   # Calculate ∆ G ° for H 2 O ( g ) + 1 2 O 2 ( g ) ⇌ H 2 O 2 ( g ) at 600. K, using the following data: H 2 ( g ) + O 2 ( g ) ⇌ H 2 O 2 ( g ) K = 2.3 × 10 6 at 600 . K 2H 2 ( g ) + O 2 ( g ) ⇌ 2 H 2 O ( g ) K = 1.8 × 10 37 at 600 . K ### 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 73E
Textbook Problem
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## Calculate ∆G° for H 2 O ( g ) + 1 2 O 2 ( g ) ⇌ H 2 O 2 ( g ) at 600. K, using the following data: H 2 ( g ) + O 2 ( g ) ⇌ H 2 O 2 ( g ) K =   2.3 ×   10 6   at   600 .   K 2H 2 ( g ) + O 2 ( g ) ⇌ 2 H 2 O ( g ) K =   1.8 ×   10 37   at   600 .   K

Interpretation Introduction

Interpretation: The reaction of formation of H2O2 and H2O , the value of K and T for these reactions is given. The value of ΔG° is to be calculated for the given reaction.

Concept introduction: Equilibrium constant, K , is defined as the ratio of the concentration of products to that of the reactants at equilibrium. If a given reaction is at equilibrium, the free energy change is,

ΔG=0Q=K

ΔG=ΔG°+RTln(K)

### Explanation of Solution

Given

The stated first reaction at 600K is,

H2(g)+O2(g)H2O2(g) (1)

The equilibrium constant for this reaction is,

K=2.3×106

The stated second reaction at 600K is,

2H2(g)+O2(g)2H2O(g) (2)

The equilibrium constant for this reaction is,

K=1.8×1037

Divide equation (2) on both sides by a factor 2 to get third equation.

12(2H2(g)+O2(g))H2O(g) (3)

The reaction for which ΔG° is to be calculated is,

H2O(g)+12O2(g)H2O2(g) (4)

Formula

ΔG1°=RTln(K)

Where,

• ΔG1° is the standard Gibbs free energy change for the first reaction.
• R is the gas constant (8.3145JK1mol1) .
• T is the absolute temperature.
• K is the equilibrium constant.

Substitute the values of R,T and K for first reaction in the above expression.

ΔG1°=RTln(K)=(8.3145J/K)(600K)ln(2.3×106)=73076.57J/mol_

ΔG2°=RTln(K)

Where,

• ΔG2° is the standard Gibbs free energy change for the second reaction.
• R is the gas constant (8.3145JK1mol1)

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