   Chapter 17, Problem 55E

Chapter
Section
Textbook Problem

# For the reaction at 298 K, 2 NO 2 ( g )   ⇌ N 2 O 4 ( g ) the values of ∆H° and ∆S° are −58.03 kJ and −176.6 J/K, respectively. What is the value of ∆G° at 298 K? Assuming that ∆H° and ∆S° do not depend on temperature, at what temperature is ∆G° = 0? Is ∆G° negative above or below this temperature?

Interpretation Introduction

Interpretation: The reaction of formation of N2O4 , and the values of ΔS , ΔH for this reaction is given. The value of ΔG at 298K is to be calculated. The temperature at which ΔG=0 is to be calculated. If ΔG is negative above or below this temperature is to be stated.

Concept introduction: The expression for ΔG is,

ΔG=ΔHTΔS

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

Explanation

Explanation

The stated reaction is,

2NO2(g)N2O4(g)

Given

The value of ΔH is 58.03kJ/mol .

The value of ΔS is 176.6J/K .

Temperature is 298K .

The conversion of joule (J) into kilo-joule (kJ) is done as,

1J=103kJ

Hence,

The conversion of 176.6J into kilo-joule (kJ) is,

176.6J=(176.6×103)kJ=176.6×103kJ

Formula

The formula of ΔG is,

ΔG=ΔHTΔS

Where,

• ΔH is the enthalpy of reaction.
• ΔG is the free energy change.
• T is the temperature.
• ΔS is the entropy of reaction.

Substitute the values of ΔS and ΔH in the above equation.

ΔG=ΔHTΔS=58.03kJ{(298K)(176.6×103kJ/K)}=5

### 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 