   # For the equilibrium A ( g ) + 2 B ( g ) ⇌ C ( g ) the initial concentrations are [A] = [B] = [C] = 0.100 atm. Once equilibrium has been established, it is found that [C] = 0.040 atm. What is ∆ G ° for this reaction at 25°C? ### 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 119IP
Textbook Problem
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## For the equilibrium A ( g ) + 2 B ( g ) ⇌ C ( g ) the initial concentrations are [A] = [B] = [C] = 0.100 atm. Once equilibrium has been established, it is found that [C] = 0.040 atm. What is ∆G° for this reaction at 25°C?

Interpretation Introduction

Interpretation: It is given that, the initial concentration of reactants [A],[B] and product [C] is same (0.100atm) , whereas concentration of product after the establishment of equilibrium is 0.040atm . The value of ΔG° is to be calculated at 25°C for the given reaction.

Concept introduction:

### Explanation of Solution

Given

Initial concentration of A(g),B(g) and C(g) is 0.100atm .

The concentration of C(g) after the establishment of equilibrium is 0.040atm .

The stated reaction is,

A(g)+2B(g)C(g)

The change in concentration of A(g) is assumed to be x .

The stated reaction will proceed in the reverse direction because final concentration of product C(g) is less than its initial concentration.

The ICE table is formed for the given reaction.

A(g)+B(g)C(g)Initial (atm)          0.1000.1000.100Change(atm)        x2xxEquilibrium(atm)0.100+x0.100+2x0.100x

The equilibrium concentration of [B] is (0.100+2x) .

The equilibrium concentration of [A] is (0.100+x) .

The equilibrium concentration of [C] is (0.100x)M . Since, the equilibrium concentration of C is 0.040atm . The value of x is calculated as,

(0.100x)=0.040x=0.060

Substitute the value of x to get the equilibrium concentration of [A] .

[A]=(0.100+x)=0.100+0.060=0.160atm_

Substitute the value of x to get the equilibrium concentration of [B] .

[B]=(0.100+2x)=0.100+2(0.060)=0.220atm_

The value of the equilibrium concentration of [A] is 0.160atm . The value of the equilibrium concentration of [B] is 0.220atm . The equilibrium concentration of C is 0.040atm

Formula

At equilibrium, the equilibrium constant expression is expressed by the formula,

K=ConcentrationofproductsConcentrationofreactants

Where,

• K is the equilibrium constant

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