Chemistry: Atoms First
Chemistry: Atoms First
3rd Edition
ISBN: 9781259923142
Author: Burdge
Publisher: MCG
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Chapter 25, Problem 25.39QP

(a)

Interpretation Introduction

Interpretation:

For the formation of NO , the values of ΔGo , Kp and Kc have to be calculated.

Concept introduction:

Gibb’s free energy: The relationship between temperature and thermodynamic properties like enthalpy and entropy is given by Gibb’s free energy.  Based on the value of Gibb’s free energy the feasibility of reactions can be explained.

G=H-TSG-Gibb'sfreeenergyH-enthalpyT-temperatureS-entropy

To calculate: Gibb’s free energy change for the formation of NO

(a)

Expert Solution
Check Mark

Answer to Problem 25.39QP

Gibb’s free energy change for the given reaction

ΔGrxno(NO)=86.7kJmol

Explanation of Solution

Formation reaction of NO

N2(g)+O2(g)2NO(g)

Given that, Gibb’s free energy change of the reaction at 298 K

ΔGo=173.4kJmol

General equation for calculation of Gibb’s free energy change of the given reaction

ΔGrxno=ΔGfo(products)-ΔGfo(reactants)ΔGrxno-freeenergychangeofthereactionΔGfo-freeenergychangeofformation

Apply the formula for the given reaction

Substances in elemental form (here oxygen and nitrogen molecule) have Gibb’s free energy of formation value is zero.

ΔGrxno=2ΔGfo(NO)-[(0+0)]173.4kJmol=2ΔGfo(NO)ΔGfo(NO)=(173.42)kJmol=86.7kJmol

Gibb’s free energy change for the formation of NO is calculated as 86.7kJmol

Gibb’s free energy change for the formation of NO molecule is calculated by the application of general equation of free energy change.

(b)

Interpretation Introduction

Interpretation:

For the formation of NO , the values of ΔGo , Kp and Kc have to be calculated.

Concept introduction:

Gibb’s free energy: The relationship between temperature and thermodynamic properties like enthalpy and entropy is given by Gibb’s free energy.  Based on the value of Gibb’s free energy the feasibility of reactions can be explained.

G=H-TSG-Gibb'sfreeenergyH-enthalpyT-temperatureS-entropy

To calculate: The Kp value for the given reaction

(b)

Expert Solution
Check Mark

Answer to Problem 25.39QP

For the given reaction, value of Kp

Kp=4×10-31

Explanation of Solution

The relationship between ΔGo and Kp is given by

ΔGo=-RTlnKpR-gas constant; T - temperature; Kp - equilibrium constant at partial pressure

Substitute the values of ΔGo , T and R

173.4×103Jmol=-(8.314Jmol)(298K)lnKpKp=4×10-31

By the use of relationship between ΔGo and Kp .  The value of Kp is calculated as 4×10-31

(c)

Interpretation Introduction

Interpretation:

For the formation of NO , the values of ΔGo , Kp and Kc have to be calculated.

Concept introduction:

Gibb’s free energy: The relationship between temperature and thermodynamic properties like enthalpy and entropy is given by Gibb’s free energy.  Based on the value of Gibb’s free energy the feasibility of reactions can be explained.

G=H-TSG-Gibb'sfreeenergyH-enthalpyT-temperatureS-entropy

To calculate: The Kc value for the given reaction

(c)

Expert Solution
Check Mark

Answer to Problem 25.39QP

For the given reaction value of Kc

Kc=4×10-31

Explanation of Solution

The relationship between Kc and Kp is given by

Kp=Kc(0.08206T)ΔnΔn-changeinnumberofmoles

Here no change in number of moles, therefore, Δn=0

Kp=Kc=4×10-31

By the use of relationship between Kc and Kp .  The value of Kc is calculated as 4×10-31

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Chapter 25 Solutions

Chemistry: Atoms First

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