   # Using data from Appendix 4 calculate ∆ H°, ∆ S° , and ∆ G° for the reaction N 2 ( g ) + O 2 ( g ) → 2NO ( g ) Why does NO form in an automobile engine but then does not readily decompose back to N 2 and O 2 in the atmosphere? ### Chemistry: An Atoms First Approach

2nd Edition
Steven S. Zumdahl + 1 other
Publisher: Cengage Learning
ISBN: 9781305079243

#### Solutions

Chapter
Section ### Chemistry: An Atoms First Approach

2nd Edition
Steven S. Zumdahl + 1 other
Publisher: Cengage Learning
ISBN: 9781305079243
Chapter 19, Problem 48E
Textbook Problem
1 views

## Using data from Appendix 4 calculate ∆H°, ∆S°, and ∆G° for the reaction N 2 ( g ) + O 2 ( g ) → 2NO ( g ) Why does NO form in an automobile engine but then does not readily decompose back to N2 and O2 in the atmosphere?

Interpretation Introduction

Interpretation: Reaction of commercial production of NO is given. The value of ΔH°,ΔG° and ΔS° is to be calculated for the given reaction. The explanation of the fact that NO formed in an automobile engine but does not readily decompose back to N2 and O2 in the atmosphere is to be stated.

Concept introduction: The expression to calculate ΔH° is,

ΔH°=npΔH°(product)nfΔH°(reactant)

The expression to calculate ΔS° is,

ΔS°=npΔS°(product)nfΔS°(reactant)

A reaction is said to be favored if the value of ΔG° is negative.

To determine: The value of ΔH°,ΔG° and ΔS° for the given reaction; the explanation of the fact that NO is formed in an automobile engine but does not readily decompose back to N2 and O2 in the atmosphere.

### Explanation of Solution

Explanation

The value of ΔH° for the given reaction is 180kJ_ .

The stated reaction is,

N2(g)+O2(g)2NO(g)

Refer to Appendix 4 .

The value of ΔH°(kJ/mol) for the given reactant and product is,

 Molecules ΔH°(kJ/mol) NO(g) 90 N2(g) 0 O2(g) 0

The formula of ΔH° is,

ΔH°=npΔH°(product)nfΔH°(reactant)

Where,

• ΔH° the standard enthalpy of reaction.
• np is the number of moles of each product.
• nr is the number of moles each reactant.
• ΔH°(product) is the standard enthalpy of product at a pressure of 1atm .
• ΔH°(reactant) is the standard enthalpy of reactant at a pressure of 1atm .

Substitute all values from the table in the above equation.

ΔH°=npΔH°(product)nfΔH°(reactant)=[2(90){(0)+(0)}]kJ=180kJ_

The value of ΔS° for the given reaction is 25J/mol_ .

Refer to Appendix 4

The stated reaction is,

N2(g)+O2(g)2NO(g)

The value of ΔS°(J/Kmol) for the given reactant and product is,

 Molecules ΔS°(J/K⋅mol) NO(g) 211 N2(g) 192 O2(g) 205

The formula of ΔS° is,

ΔS°=npΔS°(product)nfΔS°(reactant)

Where,

• ΔS° is the standard enthalpy of reaction.
• np is the number of moles of each product.
• nr is the number of moles each reactant

### Still sussing out bartleby?

Check out a sample textbook solution.

See a sample solution

#### The Solution to Your Study Problems

Bartleby provides explanations to thousands of textbook problems written by our experts, many with advanced degrees!

Get Started

Find more solutions based on key concepts
Aerobically trained muscles burn fat more readily than untrained in us. T F

Nutrition: Concepts and Controversies - Standalone book (MindTap Course List)

Understanding Nutrition (MindTap Course List)

What is physically exchanged during crossing over?

Human Heredity: Principles and Issues (MindTap Course List)

When was the first set of laws governing the allocation of ocean resources proposed?

Oceanography: An Invitation To Marine Science, Loose-leaf Versin

Liquid nitrogen has a boiling point of 77.3 K and a latent heal of vaporization of 2.01 105J/kg. A 25.0-W elec...

Physics for Scientists and Engineers, Technology Update (No access codes included)

What are the largest known structures in the Universe?

Foundations of Astronomy (MindTap Course List) 