Chemistry
Chemistry
4th Edition
ISBN: 9780078021527
Author: Julia Burdge
Publisher: McGraw-Hill Education
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Chapter 24, Problem 41QP
Interpretation Introduction

Interpretation:

The standard free energy of formation for NO, KP, and Kc for the reaction are to be determined with given ΔG value.

Concept introduction:

The ΔG for the reaction is calculated by the expression as

ΔG=ΔGf(products)ΔGf(reactant).

Here, ΔG is the change in standard Gibbs free energy, ΔGf(products) is the sum of standard Gibbs free energy of products and ΔGf(reactant) is the sum of standard Gibbs free energy of reactants.

The value of KP is calculated by using the relation given below:

lnKP=ΔGRT.

Here, ΔG is the change in standard Gibbs free energy, R is the gas constant, and T is the temperature.

The value of KP and KC are same if there is no change in the number of moles in the reaction.

The relationship between kilojoules and joules can be expressed as

1 kJ=1000 J.

To convert kilojoules to joules, conversion factor is 1000 J1 kJ.

Expert Solution & Answer
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Answer to Problem 41QP

Solution: The value of KC, KP, and ΔGf(NO) for the given reaction are 4×1031, 4×1031, and 86.7 kJ/mol, respectively.

Explanation of Solution

Given information: The given reaction is N2(g)+O2(g)2NO(g).

The value of ΔG from appendix 2 is 173.4 kJ/mol.

The value of standard free energy of formation for NO is calculated by using the relation given below:

ΔG=2ΔGf(NO)[ΔGf(N2)+ΔGf(O2)].

Here, ΔG is the change in standard Gibbs free energy, ΔGf(NO) is the change in standard Gibbs free energy of NO, ΔGf(O2) is the change in standard Gibbs free energy of O2, and ΔGf(N2) is the change in standard Gibbs free energy of N2.

The value of change in standard Gibbs free energy for atoms in their standard state is zero. In the reaction, O2 and N2 are present in their standard state. Thus, the value of ΔGf(O2) and ΔGf(N2) is zero.

Substitute 0 for ΔGf(O2), 0 for ΔGf(N2), and 173.4 kJ/mol for ΔG in theabove equation as follows:

173.4 kJ/mol=2ΔGf(NO)[0+0]ΔGf(NO)=86.7 kJ/mol.

Therefore, the value of ΔGf(NO) is 86.7 kJ/mol.

The value of ΔG is converted to J/mol by using the relation:

1 kJ=1000 J.(173.4 kJ/mol ×1000 J1 kJ)=173.4×103 J/mol.

The value of KP is calculated by using the relation given below:

lnKP=ΔGRT.

Here, ΔG is the change in standard Gibbs free energy, R is the gas constant, and T is the temperature.

Substitute 173.4 ×103 J/mol for ΔG, 298 K for T, and 8.314 J/K.mol for R in the above equation as follows:

lnKP=(173.4 ×103 J/mol)(8.314 J/K.mol)(298 K)=173.4 ×103 2477.752 KP=4×1031.

Therefore, the value of KP for the given reaction is 4×1031.

In the given reaction, the number of moles of gaseous atoms is two on both sides of the reaction, and thus there is no change in the number of moles.

Hence, the value of KP and KC is same.

Therefore, the value of KC for the given reaction is 4×1031.

Conclusion

The value of ΔGf(NO), KC, and KP for the given reaction is 4×1031.

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