EBK ORGANIC CHEMISTRY
EBK ORGANIC CHEMISTRY
6th Edition
ISBN: 8220103151757
Author: LOUDON
Publisher: MAC HIGHER
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Chapter 3, Problem 3.46AP
Interpretation Introduction

(a)

Interpretation:

The percentage of 2,2 dimethylpropane and pentane in an equilibrium mixture at 25°C is to be stated.

Concept introduction:

The product of natural log of equilibrium constant, gas constant and specific temperature gives the value of free energy for the reaction. Free energy of a reaction is expressed in J/mol or kJ/mol.

Expert Solution
Check Mark

Answer to Problem 3.46AP

The percentage of 2,2 dimethylpropane is 94.1% and the pentane is 5.9% in the mixture.

Explanation of Solution

It is given that standard free energy (ΔGο) for 2,2 dimethylpropane and pentane is 6.86kJmol1.

Assume that 2,2 dimethylpropane is A and pentane is B.

Then the equilibrium between A and B is shown below.

AB

The formula to calculate the Keq is shown below.

ΔGο=2.3RTlogKeq

Where

  • R is gas constant.
  • T is temperature.
  • ΔGο is the standard free energy.
  • Keq is the equilibrium constant.

Substitute the value of R as 8.314×103kJmol1, the value of T as 298K and ΔGο as 6.86kJmol1 in above equation.

6.86kJmol1=2.3×8.314×103×298×logKeq6.86kJmol1=5.69×logKeq

Rearrange the above equation as shown below.

6.86kJmol15.69=logKeq1.20=logKeqKeq=antilog(1.20)Keq=0.0630

The percentage of each compound in the mixture is calculated as shown below.

Keq=[B][A]

Substitute the value of Keq as 0.0630.

0.0630=[B][A]0.0630[A]=[B]

At equilibrium, consider total hydrocarbon =1. This is represented as shown below.

A+B=1

Substitute the value of B in the above equation.

A+0.0630[A]=11.0630[A]=1[A]=11.0630[A]=0.941

The value of [A] is converted into percentage as shown below.

%of[A]=0.941×100=94.1%

Therefore, the percentage of the compound [A] is 94.1% in the mixture.

The value of [B] is calculated as shown below.

%of[B]=100%94.1%=5.9%

Therefore, compound [B] is present in 5.9% in the mixture.

Conclusion

The percentage of 2,2 dimethylpropane is 94.1% and the pentane is 5.9% in the mixture.

Interpretation Introduction

(b)

Interpretation:

The percentage of anti-butane and gauche-butane in an equilibrium mixture at 25°C is to be stated.

Concept introduction:

The product of natural log of equilibrium constant, gas constant and specific temperature gives the value of free energy for the reaction. Free energy of a reaction is expressed in J/mol or kJ/mol.

Expert Solution
Check Mark

Answer to Problem 3.46AP

The percentage of anti-butane is 75.6% and the gauche-butane is 24.4% in the mixture.

Explanation of Solution

It is given that standard free energy (ΔGο) for gauche-butane and anti-butane is 2.8kJmol1.

Assume that anti-butane is A and gauche butane is B.

Then the equilibrium between A and B is shown below.

AB

The formula to calculate the Keq is shown below.

ΔGο=2.3RTlogKeq

Where

  • R is gas constant.
  • T is temperature.
  • ΔGο is the standard free energy.
  • Keq is the equilibrium constant.

Substitute the value of R as 8.314×103kJmol1, the value of T as 298K and ΔGο as 2.8kJmol1 in the above equation.

2.8kJmol1=2.3×8.314×103×298×logKeq2.8kJmol1=5.69×logKeq

Rearrange the above equation as shown below.

2.8kJmol15.69=logKeq0.492=logKeqKeq=antilog(0.492)Keq=0.322

The percentage of each compound in the mixture is calculated as shown below.

Keq=[B][A]

Substitute the value of Keq as 0.322.

0.322=[B][A]0.322[A]=[B]

At equilibrium, consider total hydrocarbon =1

A+B=1

Substitute the value of B in the above equation.

A+0.322[A]=11.322[A]=1[A]=11.322[A]=0.756

The value of [A] is converted into percentage as shown below.

%of[A]=0.756×100=75.6%

Therefore, the percentage of the compound [A] is 75.6% in the mixture.

The value of [B] is calculated as shown below.

%of[B]=100%75.6%=24.4%

Therefore, compound [B] is present in 24.4% in the mixture.

Conclusion

The percentage of anti-butane is 75.6% and the gauche-pentane is 24.4% in the mixture.

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