   # 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? ### 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 16, Problem 55E
Textbook Problem
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## 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 of Solution

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

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