   Chapter 21, Problem 63PS

Chapter
Section
Textbook Problem

Use data in Appendix L to calculate the enthalpy and free energy change for the reaction2 NO2(g) + O2(g) → 2 NO2(g)Is this reaction exothermic or endothermic? Is the reaction product- or reactant-favored at equilibrium?

Interpretation Introduction

Interpretation:

To calculate the enthalpy and free energy change for the given reaction.Check whether this reaction is exothermic or endothermic and is the reaction is product or reactant favoured at eqauilibrium.

Concept introduction:

The Gibbs free energy or the free energy change is a thermodynamic quantity represented by ΔG. The expression for the free energy change is:

ΔrG°=nΔfG°(products)nΔfG°(reactants)

The enthalpy change is expressed by the formula,

ΔrH°=nΔfH°(products)nΔfH°(reactants)

If the value of enthalpy change is negative, the reaction is exothermic.

If the value of free energy change is negative, then the reaction is product-favored at equilibrium.

Explanation

The standard free energy change and enthalpy change for the formation of N2O4(g) is calculated below.

Given:

Refer to Appendix L for the values of standard enthalpy and free energy.

The standard enthalpy change value of NO(g) is 90.29 kJ/mol.

The standard enthalpy change value of NO2(g) is 33.1 kJ/mol.

The standard enthalpy change value of O2(g) is 0 kJ/mol.

The standard free energy change value of NO(g) is 86.58 kJ/mol.

The standard free energy change value of NO2(g) is 51.23 kJ/mol.

The standard free energy change value of O2(g) is 0 kJ/mol.

The given reaction is,

2NO(g)+O2(g)2NO2(g)

The expression for enthalpy change is,

ΔrH°=nΔfH°(products)nΔfH°(reactants)=[(2 mol NO2(g)/mol-rxn)ΔfH°[NO2(g)][(1 mol O2(g)/mol-rxn)ΔfH°[O2(g)]+(2 mol NO(g)/mol-rxn)ΔfH°[NO(g)]] ]

Substitute the values,

ΔrH°=[(2 mol NO2(g)/mol-rxn)(33.1 kJ/mol)[(1 mol O2(g)/mol-rxn)(0 kJ/mol)+(2 mol NO(g)/mol-rxn)(90

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