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Carbon tetrachloride (CCl 4 ) and benzene (C 6 H 6 ) form ideal solutions. Consider an equimolar solution of CCl 4 and C 6 H 6 at 25°C. The vapor above the solution is collected and condensed. Using the following data, determine the composition in mole fraction of the condensed vapor. Substance Δ G f o C 6 H 6 ( l ) 124.50 kJ/mol C 6 H 6 ( g ) 129.66 kJ/mol CCI 4 ( l ) −65.21 kJ/mol CCI 4 ,( g ) −60.59 kJ/mol

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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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BuyFindarrow_forward

Chemistry: An Atoms First Approach

2nd Edition
Steven S. Zumdahl + 1 other
Publisher: Cengage Learning
ISBN: 9781305079243
Chapter 16, Problem 114CP
Textbook Problem
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Carbon tetrachloride (CCl4) and benzene (C6H6) form ideal solutions. Consider an equimolar solution of CCl4 and C6H6 at 25°C. The vapor above the solution is collected and condensed. Using the following data, determine the composition in mole fraction of the condensed vapor.

Substance Δ G f o
C6H6(l) 124.50 kJ/mol
C6H6(g) 129.66 kJ/mol
CCI4(l) −65.21 kJ/mol
CCI4,(g) −60.59 kJ/mol

Interpretation Introduction

Interpretation: The composition of benzene (C6H6) and carbon tetrachloride (CCl4) in mole fraction of the condensed vapor is to be calculated by using the given data.

Concept introduction: Equilibrium constant K , is the measure of concentration of reactants and products of a system at equilibrium. If the reaction is at equilibrium and the equilibrium is expressed as partial pressure (P) , then the free energy change is,

ΔG=0Q=P

The expression for free energy change is,

ΔG°=RTln(P)

The expression for standard Gibbs free energy, ΔG° , is,

ΔG°=npΔG°(product)nfΔG°(reactant)

Explanation of Solution

The reaction that takes place for benzene is,

C6H6(l)C6H6(g)

The reaction that takes place for carbon tetrachloride is,

CCl4(l)CCl4(g)

Given table,

The value of ΔG°(kJ/mol) for the given reactant and product is,

Molecules ΔG°(kJ/mol)
C6H6(l) 124.50
C6H6(g) 129.66
CCl4(l) 65.21
CCl4(g) 60.59

The formula of ΔG° is,

ΔG°=npΔG°(product)nfΔG°(reactant) (1)

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 .
  • ΔG°(reactant) is the free energy of reactant at a pressure of 1atm .

Substitute all the values from the table for C6H6 in the above equation.

ΔG°=npΔG°(product)nfΔG°(reactant)=[1(129.66)1(124.50)]kJ/mol=5.61kJ/mol_

Substitute all the values from the table for CCl4 in equation (1).

ΔG°=npΔG°(product)nfΔG°(reactant)=[1(60.59)1(65.21)]kJ/mol=4.62kJ/mol_

The value of ΔG°  for benzene is 5.61kJ/mol .

Temperature is 298K .

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

1kJ=103J

Hence,

The conversion of 5.61kJ into joule is,

5.61kJ=(5.61×103)J=5.61×103J

Formula

The expression for free energy change is,

ΔG=ΔG°+RTln(Q)

Where,

  • ΔG is the free energy change for a reaction at specified pressure.
  • R is the gas law constant (8.3145J/Kmol) .
  • T is the absolute temperature.
  • Q is the reaction constant.

Since, the reaction is at equilibrium, the free energy change,

ΔG=0Q=P1

Where,

P1 is the partial pressure of benzene

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Chapter 16 Solutions

Chemistry: An Atoms First Approach
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