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The standard free energies of formation and the standard enthalpies of formation at 298 K for difluoroacetylene (C 2 F 2 ) and hexafluorobenzene (C 6 F 6 ) are Δ G f o (KJ/mol) Δ H f o (KJ/mol) C 2 F 2 ( g ) 191.2 241.3 Hexane 78.2 132.8 For the following reaction: C 6 F 6 ( g ) ⇌ 3 C 2 F 2 ( g ) a. calculate ∆S° at 298 K. b. calculate K at 298 K. c. estimate K at 3000. K, assuming ∆H° and ∆S° do not depend on temperature.

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Chemistry: An Atoms First Approach

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

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Chapter
Section
BuyFindarrow_forward

Chemistry: An Atoms First Approach

2nd Edition
Steven S. Zumdahl + 1 other
Publisher: Cengage Learning
ISBN: 9781305079243
Chapter 16, Problem 72E
Textbook Problem
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The standard free energies of formation and the standard enthalpies of formation at 298 K for difluoroacetylene (C2F2) and hexafluorobenzene (C6F6) are

  Δ G f o (KJ/mol) Δ H f o (KJ/mol)
C2F2(g) 191.2 241.3
Hexane 78.2 132.8

For the following reaction:

C 6 F 6 ( g ) 3 C 2 F 2 ( g )

a. calculate ∆S° at 298 K.

b. calculate K at 298 K.

c. estimate K at 3000. K, assuming ∆H° and ∆S° do not depend on temperature.

(a)

Interpretation Introduction

Interpretation: The reaction corresponding to the formation of C2F2 ; the value of ΔHf° and ΔGf° for C2F2 and C6F6 is given. The value of ΔS° and K at the given temperature is to be calculated. The value of K at the given temperature, assuming that ΔH° and ΔS° are temperature independent is to be calculated.

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

The expression for free energy change is,

ΔG=ΔG°+RTln(K)

The formula of ΔH° is,

ΔH°=npΔH°(product)nfΔH°(reactant)

Explanation of Solution

Given

The value of ΔG° for C2F2 is 191.2kJ/mol .

The value of ΔG° for C6F6 is 78.2kJ/mol .

The reaction that takes place is,

C6F6(g)3C2F2(g)

The formula of ΔG° is,

ΔG°=npΔG°(product)nfΔG°(reactant)

Where,

  • np is the number of moles of each product.
  • nr is the number of moles each reactant.
  • ΔG°(product) is the free energy of product at a pressure of 1atm

(b)

Interpretation Introduction

Interpretation: The reaction corresponding to the formation of C2F2 ; the value of ΔHf° and ΔGf° for C2F2 and C6F6 is given. The value of ΔS° and K at the given temperature is to be calculated. The value of K at the given temperature, assuming that ΔH° and ΔS° are temperature independent is to be calculated.

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

The expression for free energy change is,

ΔG=ΔG°+RTln(K)

The formula of ΔH° is,

ΔH°=npΔH°(product)nfΔH°(reactant)

(c)

Interpretation Introduction

Interpretation: The reaction corresponding to the formation of C2F2 ; the value of ΔHf° and ΔGf° for C2F2 and C6F6 is given. The value of ΔS° and K at the given temperature is to be calculated. The value of K at the given temperature, assuming that ΔH° and ΔS° are temperature independent is to be calculated.

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

The expression for free energy change is,

ΔG=ΔG°+RTln(K)

The formula of ΔH° is,

ΔH°=npΔH°(product)nfΔH°(reactant)

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Chapter 16 Solutions

Chemistry: An Atoms First Approach
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