrium mixture in which \(\left[\mathrm{SO}_{2}\right]=1.5 \times 10^{-3} \mathrm{M}\) and \(\left[\mathrm{O}_{2}\right]=3.0 \times 10^{-3} \mathrm{M}\), what is \(\left[\mathrm{SO}_{3}\right] ?\)

Consider the gas-phase hydration of hexafluoroacetone, \(\left(\mathrm{CF}_{3}\right)_{2} \mathrm{CO}=\) At \(76^{\circ} \mathrm{C}\), the forward and reverse rate constants are \(k_{\mathrm{f}}=0.13 \mathrm{M}^{-1} \mathrm{~s}^{-1}\) and \(k_{\mathrm{r}}=6.2 \times 10^{-4} \mathrm{~s}^{-1}\). What is the value of the equilibrium constant \(K_{c} ?\)

For each of the following equilibria, write the equilibrium constant expression for \(K_{c}\). Where appropriate, also write the equilibrium constant expression for \(K_{\mathrm{p}^{-}}\) (a) \(\mathrm{WO}_{3}(s)+3 \mathrm{H}_{2}(g) \rightleftharpoons \mathrm{W}(s)+3 \mathrm{H}_{2} \mathrm{O}(g)\) (b) \(\mathrm{Ag}^{+}(a q)+\mathrm{Cl}^{-}(a q) \rightleftharpoons \mathrm{AgCl}(s)\) (c) \(2 \mathrm{FeCl}_{3}(s)+3 \mathrm{H}_{2} \mathrm{O}(g) \rightleftharpoons \mathrm{Fe}_{2} \mathrm{O}_{3}(g)+6 \mathrm{HCl}(g)\) (d) \(\mathrm{MgCO}_{3}(s) \rightleftharpoons \mathrm{MgO}(s)+\mathrm{CO}_{2}(g)\)

Identify the true statement about the rate of the forward and reverse reaction once a reaction has reached equilibrium. (a) The rate of the forward reaction and the reverse reaction is zero. (b) The rate of the forward reaction is greater than the rate of the reverse reaction. (c) The rate of the reverse reaction is greater than the rate of the forward reaction. (d) The rate of the forward reaction is equal to the rate of the reverse reaction.

A sample of HI (9.30 \(\times 10^{-3} \mathrm{~mol}\) ) was placed in an empty \(2.00 \mathrm{~L}\) container at \(1000 \mathrm{~K}\). After equilibrium was reached, the concentration of \(\mathrm{I}_{2}\) was \(6.29 \times 10^{-4} \mathrm{M} .\) Calculate the value of \(K_{c}\) at 1000 \(\mathrm{K}\) for the reaction \(\mathrm{H}_{2}(\mathrm{~g})+\mathrm{I}_{2}(g) \rightleftharpoons 2 \mathrm{HI}(g)\).

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.